Griffiths_D.J._Introduction_to_quantum_mechanics 2nd ed SOUTIONS
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Solution manual by David Griffiths (2005, Pearson) for the 2nd edition of his quantum mechanics textbook, a download kept in Phil's physics book collection. It gives worked solutions chapter by chapter, from the wave function and the Schrödinger equation through formalism, three dimensions, identical particles, perturbation theory, variational principle, WKB, adiabatic approximation and scattering. It ends with a linear algebra appendix and a 1st-to-2nd edition problem grid.
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Contents
P r e f a c e 2 1 The Wave Function 3
2 Time-Independent Schrödinger Equation 14 3 F o r m a l i s m 6 2 4 Quantum Mechanics in Three Dimensions 87
5 Identical Particles 132 6 Time-Independent Perturbation Theory 154
7 The Variational Principle 196
8 The WKB Approximation 219 9 Time-Dependent Perturbation Theory 236
10 The Adiabatic Approximation 254
11 Scattering 268 12 Afterword 282
Appendix Linear Algebra 283 2
nd Edition – 1st Edition Problem Correlation Grid 299
2
Preface
These are my own solutions to the problems in Introduction to Quantum Mechanics, 2nd ed. I have made every
effort to insure that they are clear and correct, but errors are bound to occur, and for this I apologize in advance.
I would like to thank the many people who pointed out mistakes in the solution manual for the first edition,and encourage anyone who finds defects in this one to alert me (griffi[email protected]). I’ll maintain a list of errataon my web page (http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in themanual itself from time to time. I also thank my students at Reed and at Smith for many useful suggestions,and above all Neelaksh Sadhoo, who did most of the typesetting.
At the end of the manual there is a grid that correlates the problem numbers in the second edition with
those in the first edition.
David Griffiths
c/circlecopyrt2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
CHAPTER 1. THE WAVE FUNCTION 3
Chapter 1
The Wave Function
Problem 1.1
(a)
/angbracketleftj/angbracketright2=2 12=441.
/angbracketleftj2/angbracketright=1
N/summationdisplay
j2N(j)=1
14/bracketleftbig
(142) + (152) + 3(162) + 2(222) + 2(242) + 5(252)/bracketrightbig
=1
14(196 + 225 + 768 + 968 + 1152 + 3125) =6434
14=459.571.
(b)j∆j=j−/angbracketleftj/angbracketright
1414−21 =−7
1515−21 =−6
1616−21 =−5
2222−21 = 1
2424−21 = 3
2525−21 = 4
σ2=1
N/summationdisplay
(∆j)2N(j)=1
14/bracketleftbig
(−7)2+(−6)2+(−5)2·3 + (1)2·2 + (3)2·2 + (4)2·5/bracketrightbig
=1
14( 4 9+3 6+7 5+2+1 8+8 0 )=260
14=18.571.
σ=√
18.571 = 4.309.
(c)
/angbracketleftj2/angbracketright−/angbracketleftj/angbracketright2= 459.571−4 4 1=1 8 .571.[Agrees with (b).]
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
4 CHAPTER 1. THE WAVE FUNCTION
Problem 1.2
(a)
/angbracketleftx2/angbracketright=/integraldisplayh
0x21
2√
hxdx=1
2√
h/parenleftbigg2
5x5/2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleh
0=h2
5.
σ2=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=h2
5−/parenleftbiggh
3/parenrightbigg2
=4
45h2⇒σ=2h
3√
5=0.2981h.
(b)
P=1−/integraldisplayx+
x−1
2√
hxdx=1−1
2√
h(2√x)/vextendsingle/vextendsingle/vextendsingle/vextendsinglex+
x−=1−1√
h/parenleftbig√x+−√x−/parenrightbig
.
x+≡/angbracketleftx/angbracketright+σ=0.3333h+0.2981h=0.6315h;x−≡/angbracketleftx/angbracketright−σ=0.3333h−0.2981h=0.0352h.
P=1−√
0.6315 +√
0.0352 = 0.393.
Problem 1.3
(a)
1=/integraldisplay∞
−∞Ae−λ(x−a)2dx. Letu≡x−a,du=dx,u:−∞→∞ .
1=A/integraldisplay∞
−∞e−λu2du=A/radicalbiggπ
λ⇒A=/radicalbigg
λ
π.
(b)
/angbracketleftx/angbracketright=A/integraldisplay∞
−∞xe−λ(x−a)2dx=A/integraldisplay∞
−∞(u+a)e−λu2du
=A/bracketleftbigg/integraldisplay∞
−∞ue−λu2du+a/integraldisplay∞
−∞e−λu2du/bracketrightbigg
=A/parenleftbigg
0+a/radicalbiggπ
λ/parenrightbigg
=a.
/angbracketleftx2/angbracketright=A/integraldisplay∞
−∞x2e−λ(x−a)2dx
=A/braceleftbigg/integraldisplay∞
−∞u2e−λu2du+2a/integraldisplay∞
−∞ue−λu2du+a2/integraldisplay∞
−∞e−λu2du/bracerightbigg
=A/bracketleftbigg1
2λ/radicalbiggπ
λ+0+a2/radicalbiggπ
λ/bracketrightbigg
=a2+1
2λ.
σ2=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=a2+1
2λ−a2=1
2λ;σ=1√
2λ.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 1. THE WAVE FUNCTION 5
(c)
A
x aρ(x)
Problem 1.4
(a)
1=|A|2
a2/integraldisplaya
0x2dx+|A|2
(b−a)2/integraldisplayb
a(b−x)2dx=|A|2/braceleftBigg
1
a2/parenleftbiggx3
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0+1
(b−a)2/parenleftbigg
−(b−x)3
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
a/bracerightBigg
=|A|2/bracketleftbigga
3+b−a
3/bracketrightbigg
=|A|2b
3⇒A=/radicalbigg
3
b.
(b)
x aA
bΨ
(c)Atx=a.
(d)
P=/integraldisplaya
0|Ψ|2dx=|A|2
a2/integraldisplaya
0x2dx=|A|2a
3=a
b./braceleftbiggP=1 i f b=a,/check
P=1/2i fb=2a./check
(e)
/angbracketleftx/angbracketright=/integraldisplay
x|Ψ|2dx=|A|2/braceleftbigg1
a2/integraldisplaya
0x3dx+1
(b−a)2/integraldisplayb
ax(b−x)2dx/bracerightbigg
=3
b/braceleftBigg
1
a2/parenleftbiggx4
4/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0+1
(b−a)2/parenleftbigg
b2x2
2−2bx3
3+x4
4/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
a/bracerightBigg
=3
4b(b−a)2/bracketleftbig
a2(b−a)2+2b4−8b4/3+b4−2a2b2+8a3b/3−a4/bracketrightbig
=3
4b(b−a)2/parenleftbiggb4
3−a2b2+2
3a3b/parenrightbigg
=1
4(b−a)2(b3−3a2b+2a3)=2a+b
4.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
6 CHAPTER 1. THE WAVE FUNCTION
Problem 1.5
(a)
1=/integraldisplay
|Ψ|2dx=2|A|2/integraldisplay∞
0e−2λxdx=2|A|2/parenleftbigge−2λx
−2λ/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0=|A|2
λ;A=√
λ.
(b)
/angbracketleftx/angbracketright=/integraldisplay
x|Ψ|2dx=|A|2/integraldisplay∞
−∞xe−2λ|x|dx=0. [Odd integrand.]
/angbracketleftx2/angbracketright=2|A|2/integraldisplay∞
0x2e−2λxdx=2λ/bracketleftbigg2
(2λ)3/bracketrightbigg
=1
2λ2.
(c)
σ2=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=1
2λ2;σ=1√
2λ.|Ψ(±σ)|2=|A|2e−2λσ=λe−2λ/√
2λ=λe−√
2=0.2431λ.
|Ψ|2
λ
σ −σ +x.24λ
Probability outside :
2/integraldisplay∞
σ|Ψ|2dx=2|A|2/integraldisplay∞
σe−2λxdx=2λ/parenleftbigge−2λx
−2λ/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
σ=e−2λσ=e−√
2=0.2431.
Problem 1.6
For integration by parts, the differentiation has to be with respect to the integration variable – in this case the
differentiation is with respect to t, but the integration variable is x. It’s true that
∂
∂t(x|Ψ|2)=∂x
∂t|Ψ|2+x∂
∂t|Ψ|2=x∂
∂t|Ψ|2,
but this does notallow us to perform the integration:
/integraldisplayb
ax∂
∂t|Ψ|2dx=/integraldisplayb
a∂
∂t(x|Ψ|2)dx/negationslash=(x|Ψ|2)/vextendsingle/vextendsingleb
a.
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CHAPTER 1. THE WAVE FUNCTION 7
Problem 1.7
From Eq. 1.33,d/angbracketleftp/angbracketright
dt=−i/planckover2pi1/integraltext∂
∂t/parenleftbig
Ψ∗∂Ψ
∂x/parenrightbig
dx. But, noting that∂2Ψ
∂x∂t=∂2Ψ
∂t∂xand using Eqs. 1.23-1.24:
∂
∂t/parenleftbigg
Ψ∗∂Ψ
∂x/parenrightbigg
=∂Ψ∗
∂t∂Ψ
∂x+Ψ∗∂
∂x/parenleftbigg∂Ψ
∂t/parenrightbigg
=/bracketleftbigg
−i/planckover2pi1
2m∂2Ψ∗
∂x2+i
/planckover2pi1VΨ∗/bracketrightbigg∂Ψ
∂x+Ψ∗∂
∂x/bracketleftbiggi/planckover2pi1
2m∂2Ψ
∂x2−i
/planckover2pi1VΨ/bracketrightbigg
=i/planckover2pi1
2m/bracketleftbigg
Ψ∗∂3Ψ
∂x3−∂2Ψ∗
∂x2∂Ψ
∂x/bracketrightbigg
+i
/planckover2pi1/bracketleftbigg
VΨ∗∂Ψ
∂x−Ψ∗∂
∂x(VΨ)/bracketrightbigg
The first term integrates to zero, using integration by parts twice, and the second term can be simplified to
VΨ∗∂Ψ
∂x−Ψ∗V∂Ψ
∂x−Ψ∗∂V
∂xΨ=−|Ψ|2∂V
∂x.So
d/angbracketleftp/angbracketright
dt=−i/planckover2pi1/parenleftbiggi
/planckover2pi1/parenrightbigg/integraldisplay
−|Ψ|2∂V
∂xdx=/angbracketleft−∂V
∂x/angbracketright.QED
Problem 1.8
Suppose Ψ satisfies the Schr¨ odinger equation without V0:i/planckover2pi1∂Ψ
∂t=−/planckover2pi12
2m∂2Ψ
∂x2+VΨ. We want to find the solution
Ψ0withV0:i/planckover2pi1∂Ψ0
∂t=−/planckover2pi12
2m∂2Ψ0
∂x2+(V+V0)Ψ0.
Claim :Ψ0=Ψe−iV0t//planckover2pi1.
Proof:i/planckover2pi1∂Ψ0
∂t=i/planckover2pi1∂Ψ
∂te−iV0t//planckover2pi1+i/planckover2pi1Ψ/parenleftbig
−iV0
/planckover2pi1/parenrightbig
e−iV0t//planckover2pi1=/bracketleftBig
−/planckover2pi12
2m∂2Ψ
∂x2+VΨ/bracketrightBig
e−iV0t//planckover2pi1+V0Ψe−iV0t//planckover2pi1
=−/planckover2pi12
2m∂2Ψ0
∂x2+(V+V0)Ψ0. QED
This has noeffect on the expectation value of a dynamical variable, since the extra phase factor, being inde-
pendent of x, cancels out in Eq. 1.36.
Problem 1.9
(a)
1=2|A|2/integraldisplay∞
0e−2amx2//planckover2pi1dx=2|A|21
2/radicalbiggπ
(2am//planckover2pi1)=|A|2/radicalbigg
π/planckover2pi1
2am;A=/parenleftbigg2am
π/planckover2pi1/parenrightbigg1/4
.
(b)
∂Ψ
∂t=−iaΨ;∂Ψ
∂x=−2amx
/planckover2pi1Ψ;∂2Ψ
∂x2=−2am
/planckover2pi1/parenleftbigg
Ψ+x∂Ψ
∂x/parenrightbigg
=−2am
/planckover2pi1/parenleftbigg
1−2amx2
/planckover2pi1/parenrightbigg
Ψ.
Plug these into the Schr¨ odinger equation, i/planckover2pi1∂Ψ
∂t=−/planckover2pi12
2m∂2Ψ
∂x2+VΨ:
VΨ=i/planckover2pi1(−ia)Ψ +/planckover2pi12
2m/parenleftbigg
−2am
/planckover2pi1/parenrightbigg/parenleftbigg
1−2amx2
/planckover2pi1/parenrightbigg
Ψ
=/bracketleftbigg
/planckover2pi1a−/planckover2pi1a/parenleftbigg
1−2amx2
/planckover2pi1/parenrightbigg/bracketrightbigg
Ψ=2a2mx2Ψ,soV(x)=2ma2x2.
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8 CHAPTER 1. THE WAVE FUNCTION
(c)
/angbracketleftx/angbracketright=/integraldisplay∞
−∞x|Ψ|2dx=0. [Odd integrand.]
/angbracketleftx2/angbracketright=2|A|2/integraldisplay∞
0x2e−2amx2//planckover2pi1dx=2|A|2 1
22(2am//planckover2pi1)/radicalbigg
π/planckover2pi1
2am=/planckover2pi1
4am.
/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=0.
/angbracketleftp2/angbracketright=/integraldisplay
Ψ∗/parenleftbigg/planckover2pi1
i∂
∂x/parenrightbigg2
Ψdx=−/planckover2pi12/integraldisplay
Ψ∗∂2Ψ
∂x2dx
=−/planckover2pi12/integraldisplay
Ψ∗/bracketleftbigg
−2am
/planckover2pi1/parenleftbigg
1−2amx2
/planckover2pi1/parenrightbigg
Ψ/bracketrightbigg
dx=2am/planckover2pi1/braceleftbigg/integraldisplay
|Ψ|2dx−2am
/planckover2pi1/integraldisplay
x2|Ψ|2dx/bracerightbigg
=2am/planckover2pi1/parenleftbigg
1−2am
/planckover2pi1/angbracketleftx2/angbracketright/parenrightbigg
=2am/planckover2pi1/parenleftbigg
1−2am
/planckover2pi1/planckover2pi1
4am/parenrightbigg
=2am/planckover2pi1/parenleftbigg1
2/parenrightbigg
=am/planckover2pi1.
(d)
σ2
x=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=/planckover2pi1
4am=⇒σx=/radicalbigg
/planckover2pi1
4am;σ2
p=/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=am/planckover2pi1=⇒σp=√
am/planckover2pi1.
σxσp=/radicalBig
/planckover2pi1
4am√
am/planckover2pi1=/planckover2pi1
2. This is(just barely) consistent with the uncertainty principle.
Problem 1.10
From Math Tables: π=3.141592653589793238462643 ···
(a)P(0) = 0 P(1) = 2/25P(2) = 3/25P(3) = 5/25P(4) = 3/25
P(5) = 3/25P(6) = 3/25P(7) = 1/25P(8) = 2/25P(9) = 3/25
In general, P(j)=N(j)
N.
(b)Most probable :3.Median : 13 are≤4, 12 are≥5, so median is 4.
Average :/angbracketleftj/angbracketright=1
25[0·0+1·2+2·3+3·5+4·3+5·3+6·3+7·1+8·2+9·3]
=1
25[ 0+2+6+1 5+1 2+1 5+1 8+7+1 6+2 7 ]=118
25=4.72.
(c)/angbracketleftj2/angbracketright=1
25[ 0+12·2+22·3+32·5+42·3+52·3+62·3+72·1+82·2+92·3]
=1
25[ 0+2+1 2+4 5+4 8+7 5+1 0 8+4 9+1 2 8+ 243] =710
25=28.4.
σ2=/angbracketleftj2/angbracketright−/angbracketleftj/angbracketright2=2 8.4−4.722=2 8.4−22.2784 = 6 .1216; σ=√
6.1216 = 2.474.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 1. THE WAVE FUNCTION 9
Problem 1.11
(a)Constant for 0 ≤θ≤π, otherwise zero. In view of Eq. 1.16, the constant is 1 /π.
ρ(θ)=/braceleftbigg
1/π,if 0≤θ≤π,
0,otherwise .
1/π
−π/2 0 π 3π/2ρ(θ)
θ
(b)
/angbracketleftθ/angbracketright=/integraldisplay
θρ(θ)dθ=1
π/integraldisplayπ
0θdθ=1
π/parenleftbiggθ2
2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=π
2[of course].
/angbracketleftθ2/angbracketright=1
π/integraldisplayπ
0θ2dθ=1
π/parenleftbiggθ3
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=π2
3.
σ2=/angbracketleftθ2/angbracketright−/angbracketleftθ/angbracketright2=π2
3−π2
4=π2
12;σ=π
2√
3.
(c)
/angbracketleftsinθ/angbracketright=1
π/integraldisplayπ
0sinθdθ=1
π(−cosθ)|π
0=1
π(1−(−1)) =2
π.
/angbracketleftcosθ/angbracketright=1
π/integraldisplayπ
0cosθdθ=1
π(sinθ)|π
0=0.
/angbracketleftcos2θ/angbracketright=1
π/integraldisplayπ
0cos2θdθ=1
π/integraldisplayπ
0(1/2)dθ=1
2.
[Because sin2θ+ cos2θ= 1, and the integrals of sin2and cos2are equal (over suitable intervals), one can
replace them by 1/2 in such cases.]
Problem 1.12
(a)x=rcosθ⇒dx=−rsinθdθ.The probability that the needle lies in range dθisρ(θ)dθ=1
πdθ, so the
probability that it’s in the range dxis
ρ(x)dx=1
πdx
rsinθ=1
πdx
r/radicalbig
1−(x/r)2=dx
π√
r2−x2.
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10 CHAPTER 1. THE WAVE FUNCTION
ρ(x)
x r2r -r -2r
∴ρ(x)=/braceleftbigg1
π√
r2−x2,if−r<x<r ,
0, otherwise .[Note: We want the magnitude ofdxhere.]
Total:/integraltextr
−r1
π√
r2−x2dx=2
π/integraltextr
01√
r2−x2dx=2
πsin−1x
r/vextendsingle/vextendsingler
0=2
πsin−1(1) =2
π·π
2=1./check
(b)
/angbracketleftx/angbracketright=1
π/integraldisplayr
−rx1√
r2−x2dx=0[odd integrand, even interval].
/angbracketleftx2/angbracketright=2
π/integraldisplayr
0x2
√
r2−x2dx=2
π/bracketleftbigg
−x
2/radicalbig
r2−x2+r2
2sin−1/parenleftBigx
r/parenrightBig/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingler
0=2
πr2
2sin−1(1) =r2
2.
σ2=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=r2/2=⇒σ=r/√
2.
To get/angbracketleftx/angbracketrightand/angbracketleftx2/angbracketrightfrom Problem 1.11(c), use x=rcosθ,s o/angbracketleftx/angbracketright=r/angbracketleftcosθ/angbracketright=0,/angbracketleftx2/angbracketright=r2/angbracketleftcos2θ/angbracketright=r2/2.
Problem 1.13
Suppose the eye end lands a distance yup from a line (0 ≤y<l), and let xbe the projection along that same
direction (−l≤x<l). The needle crosses the line above if y+x≥l(i.e.x≥l−y), and it crosses the line
below if y+x<0 (i.e.x<−y). So for a given value of y, the probability of crossing (using Problem 1.12) is
P(y)=/integraldisplay−y
−lρ(x)dx+/integraldisplayl
l−yρ(x)dx=1
π/braceleftBigg/integraldisplay−y
−l1√
l2−x2dx+/integraldisplayl
l−y1√
l2−x2dx/bracerightBigg
=1
π/braceleftbigg
sin−1/parenleftBigx
l/parenrightBig/vextendsingle/vextendsingle/vextendsingle−y
−l+ sin−1/parenleftBigx
l/parenrightBig/vextendsingle/vextendsingle/vextendsinglel
l−y/bracerightbigg
=1
π/bracketleftbig
−sin−1(y/l)+2s i n−1(1)−sin−1(1−y/l)/bracketrightbig
=1−sin−1(y/l)
π−sin−1(1−y/l)
π.
Now, all values of yare equally likely, so ρ(y)=1/l, and hence the probability of crossing is
P=1
πl/integraldisplayl
0/bracketleftbigg
π−sin−1/parenleftBigy
l/parenrightBig
−sin−1/parenleftbiggl−y
l/parenrightbigg/bracketrightbigg
dy=1
πl/integraldisplayl
0/bracketleftbig
π−2 sin−1(y/l)/bracketrightbig
dy
=1
πl/bracketleftbigg
πl−2/parenleftBig
ysin−1(y/l)+l/radicalbig
1−(y/l)2/parenrightBig/vextendsingle/vextendsingle/vextendsinglel
0/bracketrightbigg
=1−2
πl[lsin−1(1)−l]=1−1+2
π=2
π.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 1. THE WAVE FUNCTION 11
Problem 1.14
(a)Pab(t)=/integraltextb
a|Ψ(x,t)2dx, sodPab
dt=/integraltextb
a∂
∂t|Ψ|2dx.But (Eq. 1.25):
∂|Ψ|2
∂t=∂
∂x/bracketleftbiggi/planckover2pi1
2m/parenleftbigg
Ψ∗∂Ψ
∂x−∂Ψ∗
∂xΨ/parenrightbigg/bracketrightbigg
=−∂
∂tJ(x,t).
∴dPab
dt=−/integraldisplayb
a∂
∂xJ(x,t)dx=−[J(x,t)]|b
a=J(a,t)−J(b,t).QED
Probability is dimensionless, so Jhas the dimensions 1/time, and units seconds−1.
(b)Here Ψ( x,t)=f(x)e−iat, where f(x)≡Ae−amx2//planckover2pi1,s oΨ∂Ψ∗
∂x=fe−iatdf
dxeiat=fdf
dx,
and Ψ∗∂Ψ
∂x=fdf
dxtoo, so J(x,t)=0 .
Problem 1.15
(a)Eq. 1.24 now reads∂Ψ∗
∂t=−i/planckover2pi1
2m∂2Ψ∗
∂x2+i
/planckover2pi1V∗Ψ∗, and Eq. 1.25 picks up an extra term:
∂
∂t|Ψ|2=···+i
/planckover2pi1|Ψ|2(V∗−V)=···+i
/planckover2pi1|Ψ|2(V0+iΓ−V0+iΓ) =···−2Γ
/planckover2pi1|Ψ|2,
and Eq. 1.27 becomesdP
dt=−2Γ
/planckover2pi1/integraltext∞
−∞|Ψ|2dx=−2Γ
/planckover2pi1P. QED
(b)
dP
P=−2Γ
/planckover2pi1dt=⇒lnP=−2Γ
/planckover2pi1t+ constant =⇒P(t)=P(0)e−2Γt//planckover2pi1,soτ=/planckover2pi1
2Γ.
Problem 1.16
Use Eqs. [1.23] and [1.24], and integration by parts:
d
dt/integraldisplay∞
−∞Ψ∗
1Ψ2dx=/integraldisplay∞
−∞∂
∂t(Ψ∗
1Ψ2)dx=/integraldisplay∞
−∞/parenleftbigg∂Ψ∗
1
∂tΨ2+Ψ∗
1∂Ψ2
∂t/parenrightbigg
dx
=/integraldisplay∞
−∞/bracketleftbigg/parenleftbigg−i/planckover2pi1
2m∂2Ψ∗
1
∂x2+i
/planckover2pi1VΨ∗
1/parenrightbigg
Ψ2+Ψ∗
1/parenleftbiggi/planckover2pi1
2m∂2Ψ2
∂x2−i
/planckover2pi1VΨ2/parenrightbigg/bracketrightbigg
dx
=−i/planckover2pi1
2m/integraldisplay∞
−∞/parenleftbigg∂2Ψ∗
1
∂x2Ψ2−Ψ∗
1∂2Ψ2
∂x2/parenrightbigg
dx
=−i/planckover2pi1
2m/bracketleftBigg
∂Ψ∗
1
∂xΨ2/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞−/integraldisplay∞
−∞∂Ψ∗
1
∂x∂Ψ2
∂xdx−Ψ∗
1∂Ψ2
∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞+/integraldisplay∞
−∞∂Ψ∗
1
∂x∂Ψ2
∂xdx/bracketrightBigg
=0.QED
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12 CHAPTER 1. THE WAVE FUNCTION
Problem 1.17
(a)
1=|A|2/integraldisplaya
−a/parenleftbig
a2−x2/parenrightbig2dx=2|A|2/integraldisplaya
0/parenleftbig
a4−2a2x2+x4/parenrightbig
dx=2|A|2/bracketleftbigg
a4x−2a2x3
3+x5
5/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0
=2|A|2a5/parenleftbigg
1−2
3+1
5/parenrightbigg
=16
15a5|A|2,soA=/radicalbigg
15
16a5.
(b)
/angbracketleftx/angbracketright=/integraldisplaya
−ax|Ψ|2dx=0.(Odd integrand.)
(c)
/angbracketleftp/angbracketright=/planckover2pi1
iA2/integraldisplaya
−a/parenleftbig
a2−x2/parenrightbigd
dx/parenleftbig
a2−x2/parenrightbig
/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
−2xdx=0.(Odd integrand.)
Since we only know /angbracketleftx/angbracketrightatt= 0 we cannot calculate d/angbracketleftx/angbracketright/dtdirectly.
(d)
/angbracketleftx2/angbracketright=A2/integraldisplaya
−ax2/parenleftbig
a2−x2/parenrightbig2dx=2A2/integraldisplaya
0/parenleftbig
a4x2−2a2x4+x6/parenrightbig
dx
=215
16a5/bracketleftbigg
a4x3
3−2a2x5
5+x7
7/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0=15
8a5/parenleftbig
a7/parenrightbig/parenleftbigg1
3−2
5+1
7/parenrightbigg
=✚✚15a2
8/parenleftbigg35−4 2+1 5
✁3·✁5·7/parenrightbigg
=a2
8·8
7=a2
7.
(e)
/angbracketleftp2/angbracketright=−A2/planckover2pi12/integraldisplaya
−a/parenleftbig
a2−x2/parenrightbigd2
dx2/parenleftbig
a2−x2/parenrightbig
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
−2dx=2A2/planckover2pi122/integraldisplaya
0/parenleftbig
a2−x2/parenrightbig
dx
=4·15
16a5/planckover2pi12/parenleftbigg
a2x−x3
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0=15/planckover2pi12
4a5/parenleftbigg
a3−a3
3/parenrightbigg
=15/planckover2pi12
4a2·2
3=5
2/planckover2pi12
a2.
(f)
σx=/radicalbig
/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=/radicalbigg
1
7a2=a√
7.
(g)
σp=/radicalbig
/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/radicalbigg
5
2/planckover2pi12
a2=/radicalbigg
5
2/planckover2pi1
a.
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CHAPTER 1. THE WAVE FUNCTION 13
(h)
σxσp=a√
7·/radicalbigg
5
2/planckover2pi1
a=/radicalbigg
5
14/planckover2pi1=/radicalbigg
10
7/planckover2pi1
2>/planckover2pi1
2./check
Problem 1.18
h√3mkBT>d⇒T<h2
3mkBd2.
(a)Electrons ( m=9.1×10−31kg):
T<(6.6×10−34)2
3(9.1×10−31)(1.4×10−23)(3×10−10)2=1.3×105K.
Sodium nuclei ( m=2 3mp= 23(1.7×10−27)=3.9×10−26kg):
T<(6.6×10−34)2
3(3.9×10−26)(1.4×10−23)(3×10−10)2=3.0K .
(b)PV=NkBT; volume occupied by one molecule ( N=1,V=d3)⇒d=(kBT/P)1/3.
T<h2
2mkB/parenleftbiggP
kBT/parenrightbigg2/3
⇒T5/3<h2
3mP2/3
k5/3
B⇒T<1
kB/parenleftbiggh2
3m/parenrightbigg3/5
P2/5.
For helium ( m=4mp=6.8×10−27kg) at 1 atm = 1 .0×105N/m2:
T<1
(1.4×10−23)/parenleftbigg(6.6×10−34)2
3(6.8×10−27)/parenrightbigg3/5
(1.0×105)2/5=2.8 K.
For hydrogen ( m=2mp=3.4×10−27kg) with d=0.01 m:
T<(6.6×10−34)2
3(3.4×10−27)(1.4×10−23)(10−2)2=3.1×10−14K.
At 3 K it is definitely in the classical regime.
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14 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Chapter 2
Time-IndependentSchr¨odinger
Equation
Problem 2.1
(a)
Ψ(x,t)=ψ(x)e−i(E0+iΓ)t//planckover2pi1=ψ(x)eΓt//planckover2pi1e−iE0t//planckover2pi1=⇒|Ψ|2=|ψ|2e2Γt//planckover2pi1.
/integraldisplay∞
−∞|Ψ(x,t)|2dx=e2Γt//planckover2pi1/integraldisplay∞
−∞|ψ|2dx.
The second term is independent of t, so if the product is to be 1 for all time, the first term ( e2Γt//planckover2pi1) must
also be constant, and hence Γ = 0. QED
(b)Ifψsatisfies Eq. 2.5, −/planckover2pi12
2m∂2ψ
dx2+Vψ=Eψ, then (taking the complex conjugate and noting that Vand
Eare real):−/planckover2pi12
2m∂2ψ∗
dx2+Vψ∗=Eψ∗,s oψ∗alsosatisfies Eq. 2.5. Now, if ψ1andψ2satisfy Eq. 2.5, so
too does any linear combination of them ( ψ3≡c1ψ1+c2ψ2):
−/planckover2pi12
2m∂2ψ3
dx2+Vψ3=−/planckover2pi12
2m/parenleftbigg
c1∂2ψ1
dx2+c2∂2ψ2
∂x2/parenrightbigg
+V(c1ψ1+c2ψ2)
=c1/bracketleftbigg
−/planckover2pi12
2md2ψ1
dx2+Vψ1/bracketrightbigg
+c2/bracketleftbigg
−/planckover2pi12
2md2ψ2
dx2+Vψ2/bracketrightbigg
=c1(Eψ1)+c2(Eψ2)=E(c1ψ1+c2ψ2)=Eψ3.
Thus, (ψ+ψ∗) andi(ψ−ψ∗) – both of which are real– satisfy Eq. 2.5. Conclusion: From any complex
solution, we can always construct two realsolutions (of course, if ψis already real, the second one will be
zero). In particular, since ψ=1
2[(ψ+ψ∗)−i(i(ψ−ψ∗))],ψcan be expressed as a linear combination of
two real solutions. QED
(c)Ifψ(x) satisfies Eq. 2.5, then, changing variables x→−xand noting that ∂2/∂(−x)2=∂2/∂x2,
−/planckover2pi12
2m∂2ψ(−x)
dx2+V(−x)ψ(−x)=Eψ(−x);
so ifV(−x)=V(x) thenψ(−x)alsosatisfies Eq. 2.5. It follows that ψ+(x)≡ψ(x)+ψ(−x) (which is
even:ψ+(−x)=ψ+(x)) andψ−(x)≡ψ(x)−ψ(−x) (which is odd:ψ−(−x)=−ψ−(x)) both satisfy Eq.
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publisher.
CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 15
2.5. But ψ(x)=1
2(ψ+(x)+ψ−(x)), so any solution can be expressed as a linear combination of even and
odd solutions. QED
Problem 2.2
Givend2ψ
dx2=2m
/planckover2pi12[V(x)−E]ψ,i fE<V min, thenψ/prime/primeandψalways have the same sign: If ψis positive(negative),
thenψ/prime/primeis also positive(negative). This means that ψalways curves awayfrom the axis (see Figure). However,
it has got to go to zero as x→−∞ (else it would not be normalizable). At some point it’s got to depart from
zero (if it doesn’t , it’s going to be identically zero everywhere ), in (say) the positive direction. At this point its
slope is positive, and increasing ,s oψgets bigger and bigger as xincreases. It can’t ever “turn over” and head
back toward the axis, because that would requuire a negative second derivative—it always has to bend awayfrom the axis. By the same token, if it starts out heading negative, it just runs more and more negative. Inneither case is there any way for it to come back to zero, as it must (at x→∞) in order to be normalizable.
QED
xψ
Problem 2.3
Equation 2.20 saysd2ψ
dx2=−2mE
/planckover2pi12ψ; Eq. 2.23 says ψ(0) =ψ(a) = 0. If E=0 ,d2ψ/dx2=0 ,s o ψ(x)=A+Bx;
ψ(0) =A=0⇒ψ=Bx;ψ(a)=Ba=0⇒B=0 ,s o ψ=0 . I f E<0,d2ψ/dx2=κ2ψ, withκ≡√
−2mE/ /planckover2pi1
real, so ψ(x)=Aeκx+Be−κx. This time ψ(0) =A+B=0⇒B=−A,s oψ=A(eκx−e−κx), while
ψ(a)=A/parenleftbig
eκa−eiκa/parenrightbig
=0⇒eitherA=0 ,s o ψ= 0, or else eκa=e−κa,s oe2κa=1 ,s o2 κa= ln(1) = 0,
soκ= 0, and again ψ= 0. In all cases, then, the boundary conditions force ψ= 0, which is unacceptable
(non-normalizable).
Problem 2.4
/angbracketleftx/angbracketright=/integraldisplay
x|ψ|2dx=2
a/integraldisplaya
0xsin2/parenleftBignπ
ax/parenrightBig
dx. Lety≡nπ
ax,sodx=a
nπdy;y:0→nπ.
=2
a/parenleftBiga
nπ/parenrightBig2/integraldisplaynπ
0ysin2ydy=2a
n2π2/bracketleftbiggy2
4−ysin 2y
4−cos 2y
8/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglenπ
0
=2a
n2π2/bracketleftbiggn2π2
4−cos 2nπ
8+1
8/bracketrightbigg
=a
2.(Independent of n.)
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16 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
/angbracketleftx2/angbracketright=2
a/integraldisplaya
0x2sin2/parenleftBignπ
ax/parenrightBig
dx=2
a/parenleftBiga
nπ/parenrightBig3/integraldisplaynπ
0y2sin2ydy
=2a2
(nπ)3/bracketleftbiggy3
6−/parenleftbiggy3
4−1
8/parenrightbigg
sin 2y−ycos 2y
4/bracketrightbiggnπ
0
=2a2
(nπ)3/bracketleftbigg(nπ)3
6−nπcos(2nπ)
4/bracketrightbigg
=a2/bracketleftbigg1
3−1
2(nπ)2/bracketrightbigg
.
/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=0.(Note:E q.1.33 is much faster than Eq .1.35.)
/angbracketleftp2/angbracketright=/integraldisplay
ψ∗
n/parenleftbigg/planckover2pi1
id
dx/parenrightbigg2
ψndx=−/planckover2pi12/integraldisplay
ψ∗
n/parenleftbiggd2ψn
dx2/parenrightbigg
dx
=(−/planckover2pi12)/parenleftbigg
−2mEn
/planckover2pi12/parenrightbigg/integraldisplay
ψ∗
nψndx=2mEn=/parenleftbiggnπ/planckover2pi1
a/parenrightbigg2
.
σ2
x=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=a2/parenleftbigg1
3−1
2(nπ)2−1
4/parenrightbigg
=a2
4/parenleftbigg1
3−2
(nπ)2/parenrightbigg
;σx=a
2/radicalBigg
1
3−2
(nπ)2.
σ2
p=/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/parenleftbiggnπ/planckover2pi1
a/parenrightbigg2
;σp=nπ/planckover2pi1
a.∴σxσp=/planckover2pi1
2/radicalbigg
(nπ)2
3−2.
The product σxσpissmallest for n=1 ; in that case, σxσp=/planckover2pi1
2/radicalBig
π2
3−2=( 1.136)/planckover2pi1/2>/planckover2pi1/2./check
Problem 2.5
(a)
|Ψ|2=Ψ2Ψ=|A|2(ψ∗
1+ψ∗
2)(ψ1+ψ2)=|A|2[ψ∗
1ψ1+ψ∗
1ψ2+ψ∗
2ψ1+ψ∗
2ψ2].
1=/integraldisplay
|Ψ|2dx=|A|2/integraldisplay
[|ψ1|2+ψ∗
1ψ2+ψ∗
2ψ1+|ψ2|2]dx=2|A|2⇒A=1/√
2.
(b)
Ψ(x,t)=1√
2/bracketleftBig
ψ1e−iE1t//planckover2pi1+ψ2e−iE2t//planckover2pi1/bracketrightBig
(butEn
/planckover2pi1=n2ω)
=1√
2/radicalbigg
2
a/bracketleftbigg
sin/parenleftBigπ
ax/parenrightBig
e−iωt+ sin/parenleftbigg2π
ax/parenrightbigg
e−i4ωt/bracketrightbigg
=1√ae−iωt/bracketleftbigg
sin/parenleftBigπ
ax/parenrightBig
+ sin/parenleftbigg2π
ax/parenrightbigg
e−3iωt/bracketrightbigg
.
|Ψ(x,t)|2=1
a/bracketleftbigg
sin2/parenleftBigπ
ax/parenrightBig
+ sin/parenleftBigπ
ax/parenrightBig
sin/parenleftbigg2π
ax/parenrightbigg/parenleftbig
e−3iωt+e3iωt/parenrightbig
+ sin2/parenleftbigg2π
ax/parenrightbigg/bracketrightbigg
=1
a/bracketleftbigg
sin2/parenleftBigπ
ax/parenrightBig
+ sin2/parenleftbigg2π
ax/parenrightbigg
+ 2 sin/parenleftBigπ
ax/parenrightBig
sin/parenleftbigg2π
ax/parenrightbigg
cos(3ωt)/bracketrightbigg
.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 17
(c)
/angbracketleftx/angbracketright=/integraldisplay
x|Ψ(x,t)|2dx
=1
a/integraldisplaya
0x/bracketleftbigg
sin2/parenleftBigπ
ax/parenrightBig
+ sin2/parenleftbigg2π
ax/parenrightbigg
+ 2 sin/parenleftBigπ
ax/parenrightBig
sin/parenleftbigg2π
ax/parenrightbigg
cos(3ωt)/bracketrightbigg
dx
/integraldisplaya
0xsin2/parenleftBigπ
ax/parenrightBig
dx=/bracketleftBigg
x2
4−xsin/parenleftbig2π
ax/parenrightbig
4π/a−cos/parenleftbig2π
ax/parenrightbig
8(π/a)2/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0=a2
4=/integraldisplaya
0xsin2/parenleftbigg2π
ax/parenrightbigg
dx.
/integraldisplaya
0xsin/parenleftBigπ
ax/parenrightBig
sin/parenleftbigg2π
ax/parenrightbigg
dx=1
2/integraldisplaya
0x/bracketleftbigg
cos/parenleftBigπ
ax/parenrightBig
−cos/parenleftbigg3π
ax/parenrightbigg/bracketrightbigg
dx
=1
2/bracketleftbigga2
π2cos/parenleftBigπ
ax/parenrightBig
+ax
πsin/parenleftBigπ
ax/parenrightBig
−a2
9π2cos/parenleftbigg3π
ax/parenrightbigg
−ax
3πsin/parenleftbigg3π
ax/parenrightbigg/bracketrightbigga
0
=1
2/bracketleftbigga2
π2/parenleftbig
cos(π)−cos(0)/parenrightbig
−a2
9π2/parenleftbig
cos(3π)−cos(0)/parenrightbig/bracketrightbigg
=−a2
π2/parenleftbigg
1−1
9/parenrightbigg
=−8a2
9π2.
∴/angbracketleftx/angbracketright=1
a/bracketleftbigga2
4+a2
4−16a2
9π2cos(3ωt)/bracketrightbigg
=a
2/bracketleftbigg
1−32
9π2cos(3ωt)/bracketrightbigg
.
Amplitude:32
9π2/parenleftBiga
2/parenrightBig
=0.3603(a/2); angular frequency: 3ω=3π2/planckover2pi1
2ma2.
(d)
/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=m/parenleftBiga
2/parenrightBig/parenleftbigg
−32
9π2/parenrightbigg
(−3ω) sin(3ωt)=8/planckover2pi1
3asin(3ωt).
(e)You could get either E1=π2/planckover2pi12/2ma2orE2=2π2/planckover2pi12/ma2,with equal probability P1=P2=1/2.
So/angbracketleftH/angbracketright=1
2(E1+E2)=5π2/planckover2pi12
4ma2;it’s the average ofE1andE2.
Problem 2.6
From Problem 2.5, we see that
Ψ(x,t)=1√ae−iωt/bracketleftbig
sin/parenleftbigπ
ax/parenrightbig
+ sin/parenleftbig2π
ax/parenrightbig
e−3iωteiφ/bracketrightbig
;
|Ψ(x,t)|2=1
a/bracketleftbig
sin2/parenleftbigπ
ax/parenrightbig
+ sin2/parenleftbig2π
ax/parenrightbig
+ 2 sin/parenleftbigπ
ax/parenrightbig
sin/parenleftbig2π
ax/parenrightbig
cos(3ωt−φ)/bracketrightbig
;
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18 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
and hence/angbracketleftx/angbracketright=a
2/bracketleftbig
1−32
9π2cos(3ωt−φ)/bracketrightbig
.This amounts physically to starting the clock at a different time
(i.e., shifting the t= 0 point).
Ifφ=π
2,so Ψ(x,0) =A[ψ1(x)+iψ2(x)],then cos(3 ωt−φ) = sin(3 ωt);/angbracketleftx/angbracketrightstarts ata
2.
Ifφ=π,so Ψ(x,0) =A[ψ1(x)−ψ2(x)],then cos(3 ωt−φ)=−cos(3ωt);/angbracketleftx/angbracketrightstarts ata
2/parenleftbigg
1+32
9π2/parenrightbigg
.
Problem 2.7
Ψ(x,0)
xa a/2Aa/2
(a)
1=A2/integraldisplaya/2
0x2dx+A2/integraldisplaya
a/2(a−x)2dx=A2/bracketleftbiggx3
3/vextendsingle/vextendsingle/vextendsingle/vextendsinglea/2
0−(a−x)3
3/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
a/2/bracketrightbigg
=A2
3/parenleftbigga3
8+a3
8/parenrightbigg
=A2a3
12⇒A=2√
3√
a3.
(b)
cn=/radicalbigg
2
a2√
3
a√a/bracketleftbigg/integraldisplaya/2
0xsin/parenleftbiggnπ
ax/parenrightbigg
dx+/integraldisplaya
a/2(a−x) sin/parenleftbiggnπ
ax/parenrightbigg
dx/bracketrightbigg
=2√
6
a2/braceleftbigg/bracketleftbigg/parenleftbigga
nπ/parenrightbigg2
sin/parenleftbiggnπ
ax/parenrightbigg
−xa
nπcos/parenleftbiggnπ
ax/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea/2
0
+a/bracketleftbigg
−a
nπcos/parenleftbiggnπ
ax/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
a/2−/bracketleftbigg/parenleftbigga
nπ/parenrightbigg2
sin/parenleftbiggnπ
ax/parenrightbigg
−/parenleftbiggax
nπ/parenrightbigg
cos/parenleftbiggnπ
ax/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
a/2/bracerightbigg
=2√
6
a2/bracketleftbigg/parenleftbigga
nπ/parenrightbigg2
sin/parenleftbiggnπ
2/parenrightbigg
−✘✘✘✘✘✘✘ a2
2nπcos/parenleftbiggnπ
2/parenrightbigg
−✟✟✟✟✟a2
nπcosnπ+
✟✟✟✟✟✟✟a2
nπcos/parenleftbiggnπ
2/parenrightbigg
+/parenleftbigga
nπ/parenrightbigg2
sin/parenleftbiggnπ
2/parenrightbigg
+✟✟✟✟✟a2
nπcosnπ−✘✘✘✘✘✘✘ a2
2nπcos/parenleftbiggnπ
2/parenrightbigg/bracketrightbigg
=2√
6
a22 a2
(nπ)2sin/parenleftbiggnπ
2/parenrightbigg
=4√
6
(nπ)2sin/parenleftbiggnπ
2/parenrightbigg
=/braceleftBigg
0,n even,
(−1)(n−1)/24√
6
(nπ)2,nodd.
SoΨ(x,t)=4√
6
π2/radicalbigg
2
a/summationdisplay
n=1,3,5,...(−1)(n−1)/21
n2sin/parenleftbiggnπ
ax/parenrightbigg
e−Ent//planckover2pi1,whereEn=n2π2/planckover2pi12
2ma2.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 19
(c)
P1=|c1|2=16·6
π4=0.9855.
(d)
/angbracketleftH/angbracketright=/summationdisplay
|cn|2En=96
π4π2/planckover2pi12
2ma2/parenleftbigg1
1+1
32+1
52+1
72+···
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
π2/8/parenrightbigg
=48/planckover2pi12
π2ma2π2
8=6/planckover2pi12
ma2.
Problem 2.8
(a)
Ψ(x,0) =/braceleftBigg
A,0<x<a / 2;
0,otherwise .1=A2/integraldisplaya/2
0dx=A2(a/2)⇒A=/radicalbigg
2
a.
(b)From Eq. 2.37,
c1=A/radicalbigg
2
a/integraldisplaya/2
0sin/parenleftBigπ
ax/parenrightBig
dx=2
a/bracketleftBig
−a
πcos/parenleftBigπ
ax/parenrightBig/bracketrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsinglea/2
0=−2
π/bracketleftBig
cos/parenleftBigπ
2/parenrightBig
−cos 0/bracketrightBig
=2
π.
P1=|c1|2=(2/π)2=0.4053.
Problem 2.9
ˆHΨ(x,0) =−/planckover2pi12
2m∂2
∂x2[Ax(a−x)] =−A/planckover2pi12
2m∂
∂x(a−2x)=A/planckover2pi12
m.
/integraldisplay
Ψ(x,0)∗ˆHΨ(x,0)dx=A2/planckover2pi12
m/integraldisplaya
0x(a−x)dx=A2/planckover2pi12
m/parenleftbigg
ax2
2−x3
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0
=A2/planckover2pi12
m/parenleftbigga3
2−a3
3/parenrightbigg
=30
a5/planckover2pi12
ma3
6=5/planckover2pi12
ma2
(same as Example 2.3).
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20 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.10
(a)Using Eqs. 2.47 and 2.59,
a+ψ0=1√
2/planckover2pi1mω/parenleftbigg
−/planckover2pi1d
dx+mωx/parenrightbigg/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
e−mω
2/planckover2pi1x2
=1√
2/planckover2pi1mω/parenleftBigmω
π/planckover2pi1/parenrightBig1/4/bracketleftBig
−/planckover2pi1/parenleftBig
−mω
2/planckover2pi1/parenrightBig
2x+mωx/bracketrightBig
e−mω
2/planckover2pi1x2=1√
2/planckover2pi1mω/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
2mωxe−mω
2/planckover2pi1x2.
(a+)2ψ0=1
2/planckover2pi1mω/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
2mω/parenleftbigg
−/planckover2pi1d
dx+mωx/parenrightbigg
xe−mω
2/planckover2pi1x2
=1
/planckover2pi1/parenleftBigmω
π/planckover2pi1/parenrightBig1/4/bracketleftBig
−/planckover2pi1/parenleftBig
1−xmω
2/planckover2pi12x/parenrightBig
+mωx2/bracketrightBig
e−mω
2/planckover2pi1x2=/parenleftBigmω
π/planckover2pi1/parenrightBig1/4/parenleftbigg2mω
/planckover2pi1x2−1/parenrightbigg
e−mω
2/planckover2pi1x2.
Therefore, from Eq. 2.67,
ψ2=1√
2(a+)2ψ0=1√
2/parenleftBigmω
π/planckover2pi1/parenrightBig1/4/parenleftbigg2mω
/planckover2pi1x2−1/parenrightbigg
e−mω
2/planckover2pi1x2.
(b)
ψψ ψ 12 0
(c)Sinceψ0andψ2are even, whereas ψ1is odd,/integraltext
ψ∗
0ψ1dxand/integraltext
ψ∗
2ψ1dxvanish automatically. The only one
we need to check is/integraltext
ψ∗
2ψ0dx:
/integraldisplay
ψ∗
2ψ0dx=1√
2/radicalbiggmω
π/planckover2pi1/integraldisplay∞
−∞/parenleftbigg2mω
/planckover2pi1x2−1/parenrightbigg
e−mω
/planckover2pi1x2dx
=−/radicalbiggmω
2π/planckover2pi1/parenleftbigg/integraldisplay∞
−∞e−mω
/planckover2pi1x2dx−2mω
/planckover2pi1/integraldisplay∞
−∞x2e−mω
/planckover2pi1x2dx/parenrightbigg
=−/radicalbiggmω
2π/planckover2pi1/parenleftbigg/radicalbigg
π/planckover2pi1
mω−2mω
/planckover2pi1/planckover2pi1
2mω/radicalbigg
π/planckover2pi1
mω/parenrightbigg
=0./check
Problem 2.11
(a)Note that ψ0is even, and ψ1is odd. In either case |ψ|2is even, so/angbracketleftx/angbracketright=/integraltext
x|ψ|2dx=0.Therefore
/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright/dt=0.(These results hold for anystationary state of the harmonic oscillator.)
From Eqs. 2.59 and 2.62, ψ0=αe−ξ2/2,ψ1=√
2αξe−ξ2/2.S o
n=0:
/angbracketleftx2/angbracketright=α2/integraldisplay∞
−∞x2e−ξ2/2dx=α2/parenleftbigg/planckover2pi1
mω/parenrightbigg3/2/integraldisplay∞
−∞ξ2e−ξ2dξ=1√π/parenleftbigg/planckover2pi1
mω/parenrightbigg√π
2=/planckover2pi1
2mω.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 21
/angbracketleftp2/angbracketright=/integraldisplay
ψ0/parenleftbigg/planckover2pi1
id
dx/parenrightbigg2
ψ0dx=−/planckover2pi12α2/radicalbiggmω
/planckover2pi1/integraldisplay∞
−∞e−ξ2/2/parenleftbiggd2
dξ2e−ξ2/2/parenrightbigg
dξ
=−m/planckover2pi1ω√π/integraldisplay∞
−∞/parenleftbig
ξ2−1/parenrightbig
e−ξ2/2dξ=−m/planckover2pi1ω√π/parenleftbigg√π
2−√π/parenrightbigg
=m/planckover2pi1ω
2.
n=1:
/angbracketleftx2/angbracketright=2α2/integraldisplay∞
−∞x2ξ2e−ξ2dx=2α2/parenleftbigg/planckover2pi1
mω/parenrightbigg3/2/integraldisplay∞
−∞ξ4e−ξ2dξ=2/planckover2pi1√πmω3√π
4=3/planckover2pi1
2mω.
/angbracketleftp2/angbracketright=−/planckover2pi122α2/radicalbiggmω
/planckover2pi1/integraldisplay∞
−∞ξe−ξ2/2/bracketleftbiggd2
dξ2/parenleftbig
ξe−ξ2/2/parenrightbig/bracketrightbigg
dξ
=−2mω/planckover2pi1√π/integraldisplay∞
−∞/parenleftbig
ξ4−3ξ2/parenrightbig
e−ξ2dξ=−2mω/planckover2pi1√π/parenleftbigg3
4√π−3√π
2/parenrightbigg
=3m/planckover2pi1ω
2.
(b)n=0:
σx=/radicalbig
/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=/radicalbigg
/planckover2pi1
2mω;σp=/radicalbig
/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/radicalbigg
m/planckover2pi1ω
2;
σxσp=/radicalbigg
/planckover2pi1
2mω/radicalbigg
mω/planckover2pi1
2=/planckover2pi1
2.(Right atthe uncertainty limit.) /check
n=1:
σx=/radicalbigg
3/planckover2pi1
2mω;σp=/radicalbigg
3m/planckover2pi1ω
2;σxσp=3/planckover2pi1
2>/planckover2pi1
2./check
(c)
/angbracketleftT/angbracketright=1
2m/angbracketleftp2/angbracketright=
1
4/planckover2pi1ω(n=0 )
3
4/planckover2pi1ω(n=1 )
;/angbracketleftV/angbracketright=1
2mω2/angbracketleftx2/angbracketright=
1
4/planckover2pi1ω(n=0 )
3
4/planckover2pi1ω(n=1 )
.
/angbracketleftT/angbracketright+/angbracketleftV/angbracketright=/angbracketleftH/angbracketright=
1
2/planckover2pi1ω(n=0 )= E0
3
2/planckover2pi1ω(n=1 )= E1
,as expected.
Problem 2.12
From Eq. 2.69,
x=/radicalbigg
/planckover2pi1
2mω(a++a−),p=i/radicalbigg
/planckover2pi1mω
2(a+−a−),
so
/angbracketleftx/angbracketright=/radicalbigg
/planckover2pi1
2mω/integraldisplay
ψ∗
n(a++a−)ψndx.
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22 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
But (Eq. 2.66)
a+ψn=√
n+1ψn+1,a −ψn=√nψn−1.
So
/angbracketleftx/angbracketright=/radicalbigg
/planckover2pi1
2mω/bracketleftbigg√
n+1/integraldisplay
ψ∗
nψn+1dx+√n/integraldisplay
ψ∗
nψn−1dx/bracketrightbigg
=0(by orthogonality) .
/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=0.x2=/planckover2pi1
2mω(a++a−)2=/planckover2pi1
2mω/parenleftbig
a2
++a+a−+a−a++a2
−/parenrightbig
.
/angbracketleftx2/angbracketright=/planckover2pi1
2mω/integraldisplay
ψ∗
n/parenleftbig
a2
++a+a−+a−a++a2
−/parenrightbig
ψn.But
a2
+ψn=a+/parenleftbig√n+1ψn+1/parenrightbig
=√n+1√n+2ψn+2=/radicalbig
(n+ 1)(n+2 )ψn+2.
a+a−ψn=a+/parenleftbig√nψn−1/parenrightbig
=√n√nψn =nψn.
a−a+ψn=a−/parenleftbig√n+1ψn+1/parenrightbig
=/radicalbig
n+1 )√n+1ψn=(n+1 )ψn.
a2
−ψn=a−/parenleftbig√nψn−1/parenrightbig
=√n√n−1ψn−2 =/radicalbig
(n−1)nψn−2.
So
/angbracketleftx2/angbracketright=/planckover2pi1
2mω/bracketleftbigg
0+n/integraldisplay
|ψn|2dx+(n+1 )/integraldisplay
|ψn|2dx+0/bracketrightbigg
=/planckover2pi1
2mω(2n+1 )=/parenleftbigg
n+1
2/parenrightbigg/planckover2pi1
mω.
p2=−/planckover2pi1mω
2(a+−a−)2=−/planckover2pi1mω
2/parenleftbig
a2
+−a+a−−a−a++a2
−/parenrightbig
⇒
/angbracketleftp2/angbracketright=−/planckover2pi1mω
2[0−n−(n+1 )+0 ]=/planckover2pi1mω
2(2n+1 )=/parenleftbigg
n+1
2/parenrightbigg
m/planckover2pi1ω.
/angbracketleftT/angbracketright=/angbracketleftp2/2m/angbracketright=1
2/parenleftbigg
n+1
2/parenrightbigg
/planckover2pi1ω.
σx=/radicalbig
/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=/radicalbigg
n+1
2/radicalbigg
/planckover2pi1
mω;σp=/radicalbig
/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/radicalbigg
n+1
2√
m/planckover2pi1ω;σxσp=/parenleftbigg
n+1
2/parenrightbigg
/planckover2pi1≥/planckover2pi1
2./check
Problem 2.13
(a)
1=/integraldisplay
|Ψ(x,0)|2dx=|A|2/integraldisplay/parenleftbig
9|ψ0|2+1 2ψ∗
0ψ1+1 2ψ∗
1ψ0+1 6|ψ1|2/parenrightbig
dx
=|A|2( 9+0+0+1 6 )=2 5 |A|2⇒A=1/5.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 23
(b)
Ψ(x,t)=1
5/bracketleftBig
3ψ0(x)e−iE0t//planckover2pi1+4ψ1(x)e−iE1t//planckover2pi1/bracketrightBig
=1
5/bracketleftBig
3ψ0(x)e−iωt/2+4ψ1(x)e−3iωt/2/bracketrightBig
.
(Hereψ0andψ1are given by Eqs. 2.59 and 2.62; E1andE2by Eq. 2.61.)
|Ψ(x,t)|2=1
25/bracketleftBig
9ψ2
0+1 2ψ0ψ1eiωt/2e−3iωt/2+1 2ψ0ψ1e−iωt/2e3iωt/2+1 6ψ2
1/bracketrightBig
=1
25/bracketleftbig
9ψ2
0+1 6ψ2
1+2 4ψ0ψ1cos(ωt)/bracketrightbig
.
(c)
/angbracketleftx/angbracketright=1
25/bracketleftbigg
9/integraldisplay
xψ2
0dx+1 6/integraldisplay
xψ2
1dx+ 24 cos( ωt)/integraldisplay
xψ0ψ1dx/bracketrightbigg
.
But/integraltext
xψ2
0dx=/integraltext
xψ2
1dx= 0 (see Problem 2.11 or 2.12), while
/integraldisplay
xψ0ψ1dx=/radicalbiggmω
π/planckover2pi1/radicalbigg
2mω
/planckover2pi1/integraldisplay
xe−mω
2/planckover2pi1x2xe−mω
2/planckover2pi1x2dx=/radicalbigg
2
π/parenleftBigmω
/planckover2pi1/parenrightBig/integraldisplay∞
−∞x2e−mω
/planckover2pi1x2dx
=/radicalbigg
2
π/parenleftBigmω
/planckover2pi1/parenrightBig
2√π2/parenleftBigg
1
2/radicalbigg
/planckover2pi1
mω/parenrightBigg3
=/radicalbigg
/planckover2pi1
2mω.
So
/angbracketleftx/angbracketright=24
25/radicalbigg
/planckover2pi1
2mωcos(ωt);/angbracketleftp/angbracketright=md
dt/angbracketleftx/angbracketright=−24
25/radicalbigg
mω/planckover2pi1
2sin(ωt).
(Withψ2in place of ψ1the frequency would be ( E2−E0)//planckover2pi1= [(5/2)/planckover2pi1ω−(1/2)/planckover2pi1ω]//planckover2pi1=2ω.)
Ehrenfest’s theorem says d/angbracketleftp/angbracketright/dt=−/angbracketleft∂V/∂x/angbracketright. Here
d/angbracketleftp/angbracketright
dt=−24
25/radicalbigg
mω/planckover2pi1
2ωcos(ωt),V =1
2mω2x2⇒∂V
∂x=mω2x,
so
−/angbracketleftbig∂V
∂x/angbracketrightbig
=−mω2/angbracketleftx/angbracketright=−mω224
25/radicalbigg
/planckover2pi1
2mωcos(ωt)=−24
25/radicalbigg
/planckover2pi1mω
2ωcos(ωt),
so Ehrenfest’s theorem holds.
(d)You could get E0=1
2/planckover2pi1ω,with probability |c0|2=9/25,orE1=3
2/planckover2pi1ω,with probability |c1|2=16/25.
Problem 2.14
The new allowed energies are E/prime
n=(n+1
2)/planckover2pi1ω/prime=2 (n+1
2)/planckover2pi1ω=/planckover2pi1ω,3/planckover2pi1ω,5/planckover2pi1ω,.... So the probability of
getting1
2/planckover2pi1ωiszero. The probability of getting /planckover2pi1ω(the new ground state energy) is P0=|c0|2, where c0=/integraltext
Ψ(x,0)ψ/prime
0dx, with
Ψ(x,0) =ψ0(x)=/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
e−mω
2/planckover2pi1x2,ψ 0(x)/prime=/parenleftbiggm2ω
π/planckover2pi1/parenrightbigg1/4
e−m2ω
2/planckover2pi1x2.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
24 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
So
c0=21/4/radicalbiggmω
π/planckover2pi1/integraldisplay∞
−∞e−3mω
2/planckover2pi1x2dx=21/4/radicalbiggmω
π/planckover2pi12√π/parenleftBigg
1
2/radicalbigg
2/planckover2pi1
3mω/parenrightBigg
=21/4/radicalbigg
2
3.
Therefore
P0=2
3√
2=0.9428.
Problem 2.15
ψ0=/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
e−ξ2/2,soP=2/radicalbiggmω
π/planckover2pi1/integraldisplay∞
x0e−ξ2dx=2/radicalbiggmω
π/planckover2pi1/radicalbigg
/planckover2pi1
mω/integraldisplay∞
ξ0e−ξ2dξ.
Classically allowed region extends out to:1
2mω2x2
0=E0=1
2/planckover2pi1ω,orx0=/radicalBig
/planckover2pi1
mω,soξ0=1.
P=2√π/integraldisplay∞
1e−ξ2dξ= 2(1−F(√
2)) (in notation of CRC Table) = 0.157.
Problem 2.16
n=5 :j=1⇒a3=−2(5−1)
(1+1)(1+2)a1=−4
3a1;j=3⇒a5=−2(5−3)
(3+1)(3+2)a3=−1
5a3=4
15a1;j=5⇒a7=0.So
H5(ξ)=a1ξ−4
3a1ξ3+4
15a1ξ5=a1
15(15ξ−20ξ3+4ξ5).By convention the coefficient of ξ5is 25,s oa1=1 5·8,
andH5(ξ) = 120ξ−160ξ3+3 2ξ5(which agrees with Table 2.1).
n=6 :j=0⇒a2=−2(6−0)
(0+1)(0+2)a0=−6a0;j=2⇒a4=−2(6−2)
(2+1)(2+2)a2=−2
3a2=4a0;j=4⇒a6=
−2(6−4)
(4+1)(4+2)a4=−2
15a4=−8
15a0;j=6⇒a8=0.SoH6(ξ)=a0−6a0ξ2+4a0ξ4−8
15ξ6a0.The coefficient of ξ6
is 26,s o26=−8
15a0⇒a0=−15·8=−120.H6(ξ)=−120 + 720 ξ2−480ξ4+6 4ξ6.
Problem 2.17
(a)
d
dξ(e−ξ2)=−2ξe−ξ2;/parenleftbiggd
dξ/parenrightbigg2
e−ξ2=d
dξ(−2ξe−ξ2)=(−2+4ξ2)e−ξ2;
/parenleftbiggd
dξ/parenrightbigg3
e−ξ2=d
dξ/bracketleftbigg
(−2+4ξ2)e−ξ2/bracketrightbigg
=/bracketleftbigg
8ξ+(−2+4ξ2)(−2ξ)/bracketrightbigg
e−ξ2= (12ξ−8ξ3)e−ξ2;
/parenleftbiggd
dξ/parenrightbigg4
e−ξ2=d
dξ/bracketleftbigg
(12ξ−8ξ3)e−ξ2/bracketrightbigg
=/bracketleftbigg
12−24ξ2+ (12ξ−8ξ3)(−2ξ)/bracketrightbigg
e−ξ2= (12−48ξ2+1 6ξ4)e−ξ2.
H3(ξ)=−eξ2/parenleftbiggd
dξ/parenrightbigg3
e−ξ2=−12ξ+8ξ3;H4(ξ)=eξ2/parenleftbiggd
dξ/parenrightbigg4
e−ξ2=12−48ξ2+1 6ξ4.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 25
(b)
H5=2ξH4−8H3=2ξ(12−48ξ2+1 6ξ4)−8(−12ξ+8ξ3)=120ξ−160ξ3+3 2ξ5.
H6=2ξH5−10H4=2ξ(120ξ−160ξ3+3 2ξ5)−10(12−48ξ2+1 6ξ4)=−120 + 720 ξ2−480ξ4+6 4ξ6.
(c)
dH5
dξ= 120−480ξ2+ 160ξ4= 10(12−48ξ2+1 6ξ4) = (2)(5) H4./check
dH6
dξ= 1440ξ−1920ξ3+ 384ξ5= 12(120 ξ−160ξ3+3 2ξ5) = (2)(6) H5./check
(d)
d
dz(e−z2+2zξ)=(−2z+ξ)e−z2+2zξ; setting z=0,H0(ξ)=2ξ.
/parenleftbiggd
dz/parenrightbigg2
(e−z2+2zξ)=d
dz/bracketleftbigg
(−2z+2ξ)e−z2+2zξ/bracketrightbigg
=/bracketleftbigg
−2+(−2z+2ξ)2/bracketrightbigg
e−z2+2zξ; setting z=0,H1(ξ)=−2+4ξ2.
/parenleftbiggd
dz/parenrightbigg3
(e−z2+2zξ)=d
dz/braceleftbigg/bracketleftbigg
−2+(−2z+2ξ)2/bracketrightbigg
e−z2+2zξ/bracerightbigg
=/braceleftbigg
2(−2z+2ξ)(−2) +/bracketleftbigg
−2+(−2z+2ξ)2/bracketrightbigg
(−2z+2ξ)/bracerightbigg
e−z2+2zξ;
setting z=0,H2(ξ)=−8ξ+(−2+4ξ2)(2ξ)=−12ξ+8ξ3.
Problem 2.18
Aeikx+Be−ikx=A(coskx+isinkx)+B(coskx−isinkx)=(A+B) coskx+i(A−B) sinkx
=Ccoskx+Dsinkx,withC=A+B;D=i(A−B).
Ccoskx+Dsinkx=C/parenleftbiggeikx+e−ikx
2/parenrightbigg
+D/parenleftbiggeikx−e−ikx
2i/parenrightbigg
=1
2(C−iD)eikx+1
2(C+iD)e−ikx
=Aeikx+Be−ikx,withA=1
2(C−iD);B=1
2(C+iD).
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26 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.19
Equation 2.94 says Ψ = Aei(kx−/planckover2pi1k2
2mt),s o
J=i/planckover2pi1
2m/parenleftbigg
Ψ∂Ψ∗
∂x−Ψ∗∂Ψ
∂x/parenrightbigg
=i/planckover2pi1
2m|A|2/bracketleftBig
ei(kx−/planckover2pi1k2
2mt)(−ik)e−i(kx−/planckover2pi1k2
2mt)−e−i(kx−/planckover2pi1k2
2mt)(ik)ei(kx−/planckover2pi1k2
2mt)/bracketrightBig
=i/planckover2pi1
2m|A|2(−2ik)=/planckover2pi1k
m|A|2.
It flows in the positive ( x) direction (as you would expect).
Problem 2.20
(a)
f(x)=b0+∞/summationdisplay
n=1an
2i/parenleftBig
einπx/a−e−inπx/a/parenrightBig
+∞/summationdisplay
n=1bn
2/parenleftBig
einπx/a+e−inπx/a/parenrightBig
=b0+∞/summationdisplay
n=1/parenleftbiggan
2i+bn
2/parenrightbigg
einπx/a+∞/summationdisplay
n=1/parenleftbigg
−an
2i+bn
2/parenrightbigg
e−inπx/a.
Let
c0≡b0;cn=1
2(−ian+bn),forn=1,2,3,...;cn≡1
2(ia−n+b−n),forn=−1,−2,−3,....
Thenf(x)=∞/summationdisplay
n=−∞cneinπx/a.QED
(b)
/integraldisplaya
−af(x)e−imπx/adx=∞/summationdisplay
n=−∞cn/integraldisplaya
−aei(n−m)πx/adx.But for n/negationslash=m,
/integraldisplaya
−aei(n−m)πx/adx=ei(n−m)πx/a
i(n−m)π/a/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
−a=ei(n−m)π−e−i(n−m)π
i(n−m)π/a=(−1)n−m−(−1)n−m
i(n−m)π/a=0,
whereas for n=m,
/integraldisplaya
−aei(n−m)πx/adx=/integraldisplaya
−adx=2a.
So all terms except n=mare zero, and
/integraldisplaya
−af(x)e−imπx/a=2acm,socn=1
2a/integraldisplaya
−af(x)e−inπx/adx.QED
(c)
f(x)=∞/summationdisplay
n=−∞/radicalbiggπ
21
aF(k)eikx=1√
2π/summationdisplay
F(k)eikx∆k,
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 27
where ∆k≡π
ais the increment in kfromnto (n+ 1).
F(k)=/radicalbigg
2
πa1
2a/integraldisplaya
−af(x)e−ikxdx=1√
2π/integraldisplaya
−af(x)e−ikxdx.
(d)Asa→∞,kbecomes a continuous variable,
f(x)=1√
2π/integraldisplay∞
−∞F(k)eikxdk;F(k)=1√
2π/integraldisplay∞
−∞f(x)eikxdx.
Problem 2.21
(a)
1=/integraldisplay∞
−∞|Ψ(x,0)|2dx=2|A|2/integraldisplay∞
0e−2axdx=2|A|2e−2ax
−2a/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0=|A|2
a⇒A=√a.
(b)
φ(k)=A√
2π/integraldisplay∞
−∞e−a|x|e−ikxdx=A√
2π/integraldisplay∞
−∞e−a|x|(coskx−isinkx)dx.
The cosine integrand is even, and the sine is odd, so the latter vanishes and
φ(k)=2A√
2π/integraldisplay∞
0e−axcoskxdx =A√
2π/integraldisplay∞
0e−ax/parenleftbig
eikx+e−ikx/parenrightbig
dx
=A√
2π/integraldisplay∞
0/parenleftbig
e(ik−a)x+e−(ik+a)x/parenrightbig
dx=A√
2π/bracketleftbigge(ik−a)x
ik−a+e−(ik+a)x
−(ik+a)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0
=A√
2π/parenleftbigg−1
ik−a+1
ik+a/parenrightbigg
=A√
2π−ik−a+ik−a
−k2−a2=/radicalbigga
2π2a
k2+a2.
(c)
Ψ(x,t)=1√
2π2/radicalbigg
a3
2π/integraldisplay∞
−∞1
k2+a2ei(kx−/planckover2pi1k2
2mt)dk=a3/2
π/integraldisplay∞
−∞1
k2+a2ei(kx−/planckover2pi1k2
2mt)dk.
(d)Forlargea,Ψ (x,0) is a sharp narrow spike whereas φ(k)∼=/radicalbig
2/πais broad and flat; position is well-
defined but momentum is ill-defined. For smalla,Ψ (x,0) is a broad and flat whereas φ(k)∼=(/radicalbig
2a3/π)/k2
is a sharp narrow spike; position is ill-defined but momentum is well-defined.
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28 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.22
(a)
1=|A|2/integraldisplay∞
−∞e−2ax2dx=|A|2/radicalbiggπ
2a;A=/parenleftbigg2a
π/parenrightbigg1/4
.
(b)
/integraldisplay∞
−∞e−(ax2+bx)dx=/integraldisplay∞
−∞e−y2+(b2/4a)1√ady=1√aeb2/4a/integraldisplay∞
−∞e−y2dy=/radicalbiggπ
aeb2/4a.
φ(k)=1√
2πA/integraldisplay∞
−∞e−ax2e−ikxdx=1√
2π/parenleftbigg2a
π/parenrightbigg1/4/radicalbiggπ
ae−k2/4a=1
(2πa)1/4e−k2/4a.
Ψ(x,t)=1√
2π1
(2πa)1/4/integraldisplay∞
−∞e−k2/4aei(kx−/planckover2pi1k2t/2m)
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
e−[(1
4a+i/planckover2pi1t/2m)k2−ixk]dk
=1√
2π(2πa)1/4√π/radicalBig
1
4a+i/planckover2pi1t/2me−x2/4(1
4a+i/planckover2pi1t/2m)=/parenleftbigg2a
π/parenrightbigg1/4e−ax2/(1+2i/planckover2pi1at/m)
/radicalbig
1+2i/planckover2pi1at/m.
(c)
Letθ≡2/planckover2pi1at/m. Then|Ψ|2=/radicalbigg
2a
πe−ax2/(1+iθ)e−ax2/(1−iθ)
/radicalbig
(1 +iθ)(1−iθ).The exponent is
−ax2
(1 +iθ)−ax2
(1−iθ)=−ax2(1−iθ+1+iθ)
(1 +iθ)(1−iθ)=−2ax2
1+θ2;|Ψ|2=/radicalbigg
2a
πe−2ax2/(1+θ2)
√
1+θ2.
Or, with w≡/radicalbigga
1+θ2,|Ψ|2=/radicalbigg
2
πwe−2w2x2.Astincreases, the graph of |Ψ|2flattens out and broadens.
|Ψ|2|Ψ|2
xxt = 0 t > 0
(d)
/angbracketleftx/angbracketright=/integraldisplay∞
−∞x|Ψ|2dx=0(odd integrand); /angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=0.
/angbracketleftx2/angbracketright=/radicalbigg
2
πw/integraldisplay∞
−∞x2e−2w2x2dx=/radicalbigg
2
πw1
4w2/radicalbiggπ
2w2=1
4w2./angbracketleftp2/angbracketright=−/planckover2pi12/integraldisplay∞
−∞Ψ∗d2Ψ
dx2dx.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 29
Write Ψ = Be−bx2,whereB≡/parenleftbigg2a
π/parenrightbigg1/41√
1+iθandb≡a
1+iθ.
d2Ψ
dx2=Bd
dx/parenleftBig
−2bxe−bx2/parenrightBig
=−2bB(1−2bx2)e−bx2.
Ψ∗d2Ψ
dx2=−2b|B|2(1−2bx2)e−(b+b∗)x2;b+b∗=a
1+iθ+a
1−iθ=2a
1+θ2=2w2.
|B|2=/radicalbigg
2a
π1√
1+θ2=/radicalbigg
2
πw.So Ψ∗d2Ψ
dx2=−2b/radicalbigg
2
πw(1−2bx2)e−2w2x2.
/angbracketleftp2/angbracketright=2b/planckover2pi12/radicalbigg
2
πw/integraldisplay∞
−∞(1−2bx2)e−2w2x2dx
=2b/planckover2pi12/radicalbigg
2
πw/parenleftbigg/radicalbiggπ
2w2−2b1
4w2/radicalbiggπ
2w2/parenrightbigg
=2b/planckover2pi12/parenleftbigg
1−b
2w2/parenrightbigg
.
But 1−b
2w2=1−/parenleftbigga
1+iθ/parenrightbigg/parenleftbigg1+θ2
2a/parenrightbigg
=1−(1−iθ)
2=1+iθ
2=a
2b,so
/angbracketleftp2/angbracketright=2b/planckover2pi12a
2b=/planckover2pi12a. σx=1
2w;σp=/planckover2pi1√a.
(e)
σxσp=1
2w/planckover2pi1√a=/planckover2pi1
2/radicalbig
1+θ2=/planckover2pi1
2/radicalbig
1+( 2 /planckover2pi1at/m)2≥/planckover2pi1
2./check
Closest at t=0,at which time it is right atthe uncertainty limit.
Problem 2.23
(a)
(−2)3−3(−2)2+2 (−2)−1=−8−12−4−1=−25.
(b)
cos(3π)+2=−1+2= 1.
(c)
0(x= 2 is outside the domain of integration) .
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30 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.24
(a)Lety≡cx,sodx=1
cdy./braceleftbigg
Ifc>0,y:−∞→∞ .
Ifc<0,y:∞→−∞ ./bracerightbigg
/integraldisplay∞
−∞f(x)δ(cx)dx=
1
c/integraltext∞
−∞f(y/c)δ(y)dy=1
cf(0) (c>0); or
1
c/integraltext−∞
∞f(y/c)δ(y)dy=−1
c/integraltext∞
−∞f(y/c)δ(y)dy=−1
cf(0) (c<0).
In either case,/integraldisplay∞
−∞f(x)δ(cx)dx=1
|c|f(0) =/integraldisplay∞
−∞f(x)1
|c|δ(x)dx.Soδ(cx)=1
|c|δ(x)./check
(b)
/integraldisplay∞
−∞f(x)dθ
dxdx=fθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞−/integraldisplay∞
−∞df
dxθdx (integration by parts)
=f(∞)−/integraldisplay∞
0df
dxdx=f(∞)−f(∞)+f(0) =f(0) =/integraldisplay∞
−∞f(x)δ(x)dx.
Sodθ/dx =δ(x)./check[Makes sense: The θfunction is constant (so derivative is zero) except at x= 0, where
the derivative is infinite.]
Problem 2.25
ψ(x)=√mα
/planckover2pi1e−mα|x|//planckover2pi12=√mα
/planckover2pi1/braceleftBigg
e−mαx/ /planckover2pi12,(x≥0),
emαx/ /planckover2pi12,(x≤0).
/angbracketleftx/angbracketright= 0 (odd integrand) .
/angbracketleftx2/angbracketright=/integraldisplay∞
−∞x2|ψ|2dx=2mα
/planckover2pi12/integraldisplay∞
0x2e−2mαx/ /planckover2pi12dx=2mα
/planckover2pi122/parenleftbigg/planckover2pi12
2mα/parenrightbigg3
=/planckover2pi14
2m2α2;σx=/planckover2pi12
√
2mα.
dψ
dx=√mα
/planckover2pi1
−mα
/planckover2pi12e−mαx/ /planckover2pi12,(x≥0)
mα
/planckover2pi12emαx/ /planckover2pi12,(x≤0)
=/parenleftbigg√mα
/planckover2pi1/parenrightbigg3/bracketleftBig
−θ(x)e−mαx/ /planckover2pi12+θ(−x)emαx/ /planckover2pi12/bracketrightBig
.
d2ψ
dx2=/parenleftbigg√mα
/planckover2pi1/parenrightbigg3/bracketleftBig
−δ(x)e−mαx/ /planckover2pi12+mα
/planckover2pi12θ(x)e−mαx/ /planckover2pi12−δ(−x)emαx/ /planckover2pi12+mα
/planckover2pi12θ(−x)emαx/ /planckover2pi12/bracketrightBig
=/parenleftbigg√mα
/planckover2pi1/parenrightbigg3/bracketleftBig
−2δ(x)+mα
/planckover2pi12e−mα|x|//planckover2pi12/bracketrightBig
.
In the last step I used the fact that δ(−x)=δ(x) (Eq. 2.142), f(x)δ(x)=f(0)δ(x) (Eq. 2.112), and θ(−x)+
θ(x) = 1 (Eq. 2.143). Since dψ/dx is an odd function, /angbracketleftp/angbracketright=0.
/angbracketleftp2/angbracketright=−/planckover2pi12/integraldisplay∞
−∞ψd2ψ
dx2dx=−/planckover2pi12√mα
/planckover2pi1/parenleftbigg√mα
/planckover2pi1/parenrightbigg3/integraldisplay∞
−∞e−mα|x|//planckover2pi12/bracketleftBig
−2δ(x)+mα
/planckover2pi12e−mα|x|//planckover2pi12/bracketrightBig
dx
=/parenleftBigmα
/planckover2pi1/parenrightBig2/bracketleftbigg
2−2mα
/planckover2pi12/integraldisplay∞
0e−2mαx/ /planckover2pi12dx/bracketrightbigg
=2/parenleftBigmα
/planckover2pi1/parenrightBig2/bracketleftbigg
1−mα
/planckover2pi12/planckover2pi12
2mα/bracketrightbigg
=/parenleftBigmα
/planckover2pi1/parenrightBig2
.
Evidently
σp=mα
/planckover2pi1,soσxσp=/planckover2pi12
√
2mαmα
/planckover2pi1=√
2/planckover2pi1
2>/planckover2pi1
2./check
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 31
Problem 2.26
Putf(x)=δ(x) into Eq. 2.102: F(k)=1√
2π/integraldisplay∞
−∞δ(x)e−ikxdx=1√
2π.
∴f(x)=δ(x)=1√
2π/integraldisplay∞
−∞1√
2πeikxdk=1
2π/integraldisplay∞
−∞eikxdk.QED
Problem 2.27
(a)V(x)
-a a
x
(b)From Problem 2.1(c) the solutions are even or odd. Look first for even solutions :
ψ(x)=
Ae−κx(x<a),
B(eκx+e−κx)(−a<x<a ),
Aeκx(x<−a).
Continuity at a:Ae−κa=B(eκa+e−κa),orA=B(e2κa+1 ).
Discontinuous derivative at a,∆dψ
dx=−2mα
/planckover2pi12ψ(a):
−κAe−κa−B(κeκa−κe−κa)=−2mα
/planckover2pi12Ae−κa⇒A+B(e2κa−1) =2mα
/planckover2pi12κA;o r
B(e2κa−1) =A/parenleftbigg2mα
/planckover2pi12κ−1/parenrightbigg
=B(e2κa+1 )/parenleftbigg2mα
/planckover2pi12κ−1/parenrightbigg
⇒e2κa−1=e2κa/parenleftbigg2mα
/planckover2pi12κ−1/parenrightbigg
+2mα
/planckover2pi12κ−1.
1=2mα
/planckover2pi12κ−1+2mα
/planckover2pi12κe−2κa;/planckover2pi12κ
mα=1+e−2κa,ore−2κa=/planckover2pi12κ
mα−1.
This is a transcendental equation for κ(and hence for E). I’ll solve it graphically: Let z≡2κa, c≡/planckover2pi12
2amα,
soe−z=cz−1. Plot both sides and look for intersections:
1
z 1/ccz-1
e-z
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32 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
From the graph, noting that candzare both positive, we see that there is one (and only one) solution
(for even ψ). Ifα=/planckover2pi12
2ma,s oc= 1, the calculator gives z=1.278, so κ2=−2mE
/planckover2pi12=z2
(2a)2⇒E=
−(1.278)2
8/parenleftBig
/planckover2pi12
ma2/parenrightBig
=−0.204/parenleftBig
/planckover2pi12
ma2/parenrightBig
.
Now look for odd solutions:
ψ(x)=
Ae−κx(x<a),
B(eκx−e−κx)(−a<x<a ),
−Aeκx(x<−a).
Continuity at a:Ae−κa=B(eκa−e−κa),orA=B(e2κa−1).
Discontinuity in ψ/prime:−κAe−κa−B(κeκa+κe−κa)=−2mα
/planckover2pi12Ae−κa⇒B(e2κa+1 )=A/parenleftbigg2mα
/planckover2pi12κ−1/parenrightbigg
,
e2κa+1=(e2κa−1)/parenleftbigg2mα
/planckover2pi12κ−1/parenrightbigg
=e2κa/parenleftbigg2mα
/planckover2pi12κ−1/parenrightbigg
−2mα
/planckover2pi12κ+1,
1=2mα
/planckover2pi12κ−1−2mα
/planckover2pi12κe−2κa;/planckover2pi12κ
mα=1−e−2κa,e−2κa=1−/planckover2pi12κ
mα,ore−z=1−cz.
1/c 1/c1
z
This time there may or may not be a solution. Both graphs have their y-intercepts at 1, but if cis too
large (αtoo small), there may be no intersection (solid line), whereas if cis smaller (dashed line) there
will be. (Note that z=0⇒κ=0i s nota solution, since ψis then non-normalizable.) The slope of e−z
(atz=0 )i s−1; the slope of (1 −cz)i s−c. So there is an oddsolution⇔c<1, orα>/planckover2pi12/2ma.
Conclusion: Onebound state if α≤/planckover2pi12/2ma;twoifα>/planckover2pi12/2ma.
ψ ψ
x x a -a a-a
Even Odd
α=/planckover2pi12
ma⇒c=1
2./braceleftbiggEven:e−z=1
2z−1⇒z=2.21772,
Odd:e−z=1−1
2z⇒z=1.59362.
E=−0.615(/planckover2pi12/ma2);E=−0.317(/planckover2pi12/ma2).
α=/planckover2pi12
4ma⇒c=2.Only even: e−z=2z−1⇒z=0.738835; E=−0.0682( /planckover2pi12/ma2).
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 33
Problem 2.28
ψ=
Aeikx+Be−ikx(x<−a)
Ceikx+De−ikx(−a<x<a )
Feikx(x>a)
.Impose boundary conditions:
(1) Continuity at −a:Aeika+Beika=Ce−ika+Deika⇒βA+B=βC+D,whereβ≡e−2ika.
(2) Continuity at + a:Ceika+De−ika=Feika⇒F=C+βD.
(3) Discontinuity in ψ/primeat−a:ik(Ce−ika−Deika)−ik(Ae−ika−Beika)=−2mα
/planckover2pi12(Ae−ika+Beika)
⇒βC−D=β(γ+1 )A+B(γ−1),whereγ≡i2mα//planckover2pi12k.
(4) Discontinuity in ψ/primeat +a:ikFeika−ik(Ceika−De−ika)=−2mα
/planckover2pi12(Feika)
⇒C−βD=( 1−γ)F.
To solve for CandD,/braceleftbiggadd (2) and (4) : 2 C=F+( 1−γ)F⇒2C=( 2−γ)F.
subtract (2) and (4) : 2 βD=F−(1−γ)F⇒2D=(γ/β)F.
/braceleftbiggadd (1) and (3) : 2 βC=βA+B+β(γ+1 )A+B(γ−1)⇒2C=(γ+2 )A+(γ/β)B.
subtract (1) and (3) : 2 D=βA+B−β(γ+1 )A−B(γ−1)⇒2D=−γβA+( 2−γ)B.
/braceleftbiggEquate the two expressions for 2 C:( 2−γ)F=(γ+2 )A+(γ/β)B.
Equate the two expressions for 2 D:(γ/β)F=−γβA+( 2−γ)B.
Solve these for FandB, in terms of A. Multiply the first by β(2−γ), the second by γ, and subtract:
/bracketleftbig
β(2−γ)2F=β(4−γ2)A+γ(2−γ)B/bracketrightbig
;/bracketleftbig
(γ2/β)F=−βγ2A+γ(2−γ)B/bracketrightbig
.
⇒/bracketleftbig
β(2−γ)2−γ2/β/bracketrightbig
F=β/bracketleftbig
4−γ2+γ/bracketrightbig
A=4βA⇒F
A=4
(2−γ)2−γ2/β2.
Letg≡i/γ=/planckover2pi12k
2mα;φ≡4ka,soγ=i
g,β2=e−iφ.Then:F
A=4g2
(2g−i)2+eiφ.
Denominator: 4 g2−4ig−1 + cosφ+isinφ=( 4g2−1 + cosφ)+i(sinφ−4g).
|Denominator|2=( 4g2−1 + cosφ)2+ (sinφ−4g)2
=1 6g4+1+c o s2φ−8g2−2 cosφ+8g2cosφ+ sin2φ−8gsinφ+1 6g2
=1 6g4+8g2+2+( 8 g2−2) cosφ−8gsinφ.
T=/vextendsingle/vextendsingle/vextendsingle/vextendsingleF
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=8g4
(8g4+4g2+1 )+( 4 g2−1) cosφ−4gsinφ,whereg≡/planckover2pi12k
2mαandφ≡4ka.
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34 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.29
In place of Eq. 2.151, we have: ψ(x)=
Fe−κx(x>a)
Dsin(lx)( 0<x<a )
−ψ(−x)(x<0)
.
Continuity of ψ:Fe−κa=Dsin(la); continuity of ψ/prime:−Fκe−κa=Dlcos(la).
Divide:−κ=lcot(la),or−κa=lacot(la)⇒/radicalBig
z2
0−z2=−zcotz,or−cotz=/radicalbig
(z0/z)2−1.
Wide, deep well: Intersections are at π,2π,3π,etc. Same as Eq. 2.157, but now for neven. This fills in the
rest of the states for the infinite square well.
Shallow, narrow well: Ifz0<π /2, there is noodd bound state. The corresponding condition on V0is
V0<π2/planckover2pi12
8ma2⇒noodd bound state .
π2 πzz0
Problem 2.30
1=2/integraldisplay∞
0|ψ|2dx=2/parenleftbigg
|D|2/integraldisplaya
0cos2lxdx+|F|2/integraldisplay∞
ae−2κxdx/parenrightbigg
=2/bracketleftbigg
|D|2/parenleftbiggx
2+1
4lsin 2lx/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0+|F|2/parenleftbigg
−1
2κe−2κx/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
a/bracketrightbigg
=2/bracketleftbigg
|D|2/parenleftbigga
2+sin 2la
4l/parenrightbigg
+|F|2e−2κa
2κ/bracketrightbigg
.
ButF=Deκacosla(Eq. 2.152), so 1 = |D|2/parenleftbigg
a+sin(2la)
2l+cos2(la)
κ/parenrightbigg
.
Furthermore κ=ltan(la) (Eq. 2.154), so
1=|D|2/parenleftbigg
a+2 sinlacosla
2l+cos3la
lsinla/parenrightbigg
=|D|2/bracketleftbigg
a+cosla
lsinla(sin2la+ cos2la)/bracketrightbigg
=|D|2/parenleftbigg
a+1
ltanla/parenrightbigg
=|D|2/parenleftbigg
a+1
κ/parenrightbigg
.D=1/radicalbig
a+1/κ,F=eκacosla/radicalbig
a+1/κ.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 35
Problem 2.31
Equation 2.155 ⇒z0=a
/planckover2pi1√2mV0.We want α=area of potential =2aV0held constant as a→0.Therefore
V0=α
2a;z0=a
/planckover2pi1/radicalbig2mα
2a=1
/planckover2pi1√mαa→0.Soz0issmall, and the intersection in Fig. 2.18 occurs at very small
z. Solve Eq. 2.156 for very small z, by expanding tan z:
tanz∼=z=/radicalbig
(z0/z)2−1=( 1/z)/radicalBig
z2
0−z2.
Now (from Eqs. 2.146, 2.148 and 2.155) z2
0−z2=κ2a2,soz2=κa. Butz2
0−z2=z4/lessmuch1⇒z∼=z0,soκa∼=z2
0.
But we found that z0∼=1
/planckover2pi1√mαahere, so κa=1
/planckover2pi12mαa,o rκ=mα
/planckover2pi12.(At this point the a’s have canceled, and
we can go to the limit a→0.)
√
−2mE
/planckover2pi1=mα
/planckover2pi12⇒−2mE=m2α2
/planckover2pi12.E=−mα2
2/planckover2pi12(which agrees with Eq. 2.129) .
In Eq. 2.169, V0/greatermuchE⇒T−1∼=1+V2
0
4EV0sin2/parenleftbig2a
/planckover2pi1√2mV0/parenrightbig
.ButV0=α
2a, so the argument of the sine is small,
and we can replace sin FepsilonCbyFepsilonC:T−1∼=1+V0
4E/parenleftbig2a
/planckover2pi1/parenrightbig22mV0=1+( 2 aV0)2m
2/planckover2pi12E.But 2aV0=α,soT−1=1+mα2
2/planckover2pi12E,
in agreement with Eq. 2.141.
Problem 2.32
Multiply Eq. 2.165 by sin la, Eq. 2.166 by1
lcosla, and add:
Csin2la+Dsinlacosla=Feikasinla
Ccos2la−Dsinlacosla=ik
lFeikacosla/bracerightbigg
C=Feika/bracketleftbigg
sinla+ik
lcosla/bracketrightbigg
.
Multiply Eq. 2.165 by cos la, Eq. 2.166 by1
lsinla, and subtract:
Csinlacosla+Dcos2la=Feikacosla
Csinlacosla−Dsin2la=ik
lFeikasinla/bracerightbigg
D=Feika/bracketleftbigg
cosla−ik
lsinla/bracketrightbigg
.
Put these into Eq. 2.163:
(1)Ae−ika+Beika=−Feika/bracketleftbigg
sinla+ik
lcosla/bracketrightbigg
sinla+Feika/bracketleftbigg
cosla−ik
lsinla/bracketrightbigg
cosla
=Feika/bracketleftbigg
cos2la−ik
lsinlacosla−sin2la−ik
lsinlacosla/bracketrightbigg
=Feika/bracketleftbigg
cos(2la)−ik
lsin(2la)/bracketrightbigg
.
Likewise, from Eq. 2.164:
(2)Ae−ika−Beika=−il
kFeika/bracketleftbigg/parenleftbigg
sinla+ik
lcosla/parenrightbigg
cosla+/parenleftbigg
cosla−ik
lsinla/parenrightbigg
sinla/bracketrightbigg
=−il
kFeika/bracketleftbigg
sinlacosla+ik
lcos2la+ sinlacosla−ik
lsin2la/bracketrightbigg
=−il
kFeika/bracketleftbigg
sin(2la)+ik
lcos(2la)/bracketrightbigg
=Feika/bracketleftbigg
cos(2la)−il
ksin(2la)/bracketrightbigg
.
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36 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Add(1)and(2):2Ae−ika=Feika/bracketleftbigg
2 cos(2la)−i/parenleftbiggk
l+l
k/parenrightbigg
sin(2la)/bracketrightbigg
,or:
F=e−2ikaA
cos(2la)−isin(2la)
2kl(k2+l2)(confirming Eq. 2.168). Now subtract (2)from(1):
2Beika=Feika/bracketleftbigg
i/parenleftbiggl
k−k
l/parenrightbigg
sin(2la)/bracketrightbigg
⇒B=isin(2la)
2kl(l2−k2)F(confirming Eq. 2.167) .
T−1=/vextendsingle/vextendsingle/vextendsingle/vextendsingleA
F/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=/vextendsingle/vextendsingle/vextendsingle/vextendsinglecos(2la)−isin(2la)
2kl(k2+l2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
= cos2(2la)+sin2(2la)
(2lk)2(k2+l2)2.
But cos2(2la)=1−sin2(2la),so
T−1= 1 + sin2(2la)/bracketleftbigg(k2+l2)2
(2lk)2−1
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
1
(2kl)2[k4+2k2l2+l4−4k2l2]=1
(2kl)2[k4−2k2l2+l4]=(k2−l2)2
(2kl)2./bracketrightbigg
=1+(k2−l2)2
(2kl)2sin2(2la).
Butk=√
2mE
/planckover2pi1,l=/radicalbig
2m(E+V0)
/planckover2pi1;s o ( 2la)=2a
/planckover2pi1/radicalbig
2m(E+V0);k2−l2=−2mV0
/planckover2pi12,and
(k2−l2)2
(2kl)2=/parenleftbig2m
/planckover2pi12/parenrightbig2V2
0
4/parenleftbig2m
/planckover2pi12/parenrightbig2E(E+V0)=V2
0
4E(E+V0).
∴T−1=1+V2
0
4E(E+V0)sin2/parenleftbigg2a
/planckover2pi1/radicalbig
2m(E+V0)/parenrightbigg
,confirming Eq. 2.169.
Problem 2.33
E<V0.ψ =
Aeikx+Be−ikx(x<−a)
Ceκx+De−κx(−a<x<a )
Feikx(x>a)
k=√
2mE
/planckover2pi1;κ=/radicalbig
2m(V0−E)
/planckover2pi1.
(1)Continuity of ψat−a:Ae−ika+Beika=Ce−κa+Deκa.
(2)Continuity of ψ/primeat−a:ik(Ae−ika−Beika)=κ(Ce−κa−Deκa).
⇒2Ae−ika=/parenleftBig
1−iκ
k/parenrightBig
Ce−κa+/parenleftBig
1+iκ
k/parenrightBig
Deκa.
(3)Continuity of ψat +a:Ceκa+De−κa=Feika.
(4)Continuity of ψ/primeat +a:κ(Ceκa−De−κa)=ikFeika.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 37
⇒2Ceκa=/parenleftbigg
1+ik
κ/parenrightbigg
Feika;2De−κa=/parenleftbigg
1−ik
κ/parenrightbigg
Feika.
2Ae−ika=/parenleftbigg
1−iκ
k/parenrightbigg/parenleftbigg
1+ik
κ/parenrightbigg
Feikae−2κa
2+/parenleftbigg
1+iκ
k/parenrightbigg/parenleftbigg
1−ik
κ/parenrightbigg
Feikae2κa
2
=Feika
2/braceleftbigg/bracketleftbigg
1+i/parenleftbiggk
κ−κ
k/parenrightbigg
+1/bracketrightbigg
e−2κa+/bracketleftbigg
1+i/parenleftbiggκ
k−k
κ/parenrightbigg
+1/bracketrightbigg
e2κa/bracerightbigg
=Feika
2/bracketleftbigg
2/parenleftbig
e−2κa+e2κa/parenrightbig
+i(κ2−k2)
kκ/parenleftbig
e2κa−e−2κa/parenrightbig/bracketrightbigg
.
But sinh x≡ex−e−x
2,coshx≡ex+e−x
2,so
=Feika
2/bracketleftbigg
4 cosh(2 κa)+i(κ2−k2)
kκ2 sinh(2 κa)/bracketrightbigg
=2Feika/bracketleftbigg
cosh(2κa)+i(κ2−k2)
2kκsinh(2κa)/bracketrightbigg
.
T−1=/vextendsingle/vextendsingle/vextendsingle/vextendsingleA
F/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
= cosh2(2κa)+(κ2−k2)2
(2κk)2sinh2(2κa).But cosh2= 1 + sinh2,so
T−1=1+/bracketleftbigg
1+(κ2−k2)2
(2κk)2
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
⋆/bracketrightbigg
sinh2(2κa)=1+V2
0
4E(V0−E)sinh2/parenleftbigg2a
/planckover2pi1/radicalbig
2m(V0−E)/parenrightbigg
,
where⋆=4κ2k2+k4+κ4−2κ2k2
(2κk)2=(κ2+k2)2
(2κk)2=/parenleftBig
2mE
/planckover2pi12+2m(V0−E)
/planckover2pi12/parenrightBig2
42mE
/planckover2pi122m(V0−E)
/planckover2pi12=V2
0
4E(V0−E).
(You can also get this from Eq. 2.169 by switching the sign of V0and using sin( iθ)=isinhθ.)
E=V0.ψ =
Aeikx+Be−ikx(x<−a)
C+Dx (−a<x<a )
Feikx(x>a)
(In central region −/planckover2pi12
2md2ψ
dx2+V0ψ=Eψ⇒d2ψ
dx2=0,soψ=C+Dx.)
(1)Continuous ψat−a:Ae−ika+Beika=C−Da.
(2)Continuous ψat +a:Feika=C+Da.
⇒(2.5)2Da=Feika−Ae−ika−Beika.
(3)Continuous ψ/primeat−a:ik/parenleftbig
Ae−ika−Beika/parenrightbig
=D.
(4)Continuous ψ/primeat +a:ikFeika=D.
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38 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
⇒(4.5)Ae−2ika−B=F.
Use(4)to eliminate Din(2.5):Ae−2ika+B=F−2aikF=( 1−2iak)F, and add to (4.5):
2Ae−2ika=2F(1−ika),soT−1=/vextendsingle/vextendsingle/vextendsingle/vextendsingleA
F/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=1+(ka)2=1+2mE
/planckover2pi12a2.
(You can also get this from Eq. 2.169 by changing the sign of V0and taking the limit E→V0, using sin FepsilonC∼=FepsilonC.)
E>V0.This case is identical to the one in the book, only with V0→−V0.S o
T−1=1+V2
0
4E(E−V0)sin2/parenleftbigg2a
/planckover2pi1/radicalbig
2m(E−V0)/parenrightbigg
.
Problem 2.34
(a)
ψ=/braceleftbigg
Aeikx+Be−ikx(x<0)
Fe−κx(x>0)/bracerightbigg
wherek=√
2mE
/planckover2pi1;κ=/radicalbig
2m(V0−E)
/planckover2pi1.
(1)Continuity of ψ:A+B=F.
(2)Continuity of ψ/prime:ik(A−B)=−κF.
⇒A+B=−ik
κ(A−B)⇒A/parenleftbigg
1+ik
κ/parenrightbigg
=−B/parenleftbigg
1−ik
κ/parenrightbigg
.
R=/vextendsingle/vextendsingle/vextendsingle/vextendsingleB
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=|(1 +ik/κ)|2
|(1−ik/κ)|2=1+(k/κ)2
1+(k/κ)2=1.
Although the wave function penetrates into the barrier, it is eventually all reflected.
(b)
ψ=/braceleftbiggAeikx+Be−ikx(x<0)
Feilx(x>0)/bracerightbigg
wherek=√
2mE
/planckover2pi1;l=/radicalbig
2m(E−V0)
/planckover2pi1.
(1)Continuity of ψ:A+B=F.
(2)Continuity of ψ/prime:ik(A−B)=ilF.
⇒A+B=k
l(A−B);A/parenleftbigg
1−k
l/parenrightbigg
=−B/parenleftbigg
1+k
l/parenrightbigg
.
R=/vextendsingle/vextendsingle/vextendsingle/vextendsingleB
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=(1−k/l)2
(1 +k/l)2=(k−l)2
(k+l)2=(k−l)4
(k2−l2)2.
Nowk2−l2=2m
/planckover2pi12(E−E+V0)=/parenleftbigg2m
/planckover2pi12/parenrightbigg
V0;k−l=√
2m
/planckover2pi1[√
E−/radicalbig
E−V0],so
R=(√
E−√E−V0)4
V2
0.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 39
(c)
vivivtvtdt dt
From the diagram, T=Pt/Pi=|F|2vt/|A|2vi,wherePiis the probability of finding the incident particle
in the box corresponding to the time interval dt, andPtis the probability of finding the transmitted
particle in the associated box to the rightof the barrier.
Butvt
vi=√E−V0√
E(from Eq. 2.98). So T=/radicalbigg
E−V0
E/vextendsingle/vextendsingle/vextendsingle/vextendsingleF
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
.Alternatively, from Problem 2.19:
Ji=/planckover2pi1k
m|A|2;Jt=/planckover2pi1l
m|F|2;T=Jt
Ji=/vextendsingle/vextendsingle/vextendsingle/vextendsingleF
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2l
k=/vextendsingle/vextendsingle/vextendsingle/vextendsingleF
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2/radicalbigg
E−V0
E.
ForE<V 0,of course, T=0.
(d)
ForE>V 0,F=A+B=A+A/parenleftbigk
l−1/parenrightbig
/parenleftbigk
l+1/parenrightbig=A2k/l/parenleftbigk
l+1/parenrightbig=2k
k+lA.
T=/vextendsingle/vextendsingle/vextendsingle/vextendsingleF
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2l
k=/parenleftbigg2k
k+l/parenrightbigg2l
k=4kl
(k+l)2=4kl(k−l)2
(k2−l2)2=4√
E√E−V0(√
E−√E−V0)2
V2
0.
T+R=4kl
(k+l)2+(k−l)2
(k+l)2=4kl+k2−2kl+l2
(k+l)2=k2+2kl+l2
(k+l)2=(k+l)2
(k+l)2=1./check
Problem 2.35
(a)
ψ(x)=/braceleftbiggAeikx+Be−ikx(x<0)
Feilx(x>0)/bracerightbigg
wherek≡√
2mE
/planckover2pi1,l≡/radicalbig
2m(E+V0)
/planckover2pi1.
Continuity of ψ⇒A+B=F
Continuity of ψ/prime⇒ik(A−B)=ilF/bracerightbigg
=⇒
A+B=k
l(A−B);A/parenleftbigg
1−k
l/parenrightbigg
=−B/parenleftbigg
1+k
l/parenrightbigg
;B
A=−/parenleftbigg1−k/l
1+k/l/parenrightbigg
.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
40 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
R=/vextendsingle/vextendsingle/vextendsingle/vextendsingleB
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=/parenleftbiggl−k
l+k/parenrightbigg2
=/parenleftBigg√E+V0−√
E√E+V0+√
E/parenrightBigg2
=/parenleftBigg/radicalbig
1+V0/E−1/radicalbig
1+V0/E+1/parenrightBigg2
=/parenleftbigg√1+3−1√1+3+1/parenrightbigg2
=/parenleftbigg2−1
2+1/parenrightbigg2
=1
9.
(b)The cliff is two-dimensional, and even if we pretend the car drops straight down, the potential as a function
of distance along the (crooked, but now one-dimensional) pathis−mgx(withxthe vertical coordinate),
as shown.
V(x)
x
-V0
(c)HereV0/E=1 2/4 = 3, the same as in part (a), so R=1/9, and hence T=8/9 = 0.8889.
Problem 2.36
Start with Eq. 2.22: ψ(x)=Asinkx+Bcoskx. This time the boundary conditions are ψ(a)=ψ(−a)=0 :
Asinka+Bcoska=0 ;−Asinka+Bcoska=0.
/braceleftBigg
Subtract :Asinka=0⇒ka=jπorA=0,
Add: Bcoska=0⇒ka=(j−1
2)πorB=0,
(where j=1,2,3,...).
IfB= 0 (so A/negationslash= 0),k=jπ/a. In this case let n≡2j(sonis aneveninteger); then k=nπ/2a,
ψ=Asin(nπx/2a). Normalizing: 1 = |A|2/integraltexta
−asin2(nπx/2a)dx=|A|2/2⇒A=√
2.
IfA= 0 (so B/negationslash= 0),k=(j−1
2)π/a. In this case let n≡2j−1(nis anoddinteger); again k=nπ/2a,
ψ=Bcos(nπx/2a). Normalizing: 1 = |B|2/integraltexta
−acos2(nπx/2a)dx=|a|2/2⇒B=√
2.
In either case Eq. 2.21 yields E=/planckover2pi12k2
2m=n2π2/planckover2pi12
2m(2a)2(in agreement with Eq. 2.27 for a well of width 2 a).
The substitution x→(x+a)/2 takes Eq. 2.28 to
/radicalbigg
2
asin/parenleftbiggnπ
a(x+a)
2/parenrightbigg
=/radicalbigg
2
asin/parenleftBignπx
2a+nπ
2/parenrightBig
=
(−1)n/2/radicalBig
2
asin/parenleftbignπx
2a/parenrightbig
(neven),
(−1)(n−1)/2/radicalBig
2
acos/parenleftbignπx
2a/parenrightbig
(nodd).
So (apart from normalization) we recover the results above. The graphs are the same as Figure 2.2, except that
some are upside down (different normalization).
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 41
cos(πx/2a) sin(2 πx/2a) cos(3 πx/2a)
Problem 2.37
Use the trig identity sin 3 θ= 3 sinθ−4 sin3θto write
sin3/parenleftbiggπx
a/parenrightbigg
=3
4sin/parenleftbiggπx
a/parenrightbigg
−1
4sin/parenleftbigg3πx
a/parenrightbigg
.So (Eq. 2.28): Ψ( x,0) =A/radicalbigga
2/bracketleftbigg3
4ψ1(x)−1
4ψ3(x)/bracketrightbigg
.
Normalize using Eq. 2.38: |A|2a
2/parenleftbigg9
16+1
16/parenrightbigg
=5
16a|A|2=1⇒A=4√
5a.
So Ψ(x,0) =1√
10[3ψ1(x)−ψ3(x)],and hence (Eq. 2.17)
Ψ(x,t)=1√
10/bracketleftBig
3ψ1(x)e−iE1t//planckover2pi1−ψ3(x)e−iE3t//planckover2pi1/bracketrightBig
.
|Ψ(x,t)|2=1
10/bracketleftbigg
9ψ2
1+ψ2
3−6ψ1ψ3cos/parenleftbiggE3−E1
/planckover2pi1t/parenrightbigg/bracketrightbigg
;s o
/angbracketleftx/angbracketright=/integraldisplaya
0x|Ψ(x,t)|2dx=9
10/angbracketleftx/angbracketright1+1
10/angbracketleftx/angbracketright3−3
5cos/parenleftbiggE3−E1
/planckover2pi1t/parenrightbigg/integraldisplaya
0xψ1(x)ψ3(x)dx,
where/angbracketleftx/angbracketrightn=a/2 is the expectation value of xin thenth stationary state. The remaining integral is
2
a/integraldisplaya
0xsin/parenleftbiggπx
a/parenrightbigg
sin/parenleftbigg3πx
a/parenrightbigg
dx=1
a/integraldisplaya
0x/bracketleftbigg
cos/parenleftbigg2πx
a/parenrightbigg
−cos/parenleftbigg4πx
a/parenrightbigg/bracketrightbigg
dx
=1
a/bracketleftbigg/parenleftbigga
2π/parenrightbigg2
cos/parenleftbigg2πx
a/parenrightbigg
+/parenleftbiggxa
2π/parenrightbigg
sin/parenleftbigg2πx
a/parenrightbigg
−/parenleftbigga
4π/parenrightbigg2
cos/parenleftbigg4πx
a/parenrightbigg
−/parenleftbiggxa
4π/parenrightbigg
sin/parenleftbigg4πx
a/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0=0.
Evidently then,
/angbracketleftx/angbracketright=9
10/parenleftbigga
2/parenrightbigg
+1
10/parenleftbigga
2/parenrightbigg
=a
2.
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42 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.38
(a)New allowed energies: En=n2π2/planckover2pi12
2m(2a)2;Ψ (x,0) =/radicalbigg
2
asin/parenleftBigπ
ax/parenrightBig
,ψn(x)=/radicalbigg
2
2asin/parenleftBignπ
2ax/parenrightBig
.
cn=√
2
a/integraldisplaya
0sin/parenleftBigπ
ax/parenrightBig
sin/parenleftBignπ
2ax/parenrightBig
dx=√
2
2a/integraldisplaya
0/braceleftBig
cos/bracketleftBig/parenleftBign
2−1/parenrightBigπx
a/bracketrightBig
−cos/bracketleftBig/parenleftBign
2+1/parenrightBigπx
a/bracketrightBig/bracerightBig
dx.
=1√
2a/braceleftBigg
sin/bracketleftbig/parenleftbign
2−1/parenrightbigπx
a/bracketrightbig
/parenleftbign
2−1/parenrightbigπ
a−sin/bracketleftbig/parenleftbign
2+1/parenrightbigπx
a/bracketrightbig
/parenleftbign
2+1/parenrightbigπ
a/bracerightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0(forn/negationslash=2 )
=1√
2π/braceleftBigg
sin/bracketleftbig/parenleftbign
2−1/parenrightbig
π/bracketrightbig
/parenleftbign
2−1/parenrightbig−sin/bracketleftbig/parenleftbign
2+1/parenrightbig
π/bracketrightbig
/parenleftbign
2+1/parenrightbig/bracerightBigg
=sin/bracketleftbig/parenleftbign
2+1/parenrightbig
π/bracketrightbig
√
2π/bracketleftBigg
1/parenleftbign
2−1/parenrightbig−1/parenleftbign
2+1/parenrightbig/bracketrightBigg
=4√
2
πsin/bracketleftbig/parenleftbign
2+1/parenrightbig
π/bracketrightbig
(n2−4)=/braceleftBigg
0, ifnis even
±4√
2
π(n2−4),ifnis odd/bracerightBigg
.
c2=√
2
a/integraldisplaya
0sin2/parenleftBigπ
ax/parenrightBig
dx=√
2
a/integraldisplaya
01
2dx=1√
2.So the probability of getting Enis
Pn=|cn|2=
1
2, ifn=2
32
π2(n2−4)2,ifnis odd
0, otherwise
.
Most probable :E2=π2/planckover2pi12
2ma2(same as before). Probability :P2=1/2.
(b)Next most probable: E1=π2/planckover2pi12
8ma2,with probability P1=32
9π2=0.36025.
(c)/angbracketleftH/angbracketright=/integraltext
Ψ∗HΨdx=2
a/integraltexta
0sin/parenleftbigπ
ax/parenrightbig/parenleftBig
−/planckover2pi12
2md2
dx2/parenrightBig
sin/parenleftbigπ
ax/parenrightbig
dx,but this is exactly the same as before the wall
moved – for which we know the answer:π2/planckover2pi12
2ma2.
Problem 2.39
(a)According to Eq. 2.36, the most general solution to the time-dependent Schr¨ odinger equation for the
infinite square well is
Ψ(x,t)=∞/summationdisplay
n=1cnψn(x)e−i(n2π2/planckover2pi1/2ma2)t.
Nown2π2/planckover2pi1
2ma2T=n2π2/planckover2pi1
2ma24ma2
π/planckover2pi1=2πn2,s oe−i(n2π2/planckover2pi1/2ma2)(t+T)=e−i(n2π2/planckover2pi1/2ma2)te−i2πn2, and since n2is
an integer, e−i2πn2=1.Therefore Ψ( x,t+T)=Ψ (x,t).QED
(b)The classical revival time is the time it takes the particle to go down and back: Tc=2a/v, with the
velocity given by
E=1
2mv2⇒v=/radicalbigg
2E
m⇒Tc=a/radicalbigg
2m
E.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 43
(c)The two revival times are equal if
4ma2
π/planckover2pi1=a/radicalbigg
2m
E,orE=π2/planckover2pi12
8ma2=E1
4.
Problem 2.40
(a)LetV0≡32/planckover2pi12/ma2. This is just like the oddbound states for the finite square well, since they are the
ones that go to zero at the origin. Referring to the solution to Problem 2.29, the wave function is
ψ(x)=/braceleftBigg
Dsinlx, l≡/radicalbig
2m(E+V0)//planckover2pi1(0<x<a ),
Fe−κx,κ≡√
−2mE/ /planckover2pi1 (x>a),
and the boundary conditions at x=ayield
−cotz=/radicalbig
(z0/z)2−1
with
z0=√2mV0
/planckover2pi1a=/radicalbig
2m(32/planckover2pi12/ma2)
/planckover2pi1a=8.
Referring to the figure (Problem 2.29), and noting that (5 /2)π=7.85<z0<3π=9.42, we see that there
arethree bound states.
(b)Let
I1≡/integraldisplaya
0|ψ|2dx=|D|2/integraldisplaya
0sin2lxdx=|D|2/bracketleftbiggx
2−1
2lsinlxcoslx/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0=|D|2/bracketleftbigga
2−1
2lsinlzcosla/bracketrightbigg
;
I2≡/integraldisplay∞
a|ψ|2dx=|F|2/integraldisplay∞
ae−2κxdx=|F|2/bracketleftbigg
−e−2κx
2κ/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
a=|F|2e−2κa
2κ.
But continuity at x=a⇒Fe−κa=Dsinla,s oI2=|D|2sin2la
2κ.
Normalizing:
1=I1+I2=|D|2/bracketleftbigga
2−1
2lsinlacosla+sin2la
2κ/bracketrightbigg
=1
2κ|D|2/bracketleftBig
κa−κ
lsinlacosla+ sin2la/bracketrightBig
But (referring again to Problem 2 .29)κ/l=−cotla,so
=1
2κ|D|2/bracketleftbig
κa+ cotlasinlacosla+ sin2la/bracketrightbig
=|D|2(1 +κa)
2κ.
So|D|2=2κ/(1 +κa), and the probability of finding the particle outside the well is
P=I2=2κ
1+κasin2la
2κ=sin2la
1+κa.
We can express this interms of z≡laandz0:κa=/radicalbig
z2
0−z2(page 80),
sin2la= sin2z=1
1 + cot2z=1
1+(z0/z)2−1=/parenleftbiggz
z0/parenrightbigg2
⇒P=z2
z2
0(1 +/radicalbig
z2
0−z2).
So far, this is correct for anybound state. In the present case z0= 8 and zis the third solution
to−cotz=/radicalbig
(8/z)2−1, which occurs somewhere in the interval 7 .85<z< 8. Mathematica gives
z=7.9573 and P=0.54204.
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44 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.41
(a)In the standard notation ξ≡/radicalbig
mω/ /planckover2pi1x,α≡(mω/π /planckover2pi1)1/4,
Ψ(x,0) =A(1−2ξ)2e−ξ2/2=A(1−4ξ+4ξ2)e−ξ2/2.
It can be expressed as a linear combination of the first three stationary states (Eq. 2.59 and 2.62, and
Problem 2.10):
ψ0(x)=αe−ξ2/2,ψ 1(x)=√
2αξe−ξ2/2,ψ 2(x)=α√
2(2ξ2−1)e−ξ2/2.
So Ψ(x,0) =c0ψ0+c1ψ1+c2ψ2=α(c0+√
2ξc1+√
2ξ2c2−1√
2c2)e−ξ2/2with (equating like powers)
α√
2c2=4A⇒c2=2√
2A/α,
α√
2c1=−4A⇒c1=−2√
2A/α,
α(c0−c2/√
2) =A⇒c0=(A/α)+c2/√
2=( 1+2 ) A/α=3A/α.
Normalizing: 1 = |c0|2+|c1|2+|c2|2=( 8+8+9 ) ( A/α)2= 25(A/α)2⇒A=α/5.
c0=3
5,c1=−2√
2
5,c2=2√
2
5.
/angbracketleftH/angbracketright=/summationdisplay
|cn|2(n+1
2)/planckover2pi1ω=9
25/parenleftbigg1
2/planckover2pi1ω/parenrightbigg
+8
25/parenleftbigg3
2/planckover2pi1ω/parenrightbigg
+8
25/parenleftbigg5
2/planckover2pi1ω/parenrightbigg
=/planckover2pi1ω
50( 9+2 4+4 0 )=73
50/planckover2pi1ω.
(b)
Ψ(x,t)=3
5ψ0e−iωt/2−2√
2
5ψ1e−3iωt/2+2√
2
5ψ2e−5iωt/2=e−iωt/2/bracketleftBigg
3
5ψ0−2√
2
5ψ1e−iωt+2√
2
5ψ2e−2iωt/bracketrightBigg
.
To change the sign of the middle term we need e−iωT=−1 (then e−2iωT= 1); evidently ωT=π,o r
T=π/ω.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 45
Problem 2.42
Everything in Section 2.3.2 still applies, except that there is an additional boundary condition: ψ(0) = 0.This
eliminates all the evensolutions ( n=0,2,4,...), leaving only the odd solutions. So
En=/parenleftbigg
n+1
2/parenrightbigg
/planckover2pi1ω, n=1,3,5,....
Problem 2.43
(a)Normalization is the same as before: A=/parenleftbig2a
π/parenrightbig1/4.
(b)Equation 2.103 says
φ(k)=1√
2π/parenleftbigg2a
π/parenrightbigg1/4/integraldisplay∞
−∞e−ax2eilxe−ikxdx[same as before, only k→k−l]=1
(2πa)1/4e−(k−l)2/4a.
Ψ(x,t)=1√
2π1
(2πa)1/4/integraldisplay∞
−∞e−(k−l)2/4aei(kx−/planckover2pi1k2t/2m)
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
e−l2/4ae−[(1
4a+i/planckover2pi1t
2m)k2−(ix+l
2a)k]dk
=1√
2π1
(2πa)1/4e−l2/4a/radicalBigg
π/parenleftbig1
4a+i/planckover2pi1t
2m/parenrightbige(ix+l/2a)2/[4(1/4a+i/planckover2pi1t/2m)]
=/parenleftbigg2a
π/parenrightbigg1/41/radicalbig
1+2i/planckover2pi1at/me−l2/4aea(ix+l/2a)2/(1+2ia/planckover2pi1t/m).
(c)Letθ≡2/planckover2pi1at/m, as before:|Ψ|2=/radicalbigg
2a
π1√
1+θ2e−l2/2aea/bracketleftbigg
(ix+l/2a)2
(1+iθ)+(−ix+l/2a)2
(1−iθ)/bracketrightbigg
.Expand the term in
square brackets:
[]=1
1+θ2/bracketleftBigg
(1−iθ)/parenleftbigg
ix+l
2a/parenrightbigg2
+( 1+iθ)/parenleftbigg
−ix+l
2a/parenrightbigg2/bracketrightBigg
=1
1+θ2/bracketleftbigg/parenleftbigg
−x2+ixl
a+l2
4a2/parenrightbigg
+/parenleftbigg
−x2−ixl
a+l2
4a2/parenrightbigg
+iθ/parenleftbigg
x2−ixl
a−l2
4a2/parenrightbigg
+iθ/parenleftbigg
−x2−ixl
a+l2
4a2/parenrightbigg/bracketrightbigg
=1
1+θ2/bracketleftbigg
−2x2+l2
2a2+2θxl
a/bracketrightbigg
=1
1+θ2/bracketleftbigg
−2x2+2θxl
a−θ2l2
2a2+θ2l2
2a2+l2
2a2/bracketrightbigg
=−2
1+θ2/parenleftbigg
x−θl
2a/parenrightbigg2
+l2
2a2.
|Ψ(x,t)|2=/radicalbigg
2
π/radicalbigga
1+θ2e−l2/2ae−2a
1+θ2(x−θl/2a)2
el2/2a=/radicalbigg
2
πwe−2w2(x−θl/2a)2,
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46 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
wherew≡/radicalbig
a/(1 +θ2). The result is the same as before, except x→/parenleftbig
x−θl
2a/parenrightbig
=/parenleftbig
x−/planckover2pi1l
mt/parenrightbig
,s o|Ψ|2has
the same (flattening Gaussian) shape – only this time the center moves at constant speed v=/planckover2pi1l/m.
(d)
/angbracketleftx/angbracketright=/integraldisplay∞
−∞x|Ψ(x,t)|2dx. Lety≡x−θl/2a=x−vt,sox=y+vt.
=/integraldisplay∞
−∞(y+vt)/radicalbigg
2
πwe−2w2y2dy=vt.
(The first integral is trivially zero; the second is 1 by normalization.)
=/planckover2pi1l
mt;/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=/planckover2pi1l.
/angbracketleftx2/angbracketright=/integraldisplay∞
−∞(y+vt)2/radicalbigg
2
πwe−2w2y2dy=1
4w2+0+(vt)2(the first integral is same as before) .
/angbracketleftx2/angbracketright=1
4w2+/parenleftbigg/planckover2pi1lt
m/parenrightbigg2
./angbracketleftp2/angbracketright=−/planckover2pi12/integraldisplay∞
−∞Ψ∗d2Ψ
dx2dx.
Ψ=/parenleftbigg2a
π/parenrightbigg1/41√
1+iθe−l2/4aea(ix+l/2a)2/(1+iθ),sodΨ
dx=2ia/parenleftbig
ix+l
2a/parenrightbig
(1 +iθ)Ψ;
d2Ψ
dx2=/bracketleftbigg2ia(ix+l/2a)
1+iθ/bracketrightbiggdΨ
dx+2i2a
1+iθΨ=/bracketleftBigg
−4a2(ix+l/2a)2
(1 +iθ)2−2a
1+iθ/bracketrightBigg
Ψ.
/angbracketleftp2/angbracketright=4a2/planckover2pi12
(1 +iθ)2/integraldisplay∞
−∞/bracketleftBigg/parenleftbigg
ix+l
2a/parenrightbigg2
+(1 +iθ)
2a/bracketrightBigg
|Ψ|2dx
=4a2/planckover2pi12
(1 +iθ)2/integraldisplay∞
−∞/bracketleftBigg
−/parenleftbigg
y+vt−il
2a/parenrightbigg2
+(1 +iθ)
2a/bracketrightBigg
|Ψ|2dy
=4a2/planckover2pi12
(1 +iθ)2/braceleftbigg
−/integraldisplay∞
−∞y2|Ψ|2dy−2/parenleftbigg
vt−il
2a/parenrightbigg/integraldisplay∞
−∞y|Ψ|2dy
+/bracketleftBigg
−/parenleftbigg
vt−il
2a/parenrightbigg2
+(1 +iθ)
2a/bracketrightBigg/integraldisplay∞
−∞|Ψ|2dy/bracerightBigg
=4a2/planckover2pi12
(1 +iθ)2/bracketleftBigg
−1
4w2+0−/parenleftbigg
vt−il
2a/parenrightbigg2
+(1 +iθ)
2a/bracketrightBigg
=4a2/planckover2pi12
(1 +iθ)2/braceleftBigg
−1+θ2
4a−/bracketleftbigg/parenleftbigg−il
2a/parenrightbigg
(1 +iθ)/bracketrightbigg2
+(1 +iθ)
2a/bracerightBigg
=a/planckover2pi1
1+iθ/bracketleftbigg
−(1−iθ)+l2
a(1 +iθ)+2/bracketrightbigg
=a/planckover2pi12
1+iθ/bracketleftbigg
(1 +iθ)/parenleftbigg
1+l2
a/parenrightbigg/bracketrightbigg
=/planckover2pi12(a+l2).
σ2
x=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=1
4w2+/parenleftbigg/planckover2pi1lt
m/parenrightbigg2
−/parenleftbigg/planckover2pi1lt
m/parenrightbigg2
=1
4w2⇒σx=1
2w;
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 47
σ2
p=/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/planckover2pi12a+/planckover2pi12l2−/planckover2pi12l2=/planckover2pi12a,soσp=/planckover2pi1√a.
(e)σxandσpare same as before, so the uncertainty principle still holds.
Problem 2.44
Equation 2.22 ⇒ψ(x)=Asinkx+Bcoskx,0≤x≤a,withk=√
2mE/ /planckover2pi12.
Even solutions: ψ(x)=ψ(−x)=Asin(−kx)+Bcos(−kx)=−Asinkx+Bcoskx(−a≤x≤0).
Boundary
conditions
ψcontinuous at 0 : B=B(no new condition).
ψ/primediscontinuous (Eq. 2.125 with sign of αswitched): Ak+Ak=2mα
/planckover2pi12B⇒B=/planckover2pi12k
mαA.
ψ→0a tx=a:Asin(ka)+/planckover2pi12k
mαAcos(ka)=0⇒tan(ka)=−/planckover2pi12k
mα.
ψ(x)=A/parenleftbigg
sinkx+/planckover2pi12k
mαcoskx/parenrightbigg
(0≤x≤a);ψ(−x)=ψ(x).
π2 π 3π
tan(ka) -h kmα2ka
From the graph, the allowed energies are slightly above
ka=nπ
2(n=1,3,5,...)s oEn/greaterorsimilarn2π2/planckover2pi12
2m(2a)2(n=1,3,5,...).
These energies are somewhat higher than the corresponding energies for the infinite square well (Eq. 2.27, with
a→2a). Asα→0, the straight line ( −/planckover2pi12k/mα ) gets steeper and steeper, and the intersections get closer to
nπ/2; the energies then reduce to those of the ordinary infinite well. As α→∞, the straight line approaches
horizontal, and the intersections are at nπ(n=1,2,3,...),soEn→n2π2/planckover2pi12
2ma2– these are the allowed energies for
the infinite square well of width a. At this point the barrier is impenetrable, and we have two isolated infinite
square wells.
Odd solutions: ψ(x)=−ψ(−x)=−Asin(−kx)−Bcos(−kx)=Asin(kx)−Bcos(kx)(−a≤x≤0).
Boundary conditions
ψcontinuous at 0 : B=−B⇒B=0.
ψ/primediscontinuous: Ak−Ak=2mα
/planckover2pi12(0) (no new condition).
ψ(a)=0⇒Asin(ka)=0⇒ka=nπ
2(n=2,4,6,...).
ψ(x)=Asin(kx),(−a<x<a );En=n2π2/planckover2pi12
2m(2a)2(n=2,4,6,...).
These are the exact (evenn) energies (and wave functions) for the infinite square well (of width 2 a). The point
is that the oddsolutions (even n) arezeroat the origin, so they never “feel” the delta function at all.
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48 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Problem 2.45
−/planckover2pi12
2md2ψ1
dx2+Vψ1=Eψ1⇒−/planckover2pi12
2mψ2d2ψ1
dx2+Vψ1ψ2=Eψ1ψ2
−/planckover2pi12
2md2ψ2
dx2+Vψ2=Eψ2⇒−/planckover2pi12
2mψ1d2ψ2
dx2+Vψ1ψ2=Eψ1ψ2
⇒−/planckover2pi1
2
2m/bracketleftbigg
ψ2d2ψ1
dx2−ψ1d2ψ2
dx2/bracketrightbigg
=0.
Butd
dx/bracketleftbigg
ψ2dψ1
dx−ψ1dψ2
dx/bracketrightbigg
=dψ2
dxdψ1
dx+ψ2d2ψ1
dx2−dψ1
dxdψ2
dx−ψ1d2ψ2
dx2=ψ2d2ψ1
dx2−ψ1d2ψ2
dx2.Since this is
zero, it follows that ψ2dψ1
dx−ψ1dψ2
dx=K(a constant) .Butψ→0a t∞so the constant must be zero. Thus
ψ2dψ1
dx=ψ1dψ2
dx,or1
ψ1dψ1
dx=1
ψ2dψ2
dx,so lnψ1=l nψ2+ constant, or ψ1= (constant) ψ2.QED
Problem 2.46
−/planckover2pi12
2md2ψ
dx2=Eψ(where xis measured around the circumference), ord2ψ
dx2=−k2ψ,withk≡√
2mE
/planckover2pi1,s o
ψ(x)=Aeikx+Be−ikx.
Butψ(x+L)=ψ(x),sincex+Lis the same point as x,s o
AeikxeikL+Be−ikxe−ikL=Aeikx+Be−ikx,
and this is true for allx. In particular, for x=0:
(1)AeikL+Be−ikL=A+B.And for x=π
2k:
Aeiπ/2eikL+Be−iπ/2e−ikL=Aeiπ/2+Be−iπ/2,oriAeikL−iBe−ikL=iA−iB,so
(2)AeikL−Be−ikL=A−B.Add (1) and (2): 2 AeikL=2A.
EitherA= 0, or else eikL= 1, in which case kL=2nπ(n=0,±1,±2,...). But if A= 0, then Be−ikL=B,
leading to the same conclusion. So for every positive nthere are twosolutions: ψ+
n(x)=Aei(2nπx/L )and
ψ−
n(x)=Be−i(2nπx/L )(n= 0 is ok too, but in that case there is just onesolution). Normalizing:/integraltextL
0|ψ±|2dx=
1⇒A=B=1/√
L.Anyother solution (with the same energy) is a linear combination of these.
ψ±
n(x)=1√
Le±i(2nπx/L );En=2n2π2/planckover2pi12
mL2(n=0,1,2,3,...).
The theorem fails because here ψdoesnotgo to zero at ∞;xis restricted to a finite range, and we are unable
to determine the constant K(in Problem 2.45).
Problem 2.47
(a) (i) b=0⇒ordinary finite square well. Exponential decay outside; sinusoidal inside (cos for ψ1, sin for
ψ2). No nodes for ψ1, one node for ψ2.
(ii)Ground state is even. Exponential decay outside, sinusoidal inside the wells, hyperbolic cosine in
barrier. First excited state is odd– hyperbolic sine in barrier. No nodes for ψ1, one node for ψ2.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 49
x xψ ψ1 2
-a-a
aa
x x b/2 b/2-b/2
b/2+a -b/2-(b/2+a)
-(b/2+a) b/2+aψ12ψ
(iii)Forb/greatermucha, same as (ii), but wave function very small in barrier region. Essentially two isolated finite
square wells; ψ1andψ2are degenerate (in energy); they are even and odd linear combinations of the
ground states of the two separate wells.
ψ ψ12
x x-(b/2+a)
-(b/2+a)-b/2
-b/2b/2 b/2
b/2+a b/2+a
(b)From Eq. 2.157 we know that for b= 0 the energies fall slightly below
E1+V0≈π2/planckover2pi12
2m(2a)2=h
4
E2+V0≈4π2/planckover2pi12
2m(2a)2=h/bracerightBigg
whereh≡π2/planckover2pi12
2ma2.
Forb/greatermucha,the width of each (isolated) well is a,s o
E1+V0≈E2+V0≈π2/planckover2pi12
2ma2=h(again, slightly below this).
Hence the graph (next page). [Incidentally, within each well,d2ψ
dx2=−2m
/planckover2pi12(V0+E)ψ, so the more curved
the wave function, the higher the energy. This is consistent with the graphs above.]
(c)In the (even) ground state the energy is lowest in configuration (i), with b→0, so the electron tends to
draw the nuclei together, promoting bonding of the atoms. In the (odd) first excited state, by contrast,
the electron drives the nuclei apart.
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50 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
h
h/4
bE+V0
E +V
E +V0
012
Problem 2.48
(a)
dΨ
dx=2√
3
a√a·/braceleftbigg1,(0<x<a / 2)
−1,(a/2<x<a )/bracerightbigg
=2√
3
a√a/bracketleftbigg
1−2θ/parenleftbigg
x−a
2/parenrightbigg/bracketrightbigg
.
(b)
d2Ψ
dx2=2√
3
a√a/bracketleftbigg
−2δ/parenleftbigg
x−a
2/parenrightbigg/bracketrightbigg
=−4√
3
a√aδ/parenleftbigg
x−a
2/parenrightbigg
.
(c)
/angbracketleftH/angbracketright=−/planckover2pi12
2m/parenleftbigg
−4√
3
a√a/parenrightbigg/integraldisplay
Ψ∗δ/parenleftbigg
x−a
2/parenrightbigg
dx=2√
3/planckover2pi12
ma√aΨ∗/parenleftbigga
2/parenrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright√
3/a=2·3·/planckover2pi12
m·a·a=6/planckover2pi12
ma2./check
Problem 2.49
(a)
∂Ψ
∂t=/parenleftBig
−mω
2/planckover2pi1/parenrightBig/bracketleftbigga2
2/parenleftbig
−2iωe−2iωt/parenrightbig
+i/planckover2pi1
m−2ax(−iω)e−iωt/bracketrightbigg
Ψ,so
i/planckover2pi1∂Ψ
∂t=/bracketleftbigg
−1
2ma2ω2e−2iωt+1
2/planckover2pi1ω+maxω2e−iωt/bracketrightbigg
Ψ.
∂Ψ
∂x=/bracketleftBig/parenleftBig
−mω
2/planckover2pi1/parenrightBig/parenleftbig
2x−2ae−iωt/parenrightbig/bracketrightBig
Ψ=−mω
/planckover2pi1/parenleftbig
x−ae−iωt/parenrightbig
Ψ;
∂2Ψ
∂x2=−mω
/planckover2pi1Ψ−mω
/planckover2pi1/parenleftbig
x−ae−iωt/parenrightbig∂Ψ
∂x=/bracketleftbigg
−mω
/planckover2pi1+/parenleftBigmω
/planckover2pi1/parenrightBig2/parenleftbig
x−ae−iωt/parenrightbig2/bracketrightbigg
Ψ.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 51
−/planckover2pi12
2m∂2Ψ
∂x2+1
2mω2x2Ψ=−/planckover2pi12
2m/bracketleftbigg
−mω
/planckover2pi1+/parenleftBigmω
/planckover2pi1/parenrightBig2/parenleftbig
x−ae−iωt/parenrightbig2/bracketrightbigg
Ψ+1
2mω2x2Ψ
=/bracketleftbigg1
2/planckover2pi1ω−1
2mω2/parenleftbig
x2−2axe−iωt+a2e−2iωt/parenrightbig
+1
2mω2x2/bracketrightbigg
Ψ
=/bracketleftbigg1
2/planckover2pi1ω+maxω2e−iωt−1
2mω2a2e−2iωt/bracketrightbigg
Ψ
=i/planckover2pi1∂Ψ
∂t(comparing second line above). /check
(b)
|Ψ|2=/radicalbiggmω
π/planckover2pi1e−mω
2/planckover2pi1/bracketleftBig/parenleftBig
x2+a2
2(1+e2iωt)−i/planckover2pi1t
m−2axeiωt/parenrightBig
+/parenleftBig
x2+a2
2(1+e−2iωt)+i/planckover2pi1t
m−2axe−iωt/parenrightBig/bracketrightBig
=/radicalbiggmω
π/planckover2pi1e−mω
2/planckover2pi1[2x2+a2+a2cos(2 ωt)−4axcos(ωt)].Buta2[1 + cos(2 ωt)] = 2a2cos2ωt,so
=/radicalbiggmω
π/planckover2pi1e−mω
/planckover2pi1[x2−2axcos(ωt)+a2cos2(ωt)]=/radicalbiggmω
π/planckover2pi1e−mω
/planckover2pi1(x−acosωt)2.
The wave packet is a Gaussian of fixed shape, whose center oscillates back and forth sinusoidally, with
amplitude aand angular frequency ω.
(c)Note that this wave function iscorrectly normalized (compare Eq. 2.59). Let y≡x−acosωt:
/angbracketleftx/angbracketright=/integraldisplay
x|Ψ|2dx=/integraldisplay
(y+acosωt)|Ψ|2dy=0+acosωt/integraldisplay
|Ψ|2dy=acosωt.
/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt=−maωsinωt.d/angbracketleftp/angbracketright
dt=−maω2cosωt. V =1
2mω2x2=⇒dV
dx=mω2x.
/angbracketleft−dV
dx/angbracketright=−mω2/angbracketleftx/angbracketright=−mω2acosωt=d/angbracketleftp/angbracketright
dt,so Ehrenfest’s theorem issatisfied .
Problem 2.50
(a)
∂Ψ
∂t=/bracketleftbigg
−mα
/planckover2pi12∂
∂t|x−vt|−i(E+1
2mv2)
/planckover2pi1/bracketrightbigg
Ψ;∂
∂t|x−vt|=/braceleftbigg−v,ifx−vt >0
v,ifx−vt <0/bracerightbigg
.
We can write this in terms of the θ-function (Eq. 2.143):
2θ(z)−1=/braceleftbigg1,ifz>0
−1,ifz<0/bracerightbigg
,so∂
∂t|x−vt|=−v[2θ(x−vt)−1].
i/planckover2pi1∂Ψ
∂t=/braceleftbigg
imαv
/planckover2pi1[2θ(x−vt)−1] +E+1
2mv2/bracerightbigg
Ψ.[⋆]
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52 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
∂Ψ
∂x=/bracketleftbigg
−mα
/planckover2pi12∂
∂x|x−vt|+imv
/planckover2pi1/bracketrightbigg
Ψ
∂
∂x|x−vt|={1,ifx>v t ;−1,ifx<v t}=2θ(x−vt)−1.
=/braceleftbigg
−mα
/planckover2pi12[2θ(x−vt)−1] +imv
/planckover2pi1/bracerightbigg
Ψ.
∂2Ψ
∂x2=/braceleftbigg
−mα
/planckover2pi12[2θ(x−vt)−1] +imv
/planckover2pi1/bracerightbigg2
Ψ−2mα
/planckover2pi12/bracketleftbigg∂
∂xθ(x−vt)/bracketrightbigg
Ψ.
But (from Problem 2.24(b))∂
∂xθ(x−vt)=δ(x−vt), so
−/planckover2pi12
2m∂2Ψ
∂x2−αδ(x−vt)Ψ
=/parenleftBigg
−/planckover2pi12
2m/braceleftbigg
−mα
/planckover2pi12[2θ(x−vt)−1] +imv
/planckover2pi1/bracerightbigg2
+αδ(x−vt)−αδ(x−vt)/parenrightBigg
Ψ
=−/planckover2pi12
2m/braceleftbiggm2α2
/planckover2pi14[2θ(x−vt)−1]2
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
1−m2v2
/planckover2pi12−2imv
/planckover2pi1mα
/planckover2pi12[2θ(x−vt)−1]/bracerightbigg
Ψ
=/braceleftbigg
−mα2
2/planckover2pi12+1
2mv2+imvα
/planckover2pi1[2θ(x−vt)−1]/bracerightbigg
Ψ=i/planckover2pi1∂Ψ
∂t(compare [ ⋆])./check
(b)
|Ψ|2=mα
/planckover2pi12e−2mα|y|//planckover2pi12(y≡x−vt).
Check normalization: 2mα
/planckover2pi12/integraldisplay∞
0e−2mαy/ /planckover2pi12dy=2mα
/planckover2pi12/planckover2pi12
2mα=1./check
/angbracketleftH/angbracketright=/integraldisplay∞
−∞Ψ∗HΨdx.ButHΨ=i/planckover2pi1∂Ψ
∂t,which we calculated above [ ⋆].
=/integraldisplay/braceleftbiggimαv
/planckover2pi1[2θ(y)−1] +E+1
2mv2/bracerightbigg
|Ψ|2dy=E+1
2mv2.
(Note that [2 θ(y)−1] is an oddfunction of y.)Interpretation: The wave packet is dragged along (at speed
v) with the delta-function. The total energy is the energy it would have in a stationary delta-function
(E), plus kinetic energy due to the motion (1
2mv2).
Problem 2.51
(a)Figure at top of next page.
(b)dψ0
dx=−Aasech(ax) tanh(ax);d2ψ0
dx2=−Aa2/bracketleftbig
−sech(ax) tanh2(ax) + sech( ax) sech2(ax)/bracketrightbig
.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 53
V(x)
x
Hψ0=−/planckover2pi12
2md2ψ0
dx2−/planckover2pi12a2
msech2(ax)ψ0
=/planckover2pi12
2mAa2/bracketleftbig
−sech(ax) tanh2(ax) + sech3(ax)/bracketrightbig
−/planckover2pi12a2
mAsech3(ax)
=/planckover2pi12a2A
2m/bracketleftbig
−sech(ax) tanh2(ax) + sech3(ax)−2 sech3(ax)/bracketrightbig
=−/planckover2pi12a2
2mAsech(ax)/bracketleftbig
tanh2(ax) + sech2(ax)/bracketrightbig
.
But (tanh2θ+ sech2θ)=sinh2θ
cosh2θ+1
cosh2θ=sinh2θ+1
cosh2θ=1,so
=−/planckover2pi12a2
2mψ0.QED Evidently E=−/planckover2pi12a2
2m.
1=|A|2/integraldisplay∞
−∞sech2(ax)dx=|A|21
atanh(ax)/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞=2
a|A|2=⇒A=/radicalbigga
2.
xψ(x)
(c)
dψk
dx=A
ik+a/bracketleftbig
(ik−atanhax)ik−a2sech2ax/bracketrightbig
eikx.
d2ψk
dx2=A
ik+a/braceleftbig
ik/bracketleftbig
(ik−atanhax)ik−a2sech2ax/bracketrightbig
−a2iksech2ax+2a3sech2axtanhax/bracerightbig
eikx.
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54 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
−/planckover2pi12
2md2ψk
dx2+Vψk=A
ik+a/braceleftbigg−/planckover2pi12ik
2m/bracketleftbig
−k2−iaktanhax−a2sech2ax/bracketrightbig
+/planckover2pi12a2
2miksech2ax
−/planckover2pi12a3
msech2axtanhax−/planckover2pi12a2
msech2ax(ik−atanhax)/bracerightbigg
eikx
=Aeikx
ik+a/planckover2pi12
2m/parenleftbig
ik3−ak2tanhax+ia2ksech2ax+ia2ksech2ax
−2a3sech2axtanhax−2ia2ksech2ax+2a3sech2axtanhax/parenrightbig
=Aeikx
ik+a/planckover2pi12
2mk2(ik−atanhax)=/planckover2pi12k2
2mψk=Eψk.QED
Asx→+∞,tanhax→+1, so ψk(x)→A/parenleftbiggik−a
ik+a/parenrightbigg
eikx,which represents a transmitted wave.
R=0.T=/vextendsingle/vextendsingle/vextendsingle/vextendsingleik−a
ik+a/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=/parenleftbigg−ik−a
−ik+a/parenrightbigg/parenleftbiggik−a
ik+a/parenrightbigg
=1.
Problem 2.52
(a)(1) From Eq. 2.133: F+G=A+B.
(2) From Eq. 2.135: F−G=( 1+2 iβ)A−(1−2iβ)B, where β=mα//planckover2pi12k.
Subtract: 2 G=−2iβA+ 2(1−iβ)B⇒B=1
1−iβ(iβA+G).Multiply (1) by (1 −2iβ) and add:
2(1−iβ)F−2iβG=2A⇒F=1
1−iβ(A+iβG).S=1
1−iβ/parenleftbiggiβ1
1iβ/parenrightbigg
.
(b)For an evenpotential, V(−x)=V(x), scattering from the right is the same as scattering from the left, with
x↔−x,A↔G,B↔F(see Fig. 2.22): F=S11G+S12A, B =S21G+S22A.SoS11=S22,S21=S12.
(Note that the delta-well Smatrix in (a) has this property.) In the case of the finite square well, Eqs. 2.167
and 2.168 give
S21=e−2ika
cos 2la−i(k2+l2)
2klsin 2la;S11=i(l2−k2)
2klsin 2lae−2ika
cos 2la−i(k2+l2)
2klsin 2la.So
S=e−2ika
cos 2la−i(k2+l2)
2klsin 2la/parenleftBigg
i(l2−k2)
2klsin 2la 1
1 i(l2−k2)
2klsin 2la/parenrightBigg
.
Problem 2.53
(a)
B=S11A+S12G⇒G=1
S12(B−S11A)=M21A+M22B⇒M21=−S11
S12,M22=1
S12.
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CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION 55
F=S21A+S22B=S21A+S22
S12(B−S11A)=−(S11S22−S12S21)
S12A+S22
S12B=M11A+M12B.
⇒M11=−detS
S12,M12=S22
S12.M=1
S12/parenleftbigg
−det(S)S22
−S111/parenrightbigg
.Conversely:
G=M21A+M22B⇒B=1
M22(G−M21A)=S11A+S12G⇒S11=−M21
M22;S12=1
M22.
F=M11A+M12B=M11A+M12
M22(G−M21A)=(M11M22−M12M21)
M22A+M12
M22G=S21A+S22G.
⇒S21=detM
M22;S22=M12
M22.S=1
M22/parenleftbigg−M211
det(M)M12/parenrightbigg
.
[It happens that the time-reversal invariance of the Schr¨ odinger equation, plus conservation of probability,
requires M22=M∗
11,M21=M∗
12, and det( M) = 1, but I won’t use this here. See Merzbacher’s Quantum
Mechanics . Similarly, for evenpotentials S11=S22,S12=S21(Problem 2.52).]
Rl=|S11|2=/vextendsingle/vextendsingle/vextendsingle/vextendsingleM
21
M22/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
,Tl=|S21|2=/vextendsingle/vextendsingle/vextendsingle/vextendsingledet(M)
M22/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
,Rr=|S22|2=/vextendsingle/vextendsingle/vextendsingle/vextendsingleM
12
M22/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
,Tr=|S12|2=1
|M22|2.
(b)
A
BC
DF
G
x M M1 2
/parenleftbiggF
G/parenrightbigg
=M2/parenleftbiggC
D/parenrightbigg
,/parenleftbiggC
D/parenrightbigg
=M1/parenleftbiggA
B/parenrightbigg
,so/parenleftbiggF
G/parenrightbigg
=M2M1/parenleftbiggA
B/parenrightbigg
=M/parenleftbiggA
B/parenrightbigg
,withM=M2M1.QED
(c)
ψ(x)=/braceleftbiggAeikx+Be−ikx(x<a)
Feikx+Ge−ikx(x>a)/bracerightbigg
.
/braceleftbiggContinuity of ψ:Aeika+Be−ika=Feika+Ge−ika
Discontinuity of ψ/prime:ik/parenleftbig
Feika−Ge−ika/parenrightbig
−ik/parenleftbig
Aeika−Be−ika/parenrightbig
=−2mα
/planckover2pi12ψ(a)=−2mα
/planckover2pi12/parenleftbig
Aeika+Be−ika/parenrightbig
.
(1)Fe2ika+G=Ae2ika+B.
(2)Fe2ika−G=Ae2ika−B+i2mα
/planckover2pi12k/parenleftbig
Ae2ika+B/parenrightbig
.
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56 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATION
Add (1) and (2):
2Fe2ika=2Ae2ika+i2mα
/planckover2pi12k/parenleftbig
Ae2ika+B/parenrightbig
⇒F=/parenleftBig
1+imα
/planckover2pi12k/parenrightBig
A+imα
/planckover2pi12ke−2ikaB=M11A+M12B.
SoM11=( 1+iβ);M12=iβe−2ika;β≡mα
/planckover2pi12k.
Subtract (2) from (1):
2G=2B−2iβe2ikaA−2iβB⇒G=( 1−iβ)B−iβe2ikaA=M21A+M22B.
SoM21=−iβe2ika;M22=( 1−iβ). M=/parenleftbigg(1 +iβ)iβe−2ika
−iβe2ika(1−iβ)/parenrightbigg
.
(d)
M1=/parenleftbigg(1 +iβ)iβe−2ika
−iβe2ika(1−iβ)/parenrightbigg
; to get M2,just switch the sign of a:M2=/parenleftbigg(1 +iβ)iβe2ika
−iβe−2ika(1−iβ)/parenrightbigg
.
M=M2M1=/parenleftbigg[ 1+2iβ+β2(e4ika−1)] 2iβ[cos 2ka+βsin 2ka]
−2iβ[cos 2ka+βsin 2ka][ 1−2iβ+β2(e−4ika−1)]/parenrightbigg
.
T=Tl=Tr=1
|M22|2⇒
T−1=[ 1+2 iβ+β2(e4ika−1)][1−2iβ+β2(e−4ika−1)]
=1−2iβ+β2e−4ika−β2+2iβ+4β2+2iβ3e−4ika−2iβ3+β2e4ika
−β2−2iβ3e4ika+2iβ3+β4(1−e4ika−e−4ika+1 )
=1+2 β2+β2(e4ika+e−4ika)−2iβ3(e4ika−e−4ika)+2β4−β4(e4ika+e−4ika)
=1+2 β2+2β2cos 4ka−2iβ32isin 4ka+2β4−2β4cos 4ka
=1+2 β2(1 + cos 4 ka)+4β3sin 4ka+2β4(1−cos 4ka)
=1+4 β2cos22ka+8β3sin 2kacos 2ka+4β4sin22ka
T=1
1+4β2(cos 2ka+βsin 2ka)2
Problem 2.54
I’ll just show the first two graphs, and the last two. Evidently Klies between 0.9999 and 1.0001.
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57
Problem 2.55
Thecorrect values (in Eq. 2.72) are K=2n+ 1 (corresponding to En=(n+1
2)/planckover2pi1ω). I’ll start by “guessing”
2.9, 4.9, and 6.9, and tweaking the number until I’ve got 5 reliable significant digits. The results (see below)
are3.0000, 5.0000, 7.0000. (The actual energies are these numbers multiplied by1
2/planckover2pi1ω.)
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58
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59
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60
Problem 2.56
The Schr¨ odinger equation says −/planckover2pi12
2mψ/prime/prime=Eψ, or, with the correct energies (Eq. 2.27) and a=1 ,ψ/prime/prime+(nπ)2ψ=
0. I’ll start with a “guess” using 9 in place of π2(that is, I’ll use 9 for the ground state, 36 for the first excited
state, 81 for the next, and finally 144). Then I’ll tweak the parameter until the graph crosses the axis right
atx= 1. The results (see below) are, to five significant digits: 9.8696, 39.478, 88.826, 157.91. (The actual
energies are these numbers multiplied by /planckover2pi12/2ma2.)
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61
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62 CHAPTER 3. FORMALISM
Chapter 3
Formalism
Problem 3.1
(a)All conditions are trivial except Eq. A.1: we need to show that the sum of two square-integrable functions
is itself square-integrable. Let h(x)=f(x)+g(x), so that|h|2=(f+g)∗(f+g)=|f|2+|g|2+f∗g+g∗f
and hence /integraldisplay
|h|2dx=/integraldisplay
|f|2dx+/integraldisplay
|g|2dx+/integraldisplay
f∗gdx+/parenleftbigg/integraldisplay
f∗gdx/parenrightbigg∗
.
Iff(x) andg(x) are square-integrable, then the first two terms are finite, and (by Eq. 3.7) so too are the
last two. So/integraltext
|h|2dxis finite. QED
The set of all normalized functions is certainly nota vector space: it doesn’t include 0, and the sum of
two normalized functions is not (in general) normalized—in fact, if f(x) is normalized, then the square
integral of 2 f(x)i s4 .
(b)Equation A.19 is trivial:
/angbracketleftg|f/angbracketright=/integraldisplayb
ag(x)∗f(x)dx=/parenleftBigg/integraldisplayb
af(x)∗g(x)dx/parenrightBigg∗
=/angbracketleftf|g/angbracketright∗.
Equation A.20 holds (see Eq. 3.9) subject to the understanding in footnote 6. As for Eq. A.21, this is
pretty obvious:
/angbracketleftf|(b|g/angbracketright+c|h/angbracketright)=/integraldisplay
f(x)∗(bg(x)+ch(x))dx=b/integraldisplay
f∗gdx+c/integraldisplay
f∗hdx=b/angbracketleftf|g/angbracketright+c/angbracketleftf|h/angbracketright.
Problem 3.2
(a)
/angbracketleftf|f/angbracketright=/integraldisplay1
0x2νdx=1
2ν+1x2ν+1/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
0=1
2ν+1/parenleftbig
1−02ν+1/parenrightbig
.
Now 02ν+1is finite (in fact, zero) provided (2 ν+1 )>0, which is to say, ν>−1
2.If (2ν+1 )<0 the
integral definitely blows up. As for the critical case ν=−1
2, this must be handled separately:
/angbracketleftf|f/angbracketright=/integraldisplay1
0x−1dx=l nx/vextendsingle/vextendsingle1
0=l n1−ln 0 = 0 +∞.
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CHAPTER 3. FORMALISM 63
Sof(x) is in Hilbert space only for νstrictly greater than -1/2.
(b)Forν=1/2, we know from (a) that f(x)isin Hilbert space: yes.
Sincexf=x3/2,we know from (a) that it isin Hilbert space: yes.
Fordf/dx =1
2x−1/2, we know from (a) that it is notin Hilbert space: no.
[Moral: Simple operations, such as differenting (or multiplying by 1 /x), can carry a function outof Hilbert
space.]
Problem 3.3
Suppose/angbracketlefth|ˆQh/angbracketright=/angbracketleftˆQh|h/angbracketrightfor all functions h(x). Leth(x)=f(x)+cg(x) for some arbitrary constant c. Then
/angbracketlefth|ˆQh/angbracketright=/angbracketleft(f+cg)|ˆQ(f+cg)/angbracketright=/angbracketleftf|ˆQf/angbracketright+c/angbracketleftf|ˆQg/angbracketright+c∗/angbracketleftg|ˆQf/angbracketright+|c|2/angbracketleftg|ˆQg/angbracketright;
/angbracketleftˆQh|h/angbracketright=/angbracketleftˆQ(f+cg)|(f+cg)/angbracketright=/angbracketleftˆQf|f/angbracketright+c/angbracketleftˆQf|g/angbracketright+c∗/angbracketleftˆQg|f/angbracketright+|c|2/angbracketleftˆQg|g/angbracketright.
Equating the two and noting that /angbracketleftf|ˆQf/angbracketright=/angbracketleftˆQf|f/angbracketrightand/angbracketleftg|ˆQg/angbracketright=/angbracketleftˆQg|g/angbracketrightleaves
c/angbracketleftf|ˆQg/angbracketright+c∗/angbracketleftg|ˆQf/angbracketright=c/angbracketleftˆQf|g/angbracketright+c∗/angbracketleftˆQg|f/angbracketright.
In particlar, choosing c=1 :
/angbracketleftf|ˆQg/angbracketright+/angbracketleftg|ˆQf/angbracketright=/angbracketleftˆQf|g/angbracketright+/angbracketleftˆQg|f/angbracketright,
whereas if c=i:
/angbracketleftf|ˆQg/angbracketright−/angbracketleftg|ˆQf/angbracketright=/angbracketleftˆQf|g/angbracketright−/angbracketleftˆQg|f/angbracketright.
Adding the last two equations:
/angbracketleftf|ˆQg/angbracketright=/angbracketleftˆQf|g/angbracketright.QED
Problem 3.4
(a)/angbracketleftf|(ˆH+ˆK)g/angbracketright=/angbracketleftf|ˆHg/angbracketright+/angbracketleftf|ˆKg/angbracketright=/angbracketleftˆHf|g/angbracketright+/angbracketleftˆKf|g/angbracketright=/angbracketleft(ˆH+ˆK)f|g/angbracketright./check
(b)/angbracketleftf|αˆQg/angbracketright=α/angbracketleftf|ˆQg/angbracketright;/angbracketleftαˆQf|g/angbracketright=α∗/angbracketleftˆQf|g/angbracketright.Hermitian⇔αis real.
(c)/angbracketleftf|ˆHˆKg/angbracketright=/angbracketleftˆHf|ˆKg/angbracketright=/angbracketleftˆKˆHf|g/angbracketright,s oˆHˆKis hermitian⇔ˆHˆK=ˆKˆH,o r[ˆH,ˆK]=0.
(d)/angbracketleftf|ˆxg/angbracketright=/integraltext
f∗(xg)dx=/integraltext
(xf)∗gdx=/angbracketleftˆxf|g/angbracketright./check
/angbracketleftf|ˆHg/angbracketright=/integraldisplay
f∗/parenleftbigg
−/planckover2pi12
2md2
dx2+V/parenrightbigg
gdx=−/planckover2pi12
2m/integraldisplay
f∗d2g
dx2dx+/integraldisplay
f∗V gdx.
Integrating by parts (twice):
/integraldisplay∞
−∞f∗d2g
dx2dx=f∗dg
dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞−/integraldisplay∞
−∞df∗
dxdg
dxdx=f∗dg
dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞−df∗
dxg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞+/integraldisplay∞
−∞d2f∗
dx2gdx.
But for functions f(x) andg(x) in Hilbert space the boundary terms vanish, so
/integraldisplay∞
−∞f∗d2g
dx2dx=/integraldisplay∞
−∞d2f∗
dx2gdx, and hence (assuming that V(x) is real):
/angbracketleftf|ˆHg/angbracketright=/integraldisplay∞
−∞/parenleftbigg
−/planckover2pi12
2md2f
dx2+Vf/parenrightbigg∗
gdx=/angbracketleftˆHf|g/angbracketright./check
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64 CHAPTER 3. FORMALISM
Problem 3.5
(a)/angbracketleftf|xg/angbracketright=/integraltext
f∗(xg)dx=/integraltext
(xf)∗gdx=/angbracketleftxf|g/angbracketright,s ox†=x.
/angbracketleftf|ig/angbracketright=/integraltext
f∗(ig)dx=/integraltext
(−if)∗gdx=/angbracketleft−if|g/angbracketright,s oi†=−i.
/angbracketleftf|dg
dx/angbracketright=/integraldisplay∞
−∞f∗dg
dxdx=f∗g/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
−∞−/integraldisplay∞
−∞/parenleftbiggdf
dx/parenrightbigg∗
gdx=−/angbracketleftxf|g/angbracketright,s o/parenleftbiggd
dx/parenrightbigg†
=−d
dx.
(b)a+=1√
2/planckover2pi1mω(−ip+mωx).Butpandxare hermitian, and i†=−i,s o(a+)†=1√
2/planckover2pi1mω(ip+mωx),or
(a+)†=(a−).
(c)/angbracketleftf|(ˆQˆR)g/angbracketright=/angbracketleftˆQ†f|ˆRg/angbracketright=/angbracketleftˆR†ˆQ†f|g/angbracketright=/angbracketleft(ˆQˆR)†f|g/angbracketright,s o(ˆQˆR)†=ˆR†ˆQ†./check
Problem 3.6
/angbracketleftf|ˆQg/angbracketright=/integraldisplay2π
0f∗d2g
dφ2dφ=f∗dg
dφ/vextendsingle/vextendsingle/vextendsingle/vextendsingle2π
0−/integraldisplay2π
0df∗
dφdg
dφdφ=f∗dg
dφ/vextendsingle/vextendsingle/vextendsingle/vextendsingle2π
0−df∗
dφg/vextendsingle/vextendsingle/vextendsingle/vextendsingle2π
0+/integraldisplay2π
0d2f∗
dφ2gdφ.
As in Example 3.1, for periodic functions (Eq. 3.26) the boundary terms vanish, and we conclude that /angbracketleftf|ˆQg/angbracketright=
/angbracketleftˆQf|g/angbracketright,s oˆQis hermitian: yes.
ˆQf=qf⇒d2f
dφ2=qf⇒f±(φ)=Ae±√qφ.
The periodicity condition (Eq. 3.26) requires that√q(2π)=2nπi,o r√q=in, so the eigenvalues are
q=−n2,(n=0,1,2,...).The spectrum is doubly degenerate; for a given nthere are twoeigenfunctions
(the plus sign or the minus sign, in the exponent), except for the special case n= 0, which is not degenerate.
Problem 3.7
(a)Suppose ˆQf=qfandˆQg=qg.Leth(x)=af(x)+bg(x), for arbitrary constants aandb. Then
ˆQh=ˆQ(af+bg)=a(ˆQf)+b(ˆQg)=a(qf)+b(qg)=q(af+bg)=qh./check
(b)d2f
dx2=d2
dx2(ex)=d
dx(ex)=ex=f,d2g
dx2=d2
dx2/parenleftbig
e−x/parenrightbig
=d
dx/parenleftbig
−e−x/parenrightbig
=e−x=g.
So both of them are eigenfunctions, with the same eigenvalue 1. The simplest orthogonal linear combina-
tions are
sinhx=1
2/parenleftbig
ex−e−x/parenrightbig
=1
2(f−g) and cosh x=1
2/parenleftbig
ex+e−x/parenrightbig
=1
2(f+g).
(They are clearly orthogonal, since sinh xis odd while cosh xis even.)
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CHAPTER 3. FORMALISM 65
Problem 3.8
(a)The eigenvalues (Eq. 3.29) are 0 ,±1,±2,..., which are obviously real. /checkFor any two eigenfunctions,
f=Aqe−iqφandg=Aq/primee−iq/primeφ(Eq. 3.28), we have
/angbracketleftf|g/angbracketright=A∗
qAq/prime/integraldisplay2π
0eiqφe−iq/primeφdφ=A∗
qAq/primeei(q−q/prime)φ
i(q−q/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2π
0=A∗
qAq/prime
i(q−q/prime)/bracketleftBig
ei(q−q/prime)2π−1/bracketrightBig
.
Butqandq/primeareintegers ,s oei(q−q/prime)2π= 1, and hence /angbracketleftf|g/angbracketright= 0 (provided q/negationslash=q/prime, so the denominator is
nonzero). /check
(b)In Problem 3.6 the eigenvalues are q=−n2, withn=0,1,2,..., which are obviously real. /checkFor any
two eigenfunctions, f=Aqe±inφandg=Aq/primee±in/primeφ,w eh a v e
/angbracketleftf|g/angbracketright=A∗
qAq/prime/integraldisplay2π
0e∓inφe±in/primeφdφ=A∗
qAq/primee±i(n/prime−n)φ
±i(n/prime−n)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2π
0=A∗
qAq/prime
±i(n/prime−n)/bracketleftBig
e±i(n/prime−n)2π−1/bracketrightBig
=0
(provided n/negationslash=n/prime). But notice that for each eigenvalue (i.e. each value of n) there are twoeigenfunctions
(one with the plus sign and one with the minus sign), and these are notorthogonal to one another.
Problem 3.9
(a)Infinite square well (Eq. 2.19).
(b)Delta-function barrier (Fig. 2.16), or the finite rectangular barrier (Prob. 2.33).
(c)Delta-function well (Eq. 2.114), or the finite square well (Eq. 2.145) or the sech2potential (Prob. 2.51).
Problem 3.10
From Eq. 2.28, with n=1 :
ˆpψ1(x)=/planckover2pi1
id
dx/radicalbigg
2
asin/parenleftBigπ
ax/parenrightBig
=/planckover2pi1
i/radicalbigg
2
aπ
acos/parenleftBigπ
ax/parenrightBig
=/bracketleftbigg
−iπ/planckover2pi1
acot/parenleftBigπ
ax/parenrightBig/bracketrightbigg
ψ1(x).
Since ˆpψ1isnota (constant) multiple of ψ1,ψ1is not an eigenfunction of ˆ p:no.It’s true that the magnitude
of the momentum,√2mE1=π/planckover2pi1/a, is determinate, but the particle is just as likely to be found traveling to the
left (negative momentum) as to the right (positive momentum).
Problem 3.11
Ψ0(x,t)=/parenleftbiggmω
π/planckover2pi1/parenrightbigg1/4
e−mω
2/planckover2pi1x2e−iωt/2;Φ (p,t)=1√
2π/planckover2pi1/parenleftbiggmω
π/planckover2pi1/parenrightbigg1/4
e−iω/2/integraldisplay∞
−∞e−ipx//planckover2pi1e−mω
2/planckover2pi1x2dx.
From Problem 2.22(b):
Φ(p,t)=1√
2π/planckover2pi1/parenleftbiggmω
π/planckover2pi1/parenrightbigg1/4
e−iωt/2/radicalbigg
2π/planckover2pi1
mωe−p2/2mω/planckover2pi1=1
(πmω /planckover2pi1)1/4e−p2/2mω/planckover2pi1e−iωt/2.
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66 CHAPTER 3. FORMALISM
|Φ(p,t)|2=1√
πmω /planckover2pi1e−p2/mω/planckover2pi1.Maximum classical momentum:p2
2m=E=1
2/planckover2pi1ω=⇒p=√
mω/planckover2pi1.
So the probability it’s outside classical range is:
P=/integraldisplay−√
mω/planckover2pi1
−∞|Φ|2dp+/integraldisplay∞
√
mω/planckover2pi1|Φ|2dp=1−2/integraldisplay√
mω/planckover2pi1
0|Φ|2dp.Now
/integraldisplay√
mω/planckover2pi1
0|Φ|2dp=1√
πmω /planckover2pi1/integraldisplay√
mω/planckover2pi1
0e−p2/mω/planckover2pi1dp.Letz≡/radicalbigg
2
mω/planckover2pi1p,sodp=/radicalbigg
mω/planckover2pi1
2dz.
=1√
2π/integraldisplay√
2
0e−z2/2dz=F(√
2)−1
2,in CRC Table notation.
P=1−2/bracketleftbigg
(F(√
2)−1
2/bracketrightbigg
=1−2F(√
2) + 1 = 2/bracketleftBig
1−F(√
2)/bracketrightBig
=0.157.
To two digits: 0.16(compare Prob. 2.15).
Problem 3.12
From Eq. 3.55: Ψ( x,t)=1√
2π/planckover2pi1/integraldisplay∞
−∞eipx//planckover2pi1Φ(p,t)dp.
/angbracketleftx/angbracketright=/integraldisplay
Ψ∗xΨdx=/integraldisplay/bracketleftbigg1√
2π/planckover2pi1/integraldisplay
e−ip/primex//planckover2pi1Φ∗(p/prime,t)dp/prime/bracketrightbigg
x/bracketleftbigg1√
2π/planckover2pi1/integraldisplay
e+ipx//planckover2pi1Φ(p,t)dp/bracketrightbigg
dx.
Butxeipx//planckover2pi1=−i/planckover2pi1d
dp/parenleftbig
eipx//planckover2pi1/parenrightbig
,so (integrating by parts):
x/integraldisplay
eipx//planckover2pi1Φdp=/integraldisplay/planckover2pi1
id
dp/parenleftbig
eipx//planckover2pi1)Φdp=/integraldisplay
eipx//planckover2pi1/bracketleftbigg
−/planckover2pi1
i∂
∂pΦ(p,t)/bracketrightbigg
dp.
So/angbracketleftx/angbracketright=1
2π/planckover2pi1/integraldisplay/integraldisplay/integraldisplay /braceleftbigg
e−ip/primex//planckover2pi1Φ∗(p/prime,t)eipx//planckover2pi1/bracketleftbigg
−/planckover2pi1
i∂
∂pΦ(p,t)/bracketrightbigg/bracerightbigg
dp/primedpdx.
Do the xintegral first, letting y≡x//planckover2pi1:
1
2π/planckover2pi1/integraldisplay
e−ip/primex//planckover2pi1eipx//planckover2pi1dx=1
2π/integraldisplay
ei(p−p/prime)ydy=δ(p−p/prime),(Eq. 2.144), so
/angbracketleftx/angbracketright=/integraldisplay/integraldisplay
Φ∗(p/prime,t)δ(p−p/prime)/bracketleftbigg
−/planckover2pi1
i∂
∂pΦ(p,t)/bracketrightbigg
dp/primedp=/integraldisplay
Φ∗(p,t)/bracketleftbigg
−/planckover2pi1
i∂
∂pΦ(p,t)/bracketrightbigg
dp.QED
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CHAPTER 3. FORMALISM 67
Problem 3.13
(a)[AB,C ]=ABC−CAB =ABC−ACB +ACB−CAB =A[B,C]+[A,C]B./check
(b)Introducing a test function g(x), as in Eq. 2.50:
[xn,p]g=xn/planckover2pi1
idg
dx−/planckover2pi1
id
dx(xng)=xn/planckover2pi1
idg
dx−/planckover2pi1
i/parenleftbigg
nxn−1g+xndg
dx/parenrightbigg
=i/planckover2pi1nxn−1g.
So, dropping the test function, [ xn,p]=i/planckover2pi1nxn−1./check
(c)[f,p]g=f/planckover2pi1
idg
dx−/planckover2pi1
id
dx(fg)=f/planckover2pi1
idg
dx−/planckover2pi1
i/parenleftbiggdf
dxg+fdg
dx/parenrightbigg
=i/planckover2pi1df
dxg⇒[f,p]=i/planckover2pi1df
dx./check
Problem 3.14
/bracketleftbigg
x,p2
2m+V/bracketrightbigg
=1
2m/bracketleftbig
x,p2/bracketrightbig
+[x,V];/bracketleftbig
x,p2/bracketrightbig
=xp2−p2x=xp2−pxp+pxp−p2x=[x,p]p+p[x,p].
Using Eq. 2.51:/bracketleftbig
x,p2/bracketrightbig
=i/planckover2pi1p+pi/planckover2pi1=2i/planckover2pi1p.And [x,V]=0,so/bracketleftbigg
x,p2
2m+V/bracketrightbigg
=1
2m2i/planckover2pi1p=i/planckover2pi1p
m.
The generalized uncertainty principle (Eq. 3.62) says, in this case,
σ2
xσ2
H≥/parenleftbigg1
2ii/planckover2pi1
m/angbracketleftp/angbracketright/parenrightbigg2
=/parenleftbigg/planckover2pi1
2m/angbracketleftp/angbracketright/parenrightbigg2
⇒σxσH≥/planckover2pi1
2m|/angbracketleftp/angbracketright|.QED
For stationary states σH= 0 and/angbracketleftp/angbracketright= 0, so it just says 0 ≥0.
Problem 3.15
Suppose ˆPfn=λnfnandˆQfn=µnfn(that is: fn(x) is an eigenfunction both of ˆPand of ˆQ), and the set {fn}
is complete, so that any function f(x) (in Hilbert space) can be expressed as a linear combination: f=/summationtextcnfn.
Then
[ˆP,ˆQ]f=(ˆPˆQ−ˆQˆP)/summationdisplay
cnfn=ˆP/parenleftBig/summationdisplay
cnµnfn/parenrightBig
−ˆQ/parenleftBig/summationdisplay
cnλnfn/parenrightBig
=/summationdisplay
cnµnλnfn−/summationdisplay
cnλnµnfn=0.
Since this is true for anyfunction f, it follows that [ ˆP,ˆQ]=0.
Problem 3.16
dΨ
dx=i
/planckover2pi1(iax−ia/angbracketleftx/angbracketright+/angbracketleftp/angbracketright)Ψ =a
/planckover2pi1/parenleftbigg
−x+/angbracketleftx/angbracketright+i
a/angbracketleftp/angbracketright/parenrightbigg
Ψ.
dΨ
Ψ=a
/planckover2pi1/parenleftbigg
−x+/angbracketleftx/angbracketright+i/angbracketleftp/angbracketright
a/parenrightbigg
dx⇒ln Ψ =a
/planckover2pi1/parenleftbigg
−x2
2+/angbracketleftx/angbracketrightx+i/angbracketleftp/angbracketright
ax/parenrightbigg
+constant.
Letconstant =−/angbracketleftx/angbracketright2a
2/planckover2pi1+B(Ba new constant). Then ln Ψ = −a
2/planckover2pi1(x−/angbracketleftx/angbracketright)2+i/angbracketleftp/angbracketright
/planckover2pi1x+B.
Ψ=e−a
2/planckover2pi1(x−/angbracketleftx/angbracketright)2ei/angbracketleftp/angbracketrightx//planckover2pi1eB=Ae−a(x−/angbracketleftx/angbracketright)2/2/planckover2pi1ei/angbracketleftp/angbracketrightx//planckover2pi1,whereA≡eB.
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68 CHAPTER 3. FORMALISM
Problem 3.17
(a)1 commutes with everything, sod
dt/angbracketleftΨ|Ψ/angbracketright=0(this is the conservation of normalization, which we origi-
nally proved in Eq. 1.27).
(b)Anything commutes with itself, so [ H,H] = 0, and henced
dt/angbracketleftH/angbracketright=0(assuming Hhas no explicit time
dependence); this is conservation of energy, in the sense of the comment following Eq. 2.40.
(c)[H,x]=−i/planckover2pi1p
m(see Problem 3.14). Sod/angbracketleftx/angbracketright
dt=i
/planckover2pi1/parenleftbigg
−i/planckover2pi1/angbracketleftp/angbracketright
m/parenrightbigg
=/angbracketleftp/angbracketright
m(Eq. 1.33).
(d)[H,p]=/bracketleftbiggp2
2m+V,p/bracketrightbigg
=[V,p]=i/planckover2pi1dV
dx(Problem 3.13(c)). Sod/angbracketleftp/angbracketright
dt=i
/planckover2pi1/parenleftbigg
i/planckover2pi1/angbracketleftbigg∂V
∂x/angbracketrightbigg/parenrightbigg
=−/angbracketleftbigg∂V
∂x/angbracketrightbigg
.
This is Ehrenfest’s theorem (Eq. 1.38).
Problem 3.18
Ψ(x,t)=1√
2/parenleftbig
ψ1e−iE1t//planckover2pi1+ψ2e−E2t//planckover2pi1/parenrightbig
.H2Ψ=1√
2/bracketleftbig
(H2ψ1)e−E1t//planckover2pi1+(H2ψ2)e−iEnt//planckover2pi1/bracketrightbig
.
Hψ1=E1ψ1⇒H2ψ1=E1Hψ1=E2
1ψ1,andH2ψ2=E2
2ψ2,so
/angbracketleftH2/angbracketright=1
2/angbracketleft/parenleftbig
ψ1e−iE1t//planckover2pi1+ψ2e−iE2t//planckover2pi1/parenrightbig
|/parenleftbig
E2
1ψ1e−iE1t//planckover2pi1+E2
2ψ2e−iE2t//planckover2pi1/parenrightbig
/angbracketright
=1
2/parenleftbig
/angbracketleftψ1|ψ1/angbracketrighteiE1t//planckover2pi1E2
1e−iE1t//planckover2pi1+/angbracketleftψ1|ψ2/angbracketrighteiE1t//planckover2pi1E2
2e−iE2t//planckover2pi1
+/angbracketleftψ2|ψ1/angbracketrighteiE2t//planckover2pi1E2
1e−iE1t//planckover2pi1+/angbracketleftψ2|ψ2/angbracketrighteiE2t//planckover2pi1E2
2e−iE2t//planckover2pi1/parenrightbig
=1
2/parenleftbig
E2
1+E2
2/parenrightbig
.
Similarly,/angbracketleftH/angbracketright=1
2(E1+E2) (Problem 2.5(e)) .
σ2
H=/angbracketleftH2/angbracketright−/angbracketleftH/angbracketright2=1
2/parenleftbig
E2
1+E2
2/parenrightbig
−1
4(E1+E2)2=1
4/parenleftbig
2E2
1+2E2
2−E2
2−E2
1−2E1E2−E2
2/parenrightbig
=1
4/parenleftbig
E2
1−2E1E2+E2
2/parenrightbig
=1
4(E2−E1)2.σH=1
2(E2−E1).
/angbracketleftx2/angbracketright=1
2/bracketleftbig
/angbracketleftψ1|x2|ψ1/angbracketright+/angbracketleftψ2|x2|ψ2/angbracketright+/angbracketleftψ1|x2|ψ2/angbracketrightei(E1−E2)t//planckover2pi1+/angbracketleftψ2|x2|ψ1/angbracketrightei(E2−E1)t//planckover2pi1/bracketrightbig
.
/angbracketleftψn|x2|ψm/angbracketright=2
a/integraldisplaya
0x2sin/parenleftbiggnπ
ax/parenrightbigg
sin/parenleftbiggmπ
ax/parenrightbigg
dx=1
a/integraldisplaya
0x2/bracketleftbigg
cos/parenleftbiggn−m
aπx/parenrightbigg
−cos/parenleftbiggn+m
aπx/parenrightbigg/bracketrightbigg
dx.
Now/integraldisplaya
0x2cos/parenleftbiggk
aπx/parenrightbigg
dx=/braceleftbigg2a2x
k2π2cos/parenleftbiggk
aπx/parenrightbigg
+/parenleftbigga
kπ/parenrightbigg3/bracketleftbigg/parenleftbiggkπx
a/parenrightbigg2
−2/bracketrightbigg
sin/parenleftbiggk
aπx/parenrightbigg/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0
=2a3
k2π2cos(kπ)=2a3
k2π2(−1)k(fork= nonzero integer) .
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CHAPTER 3. FORMALISM 69
∴/angbracketleftψn|x2|ψm/angbracketright=2a2
π2/bracketleftbigg(−1)n−m
(n−m)2−(−1)n+m
(n+m)2/bracketrightbigg
=2a2
π2(−1)n+m4nm
(n2−m2)2.
So/angbracketleftψ1|x2|ψ2/angbracketright=/angbracketleftψ2|x2|ψ1/angbracketright=−16a2
9π2.Meanwhile, from Problem 2.4, /angbracketleftψn|x2|ψn/angbracketright=a2/bracketleftbigg1
3−1
2(nπ)2/bracketrightbigg
.
Thus/angbracketleftx2/angbracketright=1
2/braceleftbigg
a2/bracketleftbigg1
3−1
2π2/bracketrightbigg
+a2/bracketleftbigg1
3−1
8π2/bracketrightbigg
−16a2
9π2/bracketleftbigg
ei(E2−E1)t//planckover2pi1+e−i(E2−E1)t//planckover2pi1
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
2c o s(E2−E1
/planckover2pi1t)/bracketrightbigg/bracerightbigg
.
E2−E1
/planckover2pi1=(4−1)π2/planckover2pi12
2ma2/planckover2pi1=3π2/planckover2pi1
2ma2=3ω[in the notation of Problem 2.5(b)] .
/angbracketleftx2/angbracketright=a2
2/bracketleftbigg2
3−5
8π2−32
9π2cos(3ωt)/bracketrightbigg
.From Problem 2.5(c), /angbracketleftx/angbracketright=a
2/bracketleftbigg
1−32
9π2cos(3ωt)/bracketrightbigg
.
Soσ2
x=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=a2
4/bracketleftbigg4
3−5
4π2−64
9π2cos(3ωt)−1+64
9π2cos(3ωt)−/parenleftbigg32
9π2/parenrightbigg2
cos2(3ωt)/bracketrightbigg
.
σ2
x=a2
4/bracketleftbigg1
3−5
4π2−/parenleftbigg32
9π2/parenrightbigg2
cos2(3ωt)/bracketrightbigg
.And, from Problem 2.5(d):d/angbracketleftx/angbracketright
dt=8/planckover2pi1
3masin(3ωt).
Meanwhile, the energy-time uncertainty principle (Eq. 3.72) says σ2
Hσ2
x≥/planckover2pi12
4/parenleftbiggd/angbracketleftx/angbracketright
dt/parenrightbigg2
.Here
σ2
Hσ2
x=1
4(3/planckover2pi1ω)2a2
4/bracketleftbigg1
3−5
4π2−/parenleftbigg32
9π2/parenrightbigg2
cos2(3ωt)/bracketrightbigg
=(/planckover2pi1ωa)2/parenleftbigg3
4/parenrightbigg2/bracketleftBigg
1
3−5
4π2−/parenleftbigg32
9π2/parenrightbigg2
cos2(3ωt)/bracketrightBigg
.
/planckover2pi12
4/parenleftbiggd/angbracketleftx/angbracketright
dt/parenrightbigg2
=/parenleftbigg/planckover2pi1
28/planckover2pi1
3ma2/parenrightbigg2
sin2(3ωt)=/parenleftbigg8
3π2/parenrightbigg2
(/planckover2pi1ωa)2sin2(3ωt),since/planckover2pi1
ma=2aω
π.
So the uncertainty principle holds if/parenleftbigg3
4/parenrightbigg2/bracketleftbigg1
3−5
4π2−/parenleftbigg32
9π2/parenrightbigg2
cos2(3ωt)/bracketrightbigg
≥/parenleftbigg8
3π2/parenrightbigg2
sin2(3ωt),
which is to say, if
1
3−5
4π2≥/parenleftbigg32
9π2/parenrightbigg2
cos2(3ωt)+/parenleftbigg4
38
3π2/parenrightbigg2
sin2(3ωt)=/parenleftbigg32
9π2/parenrightbigg2
.
Evaluating both sides:1
3−5
4π2=0.20668;/parenleftbig32
9π2/parenrightbig2=0.12978. So it holds. (Whew!)
Problem 3.19
From Problem 2.43, we have:
/angbracketleftx/angbracketright=/planckover2pi1l
mt,sod/angbracketleftx/angbracketright
dt=/planckover2pi1l
m,σ2
x=1
4w2=1+θ2
4a,whereθ=2/planckover2pi1at
m;/angbracketleftH/angbracketright=1
2m/angbracketleftp2/angbracketright=1
2m/planckover2pi12(a+l2).
We need/angbracketleftH2/angbracketright(to get σH). Now, H=p2
2m,so
/angbracketleftH2/angbracketright=1
4m2/angbracketleftp4/angbracketright=1
4m2/integraldisplay∞
−∞p4|Φ(p,t)|2dp,where (Eq. 3.54): Φ( p,t)=1√
2π/planckover2pi1/integraldisplay∞
−∞e−ipx//planckover2pi1Ψ(x,t)dx.
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70 CHAPTER 3. FORMALISM
From Problem 2.43: Ψ( x,t)=/parenleftbigg2a
π/parenrightbigg1/41√
1+iθe−l2
4aea(ix+l
2a)2/(1+iθ).
So Φ(p,t)=1√
2π/planckover2pi1/parenleftbigg2a
π/parenrightbigg1/41√
1+iθe−l2/4a/integraldisplay∞
−∞e−ipx//planckover2pi1ea(ix+l
2a)2/(1+iθ)dx.Lety≡x−il
2a.
=1√
2π/planckover2pi1/parenleftbigg2a
π/parenrightbigg1/41√
1+iθe−l2/4aepl/2a/planckover2pi1/integraldisplay∞
−∞e−ipy//planckover2pi1e−ay2/(1+iθ)dy.
[See Prob. 2.22(a) for the integral.]
=1√
2π/planckover2pi1/parenleftbigg2a
π/parenrightbigg1/41√
1+iθe−l2/4aepl/2a/planckover2pi1/radicalbigg
π(1 +iθ)
ae−p2(1+iθ)
4a/planckover2pi12
=1√
/planckover2pi1/parenleftbigg1
2aπ/parenrightbigg1/4
e−l2
4aepl
2a/planckover2pi1e−p2(1+iθ)
4a/planckover2pi12.
|Φ(p,t)|2=1√
2aπ1
/planckover2pi1e−l2/2aepl/a/planckover2pi1e−p2/2a/planckover2pi12=1
/planckover2pi1√
2aπe1
2a/parenleftBig
l2−2pl
/planckover2pi1+p2
/planckover2pi12/parenrightBig
=1
/planckover2pi1√
2aπe−(l−p//planckover2pi1)2/2a.
/angbracketleftp4/angbracketright=1
/planckover2pi1√
2aπ/integraldisplay∞
−∞p4e−(l−p//planckover2pi1)2/2adp.Letp
/planckover2pi1−l≡z,sop=/planckover2pi1(z+l).
=1
/planckover2pi1√
2aπ/planckover2pi15/integraldisplay∞
−∞(z+l)4e−z2/2adz.Only even powers of zsurvive:
=/planckover2pi14
√
2aπ/integraldisplay∞
−∞/parenleftbig
z4+6z2l2+l4/parenrightbig
e−z2/2adz=/planckover2pi14
√
2aπ/bracketleftbigg3(2a)2
4√
2aπ+6l2(2a)
2√
2aπ+l4√
2aπ/bracketrightbigg
=/planckover2pi14/parenleftbig
3a2+6al2+l4/parenrightbig
.∴/angbracketleftH2/angbracketright=/planckover2pi14
4m2/parenleftbig
3a2+6al2+l4/parenrightbig
.
σ2
H=/angbracketleftH2/angbracketright−/angbracketleftH/angbracketright2=/planckover2pi12
4m2/parenleftbig
3a2+6al2+l4−a2−2al2−l4/parenrightbig
=/planckover2pi14
4m2/parenleftbig
2a2+4al2/parenrightbig
=/planckover2pi14a
2m2/parenleftbig
a+2l2/parenrightbig
.
σ2
Hσ2
x=/planckover2pi14a
2m2/parenleftbig
a+2l2/parenrightbig1
4a/bracketleftbigg
1+/parenleftbigg2/planckover2pi1at
m/parenrightbigg2/bracketrightbigg
=/planckover2pi14l2
4m2/parenleftbigg
1+a
2l2/parenrightbigg/bracketleftbigg
1+/parenleftbigg2/planckover2pi1at
m/parenrightbigg2/bracketrightbigg
≥/planckover2pi14l2
4m2=/planckover2pi12
4/parenleftbigg/planckover2pi1l
m/parenrightbigg2
=/planckover2pi12
4/parenleftbiggd/angbracketleftx/angbracketright
dt/parenrightbigg2
,so it works.
Problem 3.20
ForQ=x, Eq. 3.72 says σHσx≥/planckover2pi1
2/vextendsingle/vextendsingle/vextendsingle/vextendsingled/angbracketleftx/angbracketright
dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle. But/angbracketleftp/angbracketright=md/angbracketleftx/angbracketright
dt,s oσxσH≥/planckover2pi1
2m|/angbracketleftp/angbracketright|, which is the Griffiths
uncertainty principle of Problem 3.14.
Problem 3.21
P2|β/angbracketright=P(P|β/angbracketright)=P(/angbracketleftα|β/angbracketright|α/angbracketright)=/angbracketleftα|β/angbracketright(P|α/angbracketright)=/angbracketleftα|β/angbracketright/angbracketleftα|α/angbracketright/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
1|α/angbracketright=/angbracketleftα|β/angbracketright|α/angbracketright=P|β/angbracketright.
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CHAPTER 3. FORMALISM 71
SinceP2|β/angbracketright=P|β/angbracketrightforanyvector|β/angbracketright,P2=P. QED [ Note: To say two operators are equal means that
they have the same effect on all vectors.]
If|γ/angbracketrightis an eigenvector of ˆPwith eigenvalue λ, then ˆP|γ/angbracketright=λ|γ/angbracketright, and it follows that ˆP2|γ/angbracketright=λˆP|γ/angbracketright=λ2|γ/angbracketright.
ButˆP2=ˆP, and|γ/angbracketright/negationslash=0 ,s o λ2=λ, and hence the eigenvalues of ˆPare0 and 1. Any (complex) multiple of
|α/angbracketrightis an eigenvector of ˆP, with eigenvalue 1; any vector orthogonal to|α/angbracketrightis an eigenvector of ˆP, with eigenvalue
0.
Problem 3.22
(a)/angbracketleftα|=−i/angbracketleft1|−2/angbracketleft2|+i/angbracketleft3|;/angbracketleftβ|=−i/angbracketleft1|+2/angbracketleft3|.
(b)/angbracketleftα|β/angbracketright=(−i/angbracketleft1|−2/angbracketleft2|+i/angbracketleft3|)(i|1/angbracketright+2|3/angbracketright)=(−i)(i)/angbracketleft1|1/angbracketright+(i)(2)/angbracketleft3|3/angbracketright=1+2i.
/angbracketleftβ|α/angbracketright=(−i/angbracketleft1|+2/angbracketleft3|)(i|1/angbracketright−2|2/angbracketright−i|3/angbracketright)=(−i)(i)/angbracketleft1|1/angbracketright+ (2)(−i)/angbracketleft3|3/angbracketright=1−2i=/angbracketleftα|β/angbracketright∗./check
(c)
A11=/angbracketleft1|α/angbracketright/angbracketleftβ|1/angbracketright=(i)(−i)=1 ; A12=/angbracketleft1|α/angbracketright/angbracketleftβ|2/angbracketright=(i)(0) = 0; A13=/angbracketleft1|α/angbracketright/angbracketleftβ|3/angbracketright=(i)(2) = 2 i;
A21=/angbracketleft2|α/angbracketright/angbracketleftβ|1/angbracketright=(−2)(−i)= 2i;A22=/angbracketleft2|α/angbracketright/angbracketleftβ|2/angbracketright=(−2)(0) = 0; A23=/angbracketleft2|α/angbracketright/angbracketleftβ|3/angbracketright=(−2)(2) =−4;
A31=/angbracketleft3|α/angbracketright/angbracketleftβ|1/angbracketright=(−i)(−i)=−1;A32=/angbracketleft3|α/angbracketright/angbracketleftβ|2/angbracketright=(−i)(0) = 0; A33=/angbracketleft3|α/angbracketright/angbracketleftβ|3/angbracketright=(−i)(2) =−2i.
A=
102i
2i0−4
−10−2i
.No,it’snothermitian.
Problem 3.23
Write the eigenvector as |ψ/angbracketright=c1|1/angbracketright+c2|2/angbracketright,and call the eigenvalue E. The eigenvalue equation is
ˆH|ψ/angbracketright=FepsilonC(|1/angbracketright/angbracketleft1|−|2/angbracketright/angbracketleft2|+|1/angbracketright/angbracketleft2|+|2/angbracketright/angbracketleft1|)(c1|1/angbracketright+c2|2/angbracketright)=FepsilonC(c1|1/angbracketright+c1|2/angbracketright−c2|2/angbracketright+c2|1/angbracketright)
=FepsilonC[(c1+c2)|1/angbracketright+(c1−c2)|2/angbracketright]=E|ψ/angbracketright=E(c1|1/angbracketright+c2|2/angbracketright).
FepsilonC(c1+c2)=Ec1⇒c2=/parenleftbiggE
FepsilonC−1/parenrightbigg
c1;FepsilonC(c1−c2)=Ec2⇒c1=/parenleftbiggE
FepsilonC+1/parenrightbigg
c2.
c2=/parenleftbiggE
FepsilonC−1/parenrightbigg/parenleftbiggE
FepsilonC+1/parenrightbigg
c2⇒/parenleftbiggE
FepsilonC/parenrightbigg2
−1=1⇒E=±√
2FepsilonC.
The eigenvectors are: c2=(±√
2−1)c1⇒|ψ±/angbracketright=c1/bracketleftBig
|1/angbracketright+(±√
2−1)|2/angbracketright/bracketrightBig
.
The Hamiltonian matrix is H=FepsilonC/parenleftbigg11
1−1/parenrightbigg
.
Problem 3.24
|α/angbracketright=/summationdisplay
ncn|en/angbracketright⇒ˆQ|α/angbracketright=/summationdisplay
ncnˆQ|en/angbracketright=/summationdisplay
n/angbracketleften|α/angbracketrightqn|en/angbracketright=/parenleftBigg/summationdisplay
nqn|en/angbracketright/angbracketleften|/parenrightBigg
|α/angbracketright⇒ˆQ=/summationdisplay
nqn|en/angbracketright/angbracketleften|./check
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72 CHAPTER 3. FORMALISM
Problem 3.25
|e1/angbracketright=1 ;/angbracketlefte1|e1/angbracketright=/integraldisplay1
−11dx=2.So|e/prime
1/angbracketright=1√
2.
|e2/angbracketright=x;/angbracketlefte/prime
1|e2/angbracketright=1√
2/integraldisplay1
−1xdx=0 ;/angbracketlefte2|e2/angbracketright=/integraldisplay1
−1x2dx=x3
3/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
−1=2
3.So|e/prime
2/angbracketright=/radicalbigg
3
2x.
|e3/angbracketright=x2;/angbracketlefte/prime
1|e3/angbracketright=1√
2/integraldisplay1
−1x2dx=1√
22
3;/angbracketlefte/prime
2|e3/angbracketright=/radicalbigg
2
3/integraldisplay1
−1x3dx=0.
So (Problem A.4): |e/prime/prime
3/angbracketright=|e3/angbracketright−1√
22
3|e/prime
1/angbracketright=x2−1
3.
/angbracketlefte/prime/prime
3|e/prime/prime
3/angbracketright=/integraldisplay1
−1/parenleftbigg
x2−1
3/parenrightbigg2
dx=/parenleftbiggx5
5−2
3·x3
3+x
9/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
−1=2
5−4
9+2
9=8
45.So
|e/prime
3/angbracketright=/radicalbigg
45
8/parenleftbigg
x2−1
3/parenrightbigg
=/radicalbigg
5
2/parenleftbigg3
2x2−1
2/parenrightbigg
.
|e4/angbracketright=x3./angbracketlefte/prime
1|e4/angbracketright=1√
2/integraldisplay1
−1x3dx=0 ;/angbracketlefte/prime
2|e4/angbracketright=/radicalbigg
3
2/integraldisplay1
−1x4dx=/radicalbigg
3
2·2
5;
/angbracketlefte/prime
3|e4/angbracketright=/radicalbigg
5
2/integraldisplay1
−1/parenleftbigg3
2x5−1
2x3/parenrightbigg
dx=0.|e/prime/prime
4/angbracketright=|e4/angbracketright−/angbracketlefte/prime
2|e4/angbracketright|e/prime
2/angbracketright=x3−/radicalbigg
3
22
5/radicalbigg
3
2x=x3−3
5x.
/angbracketlefte/prime/prime
4|e/prime/prime
4/angbracketright=/integraldisplay1
−1/parenleftbigg
x3−3
5x/parenrightbigg2
dx=/bracketleftbiggx7
7−2·3
5x5
5+9
25x3
3/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
−1=2
7−12
25+18
75=8
7·25.
|e/prime
4/angbracketright=5
2/radicalbigg
7
2/parenleftbigg
x3−3
5x/parenrightbigg
=/radicalbigg
7
2/parenleftbigg5
2x3−3
2x/parenrightbigg
.
Problem 3.26
(a)/angbracketleftQ/angbracketright=/angbracketleftψ|ˆQψ/angbracketright=/angbracketleftˆQ†ψ|ψ/angbracketright=−/angbracketleftˆQψ|ψ/angbracketright=−(/angbracketleftψ|ˆQψ/angbracketright)∗=−/angbracketleftQ/angbracketright∗,so/angbracketleftQ/angbracketrightis imaginary ./check
(b)From Problem 3.5(c) we know that ( ˆPˆQ)†=ˆQ†ˆP†,s oi f ˆP=ˆP†andˆQ=ˆQ†then
[ˆP,ˆQ]†=(ˆPˆQ−ˆQˆP)†=ˆQ†ˆP†−ˆP†ˆQ†=ˆQˆP−ˆPˆQ=−[ˆP,ˆQ]./check
IfˆP=−ˆP†andˆQ=−ˆQ†, then [ ˆP,ˆQ]†=ˆQ†ˆP†−ˆP†ˆQ†=(−ˆQ)(−ˆP)−(−ˆP)(−ˆQ)=−[ˆP,ˆQ].
So in either case the commutator is antihermitian.
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CHAPTER 3. FORMALISM 73
Problem 3.27
(a)ψ1.
(b)b1(with probability 9/25) or b2(with probability 16/25).
(c)Right after the measurement of B:
•With probability 9/25 the particle is in state φ1=( 3ψ1+4ψ2)/5; in that case the probability of
getting a1is 9/25.
•With probability 16/25 the particle is in state φ2=( 4ψ1−3ψ2)/5; in that case the probability of
getting a1is 16/25.
So the total probability of getting a1is9
25·9
25+16
25·16
25=337
625=0.5392.
[Note: The measurment of B(even if we don’t know the outcome of that measurement) collapses the wave
function, and thereby alters the probabilities for the second measurment of A. If the graduate student
inadvertantly neglected to measure B, the second measurement of Awould be certain to reproduce the
resulta1.]
Problem 3.28
Ψn(x,t)=/radicalbigg
2
asin/parenleftBignπ
ax/parenrightBig
e−iEnt//planckover2pi1, with En=n2π2/planckover2pi12
2ma2.
Φn(p,t)=1√
2π/planckover2pi1/integraldisplay∞
−∞e−ipx//planckover2pi1Ψn(x,t)dx=1√
2π/planckover2pi1/radicalbigg
2
ae−iEnt//planckover2pi1/integraldisplaya
0e−ipx//planckover2pi1sin/parenleftBignπ
ax/parenrightBig
dx
=1√
π/planckover2pi1ae−iEnt//planckover2pi11
2i/integraldisplaya
0/bracketleftBig
ei(nπ/a−p//planckover2pi1)x−ei(−nπ/a−p//planckover2pi1)x/bracketrightBig
dx
=1√
π/planckover2pi1ae−iEnt//planckover2pi11
2i/bracketleftbiggei(nπ/a−p//planckover2pi1)x
i(nπ/a−p//planckover2pi1)−ei(−nπ/a−p//planckover2pi1)x
i(−nπ/a−p//planckover2pi1)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0
=−1
2√
π/planckover2pi1ae−iEnt//planckover2pi1/bracketleftbiggei(nπ−pa//planckover2pi1)−1
(nπ/a−p//planckover2pi1)+e−i(nπ+pa//planckover2pi1)−1
(nπ/a+p//planckover2pi1)/bracketrightbigg
=−1
2√
π/planckover2pi1ae−iEnt//planckover2pi1/bracketleftbigg(−1)ne−ipa//planckover2pi1−1
(nπ−ap//planckover2pi1)a+(−1)ne−ipa//planckover2pi1−1
(nπ+ap//planckover2pi1)a/bracketrightbigg
=−1
2/radicalbigga
π/planckover2pi1e−iEnt//planckover2pi1 2nπ
(nπ)2−(ap//planckover2pi1)2/bracketleftBig
(−1)ne−ipa//planckover2pi1−1/bracketrightBig
=/radicalbiggaπ
/planckover2pi1ne−iEnt//planckover2pi1
(nπ)2−(ap//planckover2pi1)2/bracketleftbig
1−(−1)ne−ipa//planckover2pi1/bracketrightbig
.
Noting that
1−(−1)ne−ipa//planckover2pi1=e−ipa/2/planckover2pi1/bracketleftbig
eipa/2/planckover2pi1−(−1)ne−ipa/2/planckover2pi1/bracketrightbig
=2e−ipa/2/planckover2pi1/braceleftBigg
cos(pa/2/planckover2pi1)(nodd),
isin(pa/2/planckover2pi1)(neven),
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74 CHAPTER 3. FORMALISM
we have
|Φ1(p,t)|2=4πa
/planckover2pi1cos2(pa/2/planckover2pi1)
[π2−(pa//planckover2pi1)2]2,|Φ2(p,t)|2=16πa
/planckover2pi1sin2(pa/2/planckover2pi1)
[(2π)2−(pa//planckover2pi1)2]2.
Mathematica has no trouble with the points p=±nπ/planckover2pi1/a, where the denominator vanishes. The reason is that
the numerator is also zero there, and the function as a whole is finite—in fact, the graphs show no interestingbehavior at these points.
p p|Φ |1 22 2|Φ |
/angbracketleftp2/angbracketright=/integraldisplay∞
−∞p2|Φn(p,t)|2dp=4n2πa
/planckover2pi1/integraldisplay∞
−∞p2
[(nπ)2−(ap//planckover2pi1)2]2/braceleftbiggcos2(pa/2/planckover2pi1)
sin2(pa/2/planckover2pi1)/bracerightbigg
dp[letx≡ap
nπ/planckover2pi1]
=4n/planckover2pi12
a2/integraldisplay∞
−∞x2
(1−x2)2Tn(x)dx=4n/planckover2pi12
a2In,
where
Tn(x)≡/braceleftbiggcos2(nπx/2),ifnis odd,
sin2(nπx/2),ifnis even./bracerightbigg
The integral can be evaluated by partial fractions:
x2
(x2−1)2=1
4/bracketleftbigg1
(x−1)2+1
(x+1 )2+1
(x−1)−1
(x+1 )/bracketrightbigg
⇒
In=1
4/bracketleftbigg/integraldisplay∞
−∞1
(x−1)2Tn(x)dx+/integraldisplay∞
−∞1
(x+1 )2Tn(x)dx+/integraldisplay∞
−∞1
(x−1)Tn(x)dx−/integraldisplay∞
−∞1
(x+1 )Tn(x)dx/bracketrightbigg
.
For odd n:
/integraldisplay∞
−∞1
(x±1)kcos2/parenleftBignπx
2/parenrightBig
dx=/integraldisplay∞
−∞1
ykcos2/bracketleftBignπ
2(y∓1)/bracketrightBig
dy=/integraldisplay∞
−∞1
yksin2/parenleftBignπy
2/parenrightBig
dy.
For even n:
/integraldisplay∞
−∞1
(x±1)ksin2/parenleftBignπx
2/parenrightBig
dx=/integraldisplay∞
−∞1
yksin2/bracketleftBignπ
2(y∓1)/bracketrightBig
dy=/integraldisplay∞
−∞1
yksin2/parenleftBignπy
2/parenrightBig
dy.
In either case, then,
In=1
2/integraldisplay∞
−∞1
y2sin2/parenleftBignπy
2/parenrightBig
dy=nπ
4/integraldisplay∞
−∞sin2u
u2du=nπ2
4.
Therefore
/angbracketleftp2/angbracketright=4n/planckover2pi12
a2In=4n/planckover2pi12
a2nπ2
4=/parenleftbiggnπ/planckover2pi1
a/parenrightbigg2
(same as Problem 2.4).
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CHAPTER 3. FORMALISM 75
Problem 3.29
Φ(p,0) =1√
2π/planckover2pi1/integraldisplay∞
−∞e−ipx//planckover2pi1Ψ(x,0)dx=1
2√
nπ/planckover2pi1λ/integraldisplaynλ
−nλei(2π/λ−p//planckover2pi1)xdx
=1
2√
nπ/planckover2pi1λei(2π/λ−p//planckover2pi1)x
i(2π/λ−p//planckover2pi1)/vextendsingle/vextendsingle/vextendsingle/vextendsinglenλ
−nλ=1
2√
nπ/planckover2pi1λei2πne−ipnλ/ /planckover2pi1)−e−i2πneipnλ/ /planckover2pi1)
i(2π/λ−p//planckover2pi1)
=/radicalbigg
/planckover2pi1λ
nπsin(npλ/ /planckover2pi1)
(pλ−2π/planckover2pi1).
|Ψ(x,0)|2=1
2nλ(−nλ < x < nλ );|Φ(p,0)|2=λ/planckover2pi1
nπsin2(npλ/ /planckover2pi1)
(pλ−2π/planckover2pi1)2.
|Ψ|2|Φ|2
-nλ nλ p x2πh/λ
The width of the |Ψ|2graph is wx=2nλ.The|Φ|2graph is a maximum at 2 π/planckover2pi1/λ, and goes to zero on either
side at2π/planckover2pi1
λ/parenleftbigg
1±1
2n/parenrightbigg
,s owp=2π/planckover2pi1
nλ.Asn→∞,wx→∞ andwp→0; in this limit the particle has a
well-defined momentum, but a completely indeterminate position. In general,
wxwp=( 2nλ)2π/planckover2pi1
nλ=4π/planckover2pi1>/planckover2pi1/2,
so the uncertainty principle is satisfied (using the widths as a measure of uncertainty). If we try to check the
uncertainty principle more rigorously, using standard deviation as the measure, we get an uninformative result,because
/angbracketleftp
2/angbracketright=λ/planckover2pi1
nπ/integraldisplay∞
−∞p2sin2(npλ/ /planckover2pi1)
(pλ−2π/planckover2pi1)2dp=∞.
(At large|p|the integrand is approximately (1 /λ2) sin2(npλ/ /planckover2pi1), so the integral blows up.) Meanwhile /angbracketleftp/angbracketrightis
zero, so σp=∞, and the uncertainty principle tells us nothing. The source of the problem is the discontinuity
in Ψ at the end points; here ˆ pΨ=−i/planckover2pi1dΨ/dxpicks up a delta function, and /angbracketleftΨ|ˆp2Ψ/angbracketright=/angbracketleftˆpΨ|ˆpΨ/angbracketright→∞ because
the integral of the square of the delta function blows up. In general, if you want σpto be finite, you cannot
allow discontinuities in Ψ.
Problem 3.30
(a)
1=|A|2/integraldisplay∞
−∞1
(x2+a2)2dx=2|A|2/integraldisplay∞
01
(x2+a2)2dx=2|A|21
2a2/bracketleftbiggx
x2+a2+1
atan−1/parenleftBigx
a/parenrightBig/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0
=1
a2|A|21
atan−1(∞)=π
2a3|A|2⇒A=a/radicalbigg
2a
π.
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76 CHAPTER 3. FORMALISM
(b)
/angbracketleftx/angbracketright=A2/integraldisplay∞
−∞x
(a2+x2)2dx=0.
/angbracketleftx2/angbracketright=2A2/integraldisplay∞
0x2
(a2+x2)2dx.[Lety≡x2
a2,x=a√y, dx =a
2√ydy.]
=2a2
π/integraldisplay∞
0y1/2
(1 +y)2dy=2a2
πΓ(3/2)Γ(1/2)
Γ(2)=2a2
π(√π/2)(√π)
1=a2.
σx=/radicalbig
/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=a.
(c)
Φ(p,0) =A√
2π/planckover2pi1/integraldisplay∞
−∞e−ipx//planckover2pi11
x2+a2dx.[Bute−ipx//planckover2pi1= cos/parenleftBigpx
/planckover2pi1/parenrightBig
−isin/parenleftBigpx
/planckover2pi1/parenrightBig
,and sine is odd.]
=2A√
2π/planckover2pi1/integraldisplay∞
0cos(px//planckover2pi1)
x2+a2dx=2A√
2π/planckover2pi1/parenleftBigπ
2ae−|p|a//planckover2pi1/parenrightBig
=/radicalbigga
/planckover2pi1e−|p|a//planckover2pi1.
/integraldisplay∞
−∞|Φ(p,0)|2dp=a
/planckover2pi1/integraldisplay∞
−∞e−2|p|a//planckover2pi1dp=2a
/planckover2pi1/parenleftbigge−2pa//planckover2pi1
−2a//planckover2pi1/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0=1./check
(d)
/angbracketleftp/angbracketright=a
/planckover2pi1/integraldisplay∞
−∞pe−2|p|a//planckover2pi1dp=0.
/angbracketleftp2/angbracketright=2a
/planckover2pi1/integraldisplay∞
0p2e−2pa//planckover2pi1dp=2a
/planckover2pi12/parenleftbigg/planckover2pi1
2a/parenrightbigg3
=/planckover2pi12
2a2.σp=/radicalbig
/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/planckover2pi1√
2a.
(e) σxσp=a/planckover2pi1√
2a=√
2/planckover2pi1
2>/planckover2pi1
2./check
Problem 3.31
Equation 3.71 ⇒d
dt/angbracketleftxp/angbracketright=i
/planckover2pi1/angbracketleft[H,xp]/angbracketright; Eq. 3.64⇒[H,xp]=[H,x]p+x[H,p]; Problem 3.14 ⇒[H,x]=
−i/planckover2pi1p
m; Problem 3.17(d) ⇒[H,p]=i/planckover2pi1dV
dx.So
d
dt/angbracketleftxp/angbracketright=i
/planckover2pi1/bracketleftbigg
−i/planckover2pi1
m/angbracketleftp2/angbracketright+i/planckover2pi1/angbracketleftxdV
dx/angbracketright/bracketrightbigg
=2/angbracketleftp2
2m/angbracketright−/angbracketleftxdV
dx/angbracketright=2/angbracketleftT/angbracketright−/angbracketleftxdV
dx/angbracketright.QED
In a stationary state all expectation values (at least, for operators that do not depend explicitly on t) are
time-independent (see item 1 on p. 26), so d/angbracketleftxp/angbracketright/dt= 0, and we are left with Eq. 3.97.
For the harmonic oscillator:
V=1
2mω2x2⇒dV
dx=mω2x⇒xdV
dx=mω2x2=2V⇒2/angbracketleftT/angbracketright=2/angbracketleftV/angbracketright⇒/angbracketleftT/angbracketright=/angbracketleftV/angbracketright.QED
In Problem 2.11(c) we found that /angbracketleftT/angbracketright=/angbracketleftV/angbracketright=1
4/planckover2pi1ω(forn= 0);/angbracketleftT/angbracketright=/angbracketleftV/angbracketright=3
4/planckover2pi1ω(forn= 1)./check
In Problem 2.12 we found that /angbracketleftT/angbracketright=1
2/parenleftbig
n+1
2/parenrightbig
/planckover2pi1ω, while/angbracketleftx2/angbracketright=(n+1
2)/planckover2pi1/mω,s o/angbracketleftV/angbracketright=1
2mω2/angbracketleftx2/angbracketright=1
2(n+1
2)/planckover2pi1ω,
and hence/angbracketleftT/angbracketright=/angbracketleftV/angbracketrightforallstationary states. /check
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 3. FORMALISM 77
Problem 3.32
Ψ(x,t)=1√
2/parenleftbig
ψ1e−iE1t//planckover2pi1+ψ2e−iE2t//planckover2pi1/parenrightbig
;/angbracketleftΨ(x,t)|Ψ(x,0)/angbracketright=0⇒
1
2/parenleftbig
eiE1t//planckover2pi1/angbracketleftψ1|ψ1/angbracketright+eiE1t//planckover2pi1/angbracketleftψ1|ψ2/angbracketright+eiE2t//planckover2pi1/angbracketleftψ2|ψ1/angbracketright+eiE2t//planckover2pi1/angbracketleftψ2|ψ2/angbracketright/parenrightbig
=1
2/parenleftbig
eiE1t//planckover2pi1+eiE2t//planckover2pi1/parenrightbig
=0,oreiE2t//planckover2pi1=−eiE1t//planckover2pi1,soei(E2−E1)t//planckover2pi1=−1=eiπ.
Thus (E2−E1)t//planckover2pi1=π(orthogonality also at 3 π,5π,etc., but this is the firstoccurrence).
∴∆t≡t
π=/planckover2pi1
E2−E1.But ∆E=σH=1
2(E2−E1) (Problem 3.18). So ∆ t∆E=/planckover2pi1
2./check
Problem 3.33
Equation 2.69: x=/radicalbigg
/planckover2pi1
2mω(a++a−),p=i/radicalbigg
/planckover2pi1mω
2(a+−a−); Eq.2.66 :/braceleftbigg
a+|n/angbracketright=√n+1|n+1/angbracketright,
a−|n/angbracketright=√n|n−1/angbracketright.
/angbracketleftn|x|n/prime/angbracketright=/radicalbigg
/planckover2pi1
2mω/angbracketleftn|(a++a−)|n/prime/angbracketright=/radicalbigg
/planckover2pi1
2mω/bracketleftbig√
n/prime+1/angbracketleftn|n/prime+1/angbracketright+√
n/prime/angbracketleftn|n/prime−1/angbracketright/bracketrightbig
=/radicalbigg
/planckover2pi1
2mω/parenleftbig√
n/prime+1δn,n/prime+1+√
n/primeδn,n/prime−1/parenrightbig
=/radicalbigg
/planckover2pi1
2mω/parenleftbig√nδn/prime,n−1+√
n/primeδn,n/prime−1/parenrightbig
.
/angbracketleftn|p|n/prime/angbracketright=i/radicalbigg
m/planckover2pi1ω
2/parenleftbig√nδn/prime,n−1−√
n/primeδn,n/prime−1/parenrightbig
.
Noting that nandn/primerun from zero to infinity, the matrices are:
X=/radicalbigg
/planckover2pi1
2mω
0√
1 0000√
10√
20 0 0
0√
20√
30 0.:
00√
30√
40
000√
40√
5
···
;P=i/radicalbigg
m/planckover2pi1ω
2
0−√
1 0000√
10−√
2 000
0√
20−√
30 0.:
00√
30−√
40
00 0√
4−√
5
···
.
Squaring these matrices:
X2=/planckover2pi1
2mω
10√
1·20 0 0
030√
2·30 0√
1·20 5 0√
3·40.:
0√
2·30 7 0√
4·5
···
;
P
2=−m/planckover2pi1ω
2
−10√
1·20 0 0
0−30√
2·30 0√
1·20−50√
3·40.:
0√
2·30−70√
4·5
···
.
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78 CHAPTER 3. FORMALISM
So the Hamiltonian, in matrix form, is
H=1
2mP2+mω2
2X2
=−/planckover2pi1ω
4
−10√
1·20 0 0
0−30√
2·30 0√
1·20−50√
3·40.:
0√
2·30−70√
4·5
···
+/planckover2pi1ω
4
10√
1·20 0 0
030√
2·30 0√
1·20 5 0√
3·40.:
0√
2·30 7 0√
4·5
···
=/planckover2pi1ω
2
1000
030000500007
...
.
It’s plainly diagonal, and the nonzero elements are Hnn=/parenleftbig
n+1
2/parenrightbig
/planckover2pi1ω, as they should be.
Problem 3.34
Evidently Ψ( x,t)=c0ψ0(x)e−iE0t//planckover2pi1+c1ψ1(x)e−iE1t//planckover2pi1, with|c0|2=|c1|2=1/2,soc0=eiθ0/√
2,c1=eiθ1/√
2,
for some real θ0,θ1.
/angbracketleftp/angbracketright=|c0|2/angbracketleftψ0|pψ0/angbracketright+|c1|2/angbracketleftψ1|pψ1/angbracketright+c∗
0c1ei(E0−E1)t//planckover2pi1/angbracketleftψ0|pψ1/angbracketright+c∗
1c0ei(E1−E0)t//planckover2pi1/angbracketleftψ1|pψ0/angbracketright.
ButE1−E0=(3
2/planckover2pi1ω)−(1
2/planckover2pi1ω)=/planckover2pi1ω, and (Problem 2.11) /angbracketleftψ0|pψ0/angbracketright=/angbracketleftψ1|pψ1/angbracketright= 0, while (Eqs. 2.69 and 2.66)
/angbracketleftψ0|pψ1/angbracketright=i/radicalbigg
/planckover2pi1mω
2/angbracketleftψ0|(a+−a−)ψ1/angbracketright=i/radicalbigg
/planckover2pi1mω
2/bracketleftBig
/angbracketleftψ0|√
2ψ2/angbracketright−/angbracketleftψ0|√
1ψ0/angbracketright/bracketrightBig
=−i/radicalbigg
/planckover2pi1mω
2;/angbracketleftψ1|pψ0/angbracketright=i/radicalbigg
/planckover2pi1mω
2.
/angbracketleftp/angbracketright=1√
2e−iθ01√
2eiθ1e−iωt/parenleftBigg
−i/radicalbigg
/planckover2pi1mω
2/parenrightBigg
+1√
2e−iθ11√
2eiθ0eiωt/parenleftBigg
i/radicalbigg
/planckover2pi1mω
2/parenrightBigg
=i
2/radicalbigg
/planckover2pi1mω
2/bracketleftBig
−e−i(ωt−θ1+θ0)+ei(ωt−θ1+θ0)/bracketrightBig
=−/radicalbigg
/planckover2pi1mω
2sin(ωt+θ0−θ1).
The maximum is/radicalbig
/planckover2pi1mω/2;it occurs at t=0⇔sin(θ0−θ1)=−1, orθ1=θ0+π/2. We might as well pick
θ0=0,θ1=π/2; then
Ψ(x,t)=1√
2/bracketleftBig
ψ0e−iωt/2+ψ1eiπ/2e−3iωt/2/bracketrightBig
=1√
2e−iωt/2/parenleftbig
ψ0+iψ1e−iωt/parenrightbig
.
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CHAPTER 3. FORMALISM 79
Problem 3.35
(a)/angbracketleftx/angbracketright=/angbracketleftα|xα/angbracketright=/radicalbigg
/planckover2pi1
2mω/angbracketleftα|(a++a−)α/angbracketright=/radicalbigg
/planckover2pi1
2mω(/angbracketlefta−α|α/angbracketright+/angbracketleftα|a−α/angbracketright)=/radicalbigg
/planckover2pi1
2mω(α+α∗).
x2=/planckover2pi1
2mω/parenleftbig
a2
++a+a−+a−a++a2
−/parenrightbig
.Buta−a+=[a−,a+]+a+a−=1+a+a−(Eq. 2.55) .
=/planckover2pi1
2mω/parenleftbig
a2
++2a+a−+1+a2
−/parenrightbig
.
/angbracketleftx2/angbracketright=/planckover2pi1
2mω/angbracketleftα|/parenleftbig
a2
++2a+a−+1+a2
−/parenrightbig
α/angbracketright=/planckover2pi1
2mω/parenleftbig
/angbracketlefta2
−α|α/angbracketright+2/angbracketlefta−α|a−α/angbracketright+/angbracketleftα|α/angbracketright+/angbracketleftα|a2
−α/angbracketright/parenrightbig
=/planckover2pi1
2mω/bracketleftbig
(α∗)2+2 (α∗)α+1+α2/bracketrightbig
=/planckover2pi1
2mω/bracketleftbig
1+(α+α∗)2/bracketrightbig
.
/angbracketleftp/angbracketright=/angbracketleftα|pα/angbracketright=i/radicalbigg
/planckover2pi1mω
2/angbracketleftα|(a+−a−)α/angbracketright=i/radicalbigg
/planckover2pi1mω
2(/angbracketlefta−α|α/angbracketright−/angbracketleftα|a−α/angbracketright)=−i/radicalbigg
/planckover2pi1mω
2(α−α∗).
p2=−/planckover2pi1mω
2/parenleftbig
a2
+−a+a−−a−a++a2
−/parenrightbig
=−/planckover2pi1mω
2/parenleftbig
a2
+−2a+a−−1+a2
−/parenrightbig
.
/angbracketleftp2/angbracketright=−/planckover2pi1mω
2/angbracketleftα|/parenleftbig
a2
+−2a+a−−1+a2
−/parenrightbig
α/angbracketright=−/planckover2pi1mω
2/parenleftbig
/angbracketlefta2
−α|α/angbracketright−2/angbracketlefta−α|a−α/angbracketright−/angbracketleftα|α/angbracketright+/angbracketleftα|a2
−α/angbracketright/parenrightbig
=−/planckover2pi1mω
2/bracketleftbig
(α∗)2−2(α∗)α−1+α2/bracketrightbig
=/planckover2pi1mω
2/bracketleftbig
1−(α−α∗)2/bracketrightbig
.
(b)
σ2
x=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2=/planckover2pi1
2mω/bracketleftbig
1+(α+α∗)2−(α+α∗)2/bracketrightbig
=/planckover2pi1
2mω;
σ2
p=/angbracketleftp2/angbracketright−/angbracketleftp/angbracketright2=/planckover2pi1mω
2/bracketleftbig
1−(α−α∗)2+(α−α∗)2/bracketrightbig
=/planckover2pi1mω
2.σ xσp=/radicalbigg
/planckover2pi1
2mω/radicalbigg
/planckover2pi1mω
2=/planckover2pi1
2.QED
(c)Using Eq. 2.67 for ψn:
cn=/angbracketleftψn|α/angbracketright=1√
n!/angbracketleft(a+)nψ0|α/angbracketright=1√
n!/angbracketleftψ0|(a−)nα/angbracketright=1√
n!αn/angbracketleftψ0|α/angbracketright=αn
√
n!c0./check
(d)1=∞/summationdisplay
n=0|cn|2=|c0|2∞/summationdisplay
n=0|α|2n
n!=|c0|2e|α|2⇒c0=e−|α|2/2.
(e)|α(t)/angbracketright=∞/summationdisplay
n=0cne−iEnt//planckover2pi1|n/angbracketright=∞/summationdisplay
n=0αn
√
n!e−|α|2/2e−i(n+1
2)ωt|n/angbracketright=e−iωt/2∞/summationdisplay
n=0/parenleftbig
αe−iωt/parenrightbign
√
n!e−|α|2/2|n/angbracketright.
Apart form the overall phase factor e−iωt/2(which doesn’t affect its status as an eigenfunction of a−,o r
its eigenvalue), |α(t)/angbracketrightis the same as |α/angbracketright, but with eigenvalue α(t)=e−iωtα./check
(f)Equation 2.58 says a−|ψ0/angbracketright=0 ,s o yes,itisa coherent state, with eigenvalue α=0.
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80 CHAPTER 3. FORMALISM
Problem 3.36
(a)Equation 3.60 becomes |z|2= [Re(z)]2+[Im(z)]2=/bracketleftbigg1
2(z+z∗)/bracketrightbigg2
+/bracketleftbigg1
2i(z−z∗)/bracketrightbigg2
; Eq. 3.61 generalizes to
σ2
Aσ2
B≥/bracketleftbigg1
2(/angbracketleftf|g/angbracketright+/angbracketleftg|f/angbracketright)/bracketrightbigg2
+/bracketleftbigg1
2i(/angbracketleftf|g/angbracketright−/angbracketleftg|f/angbracketright)/bracketrightbigg2
.
But/angbracketleftf|g/angbracketright−/angbracketleftg|f/angbracketright=/angbracketleft[ˆA,ˆB]/angbracketright(p. 111), and, by the same argument,
/angbracketleftf|g/angbracketright+/angbracketleftg|f/angbracketright=/angbracketleftˆAˆB/angbracketright−/angbracketleftA/angbracketright/angbracketleftB/angbracketright+/angbracketleftˆBˆA/angbracketright−/angbracketleftA/angbracketright/angbracketleftB/angbracketright=/angbracketleftˆAˆB+ˆBˆA−2/angbracketleftA/angbracketright/angbracketleftB/angbracketright/angbracketright=/angbracketleftD/angbracketright.
Soσ2
Aσ2
B≥1
4/parenleftbig
/angbracketleftD/angbracketright2+/angbracketleftC/angbracketright2/parenrightbig
./check
(b)IfˆB=ˆA, then ˆC=0,ˆD=2/parenleftBig
ˆA2−/angbracketleftA/angbracketright2/parenrightBig
;/angbracketleftD/angbracketright=2/parenleftBig
/angbracketleftˆA2/angbracketright−/angbracketleftA/angbracketright2/parenrightBig
=2σ2
A.So Eq. 3.99 says
σ2
Aσ2
A≥(1/4)4σ4
A=σ4
A, which is true, but not very informative.
Problem 3.37
First find the eigenvalues and eigenvectors of the Hamiltonian. The characteristic equation says
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(a−E)0 b
0(c−E)0
b 0(a−E)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=(a−E)(c−E)(a−E)−b
2(c−E)=(c−E)/bracketleftbig
(a−E)2−b2/bracketrightbig
=0,
EitherE=c, or else ( a−E)2=b2⇒E=a±b. So the eigenvalues are
E1=c, E 2=a+b, E 3=a−b.
To find the corresponding eigenvectors, write
a0b
0c0
b0a
α
β
γ
=En
α
β
γ
.
(1)
aα+bγ=cα⇒(a−c)α+bγ=0 ;
cβ=cβ (redundant) ;
bα+aγ=cγ⇒(a−c)γ+bα=0.
⇒/bracketleftbig
(a−c)2−b2/bracketrightbig
α=0.
So (excluding the degenerate case a−c=±b)α= 0, and hence also γ=0 .
(2)
aα+bγ=(a+b)α⇒ α−γ=0 ;
cβ=(a+b)β⇒ β=0 ;
bα+aγ=(a+b)γ(redundant) .
Soα=γandβ=0 .
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CHAPTER 3. FORMALISM 81
(3)
aα+bγ=(a−b)α⇒ α+γ=0 ;
cβ=(a−b)β⇒ β=0 ;
bα+aγ=(a−b)γ(redundant) .
Soα=−γandβ=0 .
Conclusion: The (normalized) eigenvectors of Hare
|s1/angbracketright=
0
10
,|s
2/angbracketright=1√
2
1
01
,|s
3/angbracketright=1√
2
1
0
−1
.
(a)Here|S(0)/angbracketright=|s1/angbracketright,s o
|S(t)/angbracketright=e−iE1t//planckover2pi1|s1/angbracketright=e−ict//planckover2pi1
0
10
.
(b)
|S(0)/angbracketright=1√
2(|s2/angbracketright+|s3/angbracketright).
|S(t)/angbracketright=1√
2/parenleftBig
e−iE2t//planckover2pi1|s2/angbracketright+e−iE3t//planckover2pi1|s3/angbracketright/parenrightBig
=1√
2
e−i(a+b)t//planckover2pi11√
2
1
01
+e
−i(a−b)t//planckover2pi11√
2
1
0
−1
=1
2e−iat//planckover2pi1
e−ibt//planckover2pi1+eibt//planckover2pi1
0
e−ibt//planckover2pi1−eibt//planckover2pi1
=e−iat//planckover2pi1
cos(bt//planckover2pi1)
0
−isin(bt//planckover2pi1)
.
Problem 3.38
(a)H:
E1=/planckover2pi1ω, E 2=E3=2/planckover2pi1ω;|h1/angbracketright=
1
00
,|h
2/angbracketright=
0
10
,|h
3/angbracketright=
0
01
.
A:
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−aλ 0
λ−a0
00 ( 2 λ−a)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=a
2(2λ−a)−(2λ−a)λ2=0⇒a1=2λ, a2=λ, a3=−λ.
λ
010
100002
α
β
γ
=a
α
β
γ
⇒
λβ=aα
λα=aβ
2λγ=aγ
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82 CHAPTER 3. FORMALISM
(1)
λβ=2λα⇒β=2α,
λα=2λβ⇒α=2β,
2λγ=2λγ;
α=β=0 ;|a1/angbracketright=
0
01
.
(2)
λβ=λα⇒β=α,
λα=λβ⇒α=β,
2λγ=λγ;⇒γ=0.
|a2/angbracketright=1√
2
1
10
.
(3)
λβ=−λα⇒β=−α,
λα=−λβ⇒α=−β,
2λγ=−λγ;⇒γ=0.
|a3/angbracketright=1√
2
1
−1
0
.
B:
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(2µ−b)0 0
0−bµ
0µ−b/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=b
2(2µ−b)−(2µ−b)µ2=0⇒b1=2µ, b2=µ, b3=−µ.
µ
200
001010
α
β
γ
=b
α
β
γ
⇒
2µα=bα
µγ=bβ
µβ=bγ
(1)
2µα=2µα,
µγ=2µβ⇒γ=2β,
µβ=2µγ⇒β=2γ;
β=γ=0 ;
|b1/angbracketright=
1
00
.
(2)
2µα=µα⇒α=0,
µγ=µβ⇒γ=β,
µβ=µγ;⇒β=γ.
|b2/angbracketright=1√
2
0
11
.
(3)
2µα=−µα⇒α=0,
µγ=−µβ⇒γ=−β,
µβ=−µγ;⇒β=−γ.
|b3/angbracketright=1√
2
0
1
−1
.
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CHAPTER 3. FORMALISM 83
(b)
/angbracketleftH/angbracketright=/angbracketleftS(0)|H|S(0)/angbracketright=/planckover2pi1ω/parenleftbig
c∗
1c∗2c∗3/parenrightbig
100
020002
c
1
c2
c3
=/planckover2pi1ω/parenleftbig
|c1|2+2|c2|2+2|c3|2/parenrightbig
.
/angbracketleftA/angbracketright=/angbracketleftS(0)|A|S(0)/angbracketright=λ/parenleftbig
c∗
1c∗2c∗3/parenrightbig
010
100002
c
1
c2
c3
=λ/parenleftbig
c∗
1c2+c∗
2c1+2|c3|2/parenrightbig
.
/angbracketleftB/angbracketright=/angbracketleftS(0)|B|S(0)/angbracketright=µ/parenleftbig
c∗
1c∗2c∗3/parenrightbig
200
001010
c
1
c2
c3
=µ/parenleftbig
2|c1|2+c∗
2c3+c∗
3c2/parenrightbig
.
(c)
|S(0)/angbracketright=c1|h1/angbracketright+c2|h2/angbracketright+c3|h3/angbracketright⇒
|S(t)/angbracketright=c1e−iE1t//planckover2pi1|h1/angbracketright+c2e−iE2t//planckover2pi1|h2/angbracketright+c3e−iE3t//planckover2pi1|h3/angbracketright=c1e−iωt|h1/angbracketright+c2e−2iωt|h2/angbracketright+c3e−2iωt|h3/angbracketright
=e−2iωt
c1eiωt
1
00
+c
2
0
10
+c
3
0
01
=
e−2iωt
c1eiωt
c2
c3
.
H:h1=/planckover2pi1ω,probability|c1|2;h2=h3=2/planckover2pi1ω,probability (|c2|2+|c3|2).
A:a1=2λ,/angbracketlefta1|S(t)/angbracketright=e−2iωt/parenleftbig001/parenrightbig
c1eiωt
c2
c3
=e−2iωtc3⇒probability|c3|2.
a2=λ,/angbracketlefta2|S(t)/angbracketright=e−2iωt1√
2/parenleftbig110/parenrightbig
c1eiωt
c2
c3
=1√
2e−2iωt/parenleftbig
c1eiωt+c2/parenrightbig
⇒
probability =1
2/parenleftbig
c∗
1e−iωt+c∗
2/parenrightbig/parenleftbig
c1eiωt+c2/parenrightbig
=1
2/parenleftbig
|c1|2+|c2|2+c∗
1c2e−iωt+c∗
2c1eiωt/parenrightbig
.
a3=−λ,/angbracketlefta3|S(t)/angbracketright=e−2iωt1√
2/parenleftbig1−10/parenrightbig
c1eiωt
c2
c3
=1√
2e−2iωt/parenleftbig
c1eiωt−c2/parenrightbig
⇒
probability =1
2/parenleftbig
c∗
1e−iωt−c∗
2/parenrightbig/parenleftbig
c1eiωt−c2/parenrightbig
=1
2/parenleftbig
|c1|2+|c2|2−c∗
1c2e−iωt−c∗
2c1eiωt/parenrightbig
.
Note that the sum of the probabilities is 1.
B:b1=2µ,/angbracketleftb1|S(t)/angbracketright=e−2iωt/parenleftbig100/parenrightbig
c1eiωt
c2
c3
=e−2iωtc1⇒probability|c1|2.
b2=µ,/angbracketleftb2|S(t)/angbracketright=e−2iωt1√
2/parenleftbig011/parenrightbig
c1eiωt
c2
c3
=1√
2e−2iωt(c2+c3)⇒
probability =1
2(c∗
1+c∗
2)(c1+c2)=1
2/parenleftbig
|c1|2+|c2|2+c∗
1c2+c∗
2c1/parenrightbig
.
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84 CHAPTER 3. FORMALISM
b3=−µ,/angbracketleftb3|S(t)/angbracketright=e−2iωt1√
2/parenleftbig
01−1/parenrightbig
c1eiωt
c2
c3
=1√
2e−2iωt(c2−c3)⇒
probability =1
2(c∗
2−c∗
3)(c2−c3)=1
2/parenleftbig
|c2|2+|c3|2−c∗
2c3−c∗
3c2/parenrightbig
.
Again, the sum of the probabilities is 1.
Problem 3.39
(a)
Expanding in a Taylor series: f(x+x0)=∞/summationdisplay
n=01
n!xn
0/parenleftbiggd
dx/parenrightbiggn
f(x).
Butp=/planckover2pi1
id
dx,sod
dx=ip
/planckover2pi1.Therefore f(x+x0)=∞/summationdisplay
n=01
n!xn
0/parenleftbiggip
/planckover2pi1/parenrightbiggn
f(x)=eipx0//planckover2pi1f(x).
(b)
Ψ(x,t+t0)=∞/summationdisplay
n=01
n!tn
0/parenleftbigg∂
∂t/parenrightbiggn
Ψ(x,t);i/planckover2pi1∂Ψ
∂t=HΨ.
[Note: It is emphatically notthe case that i/planckover2pi1∂
∂t=H. These two operators have the same effect onlywhen
(as here) they are acting on solutions to the (time-dependent) Schr¨ odinger equation.] Also,
/parenleftbigg
i/planckover2pi1∂
∂t/parenrightbigg2
Ψ=i/planckover2pi1∂
∂t(HΨ) =H/parenleftbigg
i/planckover2pi1∂Ψ
∂t/parenrightbigg
=H2Ψ,
provided His not explicitly dependent on t. And so on. So
Ψ(x,t+t0)=∞/summationdisplay
n=01
n!tn
0/parenleftbigg
−i
/planckover2pi1H/parenrightbiggn
Ψ=e−iHt0//planckover2pi1Ψ(x,t).
(c)
/angbracketleftQ/angbracketrightt+t0=/angbracketleftΨ(x,t+t0)|Q(x,p,t +t0)|Ψ(x,t+t0)/angbracketright.
But Ψ(x,t+t0)=e−iHt0//planckover2pi1Ψ(x,t),so, using the hermiticity of Hto write/parenleftbig
e−iHt0//planckover2pi1/parenrightbig†=eiHt0//planckover2pi1:
/angbracketleftQ/angbracketrightt+t0=/angbracketleftΨ(x,t)|eiHt0//planckover2pi1Q(x,p,t +t0)e−iHt0//planckover2pi1|Ψ(x,t)/angbracketright.
Ift0=dtis very small, expanding to first order, we have:
/angbracketleftQ/angbracketrightt+d/angbracketleftQ/angbracketright
dtdt=/angbracketleftΨ(x,t)|/parenleftbigg
1+iH
/planckover2pi1dt/parenrightbigg/bracketleftbigg
Q(x,p,t)+∂Q
∂tdt/bracketrightbigg/parenleftbigg
1−iH
/planckover2pi1dt/parenrightbigg
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
⋆|Ψ(x,t)/angbracketright
/bracketleftbigg
⋆=Q(x,p,t)+iH
/planckover2pi1dtQ−Q/parenleftbiggiH
/planckover2pi1dt/parenrightbigg
+∂Q
∂tdt=Q+i
/planckover2pi1[H,Q]dt+∂Q
∂tdt/bracketrightbigg
=/angbracketleftQ/angbracketrightt+i
/planckover2pi1/angbracketleft[H,Q]/angbracketrightdt+/angbracketleft∂Q
∂t/angbracketrightdt.
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CHAPTER 3. FORMALISM 85
∴d/angbracketleftQ/angbracketright
dt=i
/planckover2pi1/angbracketleft[H,Q]/angbracketright+/angbracketleft∂Q
∂t/angbracketright.QED
Problem 3.40
(a)For the free particle, V(x) = 0, so the time-dependent Schr¨ odinger equation reads
i/planckover2pi1∂Ψ
∂t=−/planckover2pi12
2m∂2Ψ
∂x2.Ψ(x,t)=1√
2π/planckover2pi1/integraldisplay∞
−∞eipx//planckover2pi1Φ(p,t)dp⇒
∂Ψ
∂t=1√
2π/planckover2pi1/integraldisplay∞
−∞eipx//planckover2pi1∂Φ
∂tdp,∂2Ψ
∂x2=1√
2π/planckover2pi1/integraldisplay∞
−∞/parenleftbigg
−p2
/planckover2pi12/parenrightbigg
eipx//planckover2pi1Φdp.So
1√
2π/planckover2pi1/integraldisplay∞
−∞eipx//planckover2pi1/bracketleftbigg
i/planckover2pi1∂Φ
∂t/bracketrightbigg
dp=1√
2π/planckover2pi1/integraldisplay∞
−∞eipx//planckover2pi1/bracketleftbiggp2
2mΦ/bracketrightbigg
dp.
But two functions with the same Fourier transform are equal (as you can easily prove using Plancherel’s
theorem), so
i/planckover2pi1∂Φ
∂t=p2
2mΦ.1
ΦdΦ=−ip2
2m/planckover2pi1dt⇒ Φ(p,t)=e−ip2t/2m/planckover2pi1Φ(p,0).
(b)
Ψ(x,0) =Ae−ax2eilx,A=/parenleftbigg2a
π/parenrightbigg1/4
(Problem2 .43(a)).
Φ(p,0) =1√
2π/planckover2pi1/parenleftbigg2a
π/parenrightbigg1/4/integraldisplay∞
−∞e−ipx//planckover2pi1e−ax2eilxdx=1
(2πa/planckover2pi12)1/4e−(l−p//planckover2pi1)2/4a(Problem2 .43(b)).
Φ(p,t)=1
(2πa/planckover2pi12)1/4e−(l−p//planckover2pi1)2/4ae−ip2t/2m/planckover2pi1;|Φ(p,t)|2=1√
2πa/planckover2pi1e−(l−p//planckover2pi1)2/2a.
(c)
/angbracketleftp/angbracketright=/integraldisplay∞
−∞p|Φ(p,t)|2dp=1√
2πa/planckover2pi1/integraldisplay∞
−∞pe−(l−p//planckover2pi1)2/2adp
[Lety≡(p//planckover2pi1)−l,sop=/planckover2pi1(y+l) anddp=/planckover2pi1dy.]
=/planckover2pi1√
2πa/integraldisplay∞
−∞(y+l)e−y2/2ady[but the first term is odd]
=2/planckover2pi1l√
2πa/integraldisplay∞
0e−y2/2ady=2/planckover2pi1l√
2πa/radicalbiggπa
2=/planckover2pi1l[as in Problem 2.43(d)].
/angbracketleftp2/angbracketright=/integraldisplay∞
−∞p2|Φ(p,t)|2dp=1√
2πa/planckover2pi1/integraldisplay∞
−∞p2e−(l−p//planckover2pi1)2/2adp=/planckover2pi12
√
2πa/integraldisplay∞
−∞(y2+2yl+l2)e−y2/2ady
=2/planckover2pi12
√
2πa/bracketleftbigg/integraldisplay∞
0y2e−y2/2ady+l2/integraldisplay∞
0e−y2/2ady/bracketrightbigg
=2/planckover2pi12
√
2πa/bracketleftBigg
2√π/parenleftbigg/radicalbigga
2/parenrightbigg3
+l2/radicalbiggπa
2/bracketrightBigg
=(a+l2)/planckover2pi12[as in Problem 2.43(d)].
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86 CHAPTER 3. FORMALISM
(d)H=p2
2m;/angbracketleftH/angbracketright=1
2m/angbracketleftp2/angbracketright=/planckover2pi12
2m(l2+a)=1
2m/angbracketleftp/angbracketright2+/planckover2pi12a
2m.But/angbracketleftH/angbracketright0=1
2m/angbracketleftp2/angbracketright0=/planckover2pi12a
2m(Problem 2 .22(d)).
So/angbracketleftH/angbracketright=1
2m/angbracketleftp/angbracketright2+/angbracketleftH/angbracketright0. QED Comment: The energy of the traveling gaussian is the energy of the
same gaussian at rest, plus the kinetic energy ( /angbracketleftp/angbracketright2/2m) associated with the motion of the wave packet
as a whole.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 87
Chapter 4
Quantum Mechanics in Three
Dimensions
Problem 4.1
(a)
[x,y]=xy−yx=0,etc., so [ri,rj]=0.
[px,py]f=/planckover2pi1
i∂
∂x/parenleftbigg/planckover2pi1
i∂f
∂y/parenrightbigg
−/planckover2pi1
i∂
∂y/parenleftbigg/planckover2pi1
i∂f
∂x/parenrightbigg
=−/planckover2pi12/parenleftbigg∂2f
∂x∂y−∂2f
∂y∂x/parenrightbigg
=0
(by the equality of cross-derivatives), so [pi,pj]=0.
[x,px]f=/planckover2pi1
i/parenleftbigg
x∂f
∂x−∂
∂x(xf)/parenrightbigg
=/planckover2pi1
i/parenleftbigg
x∂f
∂x−x∂f
∂x−f/parenrightbigg
=i/planckover2pi1f,
so [x,px]=i/planckover2pi1(likewise [ y,py]=i/planckover2pi1and [z,pz]=i/planckover2pi1).
[y,px]f=/planckover2pi1
i/parenleftbigg
y∂f
∂x−∂
∂x(yf)/parenrightbigg
=/planckover2pi1
i/parenleftbigg
y∂f
∂x−y∂f
∂y/parenrightbigg
= 0 (since∂y
∂x=0 ).So [y,px]=0,
and same goes for the other “mixed” commutators. Thus [ri,pj]=−[pj,ri]=i/planckover2pi1δij.
(b)The derivation of Eq. 3.71 (page 115) is identical in three dimensions, sod/angbracketleftx/angbracketright
dt=i
/planckover2pi1/angbracketleft[H,x]/angbracketright;
[H,x]=/bracketleftbiggp2
2m+V,x/bracketrightbigg
=1
2m[p2
x+p2
y+p2
z,x]=1
2m[p2
x,x]
=1
2m(px[px,x]+[px,x]px)=1
2m[(−i/planckover2pi1)px+(−i/planckover2pi1)px]=−i/planckover2pi1
mpx.
∴d/angbracketleftx/angbracketright
dt=i
/planckover2pi1/parenleftbigg
−i/planckover2pi1
m/angbracketleftpx/angbracketright/parenrightbigg
=1
m/angbracketleftpx/angbracketright.The same goes for yandz, so:d/angbracketleftr/angbracketright
dt=1
m/angbracketleftp/angbracketright.
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88 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
d/angbracketleftpx/angbracketright
dt=i
/planckover2pi1/angbracketleft[H,p x]/angbracketright;[H,p x]=/bracketleftbiggp2
2m+V,px/bracketrightbigg
=[V,px]=i/planckover2pi1∂V
∂x(Eq. 3.65)
=i
/planckover2pi1(i/planckover2pi1)/angbracketleftbigg∂V
∂x/angbracketrightbigg
=/angbracketleftbigg
−∂V
∂x/angbracketrightbigg
.Same for yandz, so:d/angbracketleftp/angbracketright
dt=/angbracketleft−∇V/angbracketright.
(c)From Eq. 3.62: σxσpx≥/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
2i/angbracketleft[x,px]/angbracketright/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
2ii/planckover2pi1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/planckover2pi1
2.Generally, σriσpj≥/planckover2pi1
2δij.
Problem 4.2
(a)Equation 4.8 ⇒−/planckover2pi12
2m/parenleftbigg∂2ψ
∂x2+∂2ψ
∂y2+∂2ψ
∂z2/parenrightbigg
=Eψ(inside the box). Separable solutions: ψ(x,y,z)=
X(x)Y(y)Z(z). Put this in, and divide by XYZ:
1
Xd2X
dx2+1
Yd2X
dy2+1
Zd2Z
dz2=−2m
/planckover2pi12E.
The three terms on the left are functions of x,y, andz, respectively, so each must be a constant. Call the
separation constants k2
x,k2
y, andk2
z(as we’ll soon seen, they must be positive).
d2X
dx2=−k2
xX;d2Y
dy2=−k2
yY;d2Z
dz2=−k2
zZ,withE=/planckover2pi12
2m(k2
x+k2
y+k2
z).
Solution:
X(x)=Axsinkxx+Bxcoskxx;Y(y)=Aysinkyy+Bycoskyy;Z(z)=Azsinkzz+Bzcoskzz.
ButX(0) = 0, so Bx=0 ;Y(0) = 0, so By=0 ;Z(0) = 0, so Bz= 0. And X(a)=0⇒sin(kxa)=0⇒
kx=nxπ/a(nx=1,2,3,...). [As before (page 31), nx/negationslash= 0, and negative values are redundant.] Likewise
ky=nyπ/aandkz=nzπ/a.S o
ψ(x,y,z)=AxAyAzsin/parenleftBignxπ
ax/parenrightBig
sin/parenleftBignyπ
ay/parenrightBig
sin/parenleftBignzπ
az/parenrightBig
,E =/planckover2pi12
2mπ2
a2(n2
x+n2
y+n2
z).
We might as well normalize X,Y, andZseparately: Ax=Ay=Az=/radicalbig
2/a.Conclusion:
ψ(x,y,z)=/parenleftbigg2
a/parenrightbigg3/2
sin/parenleftBignxπ
ax/parenrightBig
sin/parenleftBignyπ
ay/parenrightBig
sin/parenleftBignzπ
az/parenrightBig
;E=π2/planckover2pi12
2ma2(n2
x+n2
y+n2
z);nx,ny,nz=1,2,3,...
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 89
(b)
nxnynz(n2
x+n2
y+n2
z)
111 3
112 6
121 6
211 6
122 9
212 9
221 9
113 1 1
131 1 1
311 1 1
222 1 2
123 1 4
132 1 4
213 1 4
231 1 4
312 1 4
321 1 4Energy Degeneracy
E1=3π2/planckover2pi12
2ma2;d=1
E2=6π2/planckover2pi12
2ma2;d=3.
E3=9π2/planckover2pi12
2ma2;d=3.
E4=1 1π2/planckover2pi12
2ma2;d=3.
E5=1 2π2/planckover2pi12
2ma2;d=1.
E6=1 4π2/planckover2pi12
2ma2;d=6.
(c)The next combinations are: E7(322),E8(411),E9(331),E10(421),E11(332),E12(422),E13(431), and
E14(333 and 511). The degeneracy of E14is4.Simple combinatorics accounts for degeneracies of 1
(nx=ny=nz), 3 (two the same, one different), or 6 (all three different). But in the case of E14there is
a numerical “accident”: 32+32+32= 27, but 52+12+12isalso27, so the degeneracy is greater than
combinatorial reasoning alone would suggest.
Problem 4.3
Eq. 4.32⇒Y0
0=1√
4πP0
0(cosθ); Eq. 4.27⇒P0
0(x)=P0(x); Eq. 4.28⇒P0(x)=1.Y0
0=1√
4π.
Y1
2=−/radicalbigg
5
4π1
3·2eiφP1
2(cosθ);P1
2(x)=/radicalbig
1−x2d
dxP2(x);
P2(x)=1
4·2/parenleftbiggd
dx/parenrightbigg2/parenleftbig
x2−1/parenrightbig2=1
8d
dx/bracketleftbig
2(x2−1)2x/bracketrightbig
=1
2/bracketleftbig
x2−1+x(2x)/bracketrightbig
=1
2/parenleftbig
3x2−1/parenrightbig
;
P1
2(x)=/radicalbig
1−x2d
dx/bracketleftbigg3
2x2−1
2/bracketrightbigg
=/radicalbig
1−x23x;P1
2(cosθ) = 3 cos θsinθ.Y1
2=−/radicalbigg
15
8πeiφsinθcosθ.
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90 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Normalization:/integraldisplay/integraldisplay
|Y0
0|2sinθdθdφ =1
4π/bracketleftbigg/integraldisplayπ
0sinθdθ/bracketrightbigg/bracketleftbigg/integraldisplay2π
0dφ/bracketrightbigg
=1
4π(2)(2π)=1./check
/integraldisplay/integraldisplay
|Y1
2|2sinθdθdφ =15
8π/integraldisplayπ
0sin2θcos2θsinθdθ/integraldisplay2π
0dφ=15
4/integraldisplayπ
0cos2θ(1−cos2θ) sinθdθ
=15
4/bracketleftbigg
−cos3θ
3+cos5θ
5/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=15
4/bracketleftbigg2
3−2
5/bracketrightbigg
=5
2−3
2=1/check
Orthogonality:/integraldisplay/integraldisplay
Y0
0∗Y1
2sinθdθdφ =−1√
4π/radicalbigg
15
8π/bracketleftbigg/integraldisplayπ
0sinθcosθsinθdθ
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
(sin3θ)/3|π
0=0/bracketrightbigg/bracketleftbigg/integraldisplay2π
0eiφdφ
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
(eiφ)/i|2π
0=0/bracketrightbigg
=0./check
Problem 4.4
dΘ
dθ=A
tan(θ/2)1
2sec2(θ/2) =A
21
sin(θ/2) cos(θ/2)=A
sinθ.Therefored
dθ/parenleftbigg
sinθdΘ
dθ/parenrightbigg
=d
dθ(A)=0.
Withl=m=0,Eq. 4.25 reads:d
dθ/parenleftbigg
sinθdΘ
dθ/parenrightbigg
=0.SoAln[tan(θ/2)]doessatisfy Eq. 4.25 .However ,
Θ(0) = Aln(0) = A(−∞); Θ(π)=Aln/parenleftBig
tanπ
2/parenrightBig
=Aln(∞)=A(∞).Θ blows up at θ= 0 and at θ=π.
Problem 4.5
Yl
l=(−1)l/radicalBigg
(2l+1 )
4π1
(2l)!eilφPl
l(cosθ).Pl
l(x)=( 1−x2)l/2/parenleftbiggd
dx/parenrightbiggl
Pl(x).
Pl(x)=1
2ll!/parenleftbiggd
dx/parenrightbiggl
(x2−1)l,soPl
l(x)=1
2ll!(1−x2)l/2/parenleftbiggd
dx/parenrightbigg2l
(x2−1)l.
Now (x2−1)l=x2l+···,where all the other terms involve powers of xlessthan 2l, and hence give zero when
differentiated 2 ltimes. So
Pl
l(x)=1
2ll!(1−x2)l/2/parenleftbiggd
dx/parenrightbigg2l
x2l.But/parenleftbiggd
dx/parenrightbiggn
xn=n!,soPl
l=(2l)!
2ll!(1−x2)l/2.
∴Yl
l=(−1)l/radicalBigg
(2l+1 )
4π(2l)!eilφ(2l)!
2ll!(sinθ)l=1
l!/radicalbigg
(2l+ 1)!
4π/parenleftbigg
−1
2eiφsinθ/parenrightbiggl
.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 91
Y2
3=/radicalbigg
7
4π·1
5!e2iφP2
3(cosθ);P2
3(x)=( 1−x2)/parenleftbiggd
dx/parenrightbigg2
P3(x);P3(x)=1
8·3!/parenleftbiggd
dx/parenrightbigg3
(x2−1)3.
P3=1
8·3·2/parenleftbiggd
dx/parenrightbigg2/bracketleftbig
6x(x2−1)2/bracketrightbig
=1
8d
dx/bracketleftbig
(x2−1)2+4x2(x2−1)/bracketrightbig
=1
8/bracketleftbig
4x(x2−1 )+8x(x2−1 )+4x2·2x/bracketrightbig
=1
2/parenleftbig
x3−x+2x3−2x+2x3/parenrightbig
=1
2/parenleftbig
5x3−3x/parenrightbig
.
P2
3(x)=1
2/parenleftbig
1−x2/parenrightbig/parenleftbiggd
dx/parenrightbigg2/parenleftbig
5x3−3x/parenrightbig
=1
2/parenleftbig
1−x2/parenrightbigd
dx/parenleftbig
15x2−3/parenrightbig
=1
2(1−x2)30x=1 5x(1−x2).
Y2
3=/radicalbigg
7
4π1
5!15e2iφcosθsin2θ=1
4/radicalbigg
105
2πe2iφsin2θcosθ.
Check that Yl
lsatisfies Eq. 4.18: Let1
l!/radicalbigg
(2l+ 1)!
4π/parenleftbigg
−1
2/parenrightbiggl
≡A,s oYl
l=A(eiφsinθ)l.
∂Yl
l
∂θ=Aeilφl(sinθ)l−1cosθ; sinθ∂Yl
l
∂θ=lcosθYl
l;
sinθ∂
∂θ/parenleftbigg
sinθ∂Yl
l
∂θ/parenrightbigg
=lcosθ/parenleftbigg
sinθ∂Yl
l
∂θ/parenrightbigg
−lsin2θYl
l=/parenleftbig
l2cos2θ−lsin2θ/parenrightbig
Yl
l.∂2Yl
l
∂φ2=−l2Yl
l.
So the left side of Eq. 4.18 is/bracketleftbig
l2(1−sin2θ)−lsin2θ−l2/bracketrightbig
Yl
l=−l(l+1) sin2θYl
l, which matches the right side.
Check that Y2
3satisfies Eq. 4.18: Let B≡1
4/radicalbigg
105
2π,soY2
3=Be2iφsin2θcosθ.
∂Y2
3
∂θ=Be2iφ/parenleftbig
2 sinθcos2θ−sin3θ/parenrightbig
; sinθ∂
∂θ/parenleftbigg
sinθ∂Y2
3
∂θ/parenrightbigg
=Be2iφsinθ∂
∂θ/parenleftbig
2 sin2θcos2θ−sin4θ/parenrightbig
=Be2iφsinθ/parenleftbig
4 sinθcos3θ−4 sin3θcosθ−4 sin3θcosθ/parenrightbig
=4Be2iφsin2θcosθ/parenleftbig
cos2θ−2 sin2θ/parenrightbig
= 4(cos2θ−2 sin2θ)Y2
3.∂2Y2
3
∂φ2=−4Y2
3.So the left side of Eq. 4.18 is
4(cos2θ−2 sin2θ−1)Y2
3=4 (−3 sin2θ)Y2
3=−l(l+ 1) sin2θY2
3,
wherel= 3, so it fits the right side of Eq. 4.18.
Problem 4.6
/integraldisplay1
−1Pl(x)Pl/prime(x)dx=1
2ll!1
2l/primel/prime!/integraldisplay1
−1/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl
(x2−1)l/bracketrightBigg/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl/prime
(x2−1)l/prime/bracketrightBigg
dx.
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92 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Ifl/negationslash=l/prime, we may as well let lbe the larger of the two ( l>l/prime). Integrate by parts, pulling successively each
derivative off the first term onto the second:
2ll!2l/primel/prime!/integraldisplay1
−1Pl(x)Pl/prime(x)dx=/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl−1
(x2−1)l/bracketrightBigg/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl/prime
(x2−1)l/prime/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
−1
−/integraldisplay1
−1/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl−1
(x2−1)l/bracketrightBigg/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl/prime+1
(x2−1)l/prime/bracketrightBigg
dx
=...(boundary terms) ...+(−1)l/integraldisplay1
−1(x2−1)l/parenleftbiggd
dx/parenrightbiggl/prime+l
(x2−1)l/primedx.
But (d/dx)l/prime+l(x2−1)l/prime= 0, because ( x2−1)l/primeis a polynomial whose highest power is 2 l/prime, so more than 2 l/prime
derivatives will kill it, and l/prime+l>2l/prime. Now, the boundary terms are of the form:
/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl−n
(x2−1)l/bracketrightBigg/bracketleftBigg/parenleftbiggd
dx/parenrightbiggl/prime+n−1
(x2−1)l/prime/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+1
−1,n=1,2,3,...,l.
Look at the first term: ( x2−1)l=(x2−1)(x2−1)...(x2−1);lfactors. So 0 ,1,2,...,l−1 derivatives will
still leave at least one overall factor of ( x2−1). [Zero derivatives leaves lfactors; one derivative leaves l−1:
d/dx(x2−1)l=2lx(x2−1)l−1; two derivatives leaves l−2:d2/dx2(x2−1)l=2l(x2−1)l−1+2l(l−1)2x2(x2−1)l−2,
and so on.] So the boundary terms are all zero, and hence/integraltext1
−1Pl(x)Pl/prime(x)dx=0 .
This leaves only the case l=l/prime. Again the boundary terms vanish, but this time the remaining integral does
not:
(2ll!)2/integraldisplay1
−1[Pl(x)]2dx=(−1)l/integraldisplay1
−1(x2−1)l/parenleftbiggd
dx/parenrightbigg2l
(x2−1)l
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
(d/dx)2l(x2l)=(2l)!dx
=(−1)l(2l)!/integraldisplay1
−1(x2−1)ldx= 2(2l)!/integraldisplay1
0(1−x2)ldx.
Letx≡cosθ,s odx=−sinθdθ, (1−x2) = sin2θ, θ :π/2→0. Then
/integraldisplay1
0(1−x2)ldx=/integraldisplay0
π/2(sinθ)2l(−sinθ)dθ=/integraldisplayπ/2
0(sinθ)2l+1dθ
=(2)(4)···(2l)
(1)(3)(5)···(2l+1 )=(2ll!)2
1·2·3····(2l+1 )=(2ll!)2
(2l+ 1)!.
∴/integraldisplay1
−1[Pl(x)]2dx=1
(2ll!)22(2l)!(2ll!)2
(2l+ 1)!=2
2l+1.So/integraldisplay1
−1Pl(x)Pl/prime(x)dx=2
2l+1δll/prime.QED
Problem 4.7
(a)
n1(x)=−(−x)1
xd
dx/parenleftbiggcosx
x/parenrightbigg
=−cosx
x2−sinx
x.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 93
n2(x)=−(−x)2/parenleftbigg1
xd
dx/parenrightbigg2cosx
x=−x2/parenleftbigg1
xd
dx/parenrightbigg/bracketleftbigg1
xd
dx/parenleftbiggcosx
x/parenrightbigg/bracketrightbigg
=−xd
dx/parenleftbigg1
x·−xsinx−cosx
x2/parenrightbigg
=xd
dx/parenleftbiggsinx
x2+cosx
x3/parenrightbigg
=x/parenleftbiggx2cosx−2xsinx
x4+−x3sinx−3x2cosx
x6/parenrightbigg
=cosx
x−2sinx
x2−sinx
x2−3 cosx
x3=−/parenleftbigg3
x3−1
x/parenrightbigg
cosx−3
x2sinx.
(b)Letting sin x≈xand cos x≈1, and keeping only the lowest power of x:
n1(x)≈−1
x2+1
xx≈−1
x2.Asx→0, this blows up.
n2(x)≈−/parenleftbigg3
x3−1
x/parenrightbigg
−3
x2x≈−3
x3,which again blows up at the origin .
Problem 4.8
(a)
u=Arj1(kr)=A/bracketleftbiggsin(kr)
k2r−cos(kr)
k/bracketrightbigg
=A
k/bracketleftbiggsin(kr)
(kr)−cos(kr)/bracketrightbigg
.
du
dr=A
k/bracketleftbiggk2rcos(kr)−ksin(kr)
(kr)2+ksin(kr)/bracketrightbigg
=A/bracketleftbiggcos(kr)
kr−sin(kr)
(kr)2+ sin(kr)/bracketrightbigg
.
d2u
dr2=A/bracketleftbigg−k2rsin(kr)−kcos(kr)
(kr)2−k3r2cos(kr)−2k2rsin(kr)
(kr)4+kcos(kr)/bracketrightbigg
=Ak/bracketleftbigg
−sin(kr)
(kr)−cos(kr)
(kr)2−cos(kr)
(kr)2+2sin(kr)
(kr)3+ cos(kr)/bracketrightbigg
=Ak/bracketleftbigg/parenleftbigg
1−2
(kr)2/parenrightbigg
cos(kr)+/parenleftbigg2
(kr)3−1
(kr)/parenrightbigg
sin(kr)/bracketrightbigg
.
WithV= 0 and l= 1, Eq. 4.37 reads:d2u
dr2−2
r2u=−2mE
/planckover2pi12u=−k2u.In this case the left side is
Ak/bracketleftbigg/parenleftbigg
1−2
(kr)2/parenrightbigg
cos(kr)+/parenleftbigg2
(kr)3−1
(kr)/parenrightbigg
sin(kr)−2
(kr)2/parenleftbiggsin(kr)
(kr)−cos(kr)/parenrightbigg/bracketrightbigg
=Ak/bracketleftbigg
cos(kr)−sin(kr)
kr/bracketrightbigg
=−k2u.So this udoessatisfy Eq. 4.37 .
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94 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b)Equation 4.48 ⇒j1(z) = 0, where z=ka.T h u ssinz
z2−cosz
z=0 ,o r tanz=z.For high z(largen,
ifn=1,2,3,...counts the allowed energies in increasing order), the intersections occur slightly below
z=(n+1
2)π.
∴E=/planckover2pi12k2
2m=/planckover2pi12z2
2ma2=/planckover2pi12π2
2ma2/parenleftbigg
n+1
2/parenrightbigg2
.QED
π/2 3π/2z5π/2ztan z
Problem 4.9
Forr≤a,u(r)=Asin(kr), withk≡/radicalbig
2m(E+V0)//planckover2pi1.F o rr≥a, Eq. 4.37 with l=0,V= 0, and (for a bound
state)E<0⇒:
d2u
dr2=−2m
/planckover2pi12Eu=κ2u,withκ≡√
−2mE/ /planckover2pi1⇒u(r)=Ceκr+De−κr.
But the Ceκrterm blows up as r→∞,s ou(r)=De−κr.
Continuity of uatr=a:Asin(ka)=De−κa
Continuity of u/primeatr=a:Akcos(ka)=−Dκe−κa/bracerightbigg
divide:1
ktan(ka)=−1
κ,or−cotka=κ
k.
Letka≡z;κ
k=/radicalbig
2mV0a2//planckover2pi12−z2
z.Letz0≡√2mV0
/planckover2pi1a.−cotz=/radicalbig
(z0/z)2−1.This is exactly the
same transcendental equation we encountered in Problem 2.29—see graph there. There is no solution if z0<π /2,
which is to say, if 2 mV0a2//planckover2pi12<π2/4, orV0a2<π2/planckover2pi12/8m. Otherwise, the ground state energy occurs somewhere
between z=π/2 andz=π:
E+V0=/planckover2pi12k2a2
2ma2=/planckover2pi12
2ma2z2,so/planckover2pi12π2
8ma2<(E0+V0)</planckover2pi12π2
2ma2(precise value depends on V0).
Problem 4.10
R30(n=3,l= 0) : Eq. 4.62 ⇒v(ρ)=/summationtext
j=0cjρj.
Eq. 4.76⇒c1=2(1−3)
(1)(2)c0=−2c0;c2=2(2−3)
(2)(3)c1=−1
3c1=2
3c0;c3=2(3−3)
(3)(4)c2=0.
Eq. 4.73⇒ρ=r
3a; Eq. 4.75⇒R30=1
rρe−ρv(ρ)=1
rr
3ae−r/3a/bracketleftbigg
c0−2c0r
3a+2
3c0/parenleftBigr
3a/parenrightBig2/bracketrightbigg
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 95
R30=/parenleftBigc0
3a/parenrightBig/bracketleftbigg
1−2
3/parenleftBigr
a/parenrightBig
+2
27/parenleftBigr
a/parenrightBig2/bracketrightbigg
e−r/3a.
R31(n=3,l=1 ):c1=2(2−3)
(1)(4)c0=−1
2c0;c2=2(3−3)
(2)(5)c1=0.
R31=1
r/parenleftBigr
3a/parenrightBig2
e−r/3a/parenleftbigg
c0−1
2c0r
3a/parenrightbigg
=/parenleftBigc0
9a2/parenrightBig
r/bracketleftbigg
1−1
6/parenleftBigr
a/parenrightBig/bracketrightbigg
e−r/3a.
R32(n=3,l=2 ):c1=2(3−3)
(1)(6)c0=0.R 32=1
r/parenleftBigr
3a/parenrightBig3
e−r/3a(c0)=/parenleftBigc0
27a3/parenrightBig
r2e−r/3a.
Problem 4.11
(a)
Eq. 4.31⇒/integraldisplay∞
0|R|2r2dr=1.Eq. 4.82⇒R20=/parenleftBigc0
2a/parenrightBig/parenleftBig
1−r
2a/parenrightBig
e−r/2a.Letz≡r
a.
1=/parenleftBigc0
2a/parenrightBig2
a3/integraldisplay∞
0/parenleftBig
1−z
2/parenrightBig2
e−zz2dz=c2
0a
4/integraldisplay∞
0/parenleftbigg
z2−z3+1
4z4/parenrightbigg
e−zdz=c2
0a
4/parenleftbigg
2−6+24
4/parenrightbigg
=a
2c2
0.
∴c0=/radicalbigg
2
a.Eq. 4.15⇒ψ200=R20Y0
0.Table 4.3⇒Y0
0=1√
4π.
∴ψ200=1√
4π/radicalbigg
2
a1
2a/parenleftBig
1−r
2a/parenrightBig
e−r/2a⇒ψ200=1√
2πa1
2a/parenleftBig
1−r
2a/parenrightBig
e−r/2a.
(b)
R21=c0
4a2re−r/2a;1 =/parenleftBigc0
4a2/parenrightBig2
a5/integraldisplay∞
0z4e−zdz=c2
0a
1624 =3
2ac2
0,soc0=/radicalbigg
2
3a.
R21=1√
6a1
2a2re−r/2a;ψ21±1=1√
6a1
2a2re−r/2a/parenleftBigg
∓/radicalbigg
3
8πsinθe±iφ/parenrightBigg
=∓1√πa1
8a2re−r/2asinθe±iφ;
ψ210=1√
6a1
2a2re−r/2a/parenleftBigg/radicalbigg
3
4πcosθ/parenrightBigg
=1√
2πa1
4a2re−r/2acosθ.
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96 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.12
(a)
L0=exe−x=1.L1=exd
dx/parenleftbig
e−xx/parenrightbig
=ex/bracketleftbig
e−x−e−xx/bracketrightbig
=1−x.
L2=ex/parenleftbiggd
dx/parenrightbigg2/parenleftbig
e−xx2/parenrightbig
=exd
dx/parenleftbig
2xe−x−e−xx2/parenrightbig
=ex/parenleftbig
2e−x−2xe−x+e−xx2−2xe−x/parenrightbig
=2−4x+x2.
L3=ex/parenleftbiggd
dx/parenrightbigg3/parenleftbig
e−xx3/parenrightbig
=ex/parenleftbiggd
dx/parenrightbigg2/parenleftbig
−e−xx3+3x2e−x/parenrightbig
=exd
dx/parenleftbig
e−xx3−3x2e−x−3x2e−x+6xe−x/parenrightbig
=ex/parenleftbig
−e−xx3+3x2e−x+6x2e−x−12xe−x−6xe−x+6e−x/parenrightbig
=6−18x+9x2−x3.
(b)
v(ρ)=L5
2(2ρ);L5
2(x)=L5
7−5(x)=(−1)5/parenleftbiggd
dx/parenrightbigg5
L7(x).
L7(x)=ex/parenleftbiggd
dx/parenrightbigg7/parenleftbig
x7e−x/parenrightbig
=ex/parenleftbiggd
dx/parenrightbigg6/parenleftbig
7x6e−x−x7e−x/parenrightbig
=ex/parenleftbiggd
dx/parenrightbigg5/parenleftbig
42x5e−x−7x6e−x−7x6e−x+x7e−x/parenrightbig
=ex/parenleftbiggd
dx/parenrightbigg4/parenleftbig
210x4e−x−42x5e−x−84x5e−x+1 4x6e−x+7x6e−x−x7e−x/parenrightbig
=ex/parenleftbiggd
dx/parenrightbigg3/bracketleftbigg
840x3e−x−(210 + 630) x4e−x
+ (126 + 126) x5e−x−( 2 1+7 ) x6e−x+x7e−x/bracketrightbigg
=ex/parenleftbiggd
dx/parenrightbigg2/parenleftbig
2520x2e−x−(840 + 3360) x3e−x
+(840 + 1260) x4e−x−(252 + 168) x5e−x+ (28 + 7) x6e−x−x7e−x/parenrightbig
=ex/parenleftbiggd
dx/parenrightbigg/bracketleftbigg
5040xe−x−(2520 + 12600) x2e−x+ (4200 + 8400) x3e−x
−(2100 + 2100) x4e−x+ (420 + 210) x5e−x−( 3 5+7 ) x6e−x+x7e−x/bracketrightbigg
=ex/bracketleftbigg
5040e−x−(5040 + 30240) xe−x+ (15120 + 37800) x2e−x
−(12600 + 8400 + 8400) x3e−x+ (2100 + 2100 + 3150) x4e−x
−(630 + 252) x5e−x+( 4 2+7 ) x6e−x−x7e−x/bracketrightbigg
= 5040−35280x+ 52920 x2−29400x3+ 7350x4−882x5+4 9x6−x7.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 97
L5
2=−/parenleftbiggd
dx/parenrightbigg5/parenleftbig
−882x5+4 9x6−x7/parenrightbig
=−/bracketleftbig
−882(5·4·3·2) + 49(6·5·4·3·2)x−7·6·5·4·3x2/bracketrightbig
=6 0/bracketleftbig
(882×2)−(49×12)x+4 2x2/bracketrightbig
= 2520(42−14x+x2).
v(ρ) = 2520(42−28ρ+4ρ2)=5040/parenleftbig
21−14ρ+2ρ2/parenrightbig
.
(c)
Eq. 4.62⇒v(ρ)=∞/summationdisplay
j=0cjρj.Eq. 4.76⇒c1=2(3−5)
(1)(6)c0=−2
3c0.
c2=2(4−5)
(2)(7)c1=−1
7c1=2
21c0;c3=2(5−5)
(3)(8)c2=0.
v(ρ)=c0−2
3c0ρ+2
21c0ρ2=c0
21/parenleftbig
21−14ρ+2ρ2/parenrightbig
./check
Problem 4.13
(a)
ψ=1√
πa3e−r/a,so/angbracketleftrn/angbracketright=1
πa3/integraldisplay
rne−2r/a/parenleftbig
r2sinθdrdθdφ/parenrightbig
=4π
πa3/integraldisplay∞
0rn+2e−2r/adr.
/angbracketleftr/angbracketright=4
a3/integraldisplay∞
0r3e−2r/adr=4
a33!/parenleftBiga
2/parenrightBig4
=3
2a;/angbracketleftr2/angbracketright=4
a3/integraldisplay∞
0r4e−2r/adr=4
a34!/parenleftBiga
2/parenrightBig5
=3a2.
(b)
/angbracketleftx/angbracketright=0 ;/angbracketleftx2/angbracketright=1
3/angbracketleftr2/angbracketright=a2.
(c)
ψ211=R21Y1
1=−1√πa1
8a2re−r/2asinθeiφ(Problem 4.11(b)).
/angbracketleftx2/angbracketright=1
πa1
(8a2)2/integraldisplay/parenleftBig
r2e−r/asin2θ/parenrightBig/parenleftbig
r2sin2θcos2φ/parenrightbig
r2sinθdrdθdφ
=1
64πa5/integraldisplay∞
0r6e−r/adr/integraldisplayπ
0sin5θdθ/integraldisplay2π
0cos2φdφ
=1
64πa5/parenleftbig
6!a7/parenrightbig/parenleftbigg
22·4
1·3·5/parenrightbigg/parenleftbigg1
2·2π/parenrightbigg
=12a2.
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98 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.14
ψ=1√
πa3e−r/a;P=|ψ|24πr2dr=4
a3e−2r/ar2dr=p(r)dr;p(r)=4
a3r2e−2r/a.
dp
dr=4
a3/bracketleftbigg
2re−2r/a+r2/parenleftbigg
−2
ae−2r/a/parenrightbigg/bracketrightbigg
=8r
a3e−2r/a/parenleftBig
1−r
a/parenrightBig
=0⇒r=a.
Problem 4.15
(a)Ψ(r,t)=1√
2/parenleftBig
ψ211e−iE2t//planckover2pi1+ψ21−1e−iE2t//planckover2pi1/parenrightBig
=1√
2(ψ211+ψ21−1)e−iE2t//planckover2pi1;E2=E1
4=−/planckover2pi12
8ma2.
From Problem 4.11(b):
ψ211+ψ21−1=−1√πa1
8a2re−r/2asinθ/parenleftbig
eiφ−e−iφ/parenrightbig
=−i√πa4a2re−r/2asinθsinφ.
Ψ(r,t)=−i√
2πa4a2re−r/2asinθsinφe−iE2t//planckover2pi1.
(b)
/angbracketleftV/angbracketright=/integraldisplay
|Ψ|2/parenleftbigg
−e2
4πFepsilonC01
r/parenrightbigg
d3r=1
(2πa)(16a4)/parenleftbigg
−e2
4πFepsilonC0/parenrightbigg/integraldisplay/parenleftBig
r2e−r/asin2θsin2φ/parenrightBig1
rr2sinθdrdθdφ
=1
32πa5/parenleftbigg
−/planckover2pi12
ma2/parenrightbigg/integraldisplay∞
0r3e−r/adr/integraldisplayπ
0sin3θdθ/integraldisplay2π
0sin2φdφ=−/planckover2pi12
32πma6/parenleftbig
3!a4/parenrightbig/parenleftbigg4
3/parenrightbigg
(π)
=−/planckover2pi12
4ma2=1
2E1=1
2(−13.6eV) =−6.8eV (independent of t).
Problem 4.16
En(Z)=Z2En;E1(Z)=Z2E1;a(Z)=a/Z;R(Z)=Z2R.
Lyman lines range from ni=2t oni=∞(withnf= 1); the wavelengths range from
1
λ2=R/parenleftbigg
1−1
4/parenrightbigg
=3
4R⇒λ2=4
3Rdown to1
λ1=R/parenleftbigg
1−1
∞/parenrightbigg
=R⇒λ1=1
R.
ForZ=2: λ1=1
4R=1
4(1.097×107)=2.28×10−8mtoλ2=1
3R=3.04×10−8m,ultraviolet.
ForZ=3: λ1=1
9R=1.01×10−8mtoλ2=4
27R=1.35×10−8m,alsoultraviolet.
Problem 4.17
(a)V(r)=−GMm
r.Soe2
4πFepsilonC0→GMm translates hydrogen results to the gravitational analogs.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 99
(b)Equation 4.72: a=/parenleftbigg4πFepsilonC0
e2/parenrightbigg/planckover2pi12
m,soag=/planckover2pi12
GMm2
=(1.0546×10−34Js)2
(6.6726×10−11m3/kg·s2)(1.9892×1030kg)(5.98×1024kg)2=2.34×10−138m.
(c)Equation 4.70 ⇒En=−/bracketleftBigm
2/planckover2pi12(GMm )2/bracketrightBig1
n2.
Ec=1
2mv2−GMm
ro.ButGMm
r2o=mv2
ro⇒1
2mv2=GMm
2ro,so
Ec=−GMm
2ro=−/bracketleftBigm
2/planckover2pi12(GMm )2/bracketrightBig1
n2⇒n2=GMm2
/planckover2pi12ro=ro
ag⇒n=/radicalbiggro
ag.
ro= earth-sun distance = 1 .496×1011m⇒n=/radicalbigg
1.496×1011
2.34×10−138=2.53×1074.
(d)
∆E=−/bracketleftbiggG2M2m3
2/planckover2pi12/bracketrightbigg/bracketleftbigg1
(n+1 )2−1
n2/bracketrightbigg
.1
(n+1 )2=1
n2(1 + 1/n)2≈1
n2/parenleftbigg
1−2
n/parenrightbigg
.
So/bracketleftbigg1
(n+1 )2−1
n2/bracketrightbigg
≈1
n2/parenleftbigg
1−2
n−1/parenrightbigg
=−2
n3;∆E=G2M2m3
/planckover2pi12n3.
∆E=(6.67×10−11)2(1.99×1030)2(5.98×1024)3
(1.055×10−34)2(2.53×74)3=2.09×10−41J.Ep=∆E=hν=hc
λ.
λ=( 3×108)(6.63×10−34)/(2.09×10−41)=9.52×1015m.
But 1 ly = 9 .46×1015m. Is it a coincidence that λ≈1 ly? No: From part (c), n2=GMm2ro//planckover2pi12,s o
λ=ch
∆E=c2π/planckover2pi1/planckover2pi12n3
G2M2m3=c2π/planckover2pi13
G2M2m3/parenleftbiggGMm2ro
/planckover2pi12/parenrightbigg3/2
=c/parenleftBigg
2π/radicalbigg
r3o
GM/parenrightBigg
.
But (from (c)) v=/radicalbig
GM/r o=2πro/T, where Tis the period of the orbit (in this case one year), so
T=2π/radicalbig
r3o/GM, and hence λ=cT(one light year). [Incidentally, the same goes for hydrogen: The
wavelength of the photon emitted in a transition from a highly excited state to the next lower one is equalto the distance light would travel in one orbital period.]
Problem 4.18
/angbracketleftf|L±g/angbracketright=/angbracketleftf|Lxg/angbracketright±i/angbracketleftf|Lyg/angbracketright=/angbracketleftLxf|g/angbracketright±i/angbracketleftLyf|g/angbracketright=/angbracketleft(Lx∓iLy)f|g/angbracketright=/angbracketleftL∓f|g/angbracketright,so (L±)†=L∓.
Now, using Eq. 4.112, in the form L∓L±=L2−L2
z∓/planckover2pi1Lz:
/angbracketleftfm
l|L∓L±fm
l/angbracketright=/angbracketleftfm
l|(L2−L2
z∓/planckover2pi1Lz)fm
l/angbracketright=/angbracketleftfm
l|/bracketleftbig
/planckover2pi12l(l+1 )−/planckover2pi12m2∓/planckover2pi12m/bracketrightbig
fm
l/angbracketright
=/planckover2pi12[l(l+1 )−m(m±1)]/angbracketleftfm
l|fm
l/angbracketright=/planckover2pi12[l(l+1 )−m(m±1)]
=/angbracketleftL±fm
l|L±fm
l/angbracketright=/angbracketleftAm
lfm±1
l|Am
lfm±1
l/angbracketright=|Am
l|2/angbracketleftfm±1
l|fm±1
l/angbracketright=|Am
l|2.
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100 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Conclusion: Am
l=/planckover2pi1/radicalbig
l(l+1 )−m(m±1).
Problem 4.19
(a)
[Lz,x]=[xpy−ypx,x]=[xpy,x]−[ypx,x]=0−y[px,x]=i/planckover2pi1y./check
[Lz,y]=[xpy−ypx,y]=[xpy,y]−[ypx,y]=x[py,y]−0=−i/planckover2pi1x./check
[Lz,z]=[xpy−ypx,z]=[xpy,z]−[ypx,z]=0−0=0./check
[Lz,px]=[xpy−ypx,px]=[xpy,px]−[ypx,px]=py[x,px]−0=i/planckover2pi1py./check
[Lz,py]=[xpy−ypx,py]=[xpy,py]−[ypx,py]=0−px[y,py]=−i/planckover2pi1px./check
[Lz,pz]=[xpy−ypx,pz]=[xpy,pz]−[ypx,pz]=0−0=0./check
(b)
[Lz,Lx]=[Lz,ypz−zpy]=[Lz,ypz]−[Lz,zpy]=[Lz,y]pz−[Lz,py]z
=−i/planckover2pi1xpz+i/planckover2pi1pxz=i/planckover2pi1(zpx−xpz)=i/planckover2pi1Ly.
(So, by cyclic permutation of the indices, [ Lx,Ly]=i/planckover2pi1Lz.)
(c)
[Lz,r2]=[Lz,x2]+[Lz,y2]+[Lz,z2]=[Lz,x]x+x[Lz,x]+[Lz,y]y+y[Lz,y]+0
=i/planckover2pi1yx+xi/planckover2pi1y+(−i/planckover2pi1x)y+y(−i/planckover2pi1x)=0.
[Lz,p2]=[Lz,p2
x]+[Lz,p2
y]+[Lz,p2
z]=[Lz,px]px+px[Lz,px]+[Lz,py]py+py[Lz,py]+0
=i/planckover2pi1pypx+pxi/planckover2pi1py+(−i/planckover2pi1px)py+py(−i/planckover2pi1px)=0.
(d)It follows from (c) that all three components of Lcommute with r2andp2, and hence with the whole
Hamiltonian, since H=p2/2m+V(√
r2). QED
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 101
Problem 4.20
(a)
Equation 3.71 ⇒d/angbracketleftLx/angbracketright
dt=i
/planckover2pi1/angbracketleft[H,L x]/angbracketright.[H,L x]=1
2m[p2,Lx]+[V,Lx].
The first term is zero (Problem 4.19(c)); the second would be too if Vwere a function only of r=|r|, but
in general
[H,L x]=[V,yp z−zpy]=y[V,pz]−z[V,py].Now (Problem 3.13(c)):
[V,pz]=i/planckover2pi1∂V
∂zand [V,py]=i/planckover2pi1∂V
∂y.So [H,L x]=yi/planckover2pi1∂V
∂z−zi/planckover2pi1∂V
∂y=i/planckover2pi1[r×(∇V)]x.
Thusd/angbracketleftLx/angbracketright
dt=−/angbracketleft[r×(∇V)]x/angbracketright,and the same goes for the other two components:
d/angbracketleftL/angbracketright
dt=/angbracketleft[r×(−∇V)]/angbracketright=/angbracketleftN/angbracketright.QED
(b)
IfV(r)=V(r),then∇V=∂V
∂rˆr,andr׈r=0,sod/angbracketleftL/angbracketright
dt=0.QED
Problem 4.21
(a)
L+L−f=−/planckover2pi12eiφ/parenleftbigg∂
∂θ+icotθ∂
∂φ/parenrightbigg/bracketleftbigg
e−iφ/parenleftbigg∂f
∂θ−icotθ∂f
∂φ/parenrightbigg/bracketrightbigg
=−/planckover2pi12eiφ/braceleftbigg
e−iφ/bracketleftbigg∂2f
∂θ2−i/parenleftbigg
−csc2θ∂f
∂φ+ cotθ∂2f
∂θ∂φ/parenrightbigg/bracketrightbigg
+icotθ/bracketleftbigg
−ie−iφ/parenleftbigg∂f
∂θ−icotθ∂f
∂φ/parenrightbigg
+e−iφ/parenleftbigg∂2f
∂φ∂θ−icotθ∂2f
∂φ2/parenrightbigg/bracketrightbigg /bracerightbigg
=−/planckover2pi12/parenleftbigg∂2f
∂θ2+icsc2θ∂f
∂φ−icotθ∂2f
∂θ∂φ+ cotθ∂f
∂θ−icot2θ∂f
∂φ+icotθ∂2f
∂φ∂θ+ cot2θ∂2f
∂φ2/parenrightbigg
=−/planckover2pi12/bracketleftbigg∂2
∂θ2+ cotθ∂
∂θ+ cot2θ∂2
∂φ2+i(csc2θ−cot2θ)∂
∂φ/bracketrightbigg
f,so
L+L−=−/planckover2pi12/parenleftbigg∂2
∂θ2+ cotθ∂
∂θ+ cot2θ∂2
∂φ2+i∂
∂φ/parenrightbigg
.QED
(b)Equation 4.129 ⇒Lz=/planckover2pi1
i∂
∂φ,Eq. 4.112⇒L2=L+L−+L2
z−/planckover2pi1Lz, so, using (a):
L2=−/planckover2pi12/parenleftbigg∂2
∂θ2+ cotθ∂
∂θ+ cot2θ∂2
∂φ2+i∂
∂φ/parenrightbigg
−/planckover2pi12∂2
∂φ2−/planckover2pi1/parenleftbigg/planckover2pi1
i/parenrightbigg∂
∂φ
=−/planckover2pi12/parenleftbigg∂2
∂θ2+ cotθ∂
∂θ+ (cot2θ+1 )∂2
∂φ2+i∂
∂φ−i∂
∂φ/parenrightbigg
=−/planckover2pi12/parenleftbigg∂2
∂θ2+ cotθ∂
∂θ+1
sin2θ∂2
∂φ2/parenrightbigg
=−/planckover2pi12/bracketleftbigg1
sinθ∂
∂θ/parenleftbigg
sinθ∂
∂θ/parenrightbigg
+1
sin2θ∂2
∂φ2/bracketrightbigg
.QED
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102 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.22
(a)L+Yl
l=0(top of the ladder).
(b)
LzYl
l=/planckover2pi1lYl
l⇒/planckover2pi1
i∂
∂φYl
l=/planckover2pi1lYl
l,so∂Yl
l
∂φ=ilYl
l,and hence Yl
l=f(θ)eilφ.
[Note:f(θ) is the “constant” here—it’s constant with respect to φ... but still can depend on θ.]
L+Yl
l=0⇒/planckover2pi1eiφ/parenleftbigg∂
∂θ+icotθ∂
∂φ/parenrightbigg/bracketleftbig
f(θ)eilφ/bracketrightbig
=0,ordf
dθeilφ+ifcotθileilφ=0,so
df
dθ=lcotθf⇒df
f=lcotθdθ⇒/integraldisplaydf
f=l/integraldisplaycosθ
sinθdθ⇒lnf=lln(sinθ)+constant.
lnf= ln(sinlθ)+K⇒ln/parenleftbiggf
sinlθ/parenrightbigg
=K⇒f
sinlθ=constant⇒f(θ)=Asinlθ.
Yl
l(θ,φ)=A(eiφsinθ)l.
(c)
1=A2/integraldisplay
sin2lθsinθdθdφ =2πA2/integraldisplayπ
0sin(2l+1)θdθ=2πA22(2·4·6·····(2l))
1·3·5·····(2l+1 )
=4πA2(2·4·6·····2l)2
1·2·3·4·5·····(2l+1 )=4πA2(2ll!)2
(2l+ 1)!,soA=1
2l+1l!/radicalbigg
(2l+ 1)!
π,
the same as Problem 4.5, except for an overall factor of ( −1)l, which is arbitrary anyway.
Problem 4.23
L+Y1
2=/planckover2pi1eiφ/parenleftbigg∂
∂θ+icotθ∂
∂θ/parenrightbigg/bracketleftBigg
−/radicalbigg
15
8πsinθcosθeiφ/bracketrightBigg
=−/radicalbigg
15
8π/planckover2pi1eiφ/bracketleftbigg
eiφ(cos2θ−sin2θ)+icosθ
sinθsinθcosθieiφ/bracketrightbigg
=−/radicalbigg
15
8π/planckover2pi1e2iφ/parenleftbig
cos2θ−sin2θ−cos2θ/parenrightbig
=/radicalbigg
15
8π/planckover2pi1/parenleftbig
eiφsinθ/parenrightbig2
=/planckover2pi1√
2·3−1·2Y2
2=2/planckover2pi1Y2
2.∴Y2
2=1
4/radicalbigg
15
2π/parenleftbig
eiφsinθ/parenrightbig2.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 103
Problem 4.24
(a)
H=2/parenleftbigg1
2mv2/parenrightbigg
=mv2;|L|=2a
2mv=amv, soL2=a2m2v2,and hence H=L2
ma2.
But we know the eigenvalues of L2:/planckover2pi12l(l+ 1); or, since we usually label energies with n:
En=/planckover2pi12n(n+1 )
ma2(n=0,1,2,...).
(b)ψnm(θ,φ)=Ym
n(θ,φ),the ordinary spherical harmonics. The degeneracy of the nth energy level is the
number of m-values for given n:2n+1.
Problem 4.25
rc=(1.6×10−19)2
4π(8.85×10−12)(9.11×10−31)(3.0×108)2=2.81×10−15m.
L=1
2/planckover2pi1=Iω=/parenleftbigg2
5mr2/parenrightbigg/parenleftBigv
r/parenrightBig
=2
5mrv so
v=5/planckover2pi1
4mr=(5)(1.055×10−34)
(4)(9.11×10−31)(2.81×10−15)=5.15×1010m/s.
Since the speed of light is 3 ×108m/s, a point on the equator would be going more than 100 times the speed
of light. Nope : This doesn’t look like a very realistic model for spin.
Problem 4.26
(a)
[Sx,Sy]=SxSy−SySx=/planckover2pi12
4/bracketleftbigg/parenleftbigg01
10/parenrightbigg/parenleftbigg0−i
i0/parenrightbigg
−/parenleftbigg0−i
i0/parenrightbigg/parenleftbigg01
10/parenrightbigg/bracketrightbigg
=/planckover2pi12
4/bracketleftbigg/parenleftbiggi0
0−i/parenrightbigg
−/parenleftbigg−i0
0i/parenrightbigg/bracketrightbigg
=/planckover2pi12
4/parenleftbigg2i0
0−2i/parenrightbigg
=i/planckover2pi1/planckover2pi1
2/parenleftbigg10
0−1/parenrightbigg
=i/planckover2pi1Sz./check
(b)
σxσx=/parenleftbigg10
01/parenrightbigg
=1=σyσy=σzσz,soσjσj= 1 for j=x,y,orz.
σxσy=/parenleftbiggi0
0−i/parenrightbigg
=iσz;σyσz=/parenleftbigg0i
i0/parenrightbigg
=iσx;σzσx=/parenleftbigg01
−10/parenrightbigg
=iσy;
σyσx=/parenleftbigg−i0
0i/parenrightbigg
=−iσz;σzσy=/parenleftbigg0−i
−i0/parenrightbigg
=−iσx;σxσz=/parenleftbigg0−1
10/parenrightbigg
=−iσy.
Equation 4.153 packages all this in a single formula. /check
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104 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.27
(a)
χ†χ=|A|2(9 + 16) = 25|A|2=1⇒A=1/5.
(b)
/angbracketleftSx/angbracketright=χ†Sxχ=1
25/planckover2pi1
2/parenleftbig
−3i4/parenrightbig/parenleftbigg
01
10/parenrightbigg/parenleftbigg
3i
4/parenrightbigg
=/planckover2pi1
50/parenleftbig
−3i4/parenrightbig/parenleftbigg
4
3i/parenrightbigg
=/planckover2pi1
50(12i+1 2i)=0.
/angbracketleftSy/angbracketright=χ†Syχ=1
25/planckover2pi1
2/parenleftbig−3i4/parenrightbig/parenleftbigg0−i
i0/parenrightbigg/parenleftbigg3i
4/parenrightbigg
=/planckover2pi1
50/parenleftbig−3i4/parenrightbig/parenleftbigg−4i
−3/parenrightbigg
=/planckover2pi1
50(−12−12) =−12
25/planckover2pi1.
/angbracketleftSz/angbracketright=χ†Szχ=1
25/planckover2pi1
2/parenleftbig−3i4/parenrightbig/parenleftbigg10
0−1/parenrightbigg/parenleftbigg3i
4/parenrightbigg
=/planckover2pi1
50/parenleftbig−3i4/parenrightbig/parenleftbigg3i
−4/parenrightbigg
=/planckover2pi1
50(9−16) =−7
50/planckover2pi1.
(c)
/angbracketleftS2
x/angbracketright=/angbracketleftS2
y/angbracketright=/angbracketleftS2
z/angbracketright=/planckover2pi12
4(always, for spin 1/2), so σ2
Sx=/angbracketleftS2
x/angbracketright−/angbracketleftSx/angbracketright2=/planckover2pi12
4−0,σSx=/planckover2pi1
2.
σ2
Sy=/angbracketleftS2
y/angbracketright−/angbracketleftSy/angbracketright2=/planckover2pi1
4−/parenleftbigg12
25/parenrightbigg2
/planckover2pi12=/planckover2pi12
2500(625−576) =49
2500/planckover2pi12,σSy=7
50/planckover2pi1.
σ2
Sz=/angbracketleftS2
z/angbracketright−/angbracketleftSz/angbracketright2=/planckover2pi12
4−/parenleftbigg7
50/parenrightbigg2
/planckover2pi12=/planckover2pi12
2500(625−49) =576
2500/planckover2pi12,σSz=12
25/planckover2pi1.
(d)
σSxσSy=/planckover2pi1
2·7
50/planckover2pi1?
≥/planckover2pi1
2|/angbracketleftSz/angbracketright|=/planckover2pi1
2·7
50/planckover2pi1(right atthe uncertainty limit) ./check
σSyσSz=7
50/planckover2pi1·12
25/planckover2pi1?
≥/planckover2pi1
2|/angbracketleftSx/angbracketright|= 0 (trivial). /check
σSzσSx=12
25/planckover2pi1·/planckover2pi1
2?
≥/planckover2pi1
2|/angbracketleftSy/angbracketright|=/planckover2pi1
2·12
25/planckover2pi1(right atthe uncertainty limit) ./check
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 105
Problem 4.28
/angbracketleftSx/angbracketright=/planckover2pi1
2/parenleftbig
a∗b∗/parenrightbig/parenleftbigg
01
10/parenrightbigg/parenleftbigg
a
b/parenrightbigg
=/planckover2pi1
2/parenleftbig
a∗b∗/parenrightbig/parenleftbigg
b
a/parenrightbigg
=/planckover2pi1
2(a∗b+b∗a)=/planckover2pi1Re(ab∗).
/angbracketleftSy/angbracketright=/planckover2pi1
2/parenleftbig
a∗b∗/parenrightbig/parenleftbigg
0−i
i0/parenrightbigg/parenleftbigg
a
b/parenrightbigg
=/planckover2pi1
2/parenleftbig
a∗b∗/parenrightbig/parenleftbigg
−ib
ia/parenrightbigg
=/planckover2pi1
2(−ia∗b+iab∗)=/planckover2pi1
2i(ab∗−a∗b)=−/planckover2pi1Im(ab∗).
/angbracketleftSz/angbracketright=/planckover2pi1
2/parenleftbig
a∗b∗/parenrightbig/parenleftbigg
10
0−1/parenrightbigg/parenleftbigg
a
b/parenrightbigg
=/planckover2pi1
2/parenleftbig
a∗b∗/parenrightbig/parenleftbigg
a
−b/parenrightbigg
=/planckover2pi1
2(a∗a−b∗b)=/planckover2pi1
2(|a|2−|b|2).
S2
x=/planckover2pi12
4/parenleftbigg01
10/parenrightbigg/parenleftbigg01
10/parenrightbigg
=/planckover2pi12
4/parenleftbigg10
01/parenrightbigg
=/planckover2pi12
4;S2
y=/planckover2pi12
4/parenleftbigg0−i
i0/parenrightbigg/parenleftbigg0−i
i0/parenrightbigg
=/planckover2pi12
4;
S2
z=/planckover2pi12
4/parenleftbigg10
0−1/parenrightbigg/parenleftbigg10
0−1/parenrightbigg
=/planckover2pi12
4;s o/angbracketleftS2
x/angbracketright=/angbracketleftS2
y/angbracketright=/angbracketleftS2
z/angbracketright=/planckover2pi12
4.
/angbracketleftS2
x/angbracketright+/angbracketleftS2
y/angbracketright+/angbracketleftS2
z/angbracketright=3
4/planckover2pi12?=s(s+1 )/planckover2pi12=1
2(1
2+1 )/planckover2pi12=3
4/planckover2pi12=/angbracketleftS2/angbracketright./check
Problem 4.29
(a)
Sy=/planckover2pi1
2/parenleftbigg0−i
i0/parenrightbigg
;/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ−i/planckover2pi1/2
i/planckover2pi1/2−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle=λ
2−/planckover2pi12
4⇒λ=±/planckover2pi1
2(of course) .
/planckover2pi1
2/parenleftbigg0−i
i0/parenrightbigg/parenleftbiggα
β/parenrightbigg
=±/planckover2pi1
2/parenleftbiggα
β/parenrightbigg
⇒−iβ=±α;|α|2+|β|2=1⇒|α|2+|α|2=1⇒α=1√
2.
χ(y)
+=1√
2/parenleftbigg1
i/parenrightbigg
;χ(y)
−=1√
2/parenleftbigg1
−i/parenrightbigg
.
(b)
c+=/parenleftBig
χ(y)
+/parenrightBig†
χ=1√
2/parenleftbig1−i/parenrightbig/parenleftbigga
b/parenrightbigg
=1√
2(a−ib);+/planckover2pi1
2,with probability1
2|a−ib|2.
c−=/parenleftBig
χ(y)
−/parenrightBig†
χ=1√
2/parenleftbig1i/parenrightbig/parenleftbigga
b/parenrightbigg
=1√
2(a+ib);−/planckover2pi1
2,with probability1
2|a+ib|2.
P++P−=1
2[(a∗+ib∗)(a−ib)+(a∗−ib∗)(a+ib)]
=1
2/bracketleftbig
|a|2−ia∗b+iab∗+|b|2+|a|2+ia∗b−iab∗+|b|2/bracketrightbig
=|a|2+|b|2=1./check
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
106 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(c)/planckover2pi12
4,with probability 1 .
Problem 4.30
Sr=S·ˆr=Sxsinθcosφ+Sysinθsinφ+Szcosθ
=/planckover2pi1
2/bracketleftbigg/parenleftbigg
0 sin θcosφ
sinθcosφ 0/parenrightbigg
+/parenleftbigg
0−isinθsinφ
isinθsinφ 0/parenrightbigg
+/parenleftbigg
cosθ0
0−cosθ/parenrightbigg/bracketrightbigg
=/planckover2pi1
2/parenleftbigg
cosθ sinθ(cosφ−isinφ)
sinθ(cosφ+isinφ)−cosθ/parenrightbigg
=/planckover2pi1
2/parenleftbigg
cosθe−iφsinθ
eiφsinθ−cosθ/parenrightbigg
.
/vextendsingle/vextendsingle/vextendsingle/vextendsingle(
/planckover2pi1
2cosθ−λ)/planckover2pi1
2e−iφsinθ
/planckover2pi1
2eiφsinθ(−/planckover2pi1
2cosθ−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−/planckover2pi1
2
4cos2θ+λ2−/planckover2pi12
4sin2θ=0⇒
λ2=/planckover2pi12
4(sin2θ+ cos2θ)=/planckover2pi12
4⇒λ=±/planckover2pi1
2(of course) .
/planckover2pi1
2/parenleftbiggcosθe−iφsinθ
eiφsinθ−cosθ/parenrightbigg/parenleftbiggα
β/parenrightbigg
=±/planckover2pi1
2/parenleftbiggα
β/parenrightbigg
⇒αcosθ+βe−iφsinθ=±α;β=eiφ(±1−cosθ)
sinθα.
Upper sign: Use 1−cosθ= 2 sin2θ
2,sinθ= 2 sinθ
2cosθ
2. Then β=eiφsin(θ/2)
cos(θ/2)α.Normalizing:
1=|α|2+|β|2=|α2|+sin2(θ/2)
cos2(θ/2)|α|2=|α|21
cos2(θ/2)⇒α= cosθ
2,β=eiφsinθ
2,χ(r)
+=/parenleftbiggcos(θ/2)
eiφsin(θ/2)/parenrightbigg
.
Lower sign: Use 1 + cos θ= 2 cos2θ
2,β=−eiφcos(θ/2)
sin(θ/2)α;1 =|α|2+cos2(θ/2)
sin2(θ/2)|α|2=|α|21
sin2(θ/2).
Pickα=e−iφsin(θ/2); then β=−cos(θ/2),andχ(r)
−=/parenleftbigg
e−iφsin(θ/2)
−cos(θ/2)/parenrightbigg
.
Problem 4.31
There are three states: χ+=
1
00
,χ
0=
0
10
,χ
−=
0
01
.
S
zχ+=/planckover2pi1χ+,Szχ0=0,Szχ−=−/planckover2pi1χ−,⇒Sz=/planckover2pi1
10 0
00 000−1
.
From Eq. 4.136:
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 107
S+χ+=0, S+χ0=/planckover2pi1√
2χ+,S+χ−=/planckover2pi1√
2χ0
S−χ+=/planckover2pi1√
2χ0,S−χ0=/planckover2pi1√
2χ−,S−χ−=0/bracerightbigg
⇒S+=√
2/planckover2pi1
010
001000
,S
−=√
2/planckover2pi1
000
100010
.
S
x=1
2(S++S−)=/planckover2pi1√
2
010
101010
,
Sy=1
2i(S+−S−)=i/planckover2pi1√
2
0−10
10−1
01 0
.
Problem 4.32
(a)Using Eqs. 4.151 and 4.163:
c(x)
+=χ(x)†
+χ=1√
2/parenleftbig11/parenrightbig/parenleftbigg
cosα
2eiγB0t/2
sinα
2e−iγB0t/2/parenrightbigg
=1√
2/bracketleftBig
cosα
2eiγB0t/2+ sinα
2e−iγB0t/2/bracketrightBig
.
P(x)
+(t)=|c(x)
+|2=1
2/bracketleftBig
cosα
2e−iγB0t/2+ sinα
2eiγB0t/2/bracketrightBig/bracketleftBig
cosα
2eiγB0t/2+ sinα
2e−iγB0t/2/bracketrightBig
=1
2/bracketleftBig
cos2α
2+ sin2α
2+ sinα
2cosα
2/parenleftbig
eiγB0t+e−iγB0t/parenrightbig/bracketrightBig
=1
2/bracketleftBig
1+2s i nα
2cosα
2cos(γB0t)/bracketrightBig
=1
2[1 + sin αcos(γB0t)].
(b)From Problem 4.29(a): χ(y)
+=1√
2/parenleftbigg1
i/parenrightbigg
.
c(y)
+=χ(y)†
+χ=1√
2/parenleftbig1−i/parenrightbig/parenleftbiggcosα
2eiγB0t/2
sinα
2eiγB0t/2/parenrightbigg
=1√
2/bracketleftBig
cosα
2eiγB0t/2−isinα
2e−iγB0t/2/bracketrightBig
;
P(y)
+(t)=|c(y)
+|2=1
2/bracketleftBig
cosα
2e−iγB0t/2+isinα
2eiγB0t/2/bracketrightBig/bracketleftBig
cosα
2eiγB0t/2−isinα
2e−iγB0t/2/bracketrightBig
=1
2/bracketleftBig
cos2α
2+ sin2α
2+isinα
2cosα
2/parenleftbig
eiγB0t−e−iγB0t/parenrightbig/bracketrightBig
=1
2/bracketleftBig
1−2 sinα
2cosα
2sin(γB0t)/bracketrightBig
=1
2[1−sinαsin(γB0t)].
(c)
χ(z)
+=/parenleftbigg1
0/parenrightbigg
;c(z)
+=/parenleftbig10/parenrightbig/parenleftbiggcosα
2eiγB0t/2
sinα
2e−iγB0t/2/parenrightbigg
= cosα
2eiγB0t/2;P(z)
+(t)=|c(z)
+|2=cos2α
2.
Problem 4.33
(a)
H=−γB·S=−γB0cosωtSz=−γB0/planckover2pi1
2cosωt/parenleftbigg10
0−1/parenrightbigg
.
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108 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b)
χ(t)=/parenleftbigg
α(t)
β(t)/parenrightbigg
,withα(0) =β(0) =1√
2.
i/planckover2pi1∂χ
∂t=i/planckover2pi1/parenleftbigg˙α
˙β/parenrightbigg
=Hχ=−γB0/planckover2pi1
2cosωt/parenleftbigg
10
0−1/parenrightbigg/parenleftbigg
α
β/parenrightbigg
=−γB0/planckover2pi1
2cosωt/parenleftbigg
α
−β/parenrightbigg
.
˙α=i/parenleftbiggγB0
2/parenrightbigg
cosωtα⇒dα
α=i/parenleftbiggγB0
2/parenrightbigg
cosωtdt⇒lnα=iγB0
2sinωt
ω+constant.
α(t)=Aei(γB0/2ω) sinωt;α(0) =A=1√
2,soα(t)=1√
2ei(γB0/2ω) sinωt.
˙β=−i/parenleftbiggγB0
2/parenrightbigg
cosωtβ⇒β(t)=1√
2e−i(γB0/2ω) sinωt.χ(t)=1√
2/parenleftbiggei(γB0/2ω) sinωt
e−i(γB0/2ω) sinωt/parenrightbigg
.
(c)
c(x)
−=χ(x)†
−χ=1
2(1−1)/parenleftbiggei(γB0/2ω) sinωt
e−i(γB0/2ω) sinωt/parenrightbigg
=1
2/bracketleftBig
ei(γB0/2ω) sinωt−e−i(γB0/2ω) sinωt/bracketrightBig
=isin/bracketleftbiggγB0
2ωsinωt/bracketrightbigg
.P(x)
−(t)=|c(x)
−|2=sin2/bracketleftbiggγB0
2ωsinωt/bracketrightbigg
.
(d)The argument of sin2must reach π/2 (soP=1 )⇒γB0
2ω=π
2,o rB0=πω
γ.
Problem 4.34
(a)
S−|10/angbracketright=(S(1)
−+S(2)
−)1√
2(↑↓+↓↑)=1√
2[(S−↑)↓+(S−↓)↑+↑(S−↓)+↓(S−↑)].
ButS−↑=/planckover2pi1↓,S−↓= 0 (Eq. 4.143), so S−|10/angbracketright=1√
2[/planckover2pi1↓↓+ 0+0+ /planckover2pi1↓↓]=√
2/planckover2pi1↓↓=√
2/planckover2pi1|1−1/angbracketright./check
(b)
S±|00/angbracketright=(S(1)
±+S(2)
±)1√
2(↑↓−↓↑ )=1√
2[(S±↑)↓−(S±↓)↑+↑(S±↓)−↓(S±↑)].
S+|00/angbracketright=1√
2(0−/planckover2pi1↑↑+/planckover2pi1↑↑−0) = 0; S−|00/angbracketright=1√
2(/planckover2pi1↓↓−0+0−/planckover2pi1↓↓)=0./check
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 109
(c)
S2|11/angbracketright=/bracketleftBig
(S(1))2+(S(2))2+2S(1)·S(2)/bracketrightBig
↑↑
=(S2↑)↑+↑(S2↑)+2[ (Sx↑)(Sx↑)+(Sy↑)(Sy↑)+(Sz↑)(Sz↑)]
=3
4/planckover2pi12↑↑+3
4/planckover2pi12↑↑+2/bracketleftbigg/planckover2pi1
2↓/planckover2pi1
2↓+i/planckover2pi1
2↓i/planckover2pi1
2↓+/planckover2pi1
2↑/planckover2pi1
2↑/bracketrightbigg
=3
2/planckover2pi12↑↑+2/parenleftbigg/planckover2pi12
4↑↑/parenrightbigg
=2/planckover2pi12↑↑=2/planckover2pi12|11/angbracketright= (1)(1 + 1) /planckover2pi12|11/angbracketright,as itshould be.
S2|1−1/angbracketright=/bracketleftBig
(S(1))2+(S(2))2+2S(1)·S(2)/bracketrightBig
↓↓
=3/planckover2pi12
4↓↓+3/planckover2pi12
4↓↓+2 [(Sx↓)(Sx↓)+(Sy↓)(Sy↓)+(Sz↓)(Sz↓)]
=3
2/planckover2pi12↓↓+2/bracketleftbigg/parenleftbigg/planckover2pi1
2↑/parenrightbigg/parenleftbigg/planckover2pi1
2↑/parenrightbigg
+/parenleftbigg
−i/planckover2pi1
2↑/parenrightbigg/parenleftbigg
−i/planckover2pi1
2↑/parenrightbigg
+/parenleftbigg
−/planckover2pi1
2↓/parenrightbigg/parenleftbigg
−/planckover2pi1
2↓/parenrightbigg/bracketrightbigg
=3
2/planckover2pi12↓↓+2/planckover2pi12
4↓↓=2/planckover2pi12↓↓=2/planckover2pi12|1−1/angbracketright./check
Problem 4.35
(a)1/2 and 1/2 gives 1 or zero; 1/2 and 1 gives 3/2 or 1/2; 1/2 and 0 gives 1/2 only. So baryons can have
spin 3/2 or spin 1/2 (and the latter can be acheived in two distinct ways). [Incidentally, the lightest
baryons docarry spin 1/2 (proton, neutron, etc.) or 3/2 (∆ ,Ω−,etc.); heavier baryons can have higher
total spin, but this is because the quarks have orbital angular momentum as well.]
(b)1/2 and 1/2 gives spin 1 or spin 0. [Again, these arethe observed spins for the lightest mesons: π’s and
K’s have spin 0, ρ’s andω’s have spin 1.]
Problem 4.36
(a)From the 2×1 Clebsch-Gordan table we get
|31/angbracketright=/radicalbigg
1
15|22/angbracketright|1−1/angbracketright+/radicalbigg
8
15|21/angbracketright|10/angbracketright+/radicalbigg
6
15|20/angbracketright|11/angbracketright,
so you might get 2/planckover2pi1(probability 1 /15),/planckover2pi1(probability 8 /15),or (probability 6 /15).
(b)From the 1×1
2table:|10/angbracketright|1
2−1
2/angbracketright=/radicalBig
2
3|3
2−1
2/angbracketright+/radicalBig
1
3|1
2−1
2/angbracketright.So the total is 3 /2o r1/2, withl(l+1)/planckover2pi12=
15/4/planckover2pi12and 3/4/planckover2pi12, respectively. Thus you get15
4/planckover2pi12(probability 2 /3),or3
4/planckover2pi12(probability 1 /3).
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110 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.37
Using Eq. 4.179: [ S2,S(1)
z]=[S(1)2,S(1)
z]+[S(2)2,S(1)
z]+2 [S(1)·S(2),S(1)
z].But [S2,Sz] = 0 (Eq. 4.102), and
anything with superscript (2) commutes with anything with superscript (1). So
[S2,S(1)
z]=2/braceleftBig
S(2)
x[S(1)
x,S(1)
z]+S(2)
y[S(1)
y,S(1)
z]+S(2)
z[S(1)
z,S(1)
z]/bracerightBig
=2/braceleftBig
−i/planckover2pi1S(1)
yS(2)
x+i/planckover2pi1S(1)
xS(2)
y/bracerightBig
=2i/planckover2pi1(S(1)×S(2))z.
[S2,S(1)
z]=2i/planckover2pi1(S(1)
xS(2)
y−S(1)
yS(2)
x),and [S2,S(1)]=2i/planckover2pi1(S(1)×S(2)). Note that [ S2,S(2)]=2i/planckover2pi1(S(2)×S(1))=
−2i/planckover2pi1(S(1)×S(2)),so [S2,(S(1)+S(2))] = 0.]
Problem 4.38
(a)
−/planckover2pi12
2m/parenleftbigg∂2ψ
∂x2+∂2ψ
∂y2+∂2ψ
∂z2/parenrightbigg
+1
2mω2/parenleftbig
x2+y2+z2/parenrightbig
ψ=Eψ.
Letψ(x,y,z)=X(x)Y(y)Z(z); plug it in, divide by XYZ, and collect terms:
/parenleftbigg
−/planckover2pi12
2m1
Xd2X
dx2+1
2mω2x2/parenrightbigg
+/parenleftbigg
−/planckover2pi12
2m1
Yd2Y
dy2+1
2mω2y2/parenrightbigg
+/parenleftbigg
−/planckover2pi12
2m1
Zd2Z
dz2+1
2mω2z2/parenrightbigg
=E.
The first term is a function only of x, the second only of y, and the third only of z. So each is a constant
(call the constants Ex,Ey,Ez, withEx+Ey+Ez=E). Thus:
−/planckover2pi12
2md2X
dx2+1
2mω2x2X=ExX;−/planckover2pi12
2md2Y
dy2+1
2mω2y2Y=EyY;−/planckover2pi12
2md2Z
dz2+1
2mω2z2Z=EzZ.
Each of these is simply the one-dimensional harmonic oscillator (Eq. 2.44). We know the allowed energies
(Eq. 2.61):
Ex=(nx+1
2)/planckover2pi1ω;Ey=(ny+1
2)/planckover2pi1ω;Ez=(nz+1
2)/planckover2pi1ω; where nx,ny,nz=0,1,2,3,....
SoE=(nx+ny+ny+3
2)/planckover2pi1ω=(n+3
2)/planckover2pi1ω,withn≡nx+ny+nz.
(b)The question is: “How many ways can we add three non-negative integers to get sum n?”
Ifnx=n,thenny=nz=0 ;oneway.
Ifnx=n−1,thenny=0,nz=1,or elseny=1,nz=0 ;twoways.
Ifnx=n−2,thenny=0,nz=2,orny=1,nz=1,orny=2,nz=0 ;threeways.
And so on. Evidently d(n)=1+2+3+ ···+(n+1 )=(n+ 1)(n+2 )
2.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 111
Problem 4.39
Eq. 4.37:−/planckover2pi12
2md2u
dr2+/bracketleftbigg1
2mω2r2+/planckover2pi12
2ml(l+1 )
r2/bracketrightbigg
u=Eu.
Following Eq. 2.71, let ξ≡/radicalbiggmω
/planckover2pi1r.Then−/planckover2pi12
2mmω
/planckover2pi1d2u
dξ2+/bracketleftbigg1
2mω2/planckover2pi1
mωξ2+/planckover2pi12
2mmω
/planckover2pi1l(l+1 )
ξ2/bracketrightbigg
u=Eu,
ord2u
dξ2=/bracketleftbigg
ξ2+l(l+1 )
ξ2−K/bracketrightbigg
u,whereK≡2E
/planckover2pi1ω(as in Eq. 2.73) .
At large ξ,d2u
dξ2≈ξ2u,andu∼()e−ξ2/2(see Eq. 2.77) .
At small ξ,d2u
dξ2≈l(l+1 )
ξ2u,andu∼()ξl+1(see Eq. 4.59) .
So letu(ξ)≡ξl+1e−ξ2/2v(ξ).[This defines the new function v(ξ).]
du
dξ=(l+1 )ξle−ξ2/2v−ξl+2e−ξ2/2v+ξl+1e−ξ2/2v/prime.
d2u
dξ2=l(l+1 )ξl−1e−ξ2/2v−(l+1 )ξl+1e−ξ2/2v+(l+1 )ξle−ξ2/2v/prime−(l+2 )ξl+1e−ξ2/2v
+ξl+3e−ξ2/2v−ξl+2e−ξ2/2v/prime+(l+1 )ξle−ξ2/2v/prime−ξl+2e−ξ2/2v/prime+ξl+1e−ξ2/2v/prime/prime
=✭✭✭✭✭✭✭✭
l(l+1 )ξl−1e−ξ2/2v−(2l+3 )ξl+1e−ξ2/2v+✘✘✘✘✘ξl+3e−ξ2/2v+2 (l+1 )ξle−ξ2/2v/prime
−2ξl+2e−ξ2/2v/prime+ξl+1e−ξ2/2v/prime/prime=✘✘✘✘✘ξl+3e−ξ2/2v+✭✭✭✭✭✭✭✭
l(l+1 )ξl−1e−ξ2/2v−Kξl+1e−ξ2/2v.
Cancelling the indicated terms, and dividing off ξl+1e−ξ2/2, we have:
v/prime/prime+2v/prime/parenleftbiggl+1
ξ−ξ/parenrightbigg
+(K−2l−3)v=0.
Letv(ξ)≡∞/summationdisplay
j=0ajξj,sov/prime=∞/summationdisplay
j=0jajξj−1;v/prime/prime=∞/summationdisplay
j=2j(j−1)ajξj−2.Then
∞/summationdisplay
j=2j(j−1)ajξj−2+2 (l+2 )∞/summationdisplay
j=1jajξj−2−2∞/summationdisplay
j=1jajξj+(K−2l−3)∞/summationdisplay
j=0ajξj=0.
In the first two sums, let j→j+ 2 (rename the dummy index):
∞/summationdisplay
j=0(j+ 2)(j+1 )aj+2ξj+2 (l+1 )∞/summationdisplay
j=0(j+2 )aj+2ξj−2∞/summationdisplay
j=0jajξj+(K−2l−3)∞/summationdisplay
j=0ajξj=0.
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112 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Note: the second sum should start at j=−1; to eliminate this term (there is no compensating one in ξ−1)w e
must take a1= 0. Combining the terms:
∞/summationdisplay
j=0[(j+ 2)(j+2l+3 )aj+2+(K−2j−2l−3)aj]=0,soaj+2=(2j+2l+3−K)
(j+ 2)(j+2l+3 )aj.
Sincea1= 0, this gives us a single sequence: a0,a2,a4,.... But the series must terminate (else we get the
wrong behavior as ξ→∞), so there occurs some maximal (even) number jmaxsuch that ajmax+2=0 . T h u s
K=2jmax+2l+3.ButE=1
2/planckover2pi1ωK, soE=/parenleftbigg
jmax+l+3
2/parenrightbigg
/planckover2pi1ω.Or, letting jmax+l≡n,
En=(n+3
2)/planckover2pi1ω,andncan be any nonnegative integer.
[Incidentally, we can also determine the degeneracy of En. Suppose niseven; then (since jmaxis even)
l=0,2,4,...,n . For each lthere are (2 l+ 1) values for m.S o
d(n)=n/summationdisplay
l=0,2,4,...(2l+1 ).Letj=l/2; then d(n)=n/2/summationdisplay
j=0(4j+1 )=4n/2/summationdisplay
j=0j+n/2/summationdisplay
j=01
=4(n
2)(n
2+1 )
2+(n
2+1 )=(n
2+ 1)(n+1 )=(n+ 1)(n+2 )
2,as before (Problem 4.38(b)).]
Problem 4.40
(a)
d
dt/angbracketleftr·p/angbracketright=i
/planckover2pi1/angbracketleft[H,r·p]/angbracketright.
[H,r·p]=3/summationdisplay
i=1[H,ripi]=3/summationdisplay
i=1([H,ri]pi+ri[H,p i]) =3/summationdisplay
i=1/parenleftbigg1
2m[p2,ri]pi+ri[V,pi]/parenrightbigg
.
[p2,ri]=3/summationdisplay
j=1[pjpj,ri]=3/summationdisplay
j=1(pj[pj,ri]+[pj,ri]pj)=3/summationdisplay
j=1[pj(−iδij)+(−i/planckover2pi1δij)pj]=−2i/planckover2pi1pi.
[V,pi]=i/planckover2pi1∂V
∂ri(Problem 3.13(c)) .[H,r·p]=3/summationdisplay
i=1/bracketleftbigg1
2m(−2i/planckover2pi1)pipi+ri/parenleftbigg
i/planckover2pi1∂V
∂ri/parenrightbigg/bracketrightbigg
=i/planckover2pi1/parenleftbigg
−p2
m+r·∇V/parenrightbigg
.d
dt/angbracketleftr·p/angbracketright=/angbracketleftp2
m−r·∇V/angbracketright=2/angbracketleftT/angbracketright−/angbracketleftr·∇V/angbracketright.
For stationary statesd
dt/angbracketleftr·p/angbracketright=0,so 2/angbracketleftT/angbracketright=/angbracketleftr·∇V/angbracketright.QED
(b)
V(r)=−e2
4πFepsilonC01
r⇒∇V=e2
4πFepsilonC01
r2ˆr⇒r·∇V=e2
4πFepsilonC01
r=−V.So 2/angbracketleftT/angbracketright=−/angbracketleftV/angbracketright.
But/angbracketleftT/angbracketright=/angbracketleftV/angbracketright=En,so/angbracketleftT/angbracketright−2/angbracketleftT/angbracketright=En,or/angbracketleftT/angbracketright=−En;/angbracketleftV/angbracketright=2En.QED
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 113
(c)
V=1
2mω2r2⇒∇V=mω2rˆr⇒r·∇V=mω2r2=2V.So 2/angbracketleftT/angbracketright=2/angbracketleftV/angbracketright,or/angbracketleftT/angbracketright=/angbracketleftV/angbracketright.
But/angbracketleftT/angbracketright+/angbracketleftV/angbracketright=En,so/angbracketleftT/angbracketright=/angbracketleftV/angbracketright=1
2En.QED
Problem 4.41
(a)∇·J=i/planckover2pi1
2m/bracketleftbig
∇Ψ·∇Ψ∗+Ψ (∇2Ψ∗)−∇Ψ∗·∇Ψ−Ψ∗(∇2Ψ)/bracketrightbig
=i/planckover2pi1
2m/bracketleftbig
Ψ(∇2Ψ∗)−Ψ∗(∇2Ψ)/bracketrightbig
.
But the Schr¨ odinger equation says i/planckover2pi1∂Ψ
∂t=−/planckover2pi12
2m∇2Ψ+VΨ, so
∇2Ψ=2m
/planckover2pi12/parenleftbigg
VΨ−i/planckover2pi1∂Ψ
∂t/parenrightbigg
,∇2Ψ∗=2m
/planckover2pi12/parenleftbigg
VΨ∗+i/planckover2pi1∂Ψ∗
∂t/parenrightbigg
.Therefore
∇·J=i/planckover2pi1
2m2m
/planckover2pi12/bracketleftbigg
Ψ/parenleftbigg
VΨ∗+i/planckover2pi1∂Ψ∗
∂t/parenrightbigg
−Ψ∗/parenleftbigg
VΨ−i/planckover2pi1∂Ψ
∂t/parenrightbigg/bracketrightbigg
=i
/planckover2pi1i/planckover2pi1/parenleftbigg
Ψ∂Ψ∗
∂t+Ψ∗∂Ψ
∂t/parenrightbigg
=−∂
∂t(Ψ∗Ψ) =−∂
∂t|Ψ|2./check
(b)From Problem 4.11(b), Ψ 211=−1√πa1
8a2re−r/2asinθeiφe−iE2t//planckover2pi1.In spherical coordinates,
∇Ψ=∂Ψ
∂rˆr+1
r∂Ψ
∂θˆθ+1
rsinθ∂Ψ
∂φˆφ,so
∇Ψ211=−1√πa1
8a2/bracketleftbigg/parenleftBig
1−r
2a/parenrightBig
e−r/2asinθeiφe−iE2t//planckover2pi1ˆr+1
rre−r/2acosθeiφe−iE2t//planckover2pi1ˆθ
+1
rsinθre−r/2asinθieiφe−iE2t//planckover2pi1ˆφ/bracketrightbigg
=/bracketleftbigg/parenleftBig
1−r
2a/parenrightBig
ˆr+ cotθˆθ+i
sinθˆφ/bracketrightbigg1
rΨ211.
Therefore
J=i/planckover2pi1
2m/bracketleftbigg/parenleftBig
1−r
2a/parenrightBig
ˆr+ cotθˆθ−i
sinθˆφ−/parenleftBig
1−r
2a/parenrightBig
ˆr−cotθˆθ−i
sinθˆφ/bracketrightbigg1
r|Ψ211|2
=i/planckover2pi1
2m(−2i)
rsinθ|Ψ211|2ˆφ=/planckover2pi1
m1
πa1
64a4r2e−r/asin2θ
rsinθˆφ=/planckover2pi1
64πma5re−r/asinθˆφ.
(c)Nowr×J=/planckover2pi1
64πma5r2e−r/asinθ/parenleftBig
ˆr׈φ/parenrightBig
, while/parenleftBig
ˆr׈φ/parenrightBig
=−ˆθand ˆz·ˆθ=−sinθ,s o
r×Jz=/planckover2pi1
64πma5r2e−r/asin2θ, and hence
Lz=m/planckover2pi1
64πma5/integraldisplay/parenleftBig
r2e−r/asin2θ/parenrightBig
r2sinθdrdθdφ
=/planckover2pi1
64πa5/integraldisplay∞
0r4e−r/adr/integraldisplayπ
0sin3θdθ/integraldisplay2π
0dφ=/planckover2pi1
64πa5/parenleftbig
4!a5/parenrightbig/parenleftbigg4
3/parenrightbigg
(2π)= /planckover2pi1,
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114 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
as itshould be, since (Eq. 4.133) Lz=/planckover2pi1m, andm= 1 for this state.
Problem 4.42
(a)
ψ=1√
πa3e−r/a⇒φ(p)=1
(2π/planckover2pi1)3/21√
πa3/integraldisplay
e−ip·r//planckover2pi1e−r/ar2sinθdrdθdφ.
With axes as suggested, p·r=prcosθ. Doing the (trivial) φintegral:
φ(p)=2π
(2πa/planckover2pi1)3/21√π/integraldisplay∞
0r2e−r/a/bracketleftbigg/integraldisplayπ
0e−iprcosθ//planckover2pi1sinθdθ/bracketrightbigg
dr.
/integraldisplayπ
0e−iprcosθ//planckover2pi1sinθdθ=/planckover2pi1
ipre−iprcosθ//planckover2pi1/vextendsingle/vextendsingle/vextendsingleπ
0=/planckover2pi1
ipr/parenleftBig
eipr//planckover2pi1−e−ipr//planckover2pi1/parenrightBig
=2/planckover2pi1
prsin/parenleftBigpr
/planckover2pi1/parenrightBig
.
φ(p)=1
π√
21
(a/planckover2pi1)3/22/planckover2pi1
p/integraldisplay∞
0re−r/asin/parenleftBigpr
/planckover2pi1/parenrightBig
dr.
/integraldisplay∞
0re−r/asin/parenleftBigpr
/planckover2pi1/parenrightBig
dr=1
2i/bracketleftbigg/integraldisplay∞
0re−r/aeipr//planckover2pi1dr−/integraldisplay∞
0re−r/ae−ipr//planckover2pi1dr/bracketrightbigg
=1
2i/bracketleftBigg
1
(1/a−ip//planckover2pi1)2−1
(1/a+ip//planckover2pi1)2/bracketrightBigg
=1
2i(2ip/a/planckover2pi1)2
/bracketleftBig
(1/a)2+(p//planckover2pi1)2/bracketrightBig2
=(2p//planckover2pi1)a3
[ 1+(ap//planckover2pi1)2]2.
φ(p)=/radicalbigg
2
/planckover2pi11
a3/21
πp2pa3
/planckover2pi11
[1 + (ap//planckover2pi1)2]2=1
π/parenleftbigg2a
/planckover2pi1/parenrightbigg3/21
[1 + (ap//planckover2pi1)2]2.
(b)
/integraldisplay
|φ|2d3p=4π/integraldisplay∞
0p2|φ|2dp=4π1
π2/parenleftbigg2a
/planckover2pi1/parenrightbigg3/integraldisplay∞
0p2
[ 1+(ap//planckover2pi1)2]4dp.
From math tables:/integraldisplay∞
0x2
(m+x2)4dx=π
32m−5/2,so
/integraldisplay∞
0p2
[ 1+(ap//planckover2pi1)2]4dp=/parenleftbigg/planckover2pi1
a/parenrightbigg8π
32/parenleftbigg/planckover2pi1
a/parenrightbigg−5
=π
32/parenleftbigg/planckover2pi1
a/parenrightbigg3
;/integraldisplay
|φ|2d3p=32
π/parenleftBiga
/planckover2pi1/parenrightBig3π
32/parenleftbigg/planckover2pi1
a/parenrightbigg3
=1./check
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 115
(c)
/angbracketleftp2/angbracketright=/integraldisplay
p2|φ|2d3p=1
π2/parenleftbigg2a
/planckover2pi1/parenrightbigg3
4π/integraldisplay∞
0p4
[1 + (ap//planckover2pi1)2]4dp.From math tables:
/integraldisplay∞
0x4
[m+x2]4dx=/parenleftBigπ
32/parenrightBig
m−3/2.So/angbracketleftp2/angbracketright=4
π/parenleftbigg2a
/planckover2pi1/parenrightbigg3/parenleftbigg/planckover2pi1
a/parenrightbigg8π
32/parenleftbigg/planckover2pi1
a/parenrightbigg−3
=/planckover2pi12
a2.
(d)
/angbracketleftT/angbracketright=1
2m/angbracketleftp2/angbracketright=1
2m/planckover2pi12
a2=/planckover2pi12
2mm2
/planckover2pi14/parenleftbigge2
4πFepsilonC0/parenrightbigg2
=m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2
=−E1,
which isconsistent with Eq. 4.191.
Problem 4.43
(a)From Tables 4.3 and 4.7,
ψ321=R32Y1
2=4
81√
301
a3/2/parenleftBigr
a/parenrightBig2
e−r/3a/bracketleftBigg
−/radicalbigg
15
8πsinθcosθeiφ/bracketrightBigg
=−1√π1
81a7/2r2e−r/3asinθcosθeiφ.
(b)
/integraldisplay
|ψ|2d3r=1
π1
(81)2a7/integraldisplay/parenleftBig
r4e−2r/3asin2θcos2θ/parenrightBig
r2sinθdrdθdφ
=1
π(81)2a72π/integraldisplay∞
0r6e−2r/3adr/integraldisplayπ
0(1−cos2θ) cos2θsinθdθ
=2
(81)2a7/bracketleftBigg
6!/parenleftbigg3a
2/parenrightbigg7/bracketrightBigg/bracketleftbigg
−cos3θ
3+cos5θ
5/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0
=2
38a76·5·4·3·237a7
27/bracketleftbigg2
3−2
5/bracketrightbigg
=3·5
4·4
15=1./check
(c)
/angbracketleftrs/angbracketright=/integraldisplay∞
0rs|R32|2r2dr=/parenleftbigg4
81/parenrightbigg21
301
a7/integraldisplay∞
0rs+6e−2r/3adr
=8
15(81)2a7(s+ 6)!/parenleftbigg3a
2/parenrightbiggs+7
=(s+ 6)!/parenleftbigg3a
2/parenrightbigg51
720=(s+ 6)!
6!/parenleftbigg3a
2/parenrightbigg3
.
Finite for s>−7.
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116 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.44
(a)From Tables 4.3 and 4.7,
ψ433=R43Y3
3=1
768√
351
a3/2/parenleftBigr
a/parenrightBig3
e−r/4a/parenleftBigg
−/radicalbigg
35
64πsin3θcosθe3iφ/parenrightBigg
=−1
6144√πa9/2r3e−r/4asin3θe3iφ.
(b)
/angbracketleftr/angbracketright=/integraldisplay
r|ψ|2d3r=1
(6144)2πa9/integraldisplay
r/parenleftBig
r6e−r/2asin6θ/parenrightBig
r2sinθdrdθdφ
=1
(6144)2πa9/integraldisplay∞
0r9e−r/2adr/integraldisplayπ
0sin7θdθ/integraldisplay2π
0dφ
=1
(6144)2πa9/bracketleftbig
9!(2a)10/bracketrightbig/parenleftbigg
22·4·6
3·5·7/parenrightbigg
(2π)=18a.
(c)Using Eq. 4.133: L2
x+L2
y=L2−L2
z= 4(5) /planckover2pi12−(3/planckover2pi1)2=11/planckover2pi12,with probability 1 .
Problem 4.45
(a)
P=/integraldisplay
|ψ|2d3r=4π
πa3/integraldisplayb
0e−2r/ar2dr=4
a3/bracketleftbigg
−a
2r2e−2r/a+a3
4e−2r/a/parenleftbigg
−2r
a−1/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
0
=−/parenleftbigg
1+2r
a+2r2
a2/parenrightbigg
e−2r/a/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
0=1−/parenleftbigg
1+2b
a+2b2
a2/parenrightbigg
e−2b/a.
(b)
P=1−/parenleftbigg
1+FepsilonC+1
2FepsilonC2/parenrightbigg
e−ρepsilono≈1−/parenleftbigg
1+FepsilonC+1
2FepsilonC2/parenrightbigg/parenleftbigg
1−FepsilonC+FepsilonC2
2−FepsilonC3
3!/parenrightbigg
≈1−1+FepsilonC−FepsilonC2
2+FepsilonC3
6−FepsilonC+FepsilonC2−FepsilonC3
2−FepsilonC2
2+FepsilonC3
2=FepsilonC3/parenleftbigg1
6−1
2+1
2/parenrightbigg
=1
6/parenleftbigg2b
a/parenrightbigg3
=4
3/parenleftbiggb
a/parenrightbigg3
.
(c)
|ψ(0)|2=1
πa3⇒P≈4
3πb31
πa3=4
3/parenleftbiggb
a/parenrightbigg3
./check
(d)
P=4
3/parenleftbigg10−15
0.5×10−10/parenrightbigg3
=4
3/parenleftbig
2×10−5/parenrightbig3=4
3·8×10−15=32
3×10−15=1.07×10−14.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 117
Problem 4.46
(a)Equation 4.75 ⇒Rn(n−1)=1
rρne−ρv(ρ),where ρ≡r
na; Eq. 4.76⇒c1=2(n−n)
(1)(2n)c0=0.
Sov(ρ)=c0,and hence Rn(n−1)=Nnrn−1e−r/na,where Nn≡c0
(na)n.
1=/integraldisplay∞
0|R|2r2dr=(Nn)2/integraldisplay∞
0r2ne−2r/nadr=(Nn)2(2n)!/parenleftbiggna
2/parenrightbigg2n+1
;Nn=/parenleftbigg2
na/parenrightbiggn/radicalBigg
2
na(2n)!.
(b)
/angbracketleftrl/angbracketright=/integraldisplay∞
0|R|2rl+2dr=N2
n/integraldisplay∞
0r2n+le−2r/nadr.
/angbracketleftr/angbracketright=/parenleftbigg2
na/parenrightbigg2n+11
(2n)!(2n+ 1)!/parenleftbiggna
2/parenrightbigg2n+2
=/parenleftbigg
n+1
2/parenrightbigg
na.
/angbracketleftr2/angbracketright=/parenleftbigg2
na/parenrightbigg2n+11
(2n)!(2n+ 2)!/parenleftbiggna
2/parenrightbigg2n+3
=( 2n+ 2)(2n+1 )/parenleftbiggna
2/parenrightbigg2
=/parenleftbigg
n+1
2/parenrightbigg
(n+ 1)(na)2.
(c)
σ2
r=/angbracketleftr2/angbracketright−/angbracketleftr/angbracketright2=/bracketleftbigg/parenleftbigg
n+1
2/parenrightbigg
(n+ 1)(na)2−/parenleftbigg
n+1
2/parenrightbigg2
(na)2/bracketrightbigg
=1
2/parenleftbigg
n+1
2/parenrightbigg
(na)2=1
2(n+1/2)/angbracketleftr/angbracketright2;σr=/angbracketleftr/angbracketright√2n+1.
r rr 6a a 650aR10 32 26 25R R
Maxima occur at:dRn,n−1
dr=0⇒(n−1)rn−2e−r/na−1
narn−1e−r/na=0⇒r=na(n−1).
Problem 4.47
Here are a couple of examples: {32, 28}and{224,56};{221, 119}and{119, 91}. For further discussion see
D. Wyss and W. Wyss, Foundations of Physics 23, 465 (1993).
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118 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.48
(a)Using Eqs. 3.64 and 4.122: [ A,B]=[x2,Lz]=x[x,Lz]+[x,Lz]x=x(−i/planckover2pi1y)+(−i/planckover2pi1y)x=−2i/planckover2pi1xy.
Equation 3.62 ⇒σ2
Aσ2
B≥/bracketleftbigg1
2i(−2i/planckover2pi1)/angbracketleftxy/angbracketright/bracketrightbigg2
=/planckover2pi12/angbracketleftxy/angbracketright2⇒σAσB≥/planckover2pi1|/angbracketleftxy/angbracketright|.
(b)Equation 4.113 ⇒/angbracketleftB/angbracketright=/angbracketleftLz/angbracketright=m/planckover2pi1;/angbracketleftB2/angbracketright=/angbracketleftL2
z/angbracketright=m2/planckover2pi12;s o σB=m2/planckover2pi12−m2/planckover2pi12=0.
(c)Since the left side of the uncertainty principle is zero, the right side must also be: /angbracketleftxy/angbracketright=0 ,for eigenstates
ofLz.
Problem 4.49
(a)1=|A|2( 1+4+4 )=9 |A|2;A=1/3.
(b)/planckover2pi1
2, with probability5
9;−/planckover2pi1
2, with probability4
9./angbracketleftSz/angbracketright=5
9/planckover2pi1
2+4
9/parenleftbigg
−/planckover2pi1
2/parenrightbigg
=/planckover2pi1
18.
(c)From Eq. 4.151,
c(x)
+=/parenleftBig
χ(x)
+/parenrightBig†
χ=1
31√
2/parenleftbig11/parenrightbig/parenleftbigg1−2i
2/parenrightbigg
=1
3√
2(1−2i+2 )=3−2i
3√
2;|c(x)
+|2=9+4
9·2=13
18.
c(x)
−=/parenleftBig
χ(x)
−/parenrightBig†
χ=1
31√
2/parenleftbig1−1/parenrightbig/parenleftbigg
1−2i
2/parenrightbigg
=1
3√
2(1−2i−2) =−1+2i
3√
2;|c(x)
−|2=1+4
9·2=5
18.
/planckover2pi1
2, with probability13
18;−/planckover2pi1
2, with probability5
18./angbracketleftSx/angbracketright=13
18/planckover2pi1
2+5
18/parenleftbigg
−/planckover2pi1
2/parenrightbigg
=2/planckover2pi1
9.
(d)From Problem 4.29(a),
c(y)
+=/parenleftBig
χ(y)
+/parenrightBig†
χ=1
31√
2/parenleftbig1−i/parenrightbig/parenleftbigg
1−2i
2/parenrightbigg
=1
3√
2(1−2i−2i)=1−4i
3√
2;|c(y)
+|2=1+1 6
9·2=17
18.
c(y)
−=/parenleftBig
χ(y)
−/parenrightBig†
χ=1
31√
2/parenleftbig1i/parenrightbig/parenleftbigg1−2i
2/parenrightbigg
=1
3√
2(1−2i+2i)=1
3√
2;|c(y)
−|2=1
9·2=1
18.
/planckover2pi1
2, with probability17
18;−/planckover2pi1
2, with probability1
18./angbracketleftSy/angbracketright=17
18/planckover2pi1
2+1
18/parenleftbigg
−/planckover2pi1
2/parenrightbigg
=4/planckover2pi1
9.
Problem 4.50
We may as well choose axes so that ˆ alies along the zaxis and ˆbis in the xzplane. Then S(1)
a=S(1)
z,andS(2)
b=
cosθS(2)
z+ sinθS(2)
x./angbracketleft00|S(1)
aS(2)
b|00/angbracketrightis to be calculated.
S(1)
aS(2)
b|00/angbracketright=1√
2/bracketleftBig
S(1)
z(cosθS(2)
z+ sinθS(2)
x)/bracketrightBig
(↑↓−↓↑ )
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 119
=1√
2[(Sz↑)(cosθSz↓+ sinθSx↓)−(Sz↓)(cosθSz↑+ sinθSx↑)]
=1√
2/braceleftbigg/parenleftbigg/planckover2pi1
2↑/parenrightbigg/bracketleftbigg
cosθ/parenleftbigg
−/planckover2pi1
2↓/parenrightbigg
+ sinθ/parenleftbigg/planckover2pi1
2↑/parenrightbigg/bracketrightbigg
−/parenleftbigg
−/planckover2pi1
2↓/parenrightbigg/bracketleftbigg
cosθ/parenleftbigg/planckover2pi1
2↑/parenrightbigg
+ sinθ/parenleftbigg/planckover2pi1
2↓/parenrightbigg/bracketrightbigg/bracerightbigg
(using Eq. 4.145)
=/planckover2pi12
4/bracketleftbigg
cosθ1√
2(−↑↓+↓↑) + sinθ1√
2(↑↑+↓↓)/bracketrightbigg
=/planckover2pi12
4/bracketleftbigg
−cosθ|00/angbracketright+ sinθ1√
2(|11/angbracketright+|1−1/angbracketright)/bracketrightbigg
.
so/angbracketleftS(1)
aS(2)
b/angbracketright=/angbracketleft00|S(1)
aS(2)
b|00/angbracketright=/planckover2pi12
4/angbracketleft00|/bracketleftbigg
−cosθ|00/angbracketright+ sinθ1√
2(|11/angbracketright+|1−1/angbracketright)/bracketrightbigg
=−/planckover2pi12
4cosθ/angbracketleft00|00/angbracketright
(by orthogonality), and hence /angbracketleftS(1)
aS(2)
b/angbracketright=−/planckover2pi12
4cosθ.QED
Problem 4.51
(a)First note from Eqs. 4.136 and 4.144 that
Sx|sm/angbracketright=1
2[S+|sm/angbracketright+S−|sm/angbracketright]
=/planckover2pi1
2/bracketleftBig/radicalbig
s(s+1 )−m(m+1 )|sm+1/angbracketright+/radicalbig
s(s+1 )−m(m−1)|sm−1/angbracketright/bracketrightBig
Sy|sm/angbracketright=1
2i[S+|sm/angbracketright−S−|sm/angbracketright]
=/planckover2pi1
2i/bracketleftBig/radicalbig
s(s+1 )−m(m+1 )|sm+1/angbracketright−/radicalbig
s(s+1 )−m(m−1)|sm−1/angbracketright/bracketrightBig
Now, using Eqs. 4.179 and 4.147:
S2|sm/angbracketright=/bracketleftbigg
(S(1))2+(S(2))2+2(S(1)
xS(2)
x+S(1)
yS(2)
y+S(1)
zS(2)
z)/bracketrightbigg/bracketleftbigg
A|1
21
2/angbracketright|S2m−1
2/angbracketright+B|1
2−1
2/angbracketright|s2m+1
2/angbracketright/bracketrightbigg
=A/braceleftbigg/parenleftbig
S2|1
21
2/angbracketright/parenrightbig
|s2m−1
2/angbracketright+|1
21
2/angbracketright/parenleftbig
S2|s2m−1
2/angbracketright/parenrightbig
+2/bracketleftbigg/parenleftbig
Sx|1
21
2/angbracketright/parenrightbig/parenleftbig
Sx|s2m−1
2/angbracketright/parenrightbig
+/parenleftbig
Sy|1
21
2/angbracketright/parenrightbig/parenleftbig
Sy|s2m−1
2/angbracketright/parenrightbig
+/parenleftbig
Sz|1
21
2/angbracketright/parenrightbig/parenleftbig
Sz|s2m−1
2/angbracketright/parenrightbig/bracketrightbigg/bracerightbigg
+B/braceleftbigg/parenleftbig
S2|1
2−1
2/angbracketright/parenrightbig
|s2m+1
2/angbracketright+|1
2−1
2/angbracketright/parenleftbig
S2|s2m+1
2/angbracketright/parenrightbig
+2/bracketleftbigg/parenleftbig
Sx|1
2−1
2/angbracketright/parenrightbig/parenleftbig
Sx|s2m+1
2/angbracketright/parenrightbig
+/parenleftbig
Sy|1
2−1
2/angbracketright/parenrightbig/parenleftbig
Sy|s2m+1
2/angbracketright/parenrightbig
+/parenleftbig
Sz|1
2−1
2/angbracketright/parenrightbig/parenleftbig
Sz|s2m+1
2/angbracketright/parenrightbig/bracketrightbigg/bracerightbigg
=A/braceleftbigg
3
4/planckover2pi12|1
21
2/angbracketright|s2m−1
2/angbracketright+/planckover2pi12s2(s2+1 )|1
21
2/angbracketright|s2m−1
2/angbracketright
+2/bracketleftbigg
/planckover2pi1
2|1
2−1
2/angbracketright/planckover2pi1
2/parenleftbigg/radicalBig
s2(s2+1 )−(m−1
2)(m+1
2)|s2m+1
2/angbracketright
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120 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
+/radicalBig
s2(s2+1 )−(m−1
2)(m−3
2)|s2m−3
2/angbracketright/parenrightbigg
+/parenleftbigg
i/planckover2pi1
2/parenrightbigg
|1
2−1
2/angbracketright/planckover2pi1
2i/parenleftbigg/radicalBig
s2(s2+1 )−(m−1
2)(m+1
2)|s2m+1
2/angbracketright
−/radicalBig
s2(s2+1 )−(m−1
2)(m−3
2)|s2m−3
2/angbracketright/parenrightbigg
+/planckover2pi1
2|1
21
2/angbracketright/planckover2pi1(m−1
2)|s2m−1
2/angbracketright/bracketrightbigg/bracerightbigg
+B/braceleftbigg
3
4/planckover2pi12|1
2−1
2/angbracketright|s2m+1
2/angbracketright+/planckover2pi12s2(s2+1 )|1
2−1
2/angbracketright|s2m+1
2/angbracketright
+2/bracketleftbigg
/planckover2pi1
2|1
21
2/angbracketright/planckover2pi1
2/parenleftbigg/radicalBig
s2(s2+1 )−(m+1
2)(m+3
2)|s2m+3
2/angbracketright+/radicalBig
s2(s2+1 )−(m+1
2)(m−1
2)|s2m−1
2/angbracketright/parenrightbigg
+/parenleftbigg
−i/planckover2pi1
2/parenrightbigg
|1
21
2/angbracketright/planckover2pi1
2i/parenleftbigg/radicalBig
s2(s2+1 )−(m+1
2)(m+3
2)|s2m+3
2/angbracketright
−/radicalBig
s2(s2+1 )−(m+1
2)(m−1
2)|s2m−1
2/angbracketright/parenrightbigg
+/parenleftbig−/planckover2pi1
2/parenrightbig
|1
2−1
2/angbracketright/planckover2pi1(m+1
2)|s2m+1
2/angbracketright/bracketrightbigg/bracerightbigg
=/planckover2pi12/braceleftbigg
A/bracketleftbigg
3
4+s2(s2+1 )+m−1
2/bracketrightbigg
+B/radicalBig
s2(s2+1 )−m2+1
4/bracerightbigg
|1
21
2/angbracketright|s2m−1
2/angbracketright
+/planckover2pi12/braceleftbigg
B/bracketleftbigg
3
4+s2(s2+1 )−m−1
2/bracketrightbigg
+A/radicalBig
s2(s2+1 )−m2+1
4/bracerightbigg
|1
2−1
2/angbracketright|s2m+1
2/angbracketright
=/planckover2pi12s(s+1 )|sm/angbracketright=/planckover2pi12s(s+1 )/bracketleftbigg
A|1
21
2/angbracketright|s2m−1
2/angbracketright+B|1
2−1
2/angbracketright|s2m+1
2/angbracketright/bracketrightbigg
.
A/bracketleftbig
s2(s2+1 )+1
4+m/bracketrightbig
+B/radicalBig
s2(s2+1 )−m2+1
4=s(s+1 )A,
B/bracketleftbig
s2(s2+1 )+1
4−m/bracketrightbig
+A/radicalBig
s2(s2+1 )−m2+1
4=s(s+1 )B,
or
A/bracketleftbig
s2(s2+1 )−s(s+1 )+1
4+m/bracketrightbig
+B/radicalBig
s2(s2+1 )−m2+1
4=0,
B/bracketleftbig
s2(s2+1 )−s(s+1 )+1
4−m/bracketrightbig
+A/radicalBig
s2(s2+1 )−m2+1
4=0,
or/braceleftbiggA(a+m)+Bb=0
B(a−m)+Ab=0/bracerightbigg
,
wherea≡s2(s2+1 )−s(s+1 )+1
4,b≡/radicalBig
s2(s2+1 )−m2+1
4.Multiply by ( a−b) andb, then subtract:
A(a2−m2)+Bb(a−m)=0 ;Bb(a−m)+Ab2=0⇒A(a2−m2−b2)=0⇒a2−b2=m2,or:
/bracketleftbig
s2(s2+1 )−s(s+1 )+1
4/bracketrightbig2−s2(s2+1 )+m2−1
4=m2,
/bracketleftbig
s2(s2+1 )−s(s+1 )+1
4/bracketrightbig2=s2
2+s2+1
4=/parenleftbig
s2+1
2/parenrightbig2,s o
s2(s2+1 )−s(s+1 )+1
4=±/parenleftbig
s2+1
2/parenrightbig
;s(s+1 )=s2(s2+1 )∓/parenleftbig
s2+1
2/parenrightbig
+1
4.
Add1
4to both sides:
s2+s+1
4=/parenleftbig
s+1
2/parenrightbig2=s2(s2+1 )∓/parenleftbig
s2+1
2/parenrightbig
+1
2=
s2
2+s2−s2−1
2+1
2=s2
2
s22+s2+s2+1
2+1
2=(s2+1 )2
.
So
s+1
2=±s2⇒s=±s2−1
2=/braceleftbiggs2−1
2
−s2−1
2
s+1
2=±(s2+1 )⇒s=±(s2+1 )−1
2=/braceleftbiggs2+1
2
−s2−3
2
.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 121
Buts≥0, so the possibilities are s=s2±1/2.Then:
a=s2
2+s2−/parenleftbigg
s2±1
2/parenrightbigg/parenleftbigg
s2±1
2+1/parenrightbigg
+1
4
=s2
2+s2−s2
2∓1
2s2−s2∓1
2s2−1
4∓1
2+1
4=∓s2∓1
2=∓/parenleftbigg
s2+1
2/parenrightbigg
.
b=/radicalBigg/parenleftbigg
s2
2+s2+1
4/parenrightbigg
−m2=/radicalBigg/parenleftbigg
s2+1
2/parenrightbigg2
−m2=/radicalBigg/parenleftbigg
s2+1
2+m/parenrightbigg/parenleftbigg
s2+1
2−m/parenrightbigg
.
∴A/bracketleftbig
∓/parenleftbig
s2+1
2/parenrightbig
+m/bracketrightbig
=∓A/parenleftbig
s2+1
2∓m/parenrightbig
=−Bb=−B/radicalBig/parenleftbig
s2+1
2+m/parenrightbig/parenleftbig
s2+1
2−m/parenrightbig
⇒A/radicalBig
s2+1
2∓m=±B/radicalBig
s2+1
2±m.But|A|2+|B|2=1,so
|A|2+|A|2/parenleftbiggs2+1
2∓m
s2+1
2±m/parenrightbigg
=|A|2
(s2+1
2±m)/bracketleftbigg
s2+1
2±m+s2+1
2∓m/bracketrightbigg
=(2s2+1 )
(s2+1
2±m)|A|2.
⇒A=/radicalBigg
s2±m+1
2
2s2+1.B=±A/radicalBig
s2+1
2∓m
/radicalBig
s2+1
2±m=±/radicalBigg
s2∓m+1
2
2s2+1.
(b)Here are four examples:
(i) From the 1 /2×1/2 table ( s2=1/2), pick s= 1 (upper signs), m= 0. Then
A=/radicalBig
1
2+0+1
2
1+1=1√
2;B=/radicalBig
1
2−0+1
2
1+1=1√
2.
(ii) From the 1 ×1/2 table ( s2= 1), pick s=3/2 (upper signs), m=1/2. Then
A=/radicalBig
1+1
2+1
2
2+1=/radicalBig
2
3;B=/radicalBig
1−1
2+1
2
2+1=1√
3.
(iii) From the 3 /2×1/2 table ( s2=3/2), pick s= 1 (lower signs), m=−1. Then
A=/radicalBig
3
2+1+1
2
3+1=√
3
2;B=−/radicalBig
3
2−1+1
2
3+1=−1
2.
(iv) From the 2 ×1/2 table ( s2= 2), pick s=3/2 (lower signs), m=1/2. Then
A=/radicalBig
2−1
2+1
2
4+1=/radicalBig
2
5;B=−/radicalBig
2+1
2+1
2
4+1=−/radicalBig
3
5.
These all check with the values on Table 4.8, except that the signs (which are conventional) are reversed
in (iii) and (iv). Normalization does not determine the sign of A(nor, therefore, of B).
Problem 4.52
|3
23
2/angbracketright=
1
000
;|
3
21
2/angbracketright=
0
100
;|
3
2−1
2/angbracketright=
0
010
;|
3
2−3
2/angbracketright=
0
001
.Equation 4.136 ⇒
S
+|3
23
2/angbracketright=0,S +|3
21
2/angbracketright=√
3/planckover2pi1|3
23
2/angbracketright,S +|3
2−1
2/angbracketright=2/planckover2pi1|3
21
2/angbracketright,S +|3
2−3
2/angbracketright=√
3/planckover2pi1|3
2−1
2/angbracketright;
S−|3
23
2/angbracketright=√
3/planckover2pi1|3
21
2/angbracketright,S −|3
21
2/angbracketright=2/planckover2pi1|3
2−1
2/angbracketright,S −|3
2−1
2/angbracketright=√
3/planckover2pi1|3
2−3
2/angbracketright,S −|3
2−3
2/angbracketright=0.
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122 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
So:S+=/planckover2pi1
0√
30 0
0020
000√
3
0000
;S−=/planckover2pi1
0000√
30 0 0
0200
00√
30
;Sx=1
2(S++S−)=/planckover2pi1
2
0√
30 0√
30 2 0
020√
3
00√
30
.
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ√
30 0√
3−λ20
02−λ√
3
00√
3−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ20
2−λ√
3
0√
3−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−√
3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle√
32 0
0−λ√
3
0√
3−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle
=−λ/bracketleftbig
−λ
3+3λ+4λ/bracketrightbig
−√
3/bracketleftBig√
3λ2−3√
3/bracketrightBig
=λ4−7λ2−3λ2+9=0 ,
orλ4−10λ2+9=0 ; ( λ2−9)(λ2−1) = 0; λ=±3,±1.So the eigenvalues of Sxare3
2/planckover2pi1,1
2/planckover2pi1,−1
2/planckover2pi1,−3
2/planckover2pi1.
Problem 4.53
From Eq. 4.135, Sz|sm/angbracketright=/planckover2pi1m|sm/angbracketright.Sincesis fixed, here, let’s just identify the states by the value of m(which
runs from−sto +s). The matrix elements of Szare
Snm=/angbracketleftn|Sz|m/angbracketright=/planckover2pi1m/angbracketleftn|m/angbracketright=/planckover2pi1mδnm.
It’s adiagonal matrix, with elements m/planckover2pi1, ranging from m=sin the upper left corner to m=−sin the lower
right corner:
Sz=/planckover2pi1
s00···0
0s−10···0
00 s−2···0
...............
00 0 ··· −s
.
From Eq. 4.136,
S
±|sm/angbracketright=/planckover2pi1/radicalbig
s(s+1 )−m(m±1)|s(m±1)/angbracketright=/planckover2pi1/radicalbig
(s∓m)(s±m+1 )|s(m±1)/angbracketright.
(S+)nm=/angbracketleftn|S+|m/angbracketright=/planckover2pi1/radicalbig
(s−m)(s+m+1 )/angbracketleftn|m+1/angbracketright=/planckover2pi1bm+1δn(m+1)=/planckover2pi1bnδn(m+1).
All nonzero elements have row index ( n) one greater than the column index ( m), so they are on the diagonal
justabove the main diagonal (note that the indices go down, here:s,s−1,s−2...,−s):
S+=/planckover2pi1
0b
s00··· 0
00bs−10··· 0
00 0 bs−2··· 0
..................
00 0 0 ···b
−s+1
00 0 0 ··· 0
.
Similarly
(S
−)nm=/angbracketleftn|S−|m/angbracketright=/planckover2pi1/radicalbig
(s+m)(s−m+1 )/angbracketleftn|m−1/angbracketright=/planckover2pi1bmδn(m−1).
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 123
This time the nonzero elements are on the diagonal just below the main diagonal:
S−=/planckover2pi1
00 0··· 00
b
s00··· 00
0bs−10··· 00
..................
00 0···b
−s+10
.
To construct S
x=1
2(S++S−) and Sy=1
2i(S+−S−), simply add and subtract the matrices S+andS−:
Sx=/planckover2pi1
2
0b
s00··· 00
bs0bs−10··· 00
0bs−10bs−2··· 00
00 bs−20··· 00
.....................
00 0 0 ··· 0b
−s+1
00 0 0 ···b−s+10
;S
y=/planckover2pi1
2i
0b
s 00··· 00
−bs0bs−10··· 00
0−bs−10bs−2··· 00
00−bs−20··· 00
.....................
00 00 ··· 0b
−s+1
00 00 ··· −b−s+10
.
Problem 4.54
L+Ym
l=/planckover2pi1/radicalbig
l(l+1 )−m(m+1 )Ym±1
l(Eqs. 4.120 and 121). Equation 4.130 ⇒
/planckover2pi1eiφ/parenleftbigg∂
∂θ+icotθ∂
∂φ/parenrightbigg
Bm
leimφPm
l(cosθ)=/planckover2pi1/radicalbig
l(l+1 )−m(m+1 )Bm+1
lei(m+1)φPm+1
l(cosθ).
Bm
l/parenleftbiggd
dθ−mcotθ/parenrightbigg
Pm
l(cosθ)=/radicalbig
l(l+1 )−m(m+1 )Bm+1
lPm+1
l(cosθ).
Letx≡cosθ; cotθ=cosθ
sinθ=x√
1−x2;d
dθ=dx
dθd
dx=−sinθd
dx=−/radicalbig
1−x2d
dx.
Bm
l/bracketleftbigg
−/radicalbig
1−x2d
dx−mx√
1−x2/bracketrightbigg
Pm
l(x)=−Bm
l1√
1−x2/bracketleftbigg
(1−x2)dPm
l
dx+mxPm
l/bracketrightbigg
=−Bm
lPm+1
l
=/radicalbig
l(l+1 )−m(m+1 )Bm+1
lPm+1
l(x).⇒Bm+1
l=−1/radicalbig
l(l+1 )−m(m+1 )Bm
l.
Nowl(l+1 )−m(m+1 )=( l−m)(l+m+1 ),so
Bm+1
l=−1√
l−m√
l+1+mBm
l⇒B1
l=−1√
l√
l+1B0
l;B2
l=−1√
l−1√
l+2B1
l=1/radicalbig
l(l−1)/radicalbig
(l+ 1)(l+2 )B0
l;
B3
l=−1√
l−2√
l+3B2
l=−1/radicalbig
(l+ 3)(l+ 2)(l+1 )l(l−1)(l−2)B0
l,etc.
Evidently there is an overall sign factor ( −1)m, and inside the square root the quantity is [( l+m)!/(l−m)!].
Thus:Bm
l=(−1)m/radicalBigg
(l−m)!
(l+m)!C(l)(where C(l)≡B0
l), form≥0. Form<0, we have
B−1
l=−B0
l/radicalbig
(l+1 )l;B−2
l=−1/radicalbig
(l+ 2)(l−1)B−1
l=1/radicalbig
(l+ 2)(l+1 )l(l−1)B0
l,etc.
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124 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
ThusB−m
l=Bm
l, so in general: Bm
l=(−1)m/radicalBig
(l−|m|)!
(l+|m|!C(l).Now, Problem 4.22 says:
Yl
l=1
2ll!/radicalbigg
(2l+ 1)!
π(eiφsinθ)l=Bl
leilφPl
l(cosθ).But
Pl
l(x)=( 1−x2)l/2/parenleftbiggd
dx/parenrightbiggl1
2ll!/parenleftbiggd
dx/parenrightbiggl
(x2−1)l=(1−x2)l/2
2ll!/parenleftbiggd
dx/parenrightbigg2l
(x2l−...)
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
(2l)!=(2l)!
2ll!(1−x2)l/2,
soPl
l(cosθ)=(2l)!
2ll!(sinθ)l.Therefore
1
2ll!/radicalbigg
(2l+ 1)!
π(eiφsinθ)l=Bl
leilφ(2l)!
2ll!(sinθ)l⇒Bl
l=1
(2l)!/radicalbigg
(2l+ 1)!
π=/radicalBigg
(2l+1 )
π(2l)!.
ButBl
l=(−1)l/radicalBigg
1
(2l)!C(l),soC(l)=(−1)l/radicalbigg
2l+1
π,and hence Bm
l=(−1)l+m/radicalBigg
(2l+1 )
π(l−|m|)!
(l+|m|)!.
This agrees with Eq. 4.32 except for the overall sign, which of course is purely conventional.
Problem 4.55
(a)For both terms, l=1 ,s o /planckover2pi12(1)(2) = 2/planckover2pi12,P=1.
(b)0,P=1
3,or/planckover2pi1,P=2
3.
(c)3
4/planckover2pi12,P=1.
(d)/planckover2pi1
2,P=1
3,o r−/planckover2pi1
2,P=2
3.
(e)From the 1×1
2Clebsch-Gordan table (or Problem 4.51):
1√
3|1
21
2/angbracketright|10/angbracketright+/radicalBig
2
3|1
2−1
2/angbracketright|11/angbracketright=1√
3/bracketleftBig/radicalBig
2
3|3
21
2/angbracketright−1√
3|1
21
2/angbracketright/bracketrightBig
+/radicalBig
2
3/bracketleftBig
1√
3|3
21
2/angbracketright+/radicalBig
2
3|1
21
2/angbracketright/bracketrightBig
=/parenleftBig
2√
2
3/parenrightBig
|3
21
2/angbracketright+/parenleftbig1
3/parenrightbig
|1
21
2/angbracketright.Sos=3
2or1
2.15
4/planckover2pi12,P=8
9,or3
4/planckover2pi12,P=1
9.
(f)1
2/planckover2pi1,P=1.
(g)
|ψ|2=|R21|2/braceleftbigg1
3|Y0
1|2(χ†
+χ+)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
1+√
2
3/bracketleftbigg
Y0∗
1Y1
1(χ†
+χ−)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
0+Y1∗
1Y0
1(χ†
−χ+)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
0/bracketrightbigg
+2
3|Y1
1|2(χ†
−χ−)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
1/bracerightbigg
=1
3|R21|2/parenleftbig
|Y0
1|2+2|Y1
1|2/parenrightbig
=1
3·1
24·1
a3·r2
a2e−r/a/bracketleftbigg3
4πcos2θ+23
8πsin2θ/bracketrightbigg
[Tables 4.3, 4.7]
=1
3·24·a5r2e−r/a·3
4π(cos2θ+ sin2θ)=1
96πa5r2e−r/a.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 125
(h)
1
3|R21|2/integraldisplay
|Y0
1|2sin2θdθdφ =1
3|R21|2=1
3·1
24a3r2e−r/a=1
72a5r2e−r/a.
Problem 4.56
(a)Equation 4.129 says Lz=/planckover2pi1
i∂
∂φ, so this problem is identical to Problem 3.39, with ˆ p→Lzandx→φ.
(b)First note that if Mis a matrix such that M2= 1, then
eiMφ=1+iMφ+1
2(iMφ)2+1
3!(iMφ)3+···=1+iMφ−1
2φ2−iM1
3!φ3+···
=( 1−1
2φ2+1
4!φ4−···)+iM(φ−1
3!φ3+1
5!φ5−···) = cosφ+iMsinφ.
SoR=eiπσx/2= cosπ
2+iσxsinπ
2(because σ2
x= 1 – see Problem 4.26) = iσx=i/parenleftbigg01
10/parenrightbigg
.
Thus Rχ+=i/parenleftbigg01
10/parenrightbigg/parenleftbigg1
0/parenrightbigg
=i/parenleftbigg0
1/parenrightbigg
=iχ−; it converts “spin up” into “spin down” (with a factor of i).
(c)
R=eiπσy/4= cosπ
4+iσysinπ
4=1√
2(1 +iσy)=1√
2/bracketleftbigg/parenleftbigg10
01/parenrightbigg
+i/parenleftbigg0−i
i0/parenrightbigg/bracketrightbigg
=1√
2/parenleftbigg11
−11/parenrightbigg
.
Rχ+=1√
2/parenleftbigg11
−11/parenrightbigg/parenleftbigg1
0/parenrightbigg
=1√
2/parenleftbigg1
−1/parenrightbigg
=1√
2(χ+−χ−)=χ(x)
−(Eq. 4.151) .
What hadbeen spin upalongzis now spin down alongx/prime(see figure).
y
y'z
x'x
z'
(d)R=eiπσz= cosπ+iσzsinπ=−1;rotation by 360◦changes the signof the spinor. But since the sign
ofχis arbitrary, it doesn’t matter.
(e)
(σ·ˆn)2=(σxnx+σyny+σznz)(σxnx+σyny+σznz)
=σ2
xn2x+σ2
yn2y+σ2
zn2z+nxny(σxσy+σyσx)+nxnz(σxσz+σzσx)+nynz(σyσz−σzσy).
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126 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Butσ2
x=σ2
y=σ2
z=1,andσxσy+σyσx=σxσz+σzσx=σyσz+σzσy= 0 (Problem 4.26), so
(σ·ˆn)2=n2
x+n2
y+n2
z=1.Soei(σ·ˆn)φ/2= cosφ
2+i(σ·ˆn) sinφ
2.QED
Problem 4.57
(a)
[q1,q2]=1
2/bracketleftbig
x+/parenleftbig
a2//planckover2pi1/parenrightbig
py,x−/parenleftbig
a2//planckover2pi1/parenrightbig
py/bracketrightbig
=0,because [ x,py]=[x,x]=[py,py]=0.
[p1,p2]=1
2/bracketleftbig
px−/parenleftbig
/planckover2pi1/a2/parenrightbig
y, p x+/parenleftbig
/planckover2pi1/a2/parenrightbig
y/bracketrightbig
=0,because [ y,px]=[y,y]=[px,px]=0.
[q1,p1]=1
2/bracketleftbig
x+/parenleftbig
a2//planckover2pi1/parenrightbig
py,px−/parenleftbig
/planckover2pi1/a2/parenrightbig
y/bracketrightbig
=1
2([x,px]−[py,y]) =1
2[i/planckover2pi1−(−i/planckover2pi1)] =i/planckover2pi1.
[q2,p2]=1
2/bracketleftbig
x−/parenleftbig
a2//planckover2pi1/parenrightbig
py,px+/parenleftbig
/planckover2pi1/a2/parenrightbig
y/bracketrightbig
=1
2([x,px]−[py,y]) =i/planckover2pi1.
[See Eq. 4.10 for the canonical commutators.]
(b)
q2
1−q2
2=1
2/bracketleftBigg
x2+a2
/planckover2pi1(xpy+pyx)+/parenleftbigga2
/planckover2pi1/parenrightbigg2
p2
y−x2+a2
/planckover2pi1(xpy+pyx)−/parenleftbigga2
/planckover2pi1/parenrightbigg2
p2
y/bracketrightBigg
=2a
/planckover2pi1xpy.
p2
1−p2
2=1
2/bracketleftBigg
p2
x−/planckover2pi1
a2(pxy+ypx)+/parenleftbigg/planckover2pi1
a2/parenrightbigg2
y2−p2
x−/planckover2pi1
a2(pxy+ypx)−/parenleftbigg/planckover2pi1
a2/parenrightbigg2
y2/bracketrightBigg
=−2/planckover2pi1
a2ypx.
So/planckover2pi1
2a2(q2
1−q2
2)+a2
2/planckover2pi1(p2
1−p2
2)=xpy−ypx=Lz.
(c)
H=1
2mp2+1
2mω2x2=a2
2/planckover2pi1p2+/planckover2pi1
2a2x2=H(x,p).
ThenH(q1,p1)=a2
2/planckover2pi1p2
1+/planckover2pi1
2a2q2
1≡H1,H(q2,p2)=a2
2/planckover2pi1p2
2+/planckover2pi1
2a2q2
2≡H2;Lz=H1−H2.
(d)The eigenvalues of H1are (n1+1
2)/planckover2pi1, and those of H2are (n2+1
2)/planckover2pi1, so the eigenvalues of Lzare
(n1+1
2)/planckover2pi1−(n2+1
2)/planckover2pi1=(n1−n2)/planckover2pi1=m/planckover2pi1, andmis aninteger , because n1andn2are.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 127
Problem 4.58
From Problem 4.28 we know that in the generic state χ=/parenleftbigg
a
b/parenrightbigg
(with|a|2+|b|2= 1),
/angbracketleftSz/angbracketright=/planckover2pi1
2/parenleftbig
|a|2−|b|2/parenrightbig
,/angbracketleftSx/angbracketright=/planckover2pi1Re(ab∗),/angbracketleftSy/angbracketright=−/planckover2pi1Im(ab∗);/angbracketleftS2
x/angbracketright=/angbracketleftS2
y/angbracketright=/planckover2pi12
4.
Writing a=|a|eiφa,b=|b|eiφb, we have ab∗=|a||b|ei(φa−φb)=|a||b|eiθ, where θ≡φa−φbis the phase
difference between aandb. Then
/angbracketleftSx/angbracketright=/planckover2pi1Re(|a||b|eiθ)=/planckover2pi1|a||b|cosθ,/angbracketleftSy/angbracketright=−/planckover2pi1Im(|a||b|eiθ)=−/planckover2pi1|a||b|sinθ.
σ2
Sx=/angbracketleftS2
x/angbracketright−/angbracketleftSx/angbracketright2=/planckover2pi12
4−/planckover2pi12|a|2|b|2cos2θ;σ2
Sy=/angbracketleftS2
y/angbracketright−/angbracketleftSy/angbracketright2=/planckover2pi12
4−/planckover2pi12|a|2|b|2sin2θ.
We want σ2
Sxσ2
Sy=/planckover2pi12
4/angbracketleftSz/angbracketright2,o r
/planckover2pi12
4/parenleftbig
1−4|a|2|b|2cos2θ/parenrightbig/planckover2pi12
4/parenleftbig
1−4|a|2|b|2sin2θ/parenrightbig
=/planckover2pi12
4/planckover2pi12
4/parenleftbig
|a|2−|b|2/parenrightbig2.
1−4|a|2|b|2/parenleftbig
cos2θ+ sin2θ/parenrightbig
+1 6|a|4|b|4sin2θcos2θ=|a|4−2|a|2|b|2+|b|4.
1+1 6|a|4|b|4sin2θcos2θ=|a|4+2|a|2|b|2+|b|4=/parenleftbig
|a|2+|b|2/parenrightbig2=1⇒|a|2|b|2sinθcosθ=0.
So either θ=0o r π, in which case aandbare relatively real, or else θ=±π/2, in which case aandbare
relatively imaginary (these two options subsume trivially the solutions a= 0 and b= 0).
Problem 4.59
(a)
Start with Eq. 3.71:d/angbracketleftr/angbracketright
dt=i
/planckover2pi1/angbracketleft[H,r]/angbracketright.
H=1
2m(p−qA)·(p−qA)+qϕ=1
2m/bracketleftbig
p2−q(p·A+A·p)+q2A2/bracketrightbig
+qϕ.
[H,x]=1
2m[p2,x]−q
2m[(p·A+A·p),x].
[p2,x]=[ (p2
x+p2
y+p2
z),x]=[p2
x,x]=px[px,x]+[px,x]px=px(−i/planckover2pi1)+(−i/planckover2pi1)px=−2i/planckover2pi1px.
[p·A,x]=[ (pxAx+pyAy+pzAz),x]=[pxAx,x]=px[Ax,x]+[px,x]Ax=−i/planckover2pi1Ax.
[A·p,x]=[ (Axpx+Aypy+Azpz),x]=[Axpx,x]=Ax[px,x]+[Ax,x]px=−i/planckover2pi1Ax.
[H,x]=1
2m(−2i/planckover2pi1px)−q
2m(−2i/planckover2pi1Ax)=−i/planckover2pi1
m(px−qAx); [H,r]=−i/planckover2pi1
m(p−qA).
d/angbracketleftr/angbracketright
dt=1
m/angbracketleft(p−qA)/angbracketright.QED
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128 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b)
We define the operator v≡1
m(p−qA);d/angbracketleftv/angbracketright
dt=i
/planckover2pi1/angbracketleft[H,v]/angbracketright+/angbracketleft∂v
∂t/angbracketright;∂v
∂t=−q
m∂A
∂t.
H=1
2mv2+qϕ⇒[H,v]=m
2[v2,v]+q[ϕ,v]; [ϕ,v]=1
m[ϕ,p].
[ϕ,px]=i/planckover2pi1∂ϕ
∂x(Eq. 3.65), so [ ϕ,p]=i/planckover2pi1∇ϕ,and [ϕ,v]=i/planckover2pi1
m∇ϕ.
[v2,vx]=[ (v2
x+v2
y+v2
z),vx]=[v2
y,vx]+[v2
z,vx]=vy[vy,vx]+[vy,vx]vy+vz[vz,vx]+[vz,vx]vz.
[vy,vx]=1
m2[(py−qAy),(px−qAx)] =−q
m2([Ay,px]+[py,Ax])
=−q
m2/parenleftbigg
i/planckover2pi1∂Ay
∂x−i/planckover2pi1∂Ax
∂y/parenrightbigg
=−i/planckover2pi1q
m2(∇×A)z=−i/planckover2pi1q
m2Bz.
[vz,vx]=1
m2[(pz−qAz),(px−qAx)] =−q
m2([Az,px]+[pz,Ax])
=−q
m2/parenleftbigg
i/planckover2pi1∂Az
∂x−i/planckover2pi1∂Ax
∂y/parenrightbigg
=i/planckover2pi1q
m2(∇×A)y=i/planckover2pi1q
m2By.
∴[v2,vx]=i/planckover2pi1q
m2(−vyBz−Bzvy+vzBy+Byvz)=i/planckover2pi1q
m2[−(v×B)x+(B×v)x].
[v2,v]=i/planckover2pi1q
m2[(B×v)−(v×B)].Putting all this together:
d/angbracketleftv/angbracketright
dt=i
/planckover2pi1/angbracketleftbigg/bracketleftbiggm
2i/planckover2pi1q
m2(B×v−v×B)+qi/planckover2pi1
m∇ϕ/bracketrightbigg/angbracketrightbigg
−q
m/angbracketleft∂A
∂t/angbracketright.
[⋆]md/angbracketleftv/angbracketright
dt=q
2/angbracketleft(v×B)−(B×v)/angbracketright+q/angbracketleftbigg
−∇ϕ−∂A
dt/angbracketrightbigg
=q
2/angbracketleft(v×B−B×v)/angbracketright+q/angbracketleftE/angbracketright.Or, since
v×B−B×v=1
m[(p−qA)×B−B×(p−qA)] =1
m[p×B−B×p]−q
m[A×B−B×A].
[Note:pdoes not commute with B, so the order doesmatter in the first term. But Acommutes with B,
soB×A=−A×Bin the second.]
md/angbracketleftv/angbracketright
dt=q/angbracketleftE/angbracketright+q
2m/angbracketleftp×B−B×p/angbracketright−q2
m/angbracketleftA×B/angbracketright.QED
(c)Go back to Eq. ⋆, and use/angbracketleftE/angbracketright=E,/angbracketleftv×B/angbracketright=/angbracketleftv/angbracketright×B;/angbracketleftB×v/angbracketright=B×/angbracketleftv/angbracketright=−/angbracketleftv/angbracketright×B.Then
md/angbracketleftv/angbracketright
dt=q/angbracketleftv/angbracketright×B+qE.QED
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 129
Problem 4.60
(a)
E=−∇ϕ=−2Kzˆk. B=∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk
∂/∂x ∂/∂y ∂/∂z
−B
0y/2B0x/20/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=
B0ˆk.
(b)For time-independent potentials Eq. 4.205 separates in the usual way:
1
2m/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg
·/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg
ψ+qϕψ=Eψ, or
−/planckover2pi12
2m∇2ψ+iq/planckover2pi1
2m[∇·(Aψ)+A·(∇ψ)]+q2
2mA2+qϕψ=Eψ. But∇·(Aψ)=(∇·A)ψ+A·(∇ψ),so
−/planckover2pi12
2m∇2ψ+iq/planckover2pi1
2m[2A·(∇ψ)+∇·(Aψ)] +/parenleftbiggq2
2mA2+qϕ/parenrightbigg
ψ=Eψ.
This is the time-independent Schr¨ odinger equation for electrodynamics. In the present case
∇·A=0,A·(∇ψ)=B0
2/parenleftbigg
x∂ψ
∂y−y∂ψ
∂x/parenrightbigg
,A2=B2
0
4/parenleftbig
x2+y2/parenrightbig
,ϕ=Kz2.
ButLz=/planckover2pi1
i/parenleftbigg
x∂
∂y−y∂
∂x/parenrightbigg
,so−/planckover2pi12
2m∇2ψ−qB0
2mLzψ+/bracketleftbiggq2B2
0
8m/parenleftbig
x2+y2/parenrightbig
+qKz2/bracketrightbigg
ψ=Eψ.
SinceLzcommutes with H, we may as well pick simultaneous eigenfunctions of both: Lzψ=¯m/planckover2pi1ψ,
where ¯m=0,±1,±2,...(with the overbar to distinguish the magnetic quantum number from the mass).
Then /bracketleftbigg
−/planckover2pi12
2m∇2+(qB0)2
8m/parenleftbig
x2+y2/parenrightbig
+qKz2/bracketrightbigg
ψ=/parenleftbigg
E+qB0/planckover2pi1
2m¯m/parenrightbigg
ψ.
Now let ω1≡qB0/m, ω 2≡/radicalbig
2Kq/m , and use cylindrical coordinates ( r,φ,z):
−/planckover2pi12
2m/bracketleftbigg1
r∂
∂r/parenleftbigg
r∂ψ
∂r/parenrightbigg
+1
r2∂2ψ
∂φ2+∂2ψ
∂z2/bracketrightbigg
+/bracketleftbigg1
8mω2
1/parenleftbig
x2+y2/parenrightbig
+1
2mω2
2z2/bracketrightbigg
ψ=/parenleftbigg
E+1
2¯m/planckover2pi1ω1/parenrightbigg
ψ.
ButLz=/planckover2pi1
i∂
∂φ,s o∂2ψ
∂φ2=−1
/planckover2pi12L2
zψ=−1
/planckover2pi12¯m2/planckover2pi12ψ=−¯m2ψ.Use separation of variables: ψ(r,φ,z)=
R(r)Φ(φ)Z(z):
−/planckover2pi12
2m/bracketleftbigg
ΦZ1
rd
dr/parenleftbigg
rdR
dr/parenrightbigg
−¯m2
r2RΦZ+RΦd2Z
dz2/bracketrightbigg
+/parenleftbigg1
8mω2
1r2+1
2mω2
2z2/parenrightbigg
RΦZ=/parenleftbigg
E+1
2¯m/planckover2pi1ω1/parenrightbigg
RΦZ.
Divide by RΦZand collect terms:
/braceleftbigg
−/planckover2pi12
2m/bracketleftbigg1
rRd
dr/parenleftbigg
rdR
dr/parenrightbigg
−¯m2
r2/bracketrightbigg
+1
8mω2
1r2/bracerightbigg
+/braceleftbigg
−/planckover2pi12
2m1
Zd2Z
dz2+1
2mω2
2z2/bracerightbigg
=/parenleftbigg
E+1
2¯m/planckover2pi1ω1/parenrightbigg
.
The first term depends only on r, the second only on z, so they’re both constants; call them ErandEz:
−/planckover2pi12
2m/bracketleftbigg1
rd
dr/parenleftbigg
rdR
dr/parenrightbigg
−¯m2
r2R/bracketrightbigg
+1
8mω2
1r2R=ErR;−/planckover2pi12
2md2Z
dz2+1
2mω2
2z2Z=EzZ;E=Er+Ez−1
2¯m/planckover2pi1ω1.
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130 CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Thezequation is a one-dimensional harmonic oscillator, and we can read off immediately that Ez=
(n2+1/2)/planckover2pi1ω2, withn2=0,1,2,.... Therequation is actually a two-dimensional harmonic oscillator; to
getEr, letu(r)≡√rR, and follow the method of Sections 4.1.3 and 4.2.1:
R=u√r,dR
dr=u/prime
√r−u
2r3/2,rdR
dr=√ru/prime−u
2√r,d
dr/parenleftbigg
rdR
dr/parenrightbigg
=√ru/prime/prime+u
4r3/2,
1
rd
dr/parenleftbigg
rdR
dr/parenrightbigg
=u/prime/prime
√r+u
4r5/2;−/planckover2pi12
2m/parenleftbiggu/prime/prime
√r+1
4u
r21√r−¯m2
r2u√r/parenrightbigg
+1
8mω2
1r2u√r=Eru√r
−/planckover2pi12
2m/bracketleftbiggd2u
dr2+/parenleftbigg1
4−¯m2/parenrightbiggu
r2/bracketrightbigg
+1
8mω2
1r2u=Eru.
This is identical to the equation we encountered in Problem 4.39 (the three-dimentional harmonic oscil-
lator), only with ω→ω1/2,E→Er,andl(l+1 )→¯m2−1/4, which is to say, l2+l+1/4= ¯m2,o r
(l+1/2)2=¯m2,o rl=|¯m|−1/2.[Our present equation depends only on ¯ m2, and hence is the same for
either sign, but the solution to Problem 4.39 assumed l+1/2≥0 (elseuis not normalizable), so we need
|m|here.] Quoting 4.39:
E=(jmax+l+3/2)/planckover2pi1ω→Er=(jmax+|¯m|+1 )/planckover2pi1ω1/2,where jmax=0,2,4,....
E=jmax+|¯m|+1 )/planckover2pi1ω1/2+(n2+1/2)/planckover2pi1ω2−¯m/planckover2pi1ω1/2=(n1+1
2)/planckover2pi1ω1+(n2+1
2)/planckover2pi1ω2,
wheren1=0,1,2,...(if ¯m≥0, then n1=jmax/2; if ¯m<0, then n1=jmax/2−¯m).
Problem 4.61
(a)
B/prime=∇×A/prime=∇×A+∇×(∇λ)=∇×A=B.
[∇×∇λ=0,by equality of cross-derivatives: ( ∇×∇λ)x=∂
∂y/parenleftbigg∂λ
∂z/parenrightbigg
−∂
∂z/parenleftbigg∂λ
∂y/parenrightbigg
=0,etc.]
E/prime=−∇ϕ/prime−∂A/prime
∂t=−∇ϕ+∇/parenleftbigg∂Λ
∂t/parenrightbigg
−∂A
∂t−∂
∂t(∇Λ) =−∇ϕ−∂A
∂t=E.
[Again:∇/parenleftbigg∂Λ
∂t/parenrightbigg
=∂
∂t(∇Λ) by the equality of cross-derivatives .]
(b)
/bracketleftbigg/planckover2pi1
i∇−qA−q(∇Λ)/bracketrightbigg
eiqΛ//planckover2pi1Ψ=q(∇Λ)eiqΛ//planckover2pi1Ψ+/planckover2pi1
ieiqΛ//planckover2pi1∇Ψ−qAeiqΛ//planckover2pi1Ψ−q(∇Λ)eiqΛ//planckover2pi1Ψ
=/planckover2pi1
ieiqΛ//planckover2pi1∇Ψ−qAeiqΛ//planckover2pi1Ψ.
/bracketleftbigg/planckover2pi1
i∇−qA−q(∇Λ)/bracketrightbigg2
eiqΛ//planckover2pi1Ψ=/parenleftbigg/planckover2pi1
i∇−qA−q(∇Λ)/parenrightbigg/bracketleftbigg/planckover2pi1
ieiqΛ//planckover2pi1∇Ψ−qAeiqΛ//planckover2pi1Ψ/bracketrightbigg
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 131
=−/planckover2pi12/bracketleftbiggiq
/planckover2pi1(∇Λ·∇Ψ)eiqΛ//planckover2pi1+eiqΛ//planckover2pi1∇2Ψ/bracketrightbigg
−/planckover2pi1q
i(∇·A)eiqΛ//planckover2pi1Ψ−q2(A·∇Λ)eiqΛ//planckover2pi1Ψ
−q/planckover2pi1
ieiqΛ//planckover2pi1A·(∇Ψ)−q/planckover2pi1
ieiqΛ//planckover2pi1(A·∇Ψ) +q2A2eiqΛ//planckover2pi1Ψ
−q/planckover2pi1
ieiqΛ//planckover2pi1(∇Λ·∇Ψ) +q2(A·∇Λ)eiqΛ//planckover2pi1Ψ
=eiqΛ//planckover2pi1/braceleftbig/bracketleftbig
−/planckover2pi12∇2Ψ+i/planckover2pi1q(∇·A) Ψ+2iq/planckover2pi1(A·∇Ψ) +q2A2Ψ/bracketrightbig
−iq/planckover2pi1(∇Λ)·(∇Ψ)−q2(A·∇Λ)Ψ +iq/planckover2pi1(∇Λ)·(∇Ψ) +q2(A·∇Λ)Ψ/bracerightbig
=eiqΛ//planckover2pi1/bracketleftBigg/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg2
Ψ/bracketrightBigg
.
So:/bracketleftBigg
1
2m/parenleftbigg/planckover2pi1
i∇−qA/prime/parenrightbigg2
+qϕ/prime/bracketrightBigg
Ψ/prime=eiqΛ//planckover2pi1/bracketleftBigg
1
2m/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg2
+qϕ−q∂Λ
∂t/bracketrightBigg
Ψ
[using Eq. 4.205] = eiqΛ//planckover2pi1/parenleftbigg
i/planckover2pi1∂Ψ
∂t−q∂Λ
∂tΨ/parenrightbigg
=i/planckover2pi1∂
∂t/parenleftBig
eiqΛ//planckover2pi1Ψ/parenrightBig
=i/planckover2pi1∂Ψ/prime
∂t.QED
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132 CHAPTER 5. IDENTICAL PARTICLES
Chapter 5
Identical Particles
Problem 5.1
(a)
(m1+m2)R=m1r1+m2r2=m1r1+m2(r1−r)=(m1+m2)r1−m2r⇒
r1=R+m2
m1+m2r=R+µ
m1r./check
(m1+m2)R=m1(r2+r)+m2r2=(m1+m2)r2+m1r⇒r2=R−m1
m1+m2r=R−µ
m2r./check
LetR=(X,Y,Z ),r=(x,y,z).
(∇1)x=∂
∂x1=∂X
∂x1∂
∂X+∂x
∂x1∂
∂x
=/parenleftbiggm1
m1+m2/parenrightbigg∂
∂X+ (1)∂
∂x=µ
m2(∇R)x+(∇r)x,so∇1=µ
m2∇R+∇r./check
(∇2)x=∂
∂x2=∂X
∂x2∂
∂X+∂x
∂x2∂
∂x
=/parenleftbiggm2
m1+m2/parenrightbigg∂
∂X−(1)∂
∂x=µ
m1(∇R)x−(∇r)x,so∇2=µ
m1∇R−∇ r./check
(b)
∇2
1ψ=∇1·(∇1ψ)=∇1·/bracketleftbiggµ
m2∇Rψ+∇rψ/bracketrightbigg
=µ
m2∇R·/parenleftbiggµ
m2∇Rψ+∇rψ/parenrightbigg
+∇r·/parenleftbiggµ
m2∇Rψ+∇rψ/parenrightbigg
=/parenleftbiggµ
m2/parenrightbigg2
∇2
Rψ+2µ
m2(∇r·∇R)ψ+∇2
rψ.
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CHAPTER 5. IDENTICAL PARTICLES 133
Likewise,∇2
2ψ=/parenleftbiggµ
m1/parenrightbigg2
∇2
Rψ−2µ
m1(∇r·∇R)+∇2
rψ.
∴Hψ=−/planckover2pi12
2m1∇2
1ψ−/planckover2pi12
2m2∇2
2ψ+V(r1,r2)ψ
=−/planckover2pi12
2/parenleftbiggµ2
m1m2
2∇2
R+2µ
m1m2∇r·∇R+1
m1∇2
r+µ2
m2m2
1∇2
R−2µ
m2m1∇r·∇R+1
m2∇2
r/parenrightbigg
ψ
+V(r)ψ=−/planckover2pi12
2/bracketleftbiggµ2
m1m2/parenleftbigg1
m2+1
m1/parenrightbigg
∇2
R+/parenleftbigg1
m1+1
m2/parenrightbigg
∇2
r/bracketrightbigg
ψ+V(r)ψ=Eψ.
But/parenleftbigg1
m1+1
m2/parenrightbigg
=m1+m2
m1m2=1
µ,soµ2
m1m2/parenleftbigg1
m2+1
m1/parenrightbigg
=µ
m1m2=m1m2
m1m2(m1+m2)+1
m1+m2.
−/planckover2pi12
2(m1+m2)∇2
Rψ−/planckover2pi12
2µ∇2
rψ+V(r)ψ=Eψ./check
(c)Put inψ=ψr(r)ψR(R), and divide by ψrψR:
/bracketleftbigg
−/planckover2pi12
2(m1+m2)1
ψR∇2
RψR/bracketrightbigg
+/bracketleftbigg
−/planckover2pi12
2µ1
ψr∇2
rψr+V(r)/bracketrightbigg
=E.
The first term depends only on R, the second only on r, so each must be a constant; call them ERand
Er, respectively. Then:
−/planckover2pi12
2(m1+m2)∇2ψR=ERψR;−/planckover2pi12
2µ∇2ψr+V(r)ψr=Erψr,with ER+Er=E.
Problem 5.2
(a)From Eq. 4.77, E1is proportional to mass, so∆E1
E1=∆m
µ=m−µ
µ=m(m+M)
mM−M
M=m
M.
The fractional error is the ratio of the electron mass to the proton mass:
9.109×10−31kg
1.673×10−27kg=5.44×10−4. The percent error is 0.054% (pretty small).
(b)From Eq. 4.94, Ris proportional to m,s o∆(1/λ)
(1/λ)=∆R
R=∆µ
µ=−(1/λ2)∆λ
(1/λ)=−∆λ
λ.
So (in magnitude) ∆ λ/λ=∆µ/µ. Butµ=mM/(m+M), where m= electron mass, and M=
nuclear mass.
∆µ=m(2mp)
m+2mp−mmp
m+mp=mmp
(m+mp)(m+2mp)(2m+2mp−m−2mp)
=m2mp
(m+mp)(m+2mp)=mµ
m+2mp.
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134 CHAPTER 5. IDENTICAL PARTICLES
∆λ
λ=∆µ
µ=m
m+2mp≈m
2mp,so∆λ=m
2mpλh,whereλhis the hydrogen wavelength.
1
λ=R/parenleftbigg1
4−1
9/parenrightbigg5
36R⇒λ=36
5R=36
5(1.097×107)m=6.563×10−7m.
∴∆λ=9.109×10−31
2(1.673×10−27)(6.563×10−7)m = 1.79×10−10m.
(c)µ=mm
m+m=m
2, so the energy is halfwhat it would be for hydrogen: (13.6/2)eV = 6.8 eV.
(d)µ=mpmµ
mp+mµ;R∝µ,s oRis changed by a factormpmµ
mp+mµ·mp+me
mpme=mµ(mp+me)
me(mp+mµ), as compared
with hydrogen. For hydrogen, 1 /λ=R(1−1/4) =3
4R⇒λ=4/3R=4/3(1.097×107)m=1.215×10−7m,
andλ∝1/R, so for muonic hydrogen the Lyman-alpha line is at
λ=me(mp+mµ)
mµ(mp+me)(1.215×10−7m) =1
206.77(1.673×10−27+ 206.77×9.109×10−31)
(1.673×10−27+9.109×10−31)(1.215×10−7m)
=6.54×10−10m.
Problem 5.3
The energy of the emitted photon, in a transition from vibrational state nito state nf,i s
Ep=(ni+1
2)/planckover2pi1ω−(nf+1
2)/planckover2pi1ω=n/planckover2pi1ω, (where n≡ni−nf). The frequency of the photon is
ν=Ep
h=nω
2π=n
2π/radicalBigg
k
µ.The splitting of this line is given by
∆ν=/vextendsingle/vextendsingle/vextendsingle/vextendsinglen
2π√
k/parenleftbigg
−1
2µ3/2∆µ/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1
2n
2π/radicalBigg
k
µ∆µ
µ=1
2ν∆µ
µ.
Now
µ=mhmc
mh+mc=1
1
mc+1
mh⇒∆µ=−1
/parenleftBig
1
mc+1
mh/parenrightBig2/parenleftbigg
−1
m2c∆mc/parenrightbigg
=µ2
m2c∆mc.
∆ν=1
2νµ∆mc
m2c=1
2ν(∆mc/mc)/parenleftBig
1+mc
mh/parenrightBig.
Using the average value (36) for mc, we have ∆ mc/mc=2/36, and mc/mh=3 6/1, so
∆ν=1
2(1/18)
(1 + 36)ν=1
(36)(37)ν=7.51×10−4ν.
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CHAPTER 5. IDENTICAL PARTICLES 135
Problem 5.4
(a)
1=/integraldisplay
|ψ±|2d3r1d3r2
=|A|2/integraldisplay
[ψa(r1)ψb(r2)±ψb(r1)ψa(r2)]∗[ψa(r1)ψb(r2)±ψb(r1)ψa(r2)]d3r1d3r2
=|A|2/bracketleftbigg/integraldisplay
|ψa(r1)|2d3r1/integraldisplay
|ψb(r2)|2d3r2±/integraldisplay
ψa(r1)∗ψb(r1)d3r1/integraldisplay
ψb(r2)∗ψa(r2)d3r2
±/integraldisplay
ψb(r1)∗ψa(r1)d3r1/integraldisplay
ψa(r2)∗ψb(r2)d3r2+/integraldisplay
|ψb(r1)|2d3r1/integraldisplay
|ψa(r2)|2d3r2/bracketrightbigg
=|A|2(1·1±0·0±0·0+1·1) = 2|A|2=⇒A=1/√
2.
(b)
1=|A|2/integraldisplay
[2ψa(r1)ψa(r2)]∗[2ψa(r1)ψa(r2)]d3r1d3r2
=4|A|2/integraldisplay
|ψa(r1)|2d3r1/integraldisplay
|ψa(r2)|2d3r2=4|A|2.A=1/2.
Problem 5.5
(a)
−/planckover2pi12
2m∂2ψ
∂x2
1−/planckover2pi12
2m∂2ψ
∂x2
2=Eψ (for 0≤x1,x2≤a,otherwise ψ=0 ).
ψ=√
2
a/bracketleftbigg
sin/parenleftBigπx1
a/parenrightBig
sin/parenleftbigg2πx2
a/parenrightbigg
−sin/parenleftbigg2πx1
a/parenrightbigg
sin/parenleftBigπx2
a/parenrightBig/bracketrightbigg
d2ψ
dx2
1=√
2
a/bracketleftBigg
−/parenleftBigπ
a/parenrightBig2
sin/parenleftBigπx1
a/parenrightBig
sin/parenleftbigg2πx2
a/parenrightbigg
+/parenleftbigg2π
a/parenrightbigg2
sin/parenleftbigg2πx1
a/parenrightbigg
sin/parenleftBigπx2
a/parenrightBig/bracketrightBigg
d2ψ
dx2
2=√
2
a/bracketleftBigg
−/parenleftbigg2π
a/parenrightbigg2
sin/parenleftBigπx1
a/parenrightBig
sin/parenleftbigg2πx2
a/parenrightbigg
+/parenleftBigπ
a/parenrightBig2
sin/parenleftbigg2πx1
a/parenrightbigg
sin/parenleftBigπx2
a/parenrightBig/bracketrightBigg
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136 CHAPTER 5. IDENTICAL PARTICLES
/parenleftbiggd2ψ
dx2
1+d2ψ
dx2
2/parenrightbigg
=−/bracketleftBigg/parenleftBigπ
a/parenrightBig2
+/parenleftbigg2π
a/parenrightbigg2/bracketrightBigg
ψ=−5π2
a2ψ,
−/planckover2pi12
2m/parenleftbiggd2ψ
dx2
1+d2ψ
dx2
2/parenrightbigg
=5π2/planckover2pi12
2ma2ψ=Eψ, withE=5π2/planckover2pi12
2ma2=5K./check
(b) Distinguishable:
ψ22=( 2/a) sin (2πx1/a) sin (2πx2/a),withE22=8K(nondegenerate).
ψ13=( 2/a) sin (πx1/a) sin (3πx2/a)
ψ31=( 2/a) sin (3πx1/a) sin (πx2/a)/bracerightbigg
,withE13=E31=1 0K(doubly degenerate).
Identical Bosons:
ψ22=( 2/a) sin (2πx1/a) sin (2πx2/a),E22=8K(nondegenerate).
ψ13=(√
2/a) [sin (πx1/a) sin (3πx2/a) + sin (3 πx1/a) sin (πx2/a)],E13=1 0K (nondegenerate).
Identical Fermions:
ψ13=(√
2/a)/bracketleftbig
sin/parenleftbigπx1
a/parenrightbig
sin/parenleftbig3πx2
a/parenrightbig
−sin/parenleftbig3πx1
a/parenrightbig
sin/parenleftbigπx2
a/parenrightbig/bracketrightbig
,E13=1 0K (nondegenerate).
ψ23=(√
2/a)/bracketleftbig
sin/parenleftbig2πx1
a/parenrightbig
sin/parenleftbig3πx2
a/parenrightbig
−sin/parenleftbig3πx1
a/parenrightbig
sin/parenleftbig2πx2
a/parenrightbig/bracketrightbig
,E23=1 3K (nondegenerate).
Problem 5.6
(a)Use Eq. 5.19 and Problem 2.4, with /angbracketleftx/angbracketrightn=a/2 and/angbracketleftx2/angbracketrightn=a2/parenleftBig
1
3−1
2(nπ)2/parenrightBig
.
/angbracketleft(x1−x2)2/angbracketright=a2/parenleftBig
1
3−1
2(nπ)2/parenrightBig
+a2/parenleftBig
1
3−1
2(mπ)2/parenrightBig
−2·a
2·a
2=a2/bracketleftbigg1
6−1
2π2/parenleftbigg1
n2+1
m2/parenrightbigg/bracketrightbigg
.
(b)/angbracketleftx/angbracketrightmn=2
a/integraltexta
0xsin/parenleftbigmπ
ax/parenrightbig
sin/parenleftbignπ
ax/parenrightbig
dx=1
a/integraltexta
0x/bracketleftBig
cos/parenleftBig
(m−n)π
ax/parenrightBig
−cos/parenleftBig
(m+n)π
ax/parenrightBig/bracketrightBig
dx
=1
a/bracketleftbigg/parenleftBig
a
(m−n)π/parenrightBig2
cos/parenleftBig
(m−n)π
ax/parenrightBig
+/parenleftBig
ax
(m−n)π/parenrightBig
sin/parenleftBig
(m−n)π
ax/parenrightBig
−/parenleftBig
a
(m+n)π/parenrightBig2
cos/parenleftBig
(m+n)π
ax/parenrightBig
−/parenleftBig
ax
(m+n)π/parenrightBig
sin/parenleftBig
(m+n)π
ax/parenrightBig/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0
=1
a/bracketleftbigg/parenleftBig
a
(m−n)π/parenrightBig2
(cos[(m−n)π]−1)−/parenleftBig
a
(m+n)π/parenrightBig2
(cos[(m+n)π]−1)/bracketrightbigg
.
But cos[( m±n)π]=(−1)m+n,so
/angbracketleftx/angbracketrightmn=a
π2/bracketleftbig
(−1)m+n−1/bracketrightbig/parenleftbigg1
(m−n)2−1
(m+n)2/parenrightbigg
=/braceleftBigg
a(−8mn)
π2(m2−n2)2,ifmandnhave opposite parity,
0, ifmandnhave same parity.
So Eq. 5.21⇒/angbracketleft(x1−x2)2/angbracketright=a2/bracketleftbigg1
6−1
2π2/parenleftbigg1
n2+1
m2/parenrightbigg/bracketrightbigg
−128a2m2n2
π4(m2−n2)4.
(The last term is present only when m,nhave opposite parity.)
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CHAPTER 5. IDENTICAL PARTICLES 137
(c)Here Eq. 5.21 ⇒/angbracketleft(x1−x2)2/angbracketright=a2/bracketleftbigg1
6−1
2π2/parenleftbigg1
n2+1
m2/parenrightbigg/bracketrightbigg
+128a2m2n2
π4(m2−n2)4.
(Again, the last term is present only when m,nhave opposite parity.)
Problem 5.7
(a)ψ(x1,x2,x3)=ψa(x1)ψb(x2)ψc(x3).
(b)ψ(x1,x2,x3)=1√
6[ψa(x1)ψb(x2)ψc(x3)+ψa(x1)ψc(x2)ψb(x3)+ψb(x1)ψa(x2)ψc(x3)
+ψb(x1)ψc(x2)ψa(x3)+ψc(x1)ψb(x2)ψa(x3)+ψc(x1)ψa(x2)ψb(x3)].
(c)ψ(x1,x2,x3)=1√
6[ψa(x1)ψb(x2)ψc(x3)−ψa(x1)ψc(x2)ψb(x3)−ψb(x1)ψa(x2)ψc(x3)
+ψb(x1)ψc(x2)ψa(x3)−ψc(x1)ψb(x2)ψa(x3)+ψc(x1)ψa(x2)ψb(x3)].
Problem 5.8
ψ=A/bracketleftbig
ψ(r1,r2,r3,...,rZ)±ψ(r2,r1,r3,...,rZ)+ψ(r2,r3,r1,...,rZ) + etc./bracketrightbig
,
where “etc.” runs over all permutations of the arguments r1,r2,...,rZ, with a + sign for all evenpermutations
(even number of transpositions ri↔rj, starting from r1,r2,...,rZ), and±for all oddpermutations (+ for
bosons,−for fermions). At the end of the process, normalize the result to determine A. (Typically A=1/√
Z!,
but this may not be right if the starting function is already symmetric under some interchanges.)
Problem 5.9
(a)The energy of each electron is E=Z2E1/n2=4E1/4=E1=−13.6eV, so the total initial energy is
2×(−13.6) eV=−27.2 eV. One electron drops to the ground state Z2E1/1=4E1, so the other is left
with 2 E1−4E1=−2E1=27.2 eV.
(b)He+hasoneelectron; it’s a hydrogenic ion (Problem 4.16) with Z= 2, so the spectrum is
1/λ=4R/parenleftBig
1/n2
f−1/n2
i/parenrightBig
,whereRis the hydrogen Rydberg constant, and ni,nfare the initial and final
quantum numbers (1, 2, 3, ...).
Problem 5.10
(a)The ground state (Eq. 5.30) is spatially symmetric , so it goes with the symmetric (triplet) spin configura-
tion. Thus the ground state is orthohelium, and it is triply degerate. The excited states (Eq. 5.32) come
in ortho (triplet) and para (singlet) form; since the former go with the symmetric spatial wave function,the orthohelium states are higher in energy than the corresponding (nondegenerate) para states.
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138 CHAPTER 5. IDENTICAL PARTICLES
(b)The ground state (Eq. 5.30) and all excited states (Eq. 5.32) come in both ortho and para form. All are
quadruply degenerate (or at any rate we have no way a priori of knowing whether ortho or para are higher
in energy, since we don’t know which goes with the symmetric spatial configuration).
Problem 5.11
(a)
/angbracketleftbigg1
|r1−r2|/angbracketrightbigg
=/parenleftbigg8
πa3/parenrightbigg2/integraldisplay/bracketleftBigg/integraldisplaye−4(r1+r2)/a
/radicalbig
r2
1+r2
2−2r1r2cosθ2d3r2/bracketrightBigg
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
/diamondsolidd3r1
/diamondsolid=2π/integraldisplay∞
0e−4(r1+r2)/a/bracketleftBigg/integraldisplayπ
0sinθ2/radicalbig
r2
1+r2
2−2r1r2cosθ2dθ2/bracketrightBigg
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
⋆r2
2dr2
⋆=1
r1r2/radicalBig
r2
1+r2
2−2r1r2cosθ2/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=1
r1r2/bracketleftbigg/radicalBig
r2
1+r2
2+2r1r2−/radicalBig
r2
1+r2
2−2r1r2/bracketrightbigg
=1
r1r2[(r1+r2)−|r1−r2|]=/braceleftbigg2/r1(r2<r1)
2/r2(r2>r1)
/diamondsolid=4πe−4r1/a/bracketleftbigg1
r1/integraldisplayr1
0r2
2e−4r2/adr2+/integraldisplay∞
r1r2e−4r2/adr2/bracketrightbigg
.
1
r1/integraldisplayr1
0r2
2e−4r2/adr2=1
r1/bracketleftbigg
−a
4r2
2e−4r2/a+a
2/parenleftBiga
4/parenrightBig2
e−4r2/a/parenleftbigg
−4r2
a−1/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingler1
0
=−a
4r1/bracketleftbigg
r2
1e−4r1/a+ar1
2e−4r1/a+a2
8e−4r1/a−a2
8/bracketrightbigg
.
/integraldisplay∞
r1r2e−4r2/adr2=/parenleftBiga
4/parenrightBig2
e−4r2/a/parenleftbigg
−4r2
a−1/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
r1=ar1
4e−4r1/a+a2
16e−4r1/a.
/diamondsolid=4π/braceleftbigga3
32r1e−4r1/a+/bracketleftbigg
−ar1
a−a2
8−a3
32r1+ar1
4+a2
16/bracketrightbigg
e−8r1/a/bracerightbigg
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CHAPTER 5. IDENTICAL PARTICLES 139
=πa2
8/braceleftbigga
r1e−4r1/a−/parenleftbigg
2+a
r1/parenrightbigg
e−8r1/a/bracerightbigg
.
/angbracketleftbigg1
|r1−r2|/angbracketrightbigg
=8
πa4·4π/integraldisplay∞
0/bracketleftbigga
r1e−4r1/a−/parenleftbigg
2+a
r1/parenrightbigg
e−8r1/a/bracketrightbigg
r2
1dr1
=32
a4/braceleftbigg
a/integraldisplay∞
0r1e−4r1/adr1−2/integraldisplay∞
0r2
1e−8r1/adr1−a/integraldisplay∞
0r1e−8r1/adr1/bracerightbigg
=32
a4/braceleftbigg
a·/parenleftBiga
4/parenrightBig2
−2·2/parenleftBiga
8/parenrightBig3
−a·/parenleftBiga
8/parenrightBig2/bracerightbigg
=32
a/parenleftbigg1
16−1
128−1
64/parenrightbigg
=5
4a.
(b)
Vee≈e2
4πFepsilonC0/angbracketleftbigg1
|r1−r2|/angbracketrightbigg
=5
4e2
4πFepsilonC01
a=5
4m
/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2
=5
2(−E1)=5
2(13.6e V )= 34 eV.
E0+Vee=(−109 + 34)eV = −75 eV, which is pretty close to the experimental value ( −79 eV).
Problem 5.12
(a)Hydrogen: (1 s); helium: (1 s)2; lithium: (1 s)2(2s); beryllium: (1 s)2(2s)2;
boron: (1 s)2(2s)2(2p); carbon: (1 s)2(2s)2(2p)2; nitrogen: (1 s)2(2s)2(2p)3;
oxygen: (1 s)2(2s)2(2p)4; fluorine: (1 s)2(2s)2(2p)5; neon: (1 s)2(2s)2(2p)6.
These values agree with those in Table 5.1—no surprises so far.
(b)Hydrogen:2S1/2; helium:1S0; lithium:2S1/2; beryllium1S0. (These four are unambiguous,
because the orbital angular momentum is zero in all cases.) For boron, the spin (1/2) and orbital (1)
angular momenta could add to give 3/2 or 1/2, so the possibilities are2P3/2or2P1/2.For carbon, the
twopelectrons could combine for orbital angular momentum 2, 1, or 0, and the spins could add to 1 or 0:
1S0,3S1,1P1,3P2,3P1,3P0,1D2,3D3,3D2,3D1.For nitrogen, the 3 pelectrons can add to orbital angular
momentum 3, 2, 1, or 0, and the spins to 3/2 or 1/2:
2S1/2,4S3/2,2P1/2,2P3/2,4P1/2,4P3/2,4P5/2,2D3/2,2D5/2,
4D1/2,4D3/2,4D5/2,4D7/2,2F5/2,2F3/2,4F3/2,4F5/2,4F7/2,4F9/2.
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140 CHAPTER 5. IDENTICAL PARTICLES
Problem 5.13
(a)Orthohelium should have lower energy than parahelium, for corresponding states (which is true).
(b)Hund’s first rule says S= 1 for the ground state of carbon. But this (the triplet) is symmetric, so the
orbital state will have to be antisymmetric. Hund’s second rule favors L= 2, but this is symmetric, as
you can see most easily by going to the “top of the ladder”: |22/angbracketright=|11/angbracketright1||11/angbracketright2. So the ground state of
carbon will be S=1,L= 1. This leaves three possibilities:3P2,3P1, and3P0.
(c)For boron there is only one electron in the 2 psubshell (which can accommodate a total of 6), so Hund’s
third rule says the ground state will have J=|L−S|.We found in Problem 5.12(b) that L= 1 and
S=1/2, soJ=1/2, and the configuration is2P1/2.
(d)For carbon we know that S= 1 and L= 1, and there are only two electrons in the outer subshell, so
Hund’s third rule says J= 0, and the ground state configuration must be3P0.
For nitrogen Hund’s first rule says S=3/2, which is symmetric (the top of the ladder is |3
23
2/angbracketright=
|1
21
2/angbracketright1|1
21
2/angbracketright2|1
21
2/angbracketright3). Hund’s second rule favors L= 3, but this is also symmetric. In fact, the only
antisymmetric orbital configuration here is L= 0. [You can check this directly by working out the
Clebsch-Gordan coefficients, but it’s easier to reason as follows: Suppose the three outer electrons are inthe “top of the ladder” spin state, so each one has spin up ( |
1
21
2/angbracketright); then (since the spin states are all the
same) the orbital states haveto be different: |11/angbracketright,|10/angbracketright, and|1−1/angbracketright. In particular, the total z-component of
orbital angular momentum has to be zero. But the only configuration that restricts Lzto zero is L= 0.]
The outer subshell is exactly half filled (three electrons with n=2 ,l= 1), so Hund’s third rule says
J=|L−S|=|0−3
2|=3/2.Conclusion: The ground state of nitrogen is4S3/2.(Table 5.1 confirms
this.)
Problem 5.14
S=2 ;L=6 ;J=8 . (1s)2(2s)2(2p)6(3s)2(3p)6(3d)10(4s)2(4p)6
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
definite (36electrons )(4d)10(5s)2(5p)6(4f)10(6s)2
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
likely (30electrons ).
Problem 5.15
Divide Eq. 5.45 by Eq. 5.43, using Eq. 5.42:
Etot/Nq
EF=/planckover2pi12(3π2Nq)5/3
10π2mV2/31
Nq2m
/planckover2pi12(3π2Nq/V )2/3=3
5.
Problem 5.16
(a)EF=/planckover2pi12
2m(3ρπ2)2/3.ρ=Nq
V=N
V=atoms
mole×moles
gm×gm
volume=NA
M·d, where NAis Avogadro’s
number (6 .02×1023),M= atomic mass = 63 .5 gm/mol, d= density = 8 .96 gm/cm3.
ρ=(6.02×1023)(8.96 gm/cm3)
(63.5g m )=8.49×1022/cm3=8.49×1028/m3.
EF=(1.055×10−34J·s)(6.58×10−16eV·s)
(2)(9.109×10−31kg)(3π28.49×1028/m3)2/3=7.04 eV.
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CHAPTER 5. IDENTICAL PARTICLES 141
(b)
7.04 eV =1
2(0.511×106eV/c2)v2⇒v2
c2=14.08
.511×106=2.76×10−5⇒v
c=5.25×10−3,
so it’s nonrelativistic. v=( 5.25×10−3)×(3×108)=1.57×106m/s.
(c)
T=7.04 eV
8.62×10−5eV/K=8.17×104K.
(d)
P=(3π2)2/3/planckover2pi12
5mρ5/3=(3π2)2/3(1.055×10−34)2
5(9.109×10−31)(8.49×1028)5/3N/m2=3.84×1010N/m2.
Problem 5.17
P=(3π2)2/3/planckover2pi12
5m/parenleftbiggNq
V/parenrightbigg5/3
=AV−5/3⇒B=−VdP
dV=−VA/parenleftbigg−5
3/parenrightbigg
V−5/3−1=5
3AV−5/3=5
3P.
For copper, B=5
3(3.84×1010N/m2)=6.4×1010N/m2.
Problem 5.18
(a)Equations 5.59 and 5.63 ⇒ψ=Asinkx+Bcoskx;Asinka=/bracketleftbig
eiKa−coska/bracketrightbig
B.So
ψ=Asinkx+Asinka
(eiKa−coska)coskx=A
(eiKa−coska)/bracketleftbig
eiKasinkx−sinkxcoska+ coskxsinka/bracketrightbig
=C/braceleftbig
sinkx+e−iKasin[k(a−x)]/bracerightbig
,whereC≡AeiKa
eiKa−coska.
(b)Ifz=ka=jπ, then sin ka= 0, Eq. 5.64 ⇒cosKa= coska=(−1)j⇒sinKa=0 ,s o eiKa=
cosKa+isinKa=(−1)j, and the constant Cinvolves division by zero. In this case we must go back to
Eq. 5.63, which is a tautology (0=0) yielding no constraint on AorB, Eq. 5.61 holds automatically, and
Eq. 5.62 gives
kA−(−1)jk/bracketleftbig
A(−1)j−0/bracketrightbig
=2mα
/planckover2pi12B⇒B=0.Soψ=Asinkx.
Hereψiszeroat each delta spike, so the wave function never “feels” the potential at all.
Problem 5.19
We’re looking for a solution to Eq. 5.66 with β= 10 and z/lessorsimilarπ:f(z) = cosz+1 0sinz
z=1.
Mathematica gives z=2.62768. So E=/planckover2pi12k2
2m=/planckover2pi12z2
2ma2=z2
2βα
a=(2.62768)2
20eV = 0.345 eV.
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142 CHAPTER 5. IDENTICAL PARTICLES
Problem 5.20
Positive-energy solutions. These are the same as before, except that α(and hence also β) is now a negative
number.
Negative-energy solutions. On 0<x<a we have
d2ψ
dx2=κ2ψ,where κ≡√
−2mE
/planckover2pi1⇒ψ(x)=Asinhkx+Bcoshkx.
According to Bloch’s theorem the solution on −a<x< 0i s
ψ(x)=e−iKa[Asinhκ(x+a)+Bcoshκ(x+a)].
Continuity at x=0⇒
B=e−iKa[Asinhκa+Bcoshκa],orAsinhκa=B/bracketleftbig
eiKa−coshκa/bracketrightbig
. (1)
The discontinuity in ψ/prime(Eq. 2.125)⇒
κA−e−iKaκ[Acoshκa+Bsinhκa]=2mα
/planckover2pi12B,orA/bracketleftbig
1−e−iKacoshκa/bracketrightbig
=B/bracketleftbigg2mα
/planckover2pi12κ+e−iKasinhκa/bracketrightbigg
.(2)
Plugging (1) into (2) and cancelling B:
/parenleftbig
eiKa−coshκa/parenrightbig/parenleftbig
1−e−iKacoshκa/parenrightbig
=2mα
/planckover2pi12κsinhκa+e−iKasinh2κa.
eiKa−2 coshκa+e−iKacosh2κa−e−iKasinh2κa=2mα
/planckover2pi12κsinhκa.
eiKa+e−iKa= 2 cosh κa+2ma
/planckover2pi12κsinhκa, cosKa= coshκa+mα
/planckover2pi12κsinhκa.
This is the analog to Eq. 5.64. As before, we let β≡mαa/ /planckover2pi12(but remember it’s now a negative number), and
this time we define z≡−κa, extending Eq. 5.65 to negative z, where it represents negative-energy solutions.
In this region we define
f(z) = cosh z+βsinhz
z. (3)
In the Figure I have plotted f(z) forβ=−1.5, using Eq. 5.66 for postive zand (3) for negative z.A s
before, allowed energies are restricted to the range −1≤f(z)≤1, and occur at intersections of f(z) with the
Nhorizontal lines cos Ka= cos(2 πn/Na ),withn=0,1,2...N−1.Evidently the first band (partly negative,
and partly positive) contains Nstates, as do all the higher bands.
01
-1
0 π2 π 3 π 4 π
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CHAPTER 5. IDENTICAL PARTICLES 143
Problem 5.21
Equation 5.56 says K=2πn
Na⇒Ka=2πn
N; at the bottom of page 227 we found that n=0,1,2,...,N−1.
Each value of ncorresponds to a distinct state. To find the allowed energies we draw Nhorizontal lines on
Figure 5.6, at heights cos Ka= cos(2 πn/N ), and look for intersections with f(z). The point is that almost all
of these lines come in pairs—two different n’s yielding the same value of cos Ka:
N=1⇒n=0⇒cosKa=1.Nondegenerate.
N=2⇒n=0,1⇒cosKa=1,−1.Nondegenerate.
N=3⇒n=0,1,2⇒cosKa=1,−1
2,−1
2.The first is nondegenerate, the other two are degenerate.
N=4⇒n=0,1,2,3⇒cosKa=1,0,−1,0.Two are nondegenerate, the others are degenerate.
Evidently they are doubly degenerate (two different n’s give same cos Ka)except when cos Ka=±1, i.e., at
thetop or bottom of a band. The Bloch factors eiKalie at equal angles in the complex plane, starting with
1 (see Figure, drawn for the case N= 8); by symmetry, there is always one with negative imaginary part
symmetrically opposite each one with positive imaginary part; these two have the same realpart (cos Ka).
Only points which fall onthe real axis have no twins.
n=0n=1n=2
n=3
n=4
n=5
n=6n=7cos(Ka)sin(Ka)
Problem 5.22
(a)
ψ(xA,xB,xC)=1√
6/parenleftBigg/radicalbigg
2
a/parenrightBigg3/bracketleftbigg
sin/parenleftbigg5πxA
a/parenrightbigg
sin/parenleftbigg7πxB
a/parenrightbigg
sin/parenleftbigg17πxC
a/parenrightbigg
−sin/parenleftbigg5πxA
a/parenrightbigg
sin/parenleftbigg17πxB
a/parenrightbigg
sin/parenleftbigg7πxC
a/parenrightbigg
+ sin/parenleftbigg7πxA
a/parenrightbigg
sin/parenleftbigg17πxB
a/parenrightbigg
sin/parenleftbigg5πxC
a/parenrightbigg
−sin/parenleftbigg7πxA
a/parenrightbigg
sin/parenleftbigg5πxB
a/parenrightbigg
sin/parenleftbigg17πxC
a/parenrightbigg
+ sin/parenleftbigg17πxA
a/parenrightbigg
sin/parenleftbigg5πxB
a/parenrightbigg
sin/parenleftbigg7πxC
a/parenrightbigg
−sin/parenleftbigg17πxA
a/parenrightbigg
sin/parenleftbigg7πxB
a/parenrightbigg
sin/parenleftbigg5πxC
a/parenrightbigg/bracketrightbigg
.
(b)(i)
ψ=/parenleftBigg/radicalbigg
2
a/parenrightBigg3/bracketleftbigg
sin/parenleftbigg11πxA
a/parenrightbigg
sin/parenleftbigg11πxB
a/parenrightbigg
sin/parenleftbigg11πxC
a/parenrightbigg/bracketrightbigg
.
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144 CHAPTER 5. IDENTICAL PARTICLES
(ii)
ψ=1√
3/parenleftBigg/radicalbigg
2
a/parenrightBigg3/bracketleftbigg
sin/parenleftBigπxA
a/parenrightBig
sin/parenleftBigπxB
a/parenrightBig
sin/parenleftbigg19πxC
a/parenrightbigg
+ sin/parenleftBigπxA
a/parenrightBig
sin/parenleftbigg19πxB
a/parenrightbigg
sin/parenleftBigπxC
a/parenrightBig
+ sin/parenleftbigg19πxA
a/parenrightbigg
sin/parenleftBigπxB
a/parenrightBig
sin/parenleftBigπxC
a/parenrightBig/bracketrightbigg
.
(iii)
ψ=1√
6/parenleftBigg/radicalbigg
2
a/parenrightBigg3/bracketleftbigg
sin/parenleftbigg5πxA
a/parenrightbigg
sin/parenleftbigg7πxB
a/parenrightbigg
sin/parenleftbigg17πxC
a/parenrightbigg
+ sin/parenleftbigg5πxA
a/parenrightbigg
sin/parenleftbigg17πxB
a/parenrightbigg
sin/parenleftbigg7πxC
a/parenrightbigg
+ sin/parenleftbigg7πxA
a/parenrightbigg
sin/parenleftbigg17πxB
a/parenrightbigg
sin/parenleftbigg5πxC
a/parenrightbigg
+ sin/parenleftbigg7πxA
a/parenrightbigg
sin/parenleftbigg5πxB
a/parenrightbigg
sin/parenleftbigg17πxC
a/parenrightbigg
+ sin/parenleftbigg17πxA
a/parenrightbigg
sin/parenleftbigg5πxB
a/parenrightbigg
sin/parenleftbigg7πxC
a/parenrightbigg
+ sin/parenleftbigg17πxA
a/parenrightbigg
sin/parenleftbigg7πxB
a/parenrightbigg
sin/parenleftbigg5πxC
a/parenrightbigg/bracketrightbigg
.
Problem 5.23
(a)En1n2n3=(n1+n2+n3+3
2)/planckover2pi1ω=9
2/planckover2pi1ω⇒n1+n2+n3=3.(n1,n2,n3=0,1,2,3...).
State Configuration # of States
n1n2n3(N0,N1,N2...)
003
030 (2,0,0,1,0,0 ...) 3
300
012
021
102 (1,1,1,0,0,0 ...) 6
120
201
210
111 (0,3,0,0,0 ...) 1Possible single-particle energies:
E0=/planckover2pi1ω/2:P0=1 2/30 = 4/10.
E1=3/planckover2pi1ω/2:P1=9/30 = 3/10.
E2=5/planckover2pi1ω/2:P2=6/30 = 2/10.
E3=7/planckover2pi1ω/2:P3=3/30 = 1/10.
Most probable configuration: (1,1,1,0,0,0 ...).
Most probable single-particle energy: E0=1
2/planckover2pi1ω.
(b)For identical fermions the onlyconfiguration is (1,1,1,0,0,0 ...) (one state), so this is also the most
probable configuration. The possible one-particle energies are
E0(P0=1/3),E 1(P1=1/3),E 2(P2=1/3),
and they are all equally likely, so it’s a 3-way tie for the most probable energy.
(c)For identical bosons all three configurations are possible, and there is one state for each. Possible one-
particle energies: E0(P0=1/3),E1(P1=4/9),E2(P2=1/9),E3(P3=1/9).Most probable energy: E1.
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CHAPTER 5. IDENTICAL PARTICLES 145
Problem 5.24
HereN= 3, and dn= 1 for all states, so:
Eq. 5.74⇒Q=6
∞/productdisplay
n=11
Nn!(distinguishable),
Eq. 5.75⇒Q=∞/productdisplay
n=11
Nn!(1−Nn)!(fermions),
Eq. 5.77⇒Q= 1 (bosons).
(In the products, most factors are 1 /0! or 1/1!, both of which are 1, so I won’t write them.)
Configuration 1 (N11= 3, others 0):
Q=6×1
3!=1 (distinguishable) ,
Q=1
3!×1
(−2)!=0 (fermions) ,
Q=1 (bosons).
Configuration 2 (N5=1,N13= 2):
Q=6×1
1!×1
2!=3 (distinguishable) ,
Q=1
1!0!×1
2!(−1)!=0 (fermions) ,
Q=1 (bosons).
Configuration 3 (N1=2,N19= 1):
Q=6×1
2!×1
1!=3 (distinguishable) ,
Q=1
2!(−1)!×1
1!0!=0 (fermions) ,
Q=1 (bosons).
Configuration 4 (N5=N7=N17= 1):
Q=6×1
1!×1
1!×1
1!=6(distinguishable) ,
Q=1
1!0!×1
1!0!×1
1!0!=1 (fermions) ,
Q=1 (bosons).
All of these agree with what we got “by hand” at the top of page 231.
Problem 5.25
N=1: - can put the ball in any of dbaskets, so dways.
N=2:
- could put both balls in any of the dbaskets : dways, or
- could put one in one basket ( dways), the other in another( d−1) ways—but it
doesn’t matter which is which, so divide by 2.
Total:d+1
2d(d−1) =1
2d(2 +d−1) =1
2d(d+1 ) ways.
N=3:
- could put all three in one basket : dways, or
- 2 in one basket, one in another : d(d−1) ways, or
- 1 each in 3 baskets : d(d−1)(d−2)/3! ways.
Total:d+d(d−1) +d(d−1)(d−2)/6=1
6d( 6+6d−6+d2−3d+2 )=1
6d(d2+3d+2 )
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146 CHAPTER 5. IDENTICAL PARTICLES
=d(d+ 1)(d+2 )
6ways.
N=4:
- all in one basket: dways, or
- 3 in one basket, 1 in another: d(d−1) ways, or
- 2 in one basket, 2 in another: d(d−1)/2 ways, or
- 2 in one basket, one each in others: d(d−1)(d−2)/2, or
- all in different baskets: d(d−1)(d−2)(d−3)/4!
Total
:d+d(d−1) +d(d−1)/2+d(d−1)(d−2)/2+d(d−1)(d−2)(d−3)/24
=1
24( 2 4+2 4 d−2 4+1 2d−1 2+1 2d2−36d+2 4+d3−6d2+1 1d−6)
=1
24d(d3+6d2+1 1d+6 )=d(d+ 1)(d+ 2)(d+3 )
24ways.
The general formula seems to be f(N,d)=d(d+ 1)(d+2 )···(d+N−1)
N!=(d+N−1)!
N!(d−1)!=/parenleftbiggd+N−1
N/parenrightbigg
.
Proof: How many ways to put Nidentical balls in dbaskets? Call it f(N,d).
- Could put all of them in the first basket: 1 way.- Could put all but one in the first basket; there remains 1 ball for d−1 baskets: f(1,d−1) ways.
- Could put all but two in the first basket; there remain 2 for d−1 baskets: f(2,d−1) ways.
...
- Could put zero in the first basket, leaving Nford−1 baskets: f(N,d−1) ways.
Thus:f(N,d)=f(0,d−1)+f(1,d−1)+f(2,d−1)+···+f(N,d−1) =/summationtext
N
j=0f(j,d−1) (where f(0,d)≡1).
It follows that f(N,d)=/summationtextN−1
j=0f(j,d−1)+f(N,d−1) =f(N−1,d)+f(N,d−1). Use this recursion relation
to confirm the conjectured formula by induction:
/parenleftbiggd+N−1
N/parenrightbigg
?=/parenleftbiggd+N−2
N−1/parenrightbigg
+/parenleftbiggd+N−2
N/parenrightbigg
=(d+N−2)!
(N−1)!(d−1)!+(d+N−2)!
N!(d−2)!
=(d+N−2)!
N!(d−1)!(N+d−1) =(d+N−1)!
N!(d−1)!=/parenleftbiggd+N−1
d−1/parenrightbigg
./check
It works for N=0:/parenleftbigd−1
0/parenrightbig
= 1, and for d=1:/parenleftbigN
N/parenrightbig
= 1 (which is obviously correct for just one basket). QED
Problem 5.26
A(x,y)=( 2x)(2y)=4xy; maximize, subject to the constraint ( x/a)2+(y/b)2=1.
(x,y)
ab
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CHAPTER 5. IDENTICAL PARTICLES 147
G(x,y,λ )≡4xy+λ/bracketleftbig
(x/a)2+(y/b)2−1/bracketrightbig
.∂G
∂x=4y+2λx
a2=0⇒y=−λx
2a2.
∂G
∂y=4x+2λy
b2=0⇒4x=−2λ
b2/parenleftbigg
−λx
2a2/parenrightbigg
⇒4x=λ2
a2b2x⇒x= 0 (minimum), or else λ=±2ab.
Soy=∓2abx
2a2=∓b
ax.We may as well pick xandypositive, (as in the figure); then y=(b/a)x(and
λ=−2ab).∂G
∂λ=0⇒/parenleftBigx
a/parenrightBig2
+/parenleftBigy
b/parenrightBig2
= 1 (of course), sox2
a2+b2x2
a2b2=1 ,o r2
a2x2=1 ,o r x=a/√
2, and hence
y=ba/(a√
2)⇒y=b/√
2.A=4a√
2b√
2=2ab.
Problem 5.27
(a)ln(10!) = ln(3628800) = 15 .1044; 10 ln(10) −1 0=2 3 .026−1 0=1 3 .0259; 15 .1044−13.0259 =
2.0785; 2 .0785/15.1044 = 0 .1376,or 14%.
(b)The percent error is:ln(z!)−zln(z)+z
ln(z!)×100.z%
205.7
100 0.89
501.9
900.996
851.06
891.009
Since my calculator cannot compute factorials greater than 69! I used Mathematica to construct the table.
Evidently, the smallest integer for which the error is <1% is 90.
Problem 5.28
Equation 5.108 ⇒N=V
2π2/integraldisplay∞
0k2n(FepsilonC)dk, where n(FepsilonC) is given (as T→0) by Eq. 5.104.
SoN=V
2π2/integraldisplaykmax
0k2dk=V
2π2k3
max
3, where kmaxis given by/planckover2pi12k2
max
2m=µ(0) =EF⇒kmax=√2mEF
/planckover2pi1.
N=V
6π2/planckover2pi13(2mEF)3/2.Compare Eq. 5.43, which says
EF=/planckover2pi12
2m/parenleftbigg
3π2Nq
V/parenrightbigg2/3
,or(2mEF)3/2
/planckover2pi13=3π2Nq
V,orN=V
3π2q/planckover2pi13(2mEF)3/2.
Hereq= 1, and Eq. 5.108 needs an extra factor of 2 on the right, to account for spin, so the two formulas agree.
Equation 5.109 ⇒Etot=V/planckover2pi12
4π2m/integraldisplaykmax
0k4dk=V/planckover2pi12
4π2mk5
max
5⇒Etot=V
20π2m/planckover2pi13(2mEF)5/2.
Compare Eq. 5.45, which says Etot=V/planckover2pi12
10π2mk5
max. Again, Eq. 5.109 for electrons has an extra factor of 2, so
the two agree.
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148 CHAPTER 5. IDENTICAL PARTICLES
Problem 5.29
(a)Equation 5.103, n(FepsilonC)>0⇒1
e(ρepsilono−µ)/kBT−1>0⇒e(ρepsilono−µ)/kBT>1⇒(FepsilonC−µ)
kBT>0⇒FepsilonC>µ(T),for all
allowed energies FepsilonC.
(b)For a free particle gas, E=/planckover2pi12
2mk2→0 (ask→0, in the continuum limit), so µ(T)is always negative .
(Technically, the lowest energy is/planckover2pi12π2
2m/parenleftbigg1
l2x+1
l2y+1
l2z/parenrightbigg
, but we take the dimensions lxlylzto be very large
in the continuum limit.) Equation 5.108 ⇒N/V=1
2π2/integraldisplay∞
0k2
e(/planckover2pi12k2/2m−µ)/kBT−1dk. The integrand is
always positive, and the only Tdependence is in µ(T) andkBT. So, as Tdecreases, ( /planckover2pi12k2/2m)−µ(T)
must also decrease, and hence −µ(T) decreases, or µ(T)increases (always negative).
(c)N
V=1
2π2/integraldisplay∞
0k2
e/planckover2pi12k2/2mkBT−1dk.Letx≡/planckover2pi12k2
2mkBT,sok=√2mkBT
/planckover2pi1x1/2;dk=√2mkBT
/planckover2pi11
2x−1/2dx.
N
V=1
2π2/parenleftbigg2mkBT
/planckover2pi12/parenrightbigg3/21
2/integraldisplay∞
0x1/2
ex−1dx,where/integraldisplay∞
0x3/2−1
ex−1dx= Γ(3/2)ζ(3/2).
Now Γ(3 /2) =√π/2;ζ(3/2) = 2.61238,soN
V=2.612/parenleftbiggmkBT
2π/planckover2pi12/parenrightbigg3/2
;Tc=2π/planckover2pi12
mkB/parenleftbiggN
2.612V/parenrightbigg2/3
.
(d)
N
V=mass/volume
mass/atom=0.15×103kg/m3
4(1.67×10−27kg)=2.2×1028/m3.
Tc=2π(1.05×10−34J·s)2
4(1.67×10−27kg)(1.38×10−23J/K)/parenleftbigg2.2×1028
2.61 m3/parenrightbigg2/3
=3.1 K.
Problem 5.30
(a)
ω=2πν=2πc
λ,sodω=−2πc
λ2dλ, andρ(ω)=/planckover2pi1
π2c3(2πc)3
λ3(e2π/planckover2pi1c/kBTλ−1).
ρ(ω)|dω|=8π/planckover2pi11
λ3(e2π/planckover2pi1c/kBTλ−1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle−2πc
λ2dλ/vextendsingle/vextendsingle/vextendsingle/vextendsingle=
ρ(λ)dλ⇒ρ(λ)=16π2/planckover2pi1c
λ5(e2π/planckover2pi1c/kBTλ−1).
(Fordensity , we want only the sizeof the interval, not its sign.)
(b)To maximize, set dρ/dλ=0 :
0=1 6π2/planckover2pi1c/bracketleftbigg−5
λ6(e2π/planckover2pi1c/kBTλ−1)−e2π/planckover2pi1c/kBTλ(2π/planckover2pi1c/kBT)
λ5(e2π/planckover2pi1c/kBTλ−1)2/parenleftbigg
−1
λ2/parenrightbigg/bracketrightbigg
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CHAPTER 5. IDENTICAL PARTICLES 149
1 2 3 4 512345
5-x
5e-x
⇒5(e2π/planckover2pi1c/kBTλ−1) =e2π/planckover2pi1c/kBTλ/parenleftbigg2π/planckover2pi1c
kBTλ/parenrightbigg
.
Letx≡2π/planckover2pi1c/kBTλ; then 5( ex−1) =xex; or 5(1−e−x)=x,o r5e−x=5−x. From the graph, the
solution occurs slightly below x=5 .
Mathematica says x=4.966, so λmax=2π/planckover2pi1c
(4.966)kB1
T=(6.626×10−34J·s)(2.998×108m/s)
(4.966)(1.3807×10−23J/K)1
T=
2.897×10−3m·K/T.
Problem 5.31
From Eq. 5.113:
E
V=/integraldisplay∞
0ρ(ω)dω=/planckover2pi1
π2c3/integraldisplay∞
0ω3
(e/planckover2pi1ω/kBT−1)dω. Letx≡/planckover2pi1ω
kBT.Then
E
V=/planckover2pi1
π2c3/parenleftbiggkBT
/planckover2pi1/parenrightbigg4/integraldisplay∞
0x3
ex−1dx=(kBT)4
π2c3/planckover2pi13Γ(4)ζ(4) =(kBT)4
π2c3/planckover2pi13·6·π4
90=/parenleftbiggπ2k4
B
15c3/planckover2pi13/parenrightbigg
T4
=/bracketleftbiggπ2(1.3807×10−23J/K)4
15(2.998×108m/s)3(1.0546×10−34J·s)3/bracketrightbigg
T4=7.566×10−16J
m3K4T4.QED
Problem 5.32
From Problem 2.11(a),
/angbracketleftx/angbracketright0=/angbracketleftx/angbracketright1=0 ;/angbracketleftx2/angbracketright0=/planckover2pi1
2mω;/angbracketleftx2/angbracketright1=3/planckover2pi1
2mω.
From Eq. 3.98,
/angbracketleftx/angbracketright01=/integraldisplay∞
−∞xψ0(x)ψ1(x)dx=/angbracketleft0|x|1/angbracketright=/radicalbigg
/planckover2pi1
2mω/parenleftBig√
1δ00+√
0δ1−1/parenrightBig
=/radicalbigg
/planckover2pi1
2mω.
(a)Equation 5.19 ⇒/angbracketleft(x1−x2)2/angbracketrightd=/planckover2pi1
2mω+3/planckover2pi1
2mω−0=2/planckover2pi1
mω.
(b)Equation 5.21 ⇒/angbracketleft(x1−x2)2/angbracketright+=2/planckover2pi1
mω−2/planckover2pi1
2mω=/planckover2pi1
mω.
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150 CHAPTER 5. IDENTICAL PARTICLES
(c)Equation 5.21 ⇒/angbracketleft(x1−x2)2/angbracketright−=2/planckover2pi1
mω+2/planckover2pi1
2mω=3/planckover2pi1
mω.
Problem 5.33
(a)Each particle has 3 possible states: 3 ×3×3=27.
(b)All in same state: aaa, bbb, ccc ⇒3.
2 in one state: aab, aac, bba, bbc, cca, ccb ⇒6 (each symmetrized).
3 different states: abc(symmetrized) ⇒1.
Total: 10.
(c)Onlyabc(antisymmetrized) = ⇒1.
Problem 5.34
Equation 5.39 ⇒Enxny=π2/planckover2pi12
2m/parenleftBigg
n2
x
l2x+n2
y
l2y/parenrightBigg
=/planckover2pi12k2
2m, with k=/parenleftbiggπnx
lx,πny
ly/parenrightbigg
. Each state is represented by an
intersection on a grid in “ k-space”—this time a plane—and each state occupies an area π2/lxly=π2/A(where
A≡lxlyis the area of the well). Two electrons per state means
1
4πk2
F=Nq
2/parenleftbiggπ2
A/parenrightbigg
,orkF=/parenleftbigg
2πNq
A/parenrightbigg1/2
=( 2πσ)1/2,
whereσ≡Nq/A is the number of free electrons per unit area.
∴EF=/planckover2pi12k2
F
2m=/planckover2pi12
2m2πσ=π/planckover2pi12σ
m.
kky
x
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CHAPTER 5. IDENTICAL PARTICLES 151
Problem 5.35
(a)
V=4
3πR3,soE=/planckover2pi12(3π2Nq)5/3
10π2m/parenleftbigg4
3πR3/parenrightbigg−2/3
=2/planckover2pi12
15πmR2/parenleftbigg9
4πNq/parenrightbigg5/3
.
(b)Imagine building up a sphere by layers. When it has reached mass m, and radius r, the work necessary
to bring in the next increment dmis:dW=−(Gm/r )dm. In terms of the mass density ρ,m=4
3πr3ρ,
anddm=4πr2drρ, where dris the resulting increase in radius. Thus:
dW=−G4
3πr3ρ4πr2ρdr
r=−16π2
3ρ2Gr4dr,
and the totalenergy of a sphere of radius Ris therefore
Egrav=−16π2
3ρ2G/integraldisplayR
0r4dr=−16π2ρ2R5
15G.Butρ=NM
4/3πR3,so
Egrav=−16π2R5
15G9N2M2
16π2R6=−3
5GN2M2
R.
(c)
Etot=A
R2−B
R,whereA≡2/planckover2pi12
15πm/parenleftbigg9
4πNq/parenrightbigg5/3
andB≡3
5GN2M2.
dEtot
dR=−2A
R3+B
R2=0⇒2A=BR, soR=2A
B=4/planckover2pi1
15πm/parenleftbigg9
4πNq/parenrightbigg5/35
3GN2M2.
R=/bracketleftBigg/parenleftbigg4
9π/parenrightbigg/parenleftbigg9π
4/parenrightbigg5/3/bracketrightBigg/parenleftbiggN5/3
N2/parenrightbigg/planckover2pi12
GmM2q5/3=/parenleftbigg9π
4/parenrightbigg2/3/planckover2pi12
GmM2q5/3
N1/3.
R=/parenleftbigg9π
4/parenrightbigg2/3(1.055×10−34J·s)2(1/2)5/3
(6.673×10−11Nm2/kg2)(9.109×10−31kg)(1.674×10−27kg)2N−1/3
=(7.58×1025m)N−1/3.
(d)Mass of sun: 1 .989×1030kg, so N=1.989×1030
1.674×10−27=1.188×1057;N−1/3=9.44×10−20.
R=( 7.58×1025)(9.44×10−20)m = 7.16×106m(slightly larger than the earth).
(e)
From Eq. 5.43: EF=/planckover2pi12
2m/parenleftbigg
3π2Nq
4/3πR3/parenrightbigg2/3
=/planckover2pi12
2mR2/parenleftbigg9π
4Nq/parenrightbigg2/3
.Numerically:
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152 CHAPTER 5. IDENTICAL PARTICLES
EF=(1.055×10−34J·s)2
2(9.109×10−31kg)(7.16×106m)2/bracketleftbigg9π
4(1.188×1057)1
2/bracketrightbigg2/3
=3.102×10−14J,
or, in electron volts: EF=3.102×10−14
1.602×10−19eV = 1.94×105eV.
Erest=mc2=5.11×105eV, so the Fermi energy (which is the energy of the most energetic electrons) is
comparable to the rest energy, so they are getting fairly relativistic.
Problem 5.36
(a)
dE=(/planckover2pi1ck)V
π2k2dk⇒Etot=/planckover2pi1cV
π2/integraldisplaykF
0k3dk=/planckover2pi1cV
4π2k4
F;kF=/parenleftbigg3π2Nq
V/parenrightbigg1/3
.
SoEtot=/planckover2pi1c
4π2(3π2Nq)4/3V−1/3.
(b)
V=4
3πR3⇒Edeg=/planckover2pi1c
4π2R(3π2Nq)4/3/parenleftbigg4π
3/parenrightbigg−1/3
=/planckover2pi1c
3πR/parenleftbigg9
4πNq/parenrightbigg4/3
.
Adding in the gravitational energy, from Problem 5.35(b),
Etot=A
R−B
R,whereA≡/planckover2pi1c
3π/parenleftbigg9
4πNq/parenrightbigg4/3
andB≡3
5GN2M2.dEtot
dR=−(A−B)
R2=0⇒A=B,
but there is no special value of Rfor which Etotis minimal. Critical value: A=B(Etot=0 )⇒
/planckover2pi1c
3π/parenleftbigg9
4πNq/parenrightbigg4/3
=3
5GN2M2,o r
Nc=15
16√
5π/parenleftbigg/planckover2pi1c
G/parenrightbigg3/2q2
M3=15
16√
5π/parenleftbigg1.055×10−34J·s×2.998×108m/s
6.673×10−11N·m2/kg2/parenrightbigg3/2(1/2)2
(1.674×10−27kg)3
=2.04×1057.(About twice the value for the sun—Problem 5.35(d).)
(c)Same as Problem 5.35(c), with m→Mandq→1, so multiply old answer by (2)5/3m/M:
R=25/3(9.109×10−31)
(1.674×10−27)(7.58×1025m)N−1/3=( 1.31×1023m)N−1/3.UsingN=1.188×1057,
R=( 1.31×1023m)(9.44×10−20)=12.4 km. To getEF, use Problem 5.35(e) with q= 1, the new R,
and the neutron mass in place of m:
EF=22/3/parenleftbigg7.16×106
1.24×104/parenrightbigg2/parenleftbigg9.11×10−31
1.67×10−27/parenrightbigg
(1.94×105eV) = 5 .60×107eV = 56.0 MeV.
The rest energy of a neutron is 940 MeV, so a neutron star is reasonably nonrelativistic.
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CHAPTER 5. IDENTICAL PARTICLES 153
Problem 5.37
(a)From Problem 4.38: En=(n+3
2)/planckover2pi1ω, withn=0,1,2,...;dn=1
2(n+ 1)(n+ 2).
From Eq. 5.103, n(FepsilonC)=e−(ρepsilono−µ)/kBT,s oNn=1
2(n+ 1)(n+2 )e(µ−3
2/planckover2pi1ω)/kBTe−n/planckover2pi1ω/kBT.
N=∞/summationdisplay
n=0Nn=1
2e(µ−3
2/planckover2pi1ω)/kBT∞/summationdisplay
n=0(n+ 1)(n+2 )xn,wherex≡e−/planckover2pi1ω/kBT.Now
1
1−x=∞/summationdisplay
n=0xn⇒x
1−x=∞/summationdisplay
n=0xn+1⇒d
dx/parenleftbiggx
1−x/parenrightbigg
=∞/summationdisplay
n=0(n+1 )xn⇒1
(1−x)2=∞/summationdisplay
n=0(n+1 )xn.
x2
(1−x)2=∞/summationdisplay
n=0(n+1 )xn+2,and henced
dx/parenleftbiggx2
(1−x)2/parenrightbigg
=∞/summationdisplay
n=0(n+ 1)(n+2 )xn+1=2x
(1−x)3.
∞/summationdisplay
n=0(n+ 1)(n+2 )xn=2
(1−x)3.SoN=eµ/kBTe−3
2/planckover2pi1ω/kBT 1
(1−e−/planckover2pi1ω/kBT)3.
eµ/kBT=N(1−e−/planckover2pi1ω/kBT)3e3
2/planckover2pi1ω/kBT;µ=kBT/bracketleftbig
lnN+ 3 ln(1−e−/planckover2pi1ω/kBT)+3
2/planckover2pi1ω/kBT/bracketrightbig
.
E=∞/summationdisplay
n=0NnEn=1
2/planckover2pi1ωe(µ−3
2/planckover2pi1ω)/kBT∞/summationdisplay
n=0(n+3/2)(n+ 1)(n+2 )xn.From above,
2x3/2
(1−x)3=∞/summationdisplay
n=0(n+ 1)(n+2 )xn+3/2⇒d
dx/parenleftbigg2x3/2
(1−x)3/parenrightbigg
=∞/summationdisplay
n=0(n+3/2)(n+ 1)(n+2 )xn+1/2,or
∞/summationdisplay
n=0(n+3/2)(n+ 1)(n+2 )xn=1
x1/2d
dx/parenleftbigg2x3/2
(1−x)3/parenrightbigg
=2
x1/2/bracketleftBigg
3
2x1/2
(1−x)3+3x3/2
(1−x)4/bracketrightBigg
=3(1 +x)
(1−x)4.
E=1
2/planckover2pi1ωe(µ−3
2/planckover2pi1ω)/kBT3(1 +e−/planckover2pi1ω/kBT)
(1−e−/planckover2pi1ω/kBT)4.Bute(µ−3
2/planckover2pi1ω)/kBT=N(1−e−/planckover2pi1ω/kBT)3,so
E=3
2N/planckover2pi1ω/parenleftbig
1+e−/planckover2pi1ω/kBT/parenrightbig
/parenleftbig
1−e−/planckover2pi1ω/kBT/parenrightbig.
(b)kBT< < /planckover2pi1ω(low temperature) ⇒e−/planckover2pi1ω/kBT≈0,soE≈3
2N/planckover2pi1ω(µ≈3
2/planckover2pi1ω). In this limit, allparticles
are in the ground state, E0=3
2/planckover2pi1ω.
(c)kBT> > /planckover2pi1ω(high temperature) ⇒e−/planckover2pi1ω/kBT≈1−(/planckover2pi1ω/kBT),soE≈3NkBT
(µ≈kBT[lnN+ 3 ln ( /planckover2pi1ω/kBT)]).The equipartition theorem says E=N#1
2kBT, where # is the number
of degrees of freedom for each particle. In this case # /2 = 3, or #=6 (3 kinetic, 3 potential, for each
particle—one of each for each direction in space).
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154 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Chapter 6
Time-Independent Perturbation
Theory
Problem 6.1
(a)
ψ0
n(x)=/radicalbigg
2
asin/parenleftBignπ
ax/parenrightBig
,soE1
n=/angbracketleftψ0
n|H/prime|ψ0
n/angbracketright=2
aα/integraldisplaya
0sin2/parenleftBignπ
ax/parenrightBig
δ/parenleftBig
x−a
2/parenrightBig
dx.
E1
n=2α
asin2/parenleftBignπ
aa
2/parenrightBig
=2α
asin2/parenleftBignπ
2/parenrightBig
=/braceleftbigg0,ifnis even,
2α/a, ifnis odd./bracerightbigg
For even nthe wave function is zero at the location of the perturbation ( x=a/2), so it never “feels” H/prime.
(b)Heren= 1, so we need
/angbracketleftψ0
m|H/prime|ψ0
1/angbracketright=2α
a/integraldisplay
sin/parenleftBigmπ
ax/parenrightBig
δ/parenleftBig
x−a
2/parenrightBig
sin/parenleftBigπ
ax/parenrightBig
dx=2α
asin/parenleftBigmπ
2/parenrightBig
sin/parenleftBigπ
2/parenrightBig
=2α
asin/parenleftBigmπ
2/parenrightBig
.
This is zero for even m, so the first three nonzero terms will be m=3 ,m= 5, and m= 7. Meanwhile,
E0
1−E0
m=π2/planckover2pi12
2ma2(1−m2), so
ψ1
1=/summationdisplay
m=3,5,7,...(2α/a) sin(mπ/2)
E0
1−E0mψ0
m=2α
a2ma2
π2/planckover2pi12/bracketleftbigg−1
1−9ψ0
3+1
1−25ψ0
5+−1
1−49ψ0
7+.../bracketrightbigg
=4maα
π2/planckover2pi12/radicalbigg
2
a/bracketleftbigg1
8sin/parenleftbigg3π
ax/parenrightbigg
−1
24sin/parenleftbigg5π
ax/parenrightbigg
+1
48sin/parenleftbigg7π
ax/parenrightbigg
+.../bracketrightbigg
=mα
π2/planckover2pi12/radicalbigga
2/bracketleftbigg
sin/parenleftbigg3π
ax/parenrightbigg
−1
3sin/parenleftbigg5π
ax/parenrightbigg
+1
6sin/parenleftbigg7π
ax/parenrightbigg
+.../bracketrightbigg
.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 155
Problem 6.2
(a)En=(n+1
2)/planckover2pi1ω/prime, where ω/prime≡/radicalbig
k(1 +FepsilonC)/m=ω√1+FepsilonC=ω(1 +1
2FepsilonC−1
8FepsilonC2+1
16FepsilonC3···), so
En=(n+1
2)/planckover2pi1ω√1+FepsilonC=(n+1
2)/planckover2pi1ω(1 +1
2FepsilonC−1
8FepsilonC2+···).
(b)H/prime=1
2k/primex2−1
2kx2=1
2kx2(1 +FepsilonC−1) =FepsilonC(1
2kx2)=FepsilonCV, where Vis the unperturbed potential energy. So
E1
n=/angbracketleftψ0
n|H/prime|ψ0
n/angbracketright=FepsilonC/angbracketleftn|V|n/angbracketright, with/angbracketleftn|V|n/angbracketrightthe expectation value of the (unperturbed) potential energy
in thenthunperturbed state. This is most easily obtained from the virial theorem (Problem 3.31), but it
can also be derived algebraically. In this case the virial theorem says /angbracketleftT/angbracketright=/angbracketleftV/angbracketright. But/angbracketleftT/angbracketright+/angbracketleftV/angbracketright=En.S o
/angbracketleftV/angbracketright=1
2E0
n=1
2(n+1
2)/planckover2pi1ω;E1
n=ρepsilono
2(n+1
2)/planckover2pi1ω,which is precisely the FepsilonC1term in the power series from
part (a).
Problem 6.3
(a)In terms of the one-particle states (Eq. 2.28) and energies (Eq. 2.27):
Ground state :ψ0
1(x1,x2)=ψ1(x1)ψ1(x2)=2
asin/parenleftBigπx1
a/parenrightBig
sin/parenleftBigπx2
a/parenrightBig
;E0
1=2E1=π2/planckover2pi12
ma2.
First excited state :ψ0
2(x1,x2)=1√
2[ψ1(x1)ψ2(x2)+ψ2(x1)ψ1(x2)]
=√
2
a/bracketleftbigg
sin/parenleftBigπx1
a/parenrightBig
sin/parenleftbigg2πx2
a/parenrightbigg
+ sin/parenleftbigg2πx1
a/parenrightbigg
sin/parenleftBigπx2
a/parenrightBig/bracketrightbigg
;E0
2=E1+E2=5
2π2/planckover2pi12
ma2.
(b)
E1
1=/angbracketleftψ0
1|H/prime|ψ0
1/angbracketright=(−aV0)/parenleftbigg2
a/parenrightbigg2/integraldisplaya
0/integraldisplaya
0sin2/parenleftBigπx1
a/parenrightBig
sin2/parenleftBigπx2
a/parenrightBig
δ(x2−x2)dx1dx2
=−4V0
a/integraldisplaya
0sin4/parenleftBigπx
a/parenrightBig
dx=−4V0
aa
π/integraldisplayπ
0sin4ydy=−4V0
π·3π
8=−3
2V0.
E1
2=/angbracketleftψ0
2|H/prime|ψ0
2/angbracketright
=(−aV0)/parenleftbigg2
a2/parenrightbigg/integraldisplay/integraldisplaya
0/bracketleftbigg
sin/parenleftBigπx1
a/parenrightBig
sin/parenleftbigg2πx2
a/parenrightbigg
+ sin/parenleftbigg2πx1
a/parenrightbigg
sin/parenleftBigπx2
a/parenrightBig/bracketrightbigg2
δ(x1−x2)dx1dx2
=−2V0
a/integraldisplaya
0/bracketleftbigg
sin/parenleftBigπx
a/parenrightBig
sin/parenleftbigg2πx
a/parenrightbigg
+ sin/parenleftbigg2πx
a/parenrightbigg
sin/parenleftBigπx
a/parenrightBig/bracketrightbigg2
dx
=−8V0
a/integraldisplaya
0sin2/parenleftBigπx
a/parenrightBig
sin2/parenleftbigg2πx
a/parenrightbigg
dx=−8V0
a·a
π/integraldisplayπ
0sin2ysin2(2y)dy
=−8V0
π·4/integraldisplayπ
0sin2ysin2ycos2ydy=−32V0
π/integraldisplayπ
0(sin4y−sin6y)dy
=−32V0
π/parenleftbigg3π
8−5π
16/parenrightbigg
=−2V0.
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156 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.4
(a)
/angbracketleftψ0
m|H|ψ0
n/angbracketright=2
aα/integraldisplaya
0sin/parenleftBigmπ
ax/parenrightBig
δ/parenleftBig
x−a
2/parenrightBig
sin/parenleftBignπ
ax/parenrightBig
dx=2α
asin/parenleftBigmπ
2/parenrightBig
sin/parenleftBignπ
2/parenrightBig
,
which is zero unless both mandnare odd—in which case it is ±2α/a. So Eq. 6.15 says
E2
n=/summationdisplay
m/negationslash=n,odd/parenleftbigg2α
a/parenrightbigg21
(E0n−E0m).But Eq. 2.27 says E0
n=π2/planckover2pi12
2ma2n2,so
E2
n=
0, ifnis even;
2m/parenleftbigg2α
π/planckover2pi1/parenrightbigg2/summationdisplay
m/negationslash=n,odd1
(n2−m2),ifnis odd.
To sum the series, note that1
(n2−m2)=1
2n/parenleftbigg1
m+n−1
m−n/parenrightbigg
. Thus,
forn=1:/summationdisplay
=1
2/summationdisplay
3,5,7,.../parenleftbigg1
m+1−1
m−1/parenrightbigg
=1
2/parenleftbigg1
4+1
6+1
8+···−1
2−1
4−1
6−1
8···/parenrightbigg
=1
2/parenleftbigg
−1
2/parenrightbigg
=−1
4;
forn=3:/summationdisplay
=1
6/summationdisplay
1,5,7,.../parenleftbigg1
m+3−1
m−3/parenrightbigg
=1
6/parenleftbigg1
4+1
8+1
10+···+1
2−1
2−1
4−1
6−1
8−1
10···/parenrightbigg
=1
6/parenleftbigg
−1
6/parenrightbigg
=−1
36.
In general, there is perfect cancellation except for the “missing” term 1 /2nin the first sum, so the total
is1
2n/parenleftbigg
−1
2n/parenrightbigg
=−1
(2n)2. Therefore: E2
n=/braceleftbigg0, ifnis even;
−2m(α/π/planckover2pi1n)2,ifnis odd.
(b)
H/prime=1
2FepsilonCkx2;/angbracketleftψ0
m|H/prime|ψ0
n/angbracketright=1
2FepsilonCk/angbracketleftm|x2|n/angbracketright.Using Eqs. 2.66 and 2.69:
/angbracketleftm|x2|n/angbracketright=/planckover2pi1
2mω/angbracketleftm|(a2
++a+a−+a−a++a2
−)|n/angbracketright
=/planckover2pi1
2mω/bracketleftBig/radicalbig
(n+ 1)(n+2 )/angbracketleftm|n+2/angbracketright+n/angbracketleftm|n/angbracketright+(n+1 )/angbracketleftm|n/angbracketright+/radicalbig
n(n−1)/angbracketleftm|n−2/angbracketright/bracketrightBig
.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 157
So, form/negationslash=n,/angbracketleftψ0
m|H/prime|ψ0
n/angbracketright=/parenleftbigg1
2kFepsilonC/parenrightbigg/parenleftbigg/planckover2pi1
2mω/parenrightbigg/bracketleftBig/radicalbig
(n+ 1)(n+2 )δm,n+2+/radicalbig
n(n−1)δm,n−2/bracketrightBig
.
E2
n=/parenleftbiggFepsilonC/planckover2pi1ω
4/parenrightbigg2/summationdisplay
m/negationslash=n/bracketleftBig/radicalbig
(n+ 1)(n+2 )δm,n+2+/radicalbig
n(n−1)δm,n−2/bracketrightBig2
(n+1
2)/planckover2pi1ω−(m+1
2)/planckover2pi1ω
=FepsilonC2/planckover2pi1ω
16/summationdisplay
m/negationslash=n[(n+ 1)(n+2 )δm,n+2+n(n−1)δm,n−2]
(n−m)
=FepsilonC2/planckover2pi1ω
16/bracketleftbigg(n+ 1)(n+2 )
n−(n+2 )+n(n−1)
n−(n−2)/bracketrightbigg
=FepsilonC2/planckover2pi1ω
16/bracketleftbigg
−1
2(n+ 1)(n+2 )+1
2n(n−1)/bracketrightbigg
=FepsilonC2/planckover2pi1ω
32/parenleftbig
−n2−3n−2+n2−n/parenrightbig
=FepsilonC2/planckover2pi1ω
32(−4n−2) =−FepsilonC21
8/planckover2pi1ω/parenleftbigg
n+1
2/parenrightbigg
(which agrees with the FepsilonC2term in the exact solution—Problem 6.2(a)).
Problem 6.5
(a)
E1
n=/angbracketleftψ0
n|H/prime|ψ0
n/angbracketright=−qE/angbracketleftn|x|n/angbracketright=0(Problem 2.12).
From Eq. 6.15 and Problem 3.33: E2
n=(qE)2/summationdisplay
m/negationslash=n|/angbracketleftm|x|n/angbracketright|2
(n−m)/planckover2pi1ω
=(qE)2
/planckover2pi1ω/planckover2pi1
2mω/summationdisplay
m/negationslash=n[√n+1δm,n+1+√n,δm,n−1]2
(n−m)=(qE)2
2mω2/summationdisplay
m/negationslash=n[(n+1 )δm,n+1+nδm,n−1]
(n−m)
=(qE)2
2mω2/bracketleftbigg(n+1 )
n−(n+1 )+n
n−(n−1)/bracketrightbigg
=(qE)2
2mω2[−(n+1 )+n]=−(qE)2
2mω2.
(b)−/planckover2pi12
2md2ψ
dx2+/parenleftbigg1
2mω2x2−qEx/parenrightbigg
ψ=Eψ. With the suggested change of variables,
/parenleftbigg1
2mω2x2−qEx/parenrightbigg
=1
2mω2/bracketleftbigg
x/prime+/parenleftbiggqE
mω2/parenrightbigg/bracketrightbigg2
−qE/bracketleftbigg
x/prime+/parenleftbiggqE
mω2/parenrightbigg/bracketrightbigg
=1
2mω2x/prime2+mω2x/primeqE
mω2+1
2mω2(qE)2
m2ω4−qEx/prime−(qE)2
mω2=1
2mω2x/prime2−1
2(qE)2
mω2.
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158 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
So the Schr¨ odinger equation says
−/planckover2pi12
2md2ψ
dx/prime2+1
2mω2x/prime2ψ=/bracketleftbigg
E+1
2(qE)2
mω2/bracketrightbigg
ψ,
which is the Schr¨ odinger equation for a simple harmonic oscillator, in the variable x/prime. The constant on
the right must therefore be ( n+1
2)/planckover2pi1ω, and we conclude that
En=(n+1
2)/planckover2pi1ω−1
2(qE)2
mω2.
The subtracted term is exactly what we got in part (a) using perturbation theory. Evidently all the higher
corrections (like the first-order correction) are zero, in this case.
Problem 6.6
(a)
/angbracketleftψ0
+|ψ0
−/angbracketright=/angbracketleft(α+ψ0
a+β+ψ0
b)|(α−ψ0
a+β−ψ0
b)/angbracketright
=α∗
+α−/angbracketleftψ0
a|ψ0
a/angbracketright+α∗
+β−/angbracketleftψ0
a|ψ0
b/angbracketright+β∗
+α−/angbracketleftψ0
b|ψ0
a/angbracketright+β∗
+β−/angbracketleftψ0
b|ψ0
b/angbracketright
=α∗
+α−+β∗
+β−.But Eq. 6.22 ⇒β±=α±(E1
±−Waa)/Wab,so
/angbracketleftψ0
+|ψ0
−/angbracketright=α∗
+α−/bracketleftbigg
1+(E1
+−Waa)(E1
−−Waa)
Wab∗Wab/bracketrightbigg
=α∗
+α−
|Wab|2/bracketleftbig
|Wab|2+(E1
+−Waa)(E1
−−Waa)/bracketrightbig
.
The term in square brackets is:
[]=E1
+E1
−−Waa(E1
++E1
−)+|Wab|2+W2
aa.But Eq. 6.27 ⇒E1
±=1
2[(Waa+Wbb)±√], where√is
shorthand for the square root term. So E1
++E1
−=Waa+Wbb, and
E1
+E1
−=1
4/bracketleftbig
(Waa+Wbb)2−(√)2/bracketrightbig
=1
4/bracketleftbig
(Waa+Wbb)2−(Waa−Wbb)2−4|Wab|2/bracketrightbig
=WaaWbb−|Wab|2.
Thus [ ] = WaaWbb−|Wab|2−Waa(Waa+Wbb)+|Wab|2+W2
aa=0,so/angbracketleftψ0
+|ψ0
−/angbracketright=0.QED
(b)
/angbracketleftψ0
+|H/prime|ψ0
−/angbracketright=α∗
+α−/angbracketleftψ0
a|H/prime|ψ0
a/angbracketright+α∗
+β−/angbracketleftψ0
a|H/prime|ψ0
b/angbracketright+β∗
+α−/angbracketleftψ0
b|H/prime|ψ0
a/angbracketright+β∗
+β−/angbracketleftψ0
b|H/prime|ψ0
b/angbracketright
=α∗
+α−Waa+α∗
+β−Wab+β∗
+α−Wba+β∗
+β−Wbb
=α∗
+α−/bracketleftbigg
Waa+Wab(E1
−−Waa)
Wab+Wba(E1
+−Waa)
W∗
ab+Wbb(E1
+−Waa)
W∗
ab(E1
−−Waa)
Wab/bracketrightbigg
=α∗
+α−/bracketleftbigg
Waa+E1
−−Waa+E1
+−Waa+Wbb(E1
+−Waa)(E1
−−Waa)
|Wab|2/bracketrightbigg
.
But we know from (a) that(E1
+−Waa)(E1
−−Waa)
|Wab|2=−1, so
/angbracketleftψ0
+|H/prime|ψ0
−/angbracketright=α∗
+α−[E1
−+E1
+−Waa−Wbb]=0.QED
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 159
(c)
/angbracketleftψ0
±|H/prime|ψ0
±/angbracketright=α∗
±α±/angbracketleftψ0
a|H/prime|ψ0
a/angbracketright+α∗
±β±/angbracketleftψ0
a|H/prime|ψ0
b/angbracketright+β∗
±α±/angbracketleftψ0
b|H/prime|ψ0
a/angbracketright+β∗
±β±/angbracketleftψ0
b|H/prime|ψ0
b/angbracketright
=|α±|2/bracketleftbigg
Waa+Wab(E1
±−Waa)
Wab/bracketrightbigg
+|β±|2/bracketleftbigg
Wba(E1
±−Wbb)
Wba+Wbb/bracketrightbigg
(this time I used Eq. 6.24 to express αin terms of β, in the third term).
∴/angbracketleftψ0
±|H/prime|ψ0
±/angbracketright=|α±|2(E1
±)+|β±|2(E1
±)=/parenleftbig
|α±|2+|β±|2/parenrightbig
E1
±=E1
±.QED
Problem 6.7
(a)See Problem 2.46.
(b)Witha→n,b→−n, we have:
Waa=Wbb=−V0
L/integraldisplayL/2
−L/2e−x2/a2dx≈−V0
L/integraldisplay∞
−∞e−x2/a2dx=−V0
La√π.
Wab=−V0
L/integraldisplayL/2
−L/2e−x2/a2e−4πnix/Ldx≈−V0
L/integraldisplay∞
−∞e−(x2/a2+4πnix/L )dx=−V0
La√πe−(2πna/L )2.
(We did this integral in Problem 2.22.) In this case Waa=Wbb, andWabis real, so Eq. 6.26 ⇒
E1
±=Waa±|Wab|,orE1
±=−√πV0a
L/parenleftBig
1∓e−(2πna/L )2/parenrightBig
.
(c)Equation 6.22 ⇒β=α(E1
−−Waa)
Wab=α/bracketleftBigg
±√π(V0a/L)e−(2πna/L )2
−√π(V0a/L)e−(2πna/L )2/bracketrightBigg
=∓α.Evidently, the “good” linear
combinations are:
ψ+=αψn−αψ−n=1√
21√
L/bracketleftBig
ei2πnx/L−e−i2πnx/L/bracketrightBig
=i/radicalbigg
2
Lsin/parenleftbigg2πnx
L/parenrightbigg
and
ψ−=αψn+αψ−n=/radicalbigg
2
Lcos/parenleftbigg2πnx
L/parenrightbigg
.Using Eq .6.9,we have :
E1
+=/angbracketleftψ+|H/prime|ψ+/angbracketright=2
L(−V0)/integraldisplayL/2
−L/2e−x2/a2sin2/parenleftbigg2πnx
L/parenrightbigg
dx,
E1
−=/angbracketleftψ−|H/prime|ψ−/angbracketright=2
L(−V0)/integraldisplayL/2
−L/2e−x2/a2cos2/parenleftbigg2πnx
L/parenrightbigg
dx.
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160 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
But sin2θ=( 1−cos 2θ)/2, and cos2θ= (1 + cos 2 θ)/2, so
E1
±≈−V0
L/integraldisplay∞
−∞e−x2/a2/bracketleftbigg
1∓cos/parenleftbigg4πnx
L/parenrightbigg/bracketrightbigg
dx=−V0
L/bracketleftbigg/integraldisplay∞
−∞e−x2/a2dx∓/integraldisplay∞
−∞e−x2/a2cos/parenleftbigg4πnx
L/parenrightbigg
dx/bracketrightbigg
=−V0
L/bracketleftBig√πa∓a√πe−(2πna/L )2/bracketrightBig
=−√πV0a
L/bracketleftBig
1∓e−(2πna/L )2/bracketrightBig
,same as (b) .
(d)Af(x)=f(−x) (the parity operator). The eigenstates are evenfunctions (with eigenvalue +1) and odd
functions (with eigenvalue −1). The linear combinations we found in (c) are precisely the odd and even
linear combinations of ψnandψ−n.
Problem 6.8
Ground state is nondegenerate; Eqs. 6.9 and 6.31 ⇒
E1=/parenleftbigg2
a/parenrightbigg3
a3V0/integraldisplay/integraldisplay/integraldisplaya
0sin2/parenleftBigπ
ax/parenrightBig
sin2/parenleftBigπ
ay/parenrightBig
sin2/parenleftBigπ
az/parenrightBig
δ(x−a
4)δ(y−a
2)δ(z−3a
4)dxdydz
=8V0sin2/parenleftBigπ
4/parenrightBig
sin2/parenleftBigπ
2/parenrightBig
sin2/parenleftbigg3π
4/parenrightbigg
=8V0/parenleftbigg1
2/parenrightbigg
(1)/parenleftbigg1
2/parenrightbigg
=2V0.
First excited states (Eq. 6.34):
Waa=8V0/integraldisplay/integraldisplay/integraldisplay
sin2/parenleftBigπ
ax/parenrightBig
sin2/parenleftBigπ
ay/parenrightBig
sin2/parenleftbigg2π
az/parenrightbigg
δ(x−a
4)δ(y−a
2)δ(z−3a
4)dxdydz
=8V0/parenleftbigg1
2/parenrightbigg
(1)(1) = 4 V0.
Wbb=8V0/integraldisplay/integraldisplay/integraldisplay
sin2/parenleftBigπ
ax/parenrightBig
sin2/parenleftbigg2π
ay/parenrightbigg
sin2/parenleftBigπ
az/parenrightBig
δ(x−a
4)δ(y−a
2)δ(z−3a
4)dxdydz
=8V0/parenleftbigg1
2/parenrightbigg
(0)/parenleftbigg1
2/parenrightbigg
=0.
Wcc=8V0/integraldisplay/integraldisplay/integraldisplay
sin2/parenleftbigg2π
ax/parenrightbigg
sin2/parenleftBigπ
ay/parenrightBig
sin2/parenleftBigπ
az/parenrightBig
δ(x−a
4)δ(y−a
2)δ(z−3a
4)dxdydz
=8V0(1)(1)/parenleftbigg1
2/parenrightbigg
=4V0.
Wab=8V0sin2/parenleftBigπ
4/parenrightBig
sin/parenleftBigπ
2/parenrightBig
sin(π) sin/parenleftbigg3π
2/parenrightbigg
sin/parenleftbigg3π
4/parenrightbigg
=0.
Wac=8V0sin/parenleftBigπ
4/parenrightBig
sin/parenleftBigπ
2/parenrightBig
sin2/parenleftBigπ
2/parenrightBig
sin/parenleftbigg3π
2/parenrightbigg
sin/parenleftbigg3π
4/parenrightbigg
=8V0/parenleftbigg1√
2/parenrightbigg
(1)(1)(−1)/parenleftbigg1√
2/parenrightbigg
=−4V0.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 161
Wbc=8V0sin/parenleftBigπ
4/parenrightBig
sin/parenleftBigπ
2/parenrightBig
sin(π) sin/parenleftBigπ
2/parenrightBig
sin/parenleftbigg3π
4/parenrightbigg
=0.
W=4V0
10−1
000
−10 1
=4V0D; det( D−λ)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(1−λ)0−1
0−λ0
−10 ( 1−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λ(1−λ)
2+λ=0⇒
λ=0,or (1−λ)2=1⇒1−λ=±1⇒λ=0 o r λ=2.
So the first-order corrections to the energies are 0, 0, 8V0.
Problem 6.9
(a)χ1=
1
00
,
eigenvalue V0;χ2=
0
10
,
eigenvalue V0;χ3=
0
01
,
eigenvalue 2V0.
(b)Characteristic equation: det( H−λ)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle[V
0(1−FepsilonC)−λ]0 0
0[ V0−λ]FepsilonCV0
0 FepsilonCV0[2V0−λ]/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0 ;
[V
0(1−FepsilonC)−λ][(V0−λ)(2V0−λ)−(FepsilonCV0)2]=0⇒λ1=V0(1−FepsilonC).
(V0−λ)(2V0−λ)−(FepsilonCV0)2=0⇒λ2−3V0λ+( 2V2
0−FepsilonC2V2
0)=0⇒
λ=3V0±/radicalbig
9V2
0−4(2V2
0−FepsilonC2V2
0)
2=V0
2/bracketleftBig
3±/radicalbig
1+4FepsilonC2/bracketrightBig
≈V0
2/bracketleftbig
3±( 1+2FepsilonC2)/bracketrightbig
.
λ2=V0
2/parenleftBig
3−/radicalbig
1+4FepsilonC2/parenrightBig
≈V0(1−FepsilonC2);λ3=V0
2/parenleftBig
3+/radicalbig
1+4FepsilonC2/parenrightBig
≈V0(2 +FepsilonC2).
(c)
H/prime=FepsilonCV0
−100
00 101 0
;E
1
3=/angbracketleftχ3|H/prime|χ3/angbracketright=FepsilonCV0/parenleftbig001/parenrightbig
−100
00 101 0
0
01
=FepsilonCV
0/parenleftbig001/parenrightbig
0
10
=
0(no first-order correction).
E2
3=/summationdisplay
m=1,2|/angbracketleftχm|H/prime|χ3/angbracketright|2
E0
3−E0m;/angbracketleftχ1|H/prime|χ3/angbracketright=FepsilonCV0/parenleftbig100/parenrightbig
−100
00 101 0
0
01
=FepsilonCV
0/parenleftbig100/parenrightbig
0
10
=0,
/angbracketleftχ
2|H/prime|χ3/angbracketright=FepsilonCV0/parenleftbig010/parenrightbig
0
01
=FepsilonCV
0.
E0
3−E0
2=2V0−V0=V0.S o E2
3=(FepsilonCV0)2/V0=FepsilonC2V0.Through second-order, then,
E3=E0
3+E1
3+E2
3=2V0+0+FepsilonC2V0=V0(2 +FepsilonC2) (same as we got for λ3in (b)).
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162 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
(d)
Waa=/angbracketleftχ1|H/prime|χ1/angbracketright=FepsilonCV0/parenleftbig
100/parenrightbig
−100
00 101 0
1
00
=FepsilonCV
0/parenleftbig
100/parenrightbig
−1
00
=−FepsilonCV
0.
Wbb=/angbracketleftχ2|H/prime|χ2/angbracketright=FepsilonCV0/parenleftbig
010/parenrightbig
−100
00 101 0
0
10
=FepsilonCV
0/parenleftbig
010/parenrightbig
0
01
=0.
W
ab=/angbracketleftχ1|H/prime|χ2/angbracketright=FepsilonCV0/parenleftbig100/parenrightbig
−100
00 101 0
0
10
=FepsilonCV
0/parenleftbig100/parenrightbig
0
01
=0.
Plug the expressions for W
aa,Wbb, andWabinto Eq. 6.27:
E1
±=1
2/bracketleftbigg
−FepsilonCV0+0±/radicalBig
FepsilonC2V2
0+0/bracketrightbigg
=1
2(−FepsilonCV0±FepsilonCV0)={0,−FepsilonCV0}.
To first-order, then, E1=V0−FepsilonCV0,E 2=V0,and these are consistent (to first order in FepsilonC) with what
we got in (b).
Problem 6.10
Given a set of orthonornal states {ψ0
j}that are degenerate eigenfunctions of the unperturbed Hamiltonian:
Hψ0
j=E0ψ0
j,/angbracketleftψ0
j|ψ0
l/angbracketright=δjl,
construct the general linear combination,
ψ0=n/summationdisplay
j=1αjψ0
j.
It too is an eigenfunction of the unperturbed Hamiltonian, with the same eigenvalue:
H0ψ0=n/summationdisplay
j=1αjH0ψ0
j=E0n/summationdisplay
j=1αjψ0
j=E0ψ0.
We want to solve the Schr¨ odinger equation Hψ=Eψfor the perturbed Hamiltonian H=H0+λH/prime.
Expand the eigenvalues and eigenfunctions as power series in λ:
E=E0+λE1+λ2E2+..., ψ =ψ0+λψ1+λ2ψ2+....
Plug these into the Schr¨ odinger equation and collect like powers:
(H0+λH/prime)(ψ0+λψ1+λ2ψ2+...)=(E0+λE1+λ2E2+...)(ψ0+λψ1+λ2ψ2+...)⇒
H0ψ0+λ(H0ψ1+H/primeψ0)+...=E0ψ0+λ(E0ψ1+E1ψ0)+...
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 163
The zeroth-order terms cancel; to first order
H0ψ1+H/primeψ0=E0ψ1+E1ψ0.
Take the inner product with ψ0
j:
/angbracketleftψ0
j|H0ψ1/angbracketright+/angbracketleftψ0
j|H/primeψ0/angbracketright=E0/angbracketleftψ0
j|ψ1/angbracketright+E1/angbracketleftψ0
j|ψ0/angbracketright.
But/angbracketleftψ0
j|H0ψ1/angbracketright=/angbracketleftH0ψ0
j|ψ1/angbracketright=E0/angbracketleftψ0
j|ψ1/angbracketright,so the first terms cancel, leaving
/angbracketleftψ0
j|H/primeψ0/angbracketright=E1/angbracketleftψ0
j|ψ0/angbracketright.
Now use ψ0=n/summationdisplay
l=1αlψ0
l,and exploit the orthonormality of {ψ0
l}:
n/summationdisplay
l=1αl/angbracketleftψ0
j|H/prime|ψ0
l/angbracketright=E1n/summationdisplay
l=1αl/angbracketleftψ0
j|ψ0
l/angbracketright=E1αj,
or, defining
Wjl≡/angbracketleftψ0
j|H/prime|ψ0
l/angbracketright,n/summationdisplay
l=1Wjlαl=E1αl.
This (the generalization of Eq. 6.22 for the case of n-fold degeneracy) is the eigenvalue equation for the matrix
W(whose jlthelement, in the {ψ0
j}basis, is Wjl);E1is the eigenvalue, and the eigenvector (in the {ψ0
j}basis)
isχj=αj.Conclusion: The first-order corrections to the energy are the eigenvalues of W. QED
Problem 6.11
(a)From Eq. 4.70: En=−/bracketleftBigg
m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2/bracketrightBigg
1
n2=−1
2mc2/parenleftbigg1
/planckover2pi1ce2
4πFepsilonC0/parenrightbigg21
n2=−α2mc2
2n2.
(b)I have found a wonderful solution—unfortunately, there isn’t enough room on this page for the proof.
Problem 6.12
Equation 4.191 ⇒/angbracketleftV/angbracketright=2En, for hydrogen. V=−e2
4πFepsilonC01
r;En=−/bracketleftBigg
m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2/bracketrightBigg
1
n2.S o
−e2
4πFepsilonC0/angbracketleftbigg1
r/angbracketrightbigg
=−2/bracketleftBigg
m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2/bracketrightBigg
1
n2⇒/angbracketleftbigg1
r/angbracketrightbigg
=/parenleftbiggme2
4πFepsilonC0/planckover2pi12/parenrightbigg1
n2=1
an2(Eq. 4.72). QED
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164 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.13
In Problem 4.43 we found (for n=3 ,l=2 ,m= 1) that/angbracketleftrs/angbracketright=(s+ 6)!
6!/parenleftbigg3a
2/parenrightbiggs
.
s=0:/angbracketleft1/angbracketright=6!
6!(1) = 1(of course) ./check
s=−1:/angbracketleftbigg1
r/angbracketrightbigg
=5!
6!/parenleftbigg3a
2/parenrightbigg−1
=1
6·2
3a=1
9a/parenleftbigg
Eq. 6.55 says1
32a=1
9a/parenrightbigg
./check
s=−2:/angbracketleftbigg1
r2/angbracketrightbigg
=4!
6!/parenleftbigg3a
2/parenrightbigg−2
=1
6·5·4
9a2=2
135a2/parenleftbigg
Eq. 6.56 says1
(5/2)·27·a2=2
135a2/parenrightbigg
./check
s=−3:/angbracketleftbigg1
r3/angbracketrightbigg
=3!
6!/parenleftbigg3a
2/parenrightbigg−3
=1
6·5·4·8
27a3=1
405a3/parenleftbigg
Eq. 6.64 says1
2(5/2)3·27·a3=1
405a3/parenrightbigg
./check
Fors=−7 (or smaller) the integral does not converge: /angbracketleft1/r7/angbracketright=∞in this state; this is reflected in the fact
that (−1)! =∞.
Problem 6.14
Equation 6 .53⇒E1
r=−1
2mc2/bracketleftbig
E2−2E/angbracketleftV/angbracketright+/angbracketleftV2/angbracketright/bracketrightbig
.HereE=(n+1
2)/planckover2pi1ω, V =1
2mω2x2⇒
E1
r=−1
2mc2/bracketleftBigg/parenleftbigg
n+1
2/parenrightbigg2
/planckover2pi12ω2−2/parenleftbigg
n+1
2/parenrightbigg
/planckover2pi1ω1
2mω2/angbracketleftx2/angbracketright+1
4m2ω4/angbracketleftx4/angbracketright/bracketrightBigg
.
But Problem 2.12 ⇒/angbracketleftx2/angbracketright=(n+1
2)/planckover2pi1
mω,so
E1
r=−1
2mc2/bracketleftBigg/parenleftbigg
n+1
2/parenrightbigg2
/planckover2pi12ω2−/parenleftbigg
n+1
2/parenrightbigg2
/planckover2pi12ω2+1
4m2ω4/angbracketleftx4/angbracketright/bracketrightBigg
=−mω4
8c2/angbracketleftx4/angbracketright.
From Eq. 2.69: x4=/planckover2pi12
4m2ω2/parenleftbig
a2
++a+a−+a−a++a2
−/parenrightbig/parenleftbig
a2
++a+a−+a−a++a2
−/parenrightbig
,
/angbracketleftx4/angbracketright=/planckover2pi12
4m2ω2/angbracketleftn|/parenleftbig
a2
+a2−+a+a−a+a−+a+a−a−a++a−a+a+a−+a−a+a−a++a2
−a2+/parenrightbig
|n/angbracketright.
(Note that only terms with equal numbers of raising and lowering operators will survive). Using Eq. 2.66,
/angbracketleftx4/angbracketright=/planckover2pi12
4m2ω2/angbracketleftn|/bracketleftBig
a2
+/parenleftBig/radicalbig
n(n−1)|n−2/angbracketright/parenrightBig
+a+a−(n|n/angbracketright)+a+a−/parenleftbig
(n+1 )|n/angbracketright/parenrightbig
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 165
+a−a+(n|n/angbracketright)+a−a+/parenleftbig
(n+1 )|n/angbracketright/parenrightbig
+a2
−/parenleftBig/radicalbig
(n+ 1)(n+2 )|n+2/angbracketright/parenrightBig/bracketrightBig
=/planckover2pi12
4m2ω2/angbracketleftn|/bracketleftBig/radicalbig
n(n−1)/parenleftBig/radicalbig
n(n−1)|n/angbracketright/parenrightBig
+n(n|n/angbracketright)+(n+1 )(n|n/angbracketright)
+n/parenleftbig
(n+1 )|n/angbracketright/parenrightbig
+(n+1 )/parenleftbig
(n+1 )|n/angbracketright/parenrightbig
+/radicalbig
(n+ 1)(n+2 )/parenleftBig/radicalbig
(n+ 1)(n+2 )|n/angbracketright/parenrightBig/bracketrightBig
=/planckover2pi12
4m2ω2/bracketleftbig
n(n−1) +n2+(n+1 )n+n(n+1 )+( n+1 )2+(n+ 1)(n+2 )/bracketrightbig
=/parenleftbigg/planckover2pi1
2mω/parenrightbigg2
(n2−n+n2+n2+n+n2+n+n2+2n+1+n2+3n+2 )=/parenleftbigg/planckover2pi1
2mω/parenrightbigg2
(6n2+6n+3 ).
E1
r=−mω4
8c2·/planckover2pi12
4m2ω2·3(3n2+2n+1 )=−3
32/parenleftbigg/planckover2pi12ω2
mc2/parenrightbigg
(2n2+2n+ 1).
Problem 6.15
Quoting the Laplacian in spherical coordinates (Eq. 4.13), we have, for states with no dependence on θorφ:
p2=−/planckover2pi12∇2=−/planckover2pi12
r2d
dr/parenleftbigg
r2d
dr/parenrightbigg
.
Question: Is it Hermitian?
Using integration by parts (twice), and test functions f(r) andg(r):
/angbracketleftf|p2g/angbracketright=−/planckover2pi12/integraldisplay∞
0f1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg
4πr2dr=−4π/planckover2pi12/integraldisplay∞
0fd
dr/parenleftbigg
r2dg
dr/parenrightbigg
dr
=−4π/planckover2pi12/braceleftbigg
r2fdg
dr/vextendsingle/vextendsingle/vextendsingle∞
0−/integraldisplay∞
0r2df
drdg
drdr/bracerightbigg
=−4π/planckover2pi12/braceleftbigg
r2fdg
dr/vextendsingle/vextendsingle/vextendsingle∞
0−r2gdf
dr/vextendsingle/vextendsingle/vextendsingle∞
0+/integraldisplay∞
0d
dr/parenleftbigg
r2df
dr/parenrightbigg
gdr/bracerightbigg
=−4π/planckover2pi12/parenleftbigg
r2fdg
dr−r2gdf
dr/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0+/angbracketleftp2f|g/angbracketright.
The boundary term at infinity vanishes for functions f(r) andg(r) that go to zero exponentially; the boundary
term at zero is killed by the factor r2, as long as the functions (and their derivatives) are finite. So
/angbracketleftf|p2g/angbracketright=/angbracketleftp2f|g/angbracketright,
and hence p2is Hermitian.
Now we apply the same argument to
p4=/planckover2pi14
r2d
dr/braceleftbigg
r2d
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2d
dr/parenrightbigg/bracketrightbigg/bracerightbigg
,
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166 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
integrating by parts fourtimes:
/angbracketleftf|p4g/angbracketright=4π/planckover2pi14/integraldisplay∞
0fd
dr/braceleftbigg
r2d
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg/bracerightbigg
dr
=4π/planckover2pi14/braceleftbigg
r2fd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0−/integraldisplay∞
0r2df
drd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
dr/bracerightbigg
=4π/planckover2pi14/braceleftbigg/bracketleftbigg
r2fd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
−df
drd
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0+/integraldisplay∞
01
r2d
dr/parenleftbigg
r2df
dr/parenrightbiggd
dr/parenleftbigg
r2dg
dr/parenrightbigg
dr/bracerightbigg
=4π/planckover2pi14/braceleftbigg/bracketleftbigg
r2fd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
−df
drd
dr/parenleftbigg
r2dg
dr/parenrightbigg
+d
dr/parenleftbigg
r2df
dr/parenrightbiggdg
dr/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0
−/integraldisplay∞
0r2d
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2df
dr/parenrightbigg/bracketrightbiggdg
drdr/bracerightbigg
=4π/planckover2pi14/braceleftbigg/bracketleftbigg
r2fd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
−df
drd
dr/parenleftbigg
r2dg
dr/parenrightbigg
+d
dr/parenleftbigg
r2df
dr/parenrightbiggdg
dr−r2gd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2df
dr/parenrightbigg/bracketrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0
+/integraldisplay∞
0d
dr/parenleftbigg
r2d
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2df
dr/parenrightbigg/bracketrightbigg/parenrightbigg
gdr/bracerightbigg
=4π/planckover2pi14/braceleftbigg
r2fd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
−r2gd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2df
dr/parenrightbigg/bracketrightbigg/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0
−4π/planckover2pi14/braceleftbiggdf
drd
dr/parenleftbigg
r2dg
dr/parenrightbigg
−d
dr/parenleftbigg
r2df
dr/parenrightbiggdg
dr/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0+/angbracketleftp4f|g/angbracketright
This time there are fourboundary terms to worry about. Infinity is no problem; the trouble comes at r=0 .
If the functions fandgwent to zero at the origin (as they do for states with l>0) we’d be OK, but states
withl= 0 go like exp( −r/na). So let’s test the boundary terms using
f(r)=e−r/na,g (r)=e−r/ma.
In this case
r2dg
dr=−1
mar2e−r/ma
d
dr/parenleftbigg
r2dg
dr/parenrightbigg
=1
(ma)2/parenleftbig
r2−2mar/parenrightbig
e−r/ma
df
drd
dr/parenleftbigg
r2dg
dr/parenrightbigg
=−1
nae−r/na 1
(ma)2/parenleftbig
r2−2mar/parenrightbig
e−r/ma.
This goes to zero as r→0, so the second pair of boundary terms vanishes—but not the first pair:
1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg
=1
(ma)2/parenleftbigg
1−2ma
r/parenrightbigg
e−r/ma
d
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
=1
(ma)3r2/bracketleftbig
2(ma)2+2mar−r2/bracketrightbig
e−r/ma
r2fd
dr/bracketleftbigg1
r2d
dr/parenleftbigg
r2dg
dr/parenrightbigg/bracketrightbigg
=1
(ma)3/bracketleftbig
2(ma)2+2mar−r2/bracketrightbig
e−r/mae−r/na
This does notvanish as r→0; rather, it goes to 2 /ma. For these particular states, then,
/angbracketleftf|p4g/angbracketright=8π/planckover2pi14
a/parenleftbigg1
m−1
n/parenrightbigg
+/angbracketleftp4f|g/angbracketright,
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 167
or, tacking on the normalization factor,
ψn00 =1√π(na)3/2e−r/na,/angbracketleftψn00|p4ψm00/angbracketright=8/planckover2pi14
a4(n−m)
(nm)5/2+/angbracketleftp4ψn00|ψm00/angbracketright,
and hence p4is not Hermitian, for such states.
Problem 6.16
(a)
[L·S,Lx]=[LxSx+LySy+LzSz,Lx]=Sx[Lx,Lx]+Sy[Ly,Lx]+Sz[Lz,Lx]
=Sx(0) +Sy(−i/planckover2pi1Lz)+Sz(i/planckover2pi1Ly)=i/planckover2pi1(LySz−LzSy)=i/planckover2pi1(L×S)x.
Same goes for the other two components, so [L·S,L]=i/planckover2pi1(L×S).
(b)[L·S,S] is identical, only with L↔S:[L·S,S]=i/planckover2pi1(S×L).
(c)[L·S,J]=[L·S,L]+[L·S,S]=i/planckover2pi1(L×S+S×L)=0.
(d)L2commutes with all components of L(andS),s o/bracketleftbig
L·S,L2/bracketrightbig
=0 .
(e)Likewise,/bracketleftbig
L·S,S2/bracketrightbig
=0 .
(f)/bracketleftbig
L·S,J2/bracketrightbig
=/bracketleftbig
L·S,L2/bracketrightbig
+/bracketleftbig
L·S,S2/bracketrightbig
+2[L·S,L·S]=0+0+0= ⇒/bracketleftbig
L·S,J2/bracketrightbig
=0 .
Problem 6.17
With the plus sign, j=l+1/2(l=j−1/2) : Eq. 6.57 ⇒E1
r=−(En)2
2mc2/parenleftbigg4n
j−3/parenrightbigg
.
Equation 6.65 ⇒E1
so=(En)2
mc2n/bracketleftbig
j(j+1 )−(j−1
2)(j+1
2)−3
4/bracketrightbig
(j−1
2)j(j+1
2)
=(En)2
mc2n(j2+j−j2+1
4−3
4)
(j−1
2)j(j+1
2)=(En)2
mc2n
j(j+1
2).
E1
fs=E1
r+E1
so=(En)2
2mc2/parenleftbigg
−4n
j+3+2n
j(j+1
2)/parenrightbigg
=(En)2
2mc2/braceleftbigg
3+2n
j(j+1
2)/bracketleftbigg
1−2/parenleftbigg
j+1
2/parenrightbigg/bracketrightbigg/bracerightbigg
=(En)2
2mc2/parenleftbigg
3−4n
j+1
2/parenrightbigg
.
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168 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
With the minus sign, j=l−1/2(l=j+1/2) : Eq. 6.57 ⇒E1
r=−(En)2
2mc2/parenleftbigg4n
j+1−3/parenrightbigg
.
Equation 6.65 ⇒E1
so=(En)2
mc2n/bracketleftbig
j(j+1 )−(j+1
2)(j+3
2)−3
4/bracketrightbig
(j+1
2)(j+ 1)(j+3
2)
=(En)2
mc2n(j2+j−j2−2j−3
4−3
4)
(j+1
2)(j+ 1)(j+3
2)=(En)2
mc2−n
(j+ 1)(j+1
2).
E1
fs=(En)2
2mc2/bracketleftbigg
−4n
j+1−3+2n
(j+ 1)(j+1
2)/bracketrightbigg
=(En)2
2mc2/braceleftbigg
3−2n
(j+ 1)(j+1
2)/bracketleftbigg
1+2/parenleftbigg
j+1
2/parenrightbigg/bracketrightbigg/bracerightbigg
=(En)2
2mc2/parenleftbigg
3−4n
j+1
2/parenrightbigg
.For both signs, then, E1
fs=(En)2
2mc2/parenleftbigg
3−4n
j+1
2/parenrightbigg
.QED
Problem 6.18
E0
3−E0
2=hν=2π/planckover2pi1c
λ=E1/parenleftbigg1
9−1
4/parenrightbigg
=−5
36E1⇒λ=−36
52π/planckover2pi1c
E1;E1=−13.6e V;
/planckover2pi1c=1.97×10−11MeV·cm;λ=36
5(2π)(1.97×10−11×106eV·cm)
(13.6 eV)=6.55×10−5cm= 655 nm.
ν=c
λ=3.00×108m/s
6.55×10−7m=4.58×1014Hz. Equation 6.66 ⇒E1
fs=(En)2
2mc2/parenleftbigg
3−4n
j+1
2/parenrightbigg
:
Forn=2:l=0o rl=1,soj=1/2o r3/2.Thusn= 2 splits into twolevels :
j=1/2:E1
2=(E2)2
2mc2/parenleftbigg
3−8
1/parenrightbigg
=−5
2(E2)2
mc2=−5
2/parenleftbigg1
4/parenrightbigg2(E1)2
mc2=−5
32(13.6 eV)2
(.511×106eV)=−5.66×10−5eV.
j=3/2:E1
2=(E2)2
2mc2/parenleftbigg
3−8
2/parenrightbigg
=−1
2(E2)2
mc2=−1
32(3.62×10−4eV) =−1.13×10−5eV.
Forn=3:l=0,1o r2,soj=1/2,3/2o r5/2.Thusn= 3 splits into threelevels :
j=1/2:E1
3=(E3)2
2mc2/parenleftbigg
3−12
1/parenrightbigg
=−9(E3)2
mc2=−9
2/parenleftbigg1
92/parenrightbigg(E1)2
mc2=−1
18(3.62×10−4eV) =−2.01×10−5eV.
j=3/2:E1
3=(E3)2
2mc2/parenleftbigg
3−12
2/parenrightbigg
=−3
2(E3)2
mc2=−1
54(3.62×10−4eV) =−0.67×10−5eV.
j=5/2:E1
3=(E3)2
2mc2/parenleftbigg
3−12
3/parenrightbigg
=−1
2(E3)2
mc2=−1
162(3.62×10−4eV) =−0.22×10−5eV.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 169
E
E0
03
2
123
4565/2
3/2
1/2
3/2
1/2j=
j=j=j=j=
There are sixtransitions here; their energies are ( E0
3+E1
3)−(E0
2+E1
2)=(E0
3−E0
2)+∆E, where
∆E≡E1
3−E1
2. Letβ≡(E1)2/mc2=3.62×10−4eV. Then:
(1
2→3
2):∆E=/bracketleftbigg/parenleftbigg
−1
18/parenrightbigg
−/parenleftbigg
−1
32/parenrightbigg/bracketrightbigg
β=−7
288β=−8.80×10−6eV.
(3
2→3
2):∆E=/bracketleftbigg/parenleftbigg
−1
54/parenrightbigg
−/parenleftbigg
−1
32/parenrightbigg/bracketrightbigg
β=11
864β=4.61×10−6eV.
(5
2→3
2):∆E=/bracketleftbigg/parenleftbigg
−1
162/parenrightbigg
+/parenleftbigg1
32/parenrightbigg/bracketrightbigg
β=65
2592β=9.08×10−6eV.
(1
2→1
2):∆E=/bracketleftbigg/parenleftbigg5
32/parenrightbigg
−/parenleftbigg1
18/parenrightbigg/bracketrightbigg
β=29
288β=3 6.45×10−6eV.
(3
2→1
2):∆E=/bracketleftbigg/parenleftbigg
−1
54/parenrightbigg
+/parenleftbigg5
32/parenrightbigg/bracketrightbigg
β=119
864β=4 9.86×10−6eV.
(5
2→1
2):∆E=/bracketleftbigg/parenleftbigg
−1
162/parenrightbigg
+/parenleftbigg5
32/parenrightbigg/bracketrightbigg
β=389
2592β=5 4.33×10−6eV.
Conclusion: There are sixlines; one of them (1
2→3
2) has a frequency lessthan the unperturbed line, the
other five have (slightly) higher frequencies. In order they are:3
2→3
2;5
2→3
2;1
2→1
2;3
2→1
2;5
2→1
2. The
frequency spacings are:
ν2−ν1=( ∆E2−∆E1)/2π/planckover2pi1=3.23×109Hz
ν3−ν3=( ∆E3−∆E2)/2π/planckover2pi1=1.08×109Hz
ν4−ν3=( ∆E4−∆E3)/2π/planckover2pi1=6.60×109Hz
ν5−ν4=( ∆E5−∆E4)/2π/planckover2pi1=3.23×109Hz
ν6−ν5=( ∆E6−∆E5)/2π/planckover2pi1=1.08×109Hz
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170 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.19
/radicalBigg/parenleftbigg
j+1
2/parenrightbigg2
−α2=/parenleftbigg
j+1
2/parenrightbigg/radicalBigg
1−/parenleftbiggα
j+1
2/parenrightbigg2
≈/parenleftbigg
j+1
2/parenrightbigg/bracketleftBigg
1−1
2/parenleftbiggα
j+1
2/parenrightbigg2/bracketrightBigg
=(j+1
2)−α2
2(j+1
2).
α
n−(j+1
2)+/radicalBig/parenleftbig
j+1
2/parenrightbig2−α2≈α
n−/parenleftbig
j+1
2/parenrightbig
+/parenleftbig
j+1
2/parenrightbig
−α2
2(j+1
2)=α
n−α2
2(j+1
2)
=α
n/bracketleftBig
1−α2
2n(j+1
2)/bracketrightBig≈α
n/bracketleftbigg
1+α2
2n(j+1
2)/bracketrightbigg
.
1+
α
n−(j+1
2)+/radicalBig/parenleftbig
j+1
2/parenrightbig2−α2
2
−1/2
≈/bracketleftbigg
1+α2
n2/parenleftbigg
1+α2
n(j+1
2)/parenrightbigg/bracketrightbigg−1/2
≈1−1
2α2
n2/parenleftbigg
1+α2
n(j+1
2)/parenrightbigg
+3
8α4
n4=1−α2
2n2+α4
2n4/parenleftbigg−n
j+1
2+3
4/parenrightbigg
.
Enj≈mc2/bracketleftbigg
1−α2
2n2+α4
2n4/parenleftbigg−n
j+1
2+3
4/parenrightbigg
−1/bracketrightbigg
=−α2mc2
2n2/bracketleftbigg
1+α2
n2/parenleftbiggn
j+1
2−3
4/parenrightbigg/bracketrightbigg
=−13.6e V
n2/bracketleftbigg
1+α2
n2/parenleftbiggn
j+1
2−3
4/parenrightbigg/bracketrightbigg
,confirming Eq. 6.67 .
Problem 6.20
Equation 6.59 ⇒B=1
4πFepsilonC0e
mc2r3L.S a y L=/planckover2pi1,r=a; then
B=1
4πFepsilonC0e/planckover2pi1
mc2a3
=(1.60×10−19C)(1.05×10−34J·s)
4π/parenleftbig
8.9×10−12C2/N·m2/parenrightbig
(9.1×10−31kg) (3×108m/s)2(0.53×10−10m)3=12 T.
So a “strong” Zeeman field is Bext/greatermuch10 T, and a “weak” one is Bext/lessmuch10 T. Incidentally, the earth’s field
(10−4T) is definitely weak.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 171
Problem 6.21
Forn=2 ,l=0(j=1/2) orl=1(j=1/2o r3/2). The eight states are:
|1/angbracketright=|201
21
2/angbracketright
|2/angbracketright=|201
2−1
2/angbracketright
gJ=/bracketleftbigg
1+(1/2)(3/2 )+( 3/4)
2(1/2)(3/2)/bracketrightbigg
=1+3/2
3/2=2.
|3/angbracketright=|211
21
2/angbracketright
|4/angbracketright=|211
2−1
2/angbracketright
gJ=/bracketleftbigg
1+(1/2)(3/2)−(1)(2) + (3 /4)
2(1/2)(3/2)/bracketrightbigg
=1+−1/2
3/2=2/3.
In these four cases, Enj=−13.6e V
4/bracketleftbigg
1+α2
4/parenleftbigg2
1−3
4/parenrightbigg/bracketrightbigg
=−3.4e V/parenleftbigg
1+5
16α2/parenrightbigg
.
|5/angbracketright=|213
23
2/angbracketright
|6/angbracketright=|213
21
2/angbracketright
|7/angbracketright=|213
2−1
2/angbracketright
|8/angbracketright=|213
2−3
2/angbracketright
g
J=/bracketleftbigg
1+(3/2)(5/2)−(1)(2) + (3 /4)
2(3/2)(5/2)/bracketrightbigg
=1+5/2
15/2=4/3.
In these four cases, Enj=−3.4e V/bracketleftbigg
1+α2
4/parenleftbigg2
2−3
4/parenrightbigg/bracketrightbigg
=−3.4e V/parenleftbigg
1+1
16α2/parenrightbigg
.
The energies are:E1=−3.4e V/parenleftbig
1+5
16α2/parenrightbig
+µBBext.
E2=−3.4e V/parenleftbig
1+5
16α2/parenrightbig
−µBBext.
E3=−3.4e V/parenleftbig
1+5
16α2/parenrightbig
+1
3µBBext.
E4=−3.4e V/parenleftbig
1+5
16α2/parenrightbig
−1
3µBBext.
E5=−3.4e V/parenleftbig
1+1
16α2/parenrightbig
+2µBBext.
E6=−3.4e V/parenleftbig
1+1
16α2/parenrightbig
+2
3µBBext.
E7=−3.4e V/parenleftbig
1+1
16α2/parenrightbig
−2
3µBBext.
E8=−3.4e V/parenleftbig
1+1
16α2/parenrightbig
−2µBBext.
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172 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
E BextBµ
2 (slope -1)4 (slope -1/3)3 (slope 1/3)1 (slope 1)8 (slope -2)7 (slope -2/3)6 (slope 2/3)5 (slope 2)
-3.4 (1+ α /16) eV2
-3.4 (1+5 α /16) eV2
Problem 6.22
E1
fs=/angbracketleftnlm lms|(H/prime
r+H/prime
so)|nlm lms/angbracketright=−E2
n
2mc2/bracketleftbigg4n
l+1/2−3/bracketrightbigg
+e2
8πFepsilonC0m2c2/planckover2pi12mlms
l(l+1/2)(l+1 )n3a3.
Now
2E
2
n
mc2=/parenleftbigg
−2E1
mc2/parenrightbigg/parenleftbigg
−E1
n4/parenrightbigg
=α2
n4(13.6 eV). (Problem 6.11.)
e2/planckover2pi12
8πFepsilonC0m2c2a3=e2/planckover2pi12(me2)3
2·4πFepsilonC0m2c2(4πFepsilonC0/planckover2pi12)3=/bracketleftBigg
m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2/bracketrightBigg/parenleftbigge2
4πFepsilonC0/planckover2pi1c/parenrightbigg2
=α2(13.6 eV).
E1
fs=13.6e V
n3α2/braceleftbigg
−1
(l+1/2)+3
4n+mlms
l(l+1/2)(l+1 )/bracerightbigg
=13.6e V
n3α2/braceleftbigg3
4n−l(l+1 )−mlms
l(l+1/2)(l+1 )/bracerightbigg
.QED
Problem 6.23
The Bohr energy is the same for all of them: E2=−13.6e V/22=−3.4e V.The Zeeman contribution is the
second term in Eq. 6.79: µBBext(ml+2ms). The fine structure is given by Eq. 6.82: E1
fs= (13.6e V/8)α2{···} =
(1.7 eV)α2{···}. In the table below I record the 8 states, the value of ( ml+2ms), the value of {···}≡
3
8−/bracketleftbiggl(l+1 )−mlms
l(l+1/2)(l+1 )/bracketrightbigg
, and (in the last column) the total energy, −3.4e V[ 1−(α2/2){···}]+(ml+2ms)µBBext.
State =|nlm lms/angbracketright (ml+2ms){···} Total Energy
|1/angbracketright=|2001
2/angbracketright 1−5/8-3.4 eV [1 + (5 /16)α2]+µBBext
|2/angbracketright=|200−1
2/angbracketright−1−5/8-3.4 eV [1 + (5 /16)α2]−µBBext
|3/angbracketright=|2111
2/angbracketright 2−1/8-3.4 eV [1 + (1 /16)α2]+2µBBext
|4/angbracketright=|21−1−1
2/angbracketright−2−1/8-3.4 eV [1 + (1 /16)α2]−2µBBext
|5/angbracketright=|2101
2/angbracketright 1−7/24-3.4 eV [1 + (7 /48)α2]+µBBext
|6/angbracketright=|210−1
2/angbracketright−1−7/24-3.4 eV [1 + (7 /48)α2]−µBBext
|7/angbracketright=|211−1
2/angbracketright 0−11/24-3.4 eV [1 + (11 /48)α2]
|8/angbracketright=|21−11
2/angbracketright 0−11/24-3.4 eV [1 + (11 /48)α2]
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 173
Ignoring fine structure there are fivedistinct levels—corresponding to the possible values of ( ml+2ms):
2(d= 1); 1 ( d= 2); 0 ( d= 2);−1(d= 2);−2(d= 1).
Problem 6.24
Equation 6.72 ⇒E1
z=e
2mBext·/angbracketleftL+2S/angbracketright=e
2mBext2ms/planckover2pi1=2msµBBext(same as the Zeeman term in Eq. 6.79,
withml= 0). Equation 6.67 ⇒Enj=−13.6e V
n2/bracketleftbigg
1+α2
n2/parenleftbigg
n−3
4/parenrightbigg/bracketrightbigg
(sincej=1/2). So the total energy is
E=−13.6e V
n2/bracketleftbigg
1+α2
n2/parenleftbigg
n−3
4/parenrightbigg/bracketrightbigg
+2msµBBext.
Fine structure is the α2term:E1
fs=−13.6e V
n4α2/parenleftbigg
n−3
4/parenrightbigg
=13.6e V
n3α2/parenleftbigg3
4n−1/parenrightbigg
, which is the same as
Eq. 6.82, with the term in square brackets set equal to 1. QED
Problem 6.25
Equation 6 .66⇒E1
fs=E2
2
2mc2/parenleftbigg
3−8
j+1/2/parenrightbigg
=E2
1
32mc2/parenleftbigg
3−8
j+1/2/parenrightbigg
;E1
mc2=−α2
2(Problem 6.11), so
E1
fs=−E1
32/parenleftbiggα2
2/parenrightbigg/parenleftbigg
3−8
j+1/2/parenrightbigg
=13.6e V
64α2/parenleftbigg
3−8
j+1/2/parenrightbigg
=γ/parenleftbigg
3−8
j+1/2/parenrightbigg
.
Forj=1/2(ψ1,ψ2,ψ6,ψ8),H1
fs=γ(3−8) =−5γ.Forj=3/2(ψ3,ψ4,ψ5,ψ7),H1
fs=γ(3−8
2)=−γ.
This confirms all the γterms in−W(p. 281). Meanwhile, H/prime
z=(e/2m)Bext(Lz+2Sz) (Eq. 6.71); ψ1,ψ2,ψ3,ψ4
are eigenstates of LzandSz; for these there are only diagonal elements:
/angbracketleftH/prime
z/angbracketright=e/planckover2pi1
2mBext(ml+2ms)=(ml+2ms)β;/angbracketleftH/prime
z/angbracketright11=β;/angbracketleftH/prime
z/angbracketright22=−β;/angbracketleftH/prime
z/angbracketright33=2β;/angbracketleftH/prime
z/angbracketright44=−2β.
This confirms the upper left corner of −W. Finally:
(Lz+2Sz)|ψ5/angbracketright=+/planckover2pi1/radicalBig
2
3|10/angbracketright|1
21
2/angbracketright
(Lz+2Sz)|ψ6/angbracketright=−/planckover2pi1/radicalBig
1
3|10/angbracketright|1
21
2/angbracketright
(Lz+2Sz)|ψ7/angbracketright=−/planckover2pi1/radicalBig
2
3|10/angbracketright|1
2−1
2/angbracketright
(Lz+2Sz)|ψ8/angbracketright=−/planckover2pi1/radicalBig
1
3|10/angbracketright|1
2−1
2/angbracketright
so/angbracketleftH
/prime
z/angbracketright55=( 2/3)β,
/angbracketleftH/prime
z/angbracketright66=( 1/3)β,
/angbracketleftH/prime
z/angbracketright77=−(2/3)β,
/angbracketleftH/prime
z/angbracketright88=−(1/3)β,
/angbracketleftH/prime
z/angbracketright56=/angbracketleftH/prime
z/angbracketright65=−(√
2/3)β,
/angbracketleftH/prime
z/angbracketright78=/angbracketleftH/prime
z/angbracketright87=−(√
2/3)β,
which confirms the remaining elements.
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174 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.26
There are eighteen n= 3 states (in general, 2 n2).
WEAK FIELD
Equation 6 .67⇒E3j=−13.6e V
9/bracketleftbigg
1+α2
9/parenleftbigg3
j+1/2−3
4/parenrightbigg/bracketrightbigg
=−1.51 eV/bracketleftbigg
1+α2
3/parenleftbigg1
j+1/2−1
4/parenrightbigg/bracketrightbigg
.
Equation 6 .76⇒E1
z=gJmjµBBext.
State|3ljm j/angbracketright gJ(Eq. 6.75)1
3/parenleftBig
1
j+1/2−1
4/parenrightBig
Total Energy
l=0,j=1/2|301
21
2/angbracketright 2 1/4−1.51 eV/parenleftBig
1+α2
4/parenrightBig
+µBBext
l=0,j=1/2|301
2−1
2/angbracketright 2 1/4−1.51 eV/parenleftBig
1+α2
4/parenrightBig
−µBBext
l=1,j=1/2|311
21
2/angbracketright 2/3 1/4−1.51 eV/parenleftBig
1+α2
4/parenrightBig
+1
3µBBext
l=1,j=1/2|311
2−1
2/angbracketright 2/3 1/4−1.51 eV/parenleftBig
1+α2
4/parenrightBig
−1
3µBBext
l=1,j=3/2|313
23
2/angbracketright 4/3 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
+2µBBext
l=1,j=3/2|313
21
2/angbracketright 4/3 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
+2
3µBBext
l=1,j=3/2|313
2−1
2/angbracketright 4/3 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
−2
3µBBext
l=1,j=3/2|313
2−3
2/angbracketright 4/3 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
−2µBBext
l=2,j=3/2|323
23
2/angbracketright 4/5 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
+6
5µBBext
l=2,j=3/2|323
21
2/angbracketright 4/5 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
+2
5µBBext
l=2,j=3/2|323
2−1
2/angbracketright 4/5 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
−2
5µBBext
l=2,j=3/2|323
2−3
2/angbracketright 4/5 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
−6
5µBBext
l=2,j=5/2|325
25
2/angbracketright 6/5 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
+3µBBext
l=2,j=5/2|325
23
2/angbracketright 6/5 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
+9
5µBBext
l=2,j=5/2|325
21
2/angbracketright 6/5 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
+3
5µBBext
l=2,j=5/2|325
2−1
2/angbracketright 6/5 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
−3
5µBBext
l=2,j=5/2|325
2−3
2/angbracketright 6/5 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
−9
5µBBext
l=2,j=5/2|325
2−5
2/angbracketright 6/5 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
−3µBBext
STRONG FIELD
Equation 6 .79⇒−1.51 eV + ( ml+2ms)µBBext;
Equation 6 .82⇒13.6e V
27α2/braceleftbigg1
4−/bracketleftbiggl(l+1 )−mlms
l(l+1/2)(l+1 )/bracketrightbigg/bracerightbigg
=−1.51 eVα2
3/braceleftbigg/bracketleftbiggl(l+1 )−mlms
l(l+1/2)(l+1 )−1
4/bracketrightbigg/bracerightbigg
.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 175
Etot=−1.51 eV(1 + α2A)+(ml+2ms)µBBext,whereA≡1
3/braceleftbigg/bracketleftbiggl(l+1 )−mlms
l(l+1/2)(l+1 )−1
4/bracketrightbigg/bracerightbigg
.
These terms are given in the table below:
State|nlm lms/angbracketright(ml+2ms)A Total Energy
l=0|3001
2/angbracketright 1 1/4−1.51 eV/parenleftBig
1+α2
4/parenrightBig
+µBBext
l=0|300−1
2/angbracketright−1 1/4−1.51 eV/parenleftBig
1+α2
4/parenrightBig
−µBBext
l=1|3111
2/angbracketright 2 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
+2µBBext
l=1|31−1−1
2/angbracketright−2 1/12−1.51 eV/parenleftBig
1+α2
12/parenrightBig
−2µBBext
l=1|3101
2/angbracketright 1 5/36−1.51 eV/parenleftBig
1+5α2
36/parenrightBig
+µBBext
l=1|310−1
2/angbracketright−1 5/36−1.51 eV/parenleftBig
1+5α2
36/parenrightBig
−µBBext
l=1|31−11
2/angbracketright 0 7/36−1.51 eV/parenleftBig
1+7α2
36/parenrightBig
l=1|311−1
2/angbracketright 0 7/36−1.51 eV/parenleftBig
1+7α2
36/parenrightBig
l=2|3221
2/angbracketright 3 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
+3µBBext
l=2|32−2−1
2/angbracketright−3 1/36−1.51 eV/parenleftBig
1+α2
36/parenrightBig
−3µBBext
l=2|3211
2/angbracketright 2 7/180−1.51 eV/parenleftBig
1+7α2
180/parenrightBig
+2µBBext
l=2|32−1−1
2/angbracketright−2 7/180−1.51 eV/parenleftBig
1+7α2
180/parenrightBig
−2µBBext
l=2|3201
2/angbracketright 1 1/20−1.51 eV/parenleftBig
1+α2
20/parenrightBig
+µBBext
l=2|320−1
2/angbracketright−1 1/20−1.51 eV/parenleftBig
1+α2
20/parenrightBig
−µBBext
l=2|32−11
2/angbracketright 0 11/180−1.51 eV/parenleftBig
1+11α2
180/parenrightBig
l=2|321−1
2/angbracketright 0 11/180−1.51 eV/parenleftBig
1+11α2
180/parenrightBig
l=2|32−21
2/angbracketright−1 13/180−1.51 eV/parenleftBig
1+13α2
180/parenrightBig
−µBBext
l=2|322−1
2/angbracketright 1 13/180−1.51 eV/parenleftBig
1+13α2
180/parenrightBig
+µBBext
INTERMEDIATE FIELD
As in the book, I’ll use the basis |nljm j/angbracketright(same as for weak field); then the fine structure matrix elements
are diagonal: Eq. 6.66 ⇒
E1
fs=E2
3
2mc2/parenleftbigg
3−12
j+1/2/parenrightbigg
=E2
1
54mc2/parenleftbigg
1−4
j+1/2/parenrightbigg
=−E1α2
108/parenleftbigg
1−4
j+1/2/parenrightbigg
=3γ/parenleftbigg
1−4
j+1/2/parenrightbigg
,
γ≡13.6e V
324α2.Forj=1/2,E1
fs=−9γ; forj=3/2,E1
fs=−3γ; forj=5/2,E1
fs=−γ.
The Zeeman Hamiltonian is Eq. 6.71: H/prime
z=1
/planckover2pi1(Lz+2Sz)µBBext. The first eight states ( l= 0 and l= 1) are
the same as before (p. 281), so the βterms in Ware unchanged; recording just the non-zero blocks of −W:
(9γ−β),(9γ+β),(3γ−2β),(3γ+2β),/parenleftBigg
(3γ−2
3β)√
2
3β√
2
3β(9γ−1
3β)/parenrightBigg
,/parenleftBigg
(3γ+2
3β)√
2
3β√
2
3β(9γ+1
3β)/parenrightBigg
.
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176 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
The other 10 states ( l= 2) must first be decomposed into eigenstates of LzandSz:
|5
25
2/angbracketright=|22/angbracketright|1
21
2/angbracketright =⇒(γ−3β)
|5
2−5
2/angbracketright=|2−2/angbracketright|1
2−1
2/angbracketright=⇒(γ+3β)
|5
23
2/angbracketright=/radicalBig
1
5|22/angbracketright|1
2−1
2/angbracketright+/radicalBig
4
5|21/angbracketright|1
21
2/angbracketright
|3
23
2/angbracketright=/radicalBig
4
5|22/angbracketright|1
2−1
2/angbracketright−/radicalBig
1
5|21/angbracketright|1
21
2/angbracketright
=⇒/parenleftbigg
(γ−9
5β)2
5β
2
5β(3γ−6
5β)/parenrightbigg
|5
21
2/angbracketright=/radicalBig
2
5|21/angbracketright|1
2−1
2/angbracketright+/radicalBig
3
5|20/angbracketright|1
21
2/angbracketright
|3
21
2/angbracketright=/radicalBig
3
5|21/angbracketright|1
2−1
2/angbracketright−/radicalBig
2
5|20/angbracketright|1
21
2/angbracketright
=⇒/parenleftBigg
(γ−3
5β)√
6
5β√
6
5β(3γ−2
5β)/parenrightBigg
|5
2−1
2/angbracketright=/radicalBig
3
5|20/angbracketright|1
2−1
2/angbracketright+/radicalBig
2
5|2−1/angbracketright|1
21
2/angbracketright
|3
2−1
2/angbracketright=/radicalBig
2
5|20/angbracketright|1
2−1
2/angbracketright−/radicalBig
3
5|2−1/angbracketright|1
21
2/angbracketright
=⇒/parenleftBigg
(γ+3
5β)√
6
5β√
6
5β(3γ+2
5β)/parenrightBigg
|5
2−3
2/angbracketright=/radicalBig
4
5|2−1/angbracketright|1
2−1
2/angbracketright+/radicalBig
1
5|2−2/angbracketright|1
21
2/angbracketright
|3
2−3
2/angbracketright=/radicalBig
1
5|2−1/angbracketright|1
2−1
2/angbracketright−/radicalBig
4
5|2−2/angbracketright|1
21
2/angbracketright
=⇒/parenleftbigg(γ+9
5β)2
5β
2
5β(3γ+6
5β)/parenrightbigg
[Sample Calculation: For the last two, letting Q≡1
/planckover2pi1(Lz+2Sz), we have
Q|5
2−3
2/angbracketright=−2/radicalBig
4
5|2−1/angbracketright|1
2−1
2/angbracketright−/radicalBig
1
5|2−2/angbracketright|1
21
2/angbracketright;
Q|3
2−3
2/angbracketright=−2/radicalBig
1
5|2−1/angbracketright|1
2−1
2/angbracketright+/radicalBig
4
5|2−2/angbracketright|1
21
2/angbracketright.
/angbracketleft5
2−3
2|Q|5
2−3
2/angbracketright=(−2)4
5−1
5=−9
5;/angbracketleft3
2−3
2|Q|3
2−3
2/angbracketright=(−2)1
5−4
5=−6
5;
/angbracketleft5
2−3
2|Q|3
2−3
2/angbracketright=−2/radicalBig
4
5/radicalBig
1
5+/radicalBig
1
5/radicalBig
4
5=−4
5+2
5=−2
5=/angbracketleft3
2−3
2|Q|5
2−3
2/angbracketright.]
So the 18×18 matrix−Wsplits into six 1 ×1 blocks and six 2 ×2 blocks. We need the eigenvalues of the
2×2 blocks. This means solving 3 characteristic equations (the other 3 are obtained trivially by changing the
sign ofβ):
/parenleftbigg
3γ−2
3β−λ/parenrightbigg/parenleftbigg
9γ−1
3β−λ/parenrightbigg
−2
9β2=0=⇒λ2+λ(β−12γ)+γ(27γ−7β)=0.
/parenleftbigg
γ−9
5β−λ/parenrightbigg/parenleftbigg
3γ−6
5β−λ/parenrightbigg
−4
25β2=0=⇒λ2+λ(3β−4γ)+γ/parenleftbigg
3γ2−33
5γβ+2β2/parenrightbigg
=0.
/parenleftbigg
γ−3
5β−λ/parenrightbigg/parenleftbigg
3γ−2
5β−λ/parenrightbigg
−6
25β2=0=⇒λ2+λ(β−4γ)+γ/parenleftbigg
3γ−11
5β/parenrightbigg
=0.
The solutions are:
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 177
λ=−β/2+6γ±/radicalbig
(β/2)2+βγ+9γ2
λ=−3β/2+2γ±/radicalBig
(β/2)2+3
5βγ+γ2
λ=−β/2+2γ±/radicalBig
(β/2)2+1
5βγ+γ2⇒FepsilonC1=E3−9γ+β
FepsilonC2=E3−3γ+2β
FepsilonC3=E3−γ+3β
FepsilonC4=E3−6γ+β/2+/radicalbig
9γ2+βγ+β2/4
FepsilonC5=E3−6γ+β/2−/radicalbig
9γ2+βγ+β2/4
FepsilonC6=E3−2γ+3β/2+/radicalBig
γ2+3
5βγ+β2/4
FepsilonC7=E3−2γ+3β/2−/radicalBig
γ2+3
5βγ+β2/4
FepsilonC8=E3−2γ+β/2+/radicalBig
γ2+1
5βγ+β2/4
FepsilonC9=E3−2γ+β/2−/radicalBig
γ2+1
5βγ+β2/4
(The other 9 FepsilonC’s are the same, but with β→−β.) Here γ=13.6e V
324α2, andβ=µBBext.
In the weak-field limit ( β/lessmuchγ):
FepsilonC4≈E3−6γ+β/2+3γ/radicalbig
1+β/9γ≈E3−6γ+β/2+3γ(1 +β/18γ)=E3−3γ+2
3β.
FepsilonC5≈E3−6γ+β/2−3γ(1 +β/18γ)=E3−9γ+1
3β.
FepsilonC6≈E3−2γ+3β/2+γ( 1+3β/10γ)=E3−γ+9
5β.
FepsilonC7≈E3−2γ+3β/2−γ( 1+3β/10γ)=E3−3γ+6
5β.
FepsilonC8≈E3−2γ+β/2+γ(1 +β/10γ)=E3−γ+3
5β.
FepsilonC9≈E3−2γ+β/2−γ(1 +β/10γ)=E3−3γ+2
5β.
Noting that γ=−(E3/36)α2=1.51 eV
36α2, we see that the weak field energies are recovered as in the first table.
In the strong-field limit ( β/greatermuchγ):
FepsilonC4≈E3−6γ+β/2+β/2/radicalbig
1+4γ/β≈E3−6γ+β/2+β/2 ( 1+2γ/β)=E3−5γ+β.
FepsilonC5≈E3−6γ+β/2−β/2 ( 1+2γ/β)=E3−7γ.
FepsilonC6≈E3−2γ+3β/2+β/2 ( 1+6γ/5β)=E3−7
5γ+2β.
FepsilonC7≈E3−2γ+3β/2−β/2 ( 1+6γ/5β)=E3−13
5γ+β.
FepsilonC8≈E3−2γ+β/2+β/2 ( 1+2γ/5β)=E3−9
5γ+β.
FepsilonC9≈E3−2γ+β/2−β/2 ( 1+2γ/5β)=E3−11
5γ.
Again, these reproduce the strong-field results in the second table.
In the figure below each line is labeled by the level number and (in parentheses) the starting and ending
slope; for each line there is a corresponding one starting from the same point but sloping down.
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178 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
E
Ε3
Ε3Ε3−3γ−γΕ3
−9γ3(3)
6(9/5−> 2)
8(3/5−>1)
2(2)
7(6/5−>1)
4(2/5−>1)
9(2/5−>0)
1(1)
5(1/3−>0)
Problem 6.27
I≡/integraltext
(a·ˆr)(b·ˆr) sinθdθdφ
=/integraltext
(axsinθcosφ+aysinθsinφ+azcosθ)(bxsinθcosφ+bysinθsinφ+bzcosθ) sinθdθdφ.
But/integraldisplay2π
0sinφdφ=/integraldisplay2π
0cosφdφ=/integraldisplay2π
0sinφcosφdφ=0,so only three terms survive :
I=/integraldisplay
(axbxsin2θcos2φ+aybysin2θsin2φ+azbzcos2θ) sinθdθdφ.
But/integraldisplay2π
0sin2φdφ=/integraldisplay2π
0cos2φdφ=π,/integraldisplay2π
0dφ=2π,so
I=/integraldisplayπ
0/bracketleftbig
π(axbx+ayby) sin2θ+2πazbzcos2θ/bracketrightbig
sinθdθ. But/integraldisplayπ
0sin3θdθ=4
3,/integraldisplayπ
0cos2θsinθdθ=2
3,
soI=π(axbx+ayby)4
3+2πazbz2
3=4π
3(axbx+ayby+azbz)=4π
3(a·b).QED
[Alternatively, noting that Ihas to be a scalar bilinear in aandb, we know immediately that I=A(a·b), where
Ais some constant (same for all aandb). To determine A, picka=b=ˆk; thenI=A=/integraltext
cos2θsinθdθdφ =
4π/3.]
For states with l= 0, the wave function is independent of θandφ(Y0
0=1/√
4π), so
/angbracketleftbigg3(Sp·ˆr)(Se·ˆr)−Sp·Se
r3/angbracketrightbigg
=/braceleftbigg/integraldisplay∞
01
r3|ψ(r)|2r2dr/bracerightbigg/integraldisplay
[3(Sp·ˆr)(Se·ˆr)] sinθdθdφ.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 179
The first angular integral is 3(4 π/3)(Sp·Se)=4π(Sp·Se), while the second is −(Sp·Se)/integraltext
sinθdθdφ =
−4π(Sp·Se), so the two cancel, and the result is zero. QED [Actually, there is a little sleight-of-hand here,
since for l=0 ,ψ→constant as r→0, and hence the radial integral diverges logarithmically at the origin.
Technically, the first term in Eq. 6.86 is the field outside an infinitesimal sphere ; the delta-function gives the
fieldinside . For this reason it is correct to do the angular integral first (getting zero) and not worry about the
radial integral.]
Problem 6.28
From Eq. 6.89 we see that ∆ E∝/parenleftbiggg
mpmea3/parenrightbigg
; we want reduced mass in a, butnotinmpme(which come from
Eq. 6.85); the notation in Eq. 6.93 obscures this point.
(a)gandmpare unchanged; me→mµ= 207me, anda→aµ. From Eq. 4.72, a∝1/m,s o
a
aµ=mµ(reduced)
me=mµmp
mµ+mp·1
me=207
1 + 207( me/mp)=207
1 + 207(9.11×10−31)
1.67×10−27)=207
1.11= 186.
∆E=( 5.88×10−6eV) (1/207) (186)3=0.183 eV.
(b)g:5.59→2;mp→me;a
ap=mp(reduced)
me=m2
e
me+me·1
me=1
2.
∆E=( 5.88×10−6eV)/parenleftbigg2
5.59/parenrightbigg/parenleftbigg1.67×10−27
9.11×10−31/parenrightbigg/parenleftbigg1
2/parenrightbigg3
=4.82×10−4eV.
(c)g:5.59→2;mp→mµ;a
am=mm(reduced)
me=memµ
me+mµ·1
me=207
208.
∆E=( 5.88×10−6)/parenleftbigg2
5.59/parenrightbigg/parenleftbigg1.67×10−27
(207)(9.11×10−31)/parenrightbigg/parenleftbigg207
208/parenrightbigg3
=1.84×10−5eV.
Problem 6.29
Use perturbation theory:
H/prime=−e2
4πFepsilonC0/parenleftbigg1
b−1
r/parenrightbigg
,for 0<r<b . ∆E=/angbracketleftψ|H/prime|ψ/angbracketright,withψ≡1√
πa3e−r/a.
∆E=−e2
4πFepsilonC01
πa34π/integraldisplayb
0/parenleftbigg1
b−1
r/parenrightbigg
e−2r/ar2dr=−e2
πFepsilonC0a3/parenleftbigg1
b/integraldisplayb
0r2e−2r/adr−/integraldisplayb
0re−2r/adr/parenrightbigg
=−e2
πFepsilonC0a3/braceleftbigg1
b/bracketleftbigg
−a
2r2e−2r/a+a/parenleftbigga
2/parenrightbigg2
e−2r/a/parenleftbigg
−2r
a−1/parenrightbigg/bracketrightbigg
−/bracketleftbigg/parenleftbigga
2/parenrightbigg2
e−2r/a/parenleftbigg
−2r
a−1/parenrightbigg/bracketrightbigg/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
0
=−e2
πFepsilonC0a3/bracketleftbigg
−a
2bb2e−2b/a+a3
4be−2b/a/parenleftbigg
−2b
a−1/parenrightbigg
−a2
4e−2b/a/parenleftbigg
−2b
a−1/parenrightbigg
+a3
4b−a2
4/bracketrightbigg
=−e2
πFepsilonC0a3/bracketleftbigg
e−2b/a/parenleftbigg
−ab
2−a2
2−a3
4b+ab
2+a2
4/parenrightbigg
+a2
4/parenleftbigga
b−1/parenrightbigg/bracketrightbigg
=−e2
πFepsilonC0a3/bracketleftbigg
e−2b/a/parenleftbigg
−a2
4/parenrightbigg/parenleftbigga
b+1/parenrightbigg
+a2
4/parenleftbigga
b−1/parenrightbigg/bracketrightbigg
=e2
4πFepsilonC0a/bracketleftbigg/parenleftbigg
1−a
b/parenrightbigg
+/parenleftbigg
1+a
b/parenrightbigg
e−2b/a/bracketrightbigg
.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
180 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Let +2b/a=FepsilonC(very small). Then the term in square brackets is:
/parenleftbigg
1−2
FepsilonC/parenrightbigg
+/parenleftbigg
1+2
FepsilonC/parenrightbigg/parenleftbigg
1−FepsilonC+FepsilonC2
2−FepsilonC3
6+···/parenrightbigg
=✁1−✄✄✄2
FepsilonC+✁1+✄✄✄2
FepsilonC−✁FepsilonC−✁2+FepsilonC2
2+✁FepsilonC−FepsilonC3
6−FepsilonC2
3+()FepsilonC3+···=FepsilonC2
6+()FepsilonC3+()FepsilonC4···
To leading order, then, ∆ E=e2
4πFepsilonC01
a4b2
6a2.
E=E1=−me4
2(4πFepsilonC0)2/planckover2pi12;a=4πFepsilonC0/planckover2pi12
me2;s o Ea=−e2
2(4πFepsilonC0).
∆E
E=e2
4πFepsilonC0/parenleftbigg
−2(4πFepsilonC0)
e2/parenrightbigg2b2
3a2=−4
3/parenleftbiggb
a/parenrightbigg2
.
Putting in a=5×10−11m:
∆E
E=−4
3/parenleftbigg10−15
5×10−11/parenrightbigg
=−16
3×10−10≈−5×10−10.
By contrast,/braceleftbiggfine structure: ∆ E/E≈α2=( 1/137)2=5×10−5,
hyperfine structure: ∆ E/E≈(me/mp)α2=( 1/1800)(1 /137)2=3×10−8.
So the correction for the finite size of the nucleus is much smaller (about 1% of hyperfine).
Problem 6.30
(a)In terms of the one-dimensional harmonic oscillator states {ψn(x)}, the unperturbed ground state is
|0/angbracketright=ψ0(x)ψ0(y)ψ0(z).
E1
0=/angbracketleft0|H/prime|0/angbracketright=/angbracketleftψ0(x)ψ0(y)ψ0(z)|λx2yz|ψ0(x)ψ0(y)ψ0(z)/angbracketright=λ/angbracketleftx2/angbracketright0/angbracketlefty/angbracketright0/angbracketleftz/angbracketright0.
But/angbracketlefty/angbracketright0=/angbracketleftz/angbracketright0= 0. So there is nochange, in first order.
(b)The (triply degenerate) first excited states are
|1/angbracketright=ψ0(x)ψ0(y)ψ1(z)
|2/angbracketright=ψ0(x)ψ1(y)ψ0(z)
|3/angbracketright=ψ1(x)ψ0(y)ψ0(z)
In this basis the perturbation matrix is Wij=/angbracketlefti|H/prime|j/angbracketright,i=1,2,3.
/angbracketleft1|H/prime|1/angbracketright=/angbracketleftψ0(x)ψ0(y)ψ1(z)|λx2yz|ψ0(x)ψ0(y)ψ1(z)/angbracketright=λ/angbracketleftx2/angbracketright0/angbracketlefty/angbracketright0/angbracketleftz/angbracketright1=0,
/angbracketleft2|H/prime|2/angbracketright=/angbracketleftψ0(x)ψ1(y)ψ0(z)|λx2yz|ψ0(x)ψ1(y)ψ0(z)/angbracketright=λ/angbracketleftx2/angbracketright0/angbracketlefty/angbracketright1/angbracketleftz/angbracketright0=0,
/angbracketleft3|H/prime|3/angbracketright=/angbracketleftψ1(x)ψ0(y)ψ0(z)|λx2yz|ψ1(x)ψ0(y)ψ0(z)/angbracketright=λ/angbracketleftx2/angbracketright1/angbracketlefty/angbracketright0/angbracketleftz/angbracketright0=0,
/angbracketleft1|H/prime|2/angbracketright=/angbracketleftψ0(x)ψ0(y)ψ1(z)|λx2yz|ψ0(x)ψ1(y)ψ0(z)/angbracketright=λ/angbracketleftx2/angbracketright0/angbracketleft0|y|1/angbracketright/angbracketleft1|z|0/angbracketright
=λ/planckover2pi1
2mω|/angbracketleft0|x|1/angbracketright|2=λ/parenleftbigg/planckover2pi1
2mω/parenrightbigg2
[using Problems 2.11 and 3.33] .
/angbracketleft1|H/prime|3/angbracketright=/angbracketleftψ0(x)ψ0(y)ψ1(z)|λx2yz|ψ1(x)ψ0(y)ψ0(z)/angbracketright=λ/angbracketleft0|x2|1/angbracketright/angbracketlefty/angbracketright0/angbracketleft1|z|0/angbracketright=0,
/angbracketleft2|H/prime|3/angbracketright=/angbracketleftψ0(x)ψ1(y)ψ0(z)|λx2yz|ψ1(x)ψ0(y)ψ0(z)/angbracketright=λ/angbracketleft0|x2|1/angbracketright/angbracketleft1|y|0/angbracketright0/angbracketleftz/angbracketright0=0.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 181
W=
0a0
a00
000
,where a≡λ/parenleftbigg/planckover2pi1
2mω/parenrightbigg2
.
Eigenvalues of W:/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−Ea 0
a−E0
00−E/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−E
3+Ea2=0⇒E={0,±a}=0,±λ/parenleftbigg/planckover2pi1
2mω/parenrightbigg2
.
Problem 6.31
(a)The first term is the nucleus/nucleus interaction, the second is the interaction between the nucleus of
atom 2 and the electron in atom 1, the third is between nucleus 1 and electron 2, and the last term is theinteraction between the electrons.
1
R−x=1
R/parenleftBig
1−x
R/parenrightBig−1
=1
R/bracketleftbigg
1+/parenleftBigx
R/parenrightBig
+/parenleftBigx
R/parenrightBig2
+.../bracketrightbigg
,
so
H/prime∼=1
4πFepsilonC0e2
R/braceleftBigg
1−/bracketleftbigg
1+/parenleftBigx1
R/parenrightBig
+/parenleftBigx1
R/parenrightBig2/bracketrightbigg
−/bracketleftbigg
1−/parenleftBigx2
R/parenrightBig
+/parenleftBigx2
R/parenrightBig2/bracketrightbigg
+/bracketleftBigg
1+/parenleftbiggx1−x2
R/parenrightbigg
+/parenleftbiggx1−x2
R/parenrightbigg2/bracketrightBigg/bracerightBigg
≈1
4πFepsilonC0e2
R/parenleftbigg
−2x1x2
R2/parenrightbigg
=−e2x1x2
2πFepsilonC0R3./check
(b)Expanding Eq. 6.99:
H=1
2m/parenleftbig
p2
++p2
−/parenrightbig
+1
2k/parenleftbig
x2
++x2
−/parenrightbig
−e2
4πFepsilonC0R3/parenleftbig
x2
+−x2
−/parenrightbig
=1
2m/parenleftbig
p2
1+p2
2/parenrightbig
+1
2k/parenleftbig
x2
1+x2
2/parenrightbig
−e2
4πFepsilonC0R3(2x1x2)=H0+H/prime(Eqs. 6.96 and 6.98).
(c)
ω±=/radicalbigg
k
m/parenleftbigg
1∓e2
2πFepsilonC0R3k/parenrightbigg1/2
∼=ω0/bracketleftBigg
1∓1
2/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg
−1
8/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg2
+.../bracketrightBigg
.
∆V∼=1
2/planckover2pi1ω0/bracketleftbigg
1−1
2/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg
−1
8/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg2
+
1+1
2/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg
−1
8/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg2/bracketrightbigg
−/planckover2pi1ω0
=1
2/planckover2pi1ω0/parenleftbigg
−1
4/parenrightbigg/parenleftbigge2
2πFepsilonC0R3mω2
0/parenrightbigg2
=−1
8/planckover2pi1
m2ω3
0/parenleftbigge2
2πFepsilonC0/parenrightbigg21
R6./check
(d)In first order:
E1
0=/angbracketleft0|H/prime|0/angbracketright=−e2
2πFepsilonC0R3/angbracketleftψ0(x1)ψ0(x2)|x1x2|ψ0(x1)ψ0(x2)/angbracketright=−e2
2πFepsilonC0R3/angbracketleftx/angbracketright0/angbracketleftx/angbracketright0=0.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
182 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
In second order:
E2
0=∞/summationdisplay
n=1|/angbracketleftψn|H/prime|ψ0/angbracketright|2
E0−En.Here|ψ0/angbracketright=|0/angbracketright|0/angbracketright,|ψn/angbracketright=|n1/angbracketright|n2/angbracketright,so
=/parenleftbigge2
2πFepsilonC0R3/parenrightbigg2∞/summationdisplay
n1=1∞/summationdisplay
n2=1|/angbracketleftn1|x1|0/angbracketright|2|/angbracketleftn2|x2|0/angbracketright|2
E0,0−En1,n2[use Problem 3.33]
=/parenleftbigge2
2πFepsilonC0R3/parenrightbigg2|/angbracketleft1|x|0/angbracketright|2|/angbracketleft1|x|0/angbracketright|2
(1
2/planckover2pi1ω0+1
2/planckover2pi1ω0)−(3
2/planckover2pi1ω0+3
2/planckover2pi1ω0)[zero unless n1=n2=1 ]
=/parenleftbigge2
2πFepsilonC0R3/parenrightbigg2/parenleftbigg
−1
2/planckover2pi1ω0/parenrightbigg/parenleftbigg/planckover2pi1
2mω0/parenrightbigg2
=−/planckover2pi1
8m2ω3
0/parenleftbigge2
2πFepsilonC0/parenrightbigg21
R6./check
Problem 6.32
(a)Let the unperturbed Hamiltonian be H(λ0), for some fixed value λ0. Now tweak λtoλ0+dλ. The
perturbing Hamiltonian is H/prime=H(λ0+dλ)−H(λ0)=(∂H/∂λ )dλ(derivative evaluated at λ0).
The change in energy is given by Eq. 6.9:
dEn=E1
n=/angbracketleftψ0
n|H/prime|ψ0
n/angbracketright=/angbracketleftψn|∂H
∂λ|ψn/angbracketrightdλ(all evaluated at λ0); so∂En
∂λ=/angbracketleftψn|∂H
∂λ|ψn/angbracketright.
[Note: Even though we used perturbation theory, the result is exact, since all we needed (to calculate the
derivative) was the infinitesimal change in En.]
(b)En=(n+1
2)/planckover2pi1ω;H=−/planckover2pi12
2md2
dx2+1
2mω2x2.
(i)
∂En
∂ω=(n+1
2)/planckover2pi1;∂H
∂ω=mωx2; so F-H⇒(n+1
2)/planckover2pi1=/angbracketleftn|mωx2|n/angbracketright.But
V=1
2mω2x2,so/angbracketleftV/angbracketright=/angbracketleftn|1
2mω2x2|n/angbracketright=1
2ω(n+1
2)/planckover2pi1;/angbracketleftV/angbracketright=1
2(n+1
2)/planckover2pi1ω.
(ii)
∂En
∂/planckover2pi1=(n+1
2)ω;∂H
∂/planckover2pi1=−/planckover2pi1
md2
dx2=2
/planckover2pi1/parenleftbigg
−/planckover2pi12
2md2
dx2/parenrightbigg
=2
/planckover2pi1T;
so F-H⇒(n+1
2)ω=2
/planckover2pi1/angbracketleftn|T|n/angbracketright,or/angbracketleftT/angbracketright=1
2(n+1
2)/planckover2pi1ω.
(iii)
∂En
∂m=0 ;∂H
∂m=/planckover2pi12
2m2d2
dx2+1
2ω2x2=−1
m/parenleftbigg
−/planckover2pi12
2md2
dx2/parenrightbigg
+1
m/parenleftbigg1
2mω2x2/parenrightbigg
=−1
mT+1
mV.
So F-H⇒0=−1
m/angbracketleftT/angbracketright+1
m/angbracketleftV/angbracketright,o r/angbracketleftT/angbracketright=/angbracketleftV/angbracketright.These results are consistent with what we found in
Problems 2.12 and 3.31.
c/circlecopyrt2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 183
Problem 6.33
(a)
∂En
∂e=−4me3
32π2FepsilonC2
0/planckover2pi12(jmax+l+1 )2=4
eEn;∂H
∂e=−2e
4πFepsilonC01
r.So the F-H theorem says:
4
eEn=−e
2πFepsilonC0/angbracketleftbigg1
r/angbracketrightbigg
,or/angbracketleftbigg1
r/angbracketrightbigg
=−8πFepsilonC0
e2En=−8πFepsilonC0E1
e2n2=−8πFepsilonC0
e2/bracketleftBigg
−m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg2/bracketrightBigg
1
n2=e2m
4πFepsilonC0/planckover2pi121
n2.
But4πFepsilonC0/planckover2pi12
me2=a(by Eq.4.72),so/angbracketleftbigg1
r/angbracketrightbigg
=1
n2a.(Agrees with Eq. 6.55.)
(b)
∂En
∂l=2me4
32π2FepsilonC2
0/planckover2pi12(jmax+l+1 )3=−2En
n;∂H
∂l=/planckover2pi12
2mr2(2l+ 1); so F-H says
−2En
n=/planckover2pi12(2l+1 )
2m/angbracketleftbigg1
r2/angbracketrightbigg
,or/angbracketleftbigg1
r2/angbracketrightbigg
=−4mEn
n(2l+1 )/planckover2pi12=−4mE1
n3(2l+1 )/planckover2pi12.
But−4mE1
/planckover2pi12=2
a2,so/angbracketleftbigg1
r2/angbracketrightbigg
=1
n3(l+1
2)a2.(Agrees with Eq. 6.56.)
Problem 6.34
Equation 4 .53⇒u/prime/prime=/bracketleftbiggl(l+1 )
r2−2mEn
/planckover2pi12−2m
/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg1
r/bracketrightbigg
u.
Butme2
4πFepsilonC0/planckover2pi12=1
a(Eq.4.72),and−2mEn
/planckover2pi12=2m
/planckover2pi12m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg21
n2=1
a2n2.So
⋆u/prime/prime=/bracketleftbiggl(l+1 )
r2−2
ar+1
n2a2/bracketrightbigg
u.
∴/integraldisplay
(ursu/prime/prime)dr=/integraldisplay
urs/bracketleftbiggl(l+1 )
r2−2
ar+1
n2a2/bracketrightbigg
udr=l(l+1 )/angbracketleftrs−2/angbracketright−2
a/angbracketleftrs−1/angbracketright+1
n2a2/angbracketleftrs/angbracketright
/diamondsolid =−/integraldisplayd
dr(urs)u/primedr=−/integraldisplay
(u/primersu/prime)dr−s/integraldisplay
(urs−1u/prime)dr.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
184 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Lemma 1 :/integraldisplay
(ursu/prime)dr=−/integraldisplayd
dr(urs)udr=−/integraldisplay
(u/primersu)dr−s/integraldisplay
urs−1udr⇒
2/integraldisplay
(ursu/prime)dr=−s/angbracketleftrs−1/angbracketright,or/integraldisplay
(ursu/prime)dr=−s
2/angbracketleftrs−1/angbracketright.
Lemma 2 :/integraldisplay
(u/prime/primers+1u/prime)dr=−/integraldisplay
u/primed
dr(rs+1u/prime)dr=−(s+1 )/integraldisplay
(u/primersu/prime)dr−/integraldisplay
(u/primers+1u/prime/prime)dr.
2/integraldisplay
(u/prime/primers+1u/prime)dr=−(s+1 )/integraldisplay
(u/primersu/prime)dr,or:/integraldisplay
(u/primersu/prime)dr=−2
s+1/integraldisplay
(u/prime/primers+1u/prime)dr.
Lemma 3 : Use⋆in Lemma 2, and exploit Lemma 1:
/integraldisplay
(u/primersu/prime)dr=−2
s+1/integraldisplay/bracketleftbiggl(l+1 )
r2−2
ar+1
n2a2/bracketrightbigg
(urs+1u/prime)dr
=−2
s+1/bracketleftbigg
l(l+1 )/integraldisplay
(urs−1u/prime)dr−2
a/integraldisplay
(ursu/prime)dr+1
n2a2/integraldisplay
(urs+1u/prime)dr/bracketrightbigg
=−2
s+1/bracketleftbigg
l(l+1 )/parenleftbigg
−s−1
2/angbracketleftrs−2/angbracketright/parenrightbigg
−2
a/parenleftBig
−s
2/angbracketleftrs−1/angbracketright/parenrightBig
+1
n2a2/parenleftbigg
−s+1
2/angbracketleftrs/angbracketright/parenrightbigg/bracketrightbigg
=l(l+1 )/parenleftbiggs−1
s+1/parenrightbigg
/angbracketleftrs−2/angbracketright−2
a/parenleftbiggs
s+1/parenrightbigg
/angbracketleftrs−1/angbracketright+1
n2a2/angbracketleftrs/angbracketright.
Plug Lemmas 1 and 3 into /diamondsolid:
l(l+1 )/angbracketleftrs−2/angbracketright−2
a/angbracketleftrs−1/angbracketright+1
n2a2/angbracketleftrs/angbracketright
=−l(l+1 )/parenleftbiggs−1
s+1/parenrightbigg
/angbracketleftrs−2/angbracketright+2
a/parenleftbiggs
s+1/parenrightbigg
/angbracketleftrs−1/angbracketright−1
n2a2/angbracketleftrs/angbracketright+s(s−1)
2/angbracketleftrs−2/angbracketright.
2
n2a2/angbracketleftrs/angbracketright−2
a/bracketleftbigg
1+s
s+1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
2s+1
s+1/angbracketleftrs−1/angbracketright+/braceleftbigg
l(l+1 )/bracketleftbigg
1+s−1
s+1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
2s
s+1−s(s−1)
2/bracerightbigg
/angbracketleftrs−2/angbracketright=0.
2(s+1 )
n2a2/angbracketleftrs/angbracketright−2
a(2s+1 )/angbracketleftrs−1/angbracketright+2s/bracketleftbigg
l2+l−(s2−1)
4/bracketrightbigg
/angbracketleftrs−2/angbracketright=0,or, finally,
(s+1 )
n2/angbracketleftrs/angbracketright−a(2s+1 )/angbracketleftrs−1/angbracketright+sa2
4(4l2+4l+1/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
(2l+1)2−s2)/angbracketleftrs−2/angbracketright=0.QED
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 185
Problem 6.35
(a)
1
n2/angbracketleft1/angbracketright−a/angbracketleftbigg1
r/angbracketrightbigg
+0=0⇒/angbracketleftbigg1
r/angbracketrightbigg
=1
n2a.
2
n2/angbracketleftr/angbracketright−3a/angbracketleft1/angbracketright+1
4/bracketleftbig
(2l+1 )2−1/bracketrightbig
a2/angbracketleftbigg1
r/angbracketrightbigg
=0⇒2
n2/angbracketleftr/angbracketright=3a−l(l+1 )a21
n2a=a
n2/bracketleftbig
3n2−l(l+1 )/bracketrightbig
.
/angbracketleftr/angbracketright=a
2/bracketleftbig
3n2−l(l+1 )/bracketrightbig
.
3
n2/angbracketleftr2/angbracketright−5a/angbracketleftr/angbracketright+1
2/bracketleftbig
(2l+1 )2−4/bracketrightbig
a2=0⇒3
n2/angbracketleftr2/angbracketright=5aa
2/bracketleftbig
3n2−l(l+1 )/bracketrightbig
−a2
2/bracketleftbig
(2l+1 )2−4/bracketrightbig
3
n2/angbracketleftr2/angbracketright=a2
2/bracketleftbig
15n2−5l(l+1 )−4l(l+1 )−1+4/bracketrightbig
=a2
2/bracketleftbig
15n2−9l(l+1 )+3/bracketrightbig
=3a2
2/bracketleftbig
5n2−3l(l+1 )+1/bracketrightbig
;/angbracketleftr2/angbracketright=n2a2
2/bracketleftbig
5n2−3l(l+1 )+1/bracketrightbig
.
4
n2/angbracketleftr3/angbracketright−7a/angbracketleftr2/angbracketright+3
4/bracketleftbig
(2l+1 )2−9/bracketrightbig
a2/angbracketleftr/angbracketright=0=⇒
4
n2/angbracketleftr3/angbracketright=7an2a2
2/bracketleftbig
5n2−3l(l+1 )+1/bracketrightbig
−3
4[4l(l+1 )−8]a2a
2/bracketleftbig
3n2−l(l+1 )/bracketrightbig
=a3
2/braceleftbig
35n4−21l(l+1 )n2+7n2−[3l(l+1 )−6]/bracketleftbig
3n2−l(l+1 )/bracketrightbig/bracerightbig
=a3
2/bracketleftbig
35n4−21l(l+1 )n2+7n2−9l(l+1 )n2+3l2(l+1 )2+1 8n2−6l(l+1 )/bracketrightbig
=a3
2/bracketleftbig
35n4+2 5n2−30l(l+1 )n2+3l2(l+1 )2−6l(l+1 )/bracketrightbig
.
/angbracketleftr3/angbracketright=n2a3
8/bracketleftbig
35n4+2 5n2−30l(l+1 )n2+3l2(l+1 )2−6l(l+1 )/bracketrightbig
.
(b)
0+a/angbracketleftbigg1
r2/angbracketrightbigg
−1
4/bracketleftbig
(2l+1 )2−1/bracketrightbig
a2/angbracketleftbigg1
r3/angbracketrightbigg
=0⇒/angbracketleftbigg1
r2/angbracketrightbigg
=al(l+1 )/angbracketleftbigg1
r3/angbracketrightbigg
.
(c)
al(l+1 )/angbracketleftbigg1
r3/angbracketrightbigg
=1
(l+1/2)n3a2⇒/angbracketleftbigg1
r3/angbracketrightbigg
=1
l(l+1/2)(l+1 )n3a3.Agrees with Eq. 6.64.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
186 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.36
(a)
|100/angbracketright=1√
πa3e−r/a(Eq.4.80),E1
s=/angbracketleft100|H/prime|100/angbracketright=eEext1
πa3/integraldisplay
e−2r/a(rcosθ)r2sinθdrdθdφ.
But the θintegral is zero:/integraldisplayπ
0cosθsinθdθ=sin2θ
2/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=0.SoE1
s=0.QED
(b)From Problem 4.11:
|1/angbracketright=ψ
200=1√
2πa1
2a/parenleftBig
1−r
2a/parenrightBig
e−r/2a
|2/angbracketright=ψ211=−1√πa1
8a2re−r/2asinθeiφ
|3/angbracketright=ψ210=1√
2πa1
4a2re−r/2acosθ
|4/angbracketright=ψ21−1=1√πa1
8a2re−r/2asinθe−iφ
/angbracketleft1|H/prime
s|1/angbracketright={...}/integraldisplayπ
0cosθsinθdθ=0
/angbracketleft2|H/prime
s|2/angbracketright={...}/integraldisplayπ
0sin2θcosθsinθdθ=0
/angbracketleft3|H/prime
s|3/angbracketright={...}/integraldisplayπ
0cos2θcosθsinθdθ=0
/angbracketleft4|H/prime
s|4/angbracketright={...}/integraldisplayπ
0sin2θcosθsinθdθ=0
/angbracketleft1|H/prime
s|2/angbracketright={...}/integraldisplay2π
0eiφdφ=0
/angbracketleft1|H/prime
s|4/angbracketright={...}/integraldisplay2π
0e−iφdφ=0
/angbracketleft2|H/prime
s|3/angbracketright={...}/integraldisplay2π
0e−iφdφ=0
/angbracketleft2|H/prime
s|4/angbracketright={...}/integraldisplay2π
0e−2iφdφ=0
/angbracketleft3|H/prime
s|4/angbracketright={...}/integraldisplay2π
0e−iφdφ=0
All matrix elements of H
/prime
sare zero
except/angbracketleft1|H/prime
s|3/angbracketrightand/angbracketleft3|H/prime
s|1/angbracketright
(which are complex conjugates,so only one needs to be evaluated).
/angbracketleft1|H
/prime
s|3/angbracketright=eEext1√
2πa1
2a1√
2πa1
4a2/integraldisplay/parenleftBig
1−r
2a/parenrightBig
e−r/2are−r/2acosθ(rcosθ)r2sinθdrdθdφ
=eEext
2πa8a3(2π)/bracketleftbigg/integraldisplayπ
0cos2θsinθdθ/bracketrightbigg/integraldisplay∞
0/parenleftBig
1−r
2a/parenrightBig
e−r/ar4dr
=eEext
8a42
3/braceleftbigg/integraldisplay∞
0r4e−r/adr−1
2a/integraldisplay∞
0r5e−r/adr/bracerightbigg
=eEext
12a4/parenleftbigg
4!a5−1
2a5!a6/parenrightbigg
=eEext
12a424a5/parenleftbigg
1−5
2/parenrightbigg
=eaEext(−3) =−3aeEext.
W=−3aeEext
0010
000010000000
.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 187
We need the eigenvalues of this matrix. The characteristic equation is:
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ010
0−λ00
10−λ0
000−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ00
0−λ0
00−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle0−λ0
10 000−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λ(−λ)
3+(−λ2)=λ2(λ2−1) = 0.
The eigenvalues are 0, 0, 1, and −1, so the perturbed energies are
E2,E2,E2+3aeEext,E2−3aeEext. Three levels.
(c)The eigenvectors with eigenvalue 0 are |2/angbracketright=
0
100
and|4/angbracketright=
0
001
; the eigenvectors with eigenvalues ±1
are|±/angbracketright≡1
√
2
1
0
±1
0
. So the “good” states are ψ211,ψ21−1,1√
2(ψ200+ψ210),1√
2(ψ200−ψ210).
/angbracketleftpe/angbracketright4=−e1
πa1
64a4/integraldisplay
r2e−r/asin2θ/bracketleftBig
rsinθcosφˆi+rsinθsinφˆj+rcosθˆk/bracketrightBig
r2sinθdrdθdφ.
But/integraldisplay2π
0cosφdφ=/integraldisplay2π
0sinφdφ=0,/integraldisplayπ
0sin3θcosθdθ=/vextendsingle/vextendsingle/vextendsingle/vextendsinglesin
4θ
4/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=0,so
/angbracketleftpe/angbracketright4= 0. Likewise /angbracketleftpe/angbracketright2=0 .
/angbracketleftpe/angbracketright±=−1
2e/integraldisplay
(ψ1±ψ3)2(r)r2sinθdrdθdφ
=−1
2e1
2πa1
4a2/integraldisplay/bracketleftBig/parenleftBig
1−r
2a/parenrightBig
±r
2acosθ/bracketrightBig2
e−r/ar(sinθcosφˆi+ sinθsinφˆj+ cosθˆk)r2sinθdrdθdφ
=−e
2ˆk
2πa1
4a22π/integraldisplay/bracketleftBig/parenleftBig
1−r
2a/parenrightBig
±r
2acosθ/bracketrightBig2
r3e−r/acosθsinθdrdθ.
But/integraltextπ
0cosθsinθdθ=/integraltextπ
0cos3θsinθdθ= 0, so only the cross-term survives:
/angbracketleftpe/angbracketright±=−e
8a3ˆk/parenleftbigg
±1
a/parenrightbigg/integraldisplay/parenleftBig
1−r
2a/parenrightBig
rcosθr3e−r/acosθsinθdrdθ
=∓/parenleftBige
8a4ˆk/parenrightBig/bracketleftbigg/integraldisplayπ
0cos2θsinθdθ/bracketrightbigg/integraldisplay∞
0/parenleftBig
1−r
2a/parenrightBig
r4e−r/adr=∓/parenleftBige
8a4ˆk/parenrightBig2
3/bracketleftbigg
4!a5−1
2a5!a6/bracketrightbigg
=∓eˆk/parenleftbigg1
12a4/parenrightbigg
24a5/parenleftbigg
1−5
2/parenrightbigg
=±3aeˆk.
Problem 6.37
(a)The nine states are:
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188 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
l=0
:|300/angbracketright=R30Y0
0
l=1:|311/angbracketright=R31Y1
1
|310/angbracketright=R31Y0
1
|31−1/angbracketright=R31Y−1
1
l=2:|322/angbracketright=R32Y2
2
|321/angbracketright=R32Y1
2
|320/angbracketright=R32Y0
2
|32−1/angbracketright=R32Y−1
2
|32−2/angbracketright=R32Y−2
2
H/prime
scontains no φdependence, so the φintegral will be:
/angbracketleftnlm|H/prime
s|n/primel/primem/prime/angbracketright={···}/integraldisplay2π
0e−imφeim/primeφdφ, which is zero unless m/prime=m.
For diagonal elements: /angbracketleftnlm|H/prime
s|nlm/angbracketright={···}/integraltextπ
0[Pm
l(cosθ)]2cosθsinθdθ. But (p. 137 in the text)
Pm
lis a polynomial (even or odd) in cos θ, multiplied (if mis odd) by sin θ. Since sin2θ=1−cos2θ,
[Pm
l(cosθ)]2is a polynomial in even powers of cos θ. So the θintegral is of the form
/integraldisplayπ
0(cosθ)2j+1sinθdθ=−(cosθ)2j+2
(2j+2 )/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=0.All diagonal elements are zero .
There remain just 4 elements to calculate:
m=m/prime=0:/angbracketleft300|H/prime
s|310/angbracketright,/angbracketleft300|H/prime
s|320/angbracketright,/angbracketleft310|H/prime
s|320/angbracketright;m=m/prime=±1:/angbracketleft31±1|H/prime
s|32±1/angbracketright.
/angbracketleft300|H/prime
s|310/angbracketright=eEext/integraldisplay
R30R31r3dr/integraldisplay
Y0
0Y0
1cosθsinθdθdφ. From Table 4.7 :
/integraldisplay
R30R31r3dr=2√
271
a3/28
27√
61
a3/21
a/integraldisplay/parenleftbigg
1−2r
3a+2r2
27a2/parenrightbigg
e−r/3a/parenleftBig
1−r
6a/parenrightBig
re−r/3ar3dr.
Letx≡2r/3a:
/integraldisplay
R30R31r3dr=24
35√
2a4/parenleftbigg3a
2/parenrightbigg5/integraldisplay∞
0/parenleftbigg
1−x+x2
6/parenrightbigg/parenleftBig
1−x
4/parenrightBig
x4e−xdx
=a
2√
2/integraldisplay∞
0/parenleftbigg
1−5
4x+5
12x2−1
24x3/parenrightbigg
x4e−xdx=a
2√
2/parenleftbigg
4!−5
45! +5
126!−1
247!/parenrightbigg
=−9√
2a.
/integraldisplay
Y0
0Y0
1cosθsinθdθdφ =1√
4π/radicalbigg
3
4π/integraldisplay
cosθcosθsinθdθdφ =√
3
4π2π/integraldisplayπ
0cos3θsinθdθ=√
3
22
3=√
3
3.
/angbracketleft300|H/prime
s|310/angbracketright=eEext(−9√
2a)/parenleftBigg√
3
3/parenrightBigg
=−3√
6aeEext.
/angbracketleft300|H/prime
s|320/angbracketright=eEext/integraldisplay
R30R31r3dr/integraldisplay
Y0
0Y0
2cosθsinθdθdφ.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 189
/integraldisplay
Y0
0Y0
2cosθsinθdθdφ =1√
4π/radicalbigg
5
16π/integraldisplay
(3 cos2θ−1) cosθsinθdθdφ =0./angbracketleft300|H/prime
s|320/angbracketright=0 .
/angbracketleft310|H/prime
s|320/angbracketright=eEext/integraldisplay
R31R32r3dr/integraldisplay
Y0
1Y0
2cosθsinθdθdφ.
/integraldisplay
R31R32r3dr=8
27√
61
a3/21
a4
81√
301
a3/21
a2/integraldisplay/parenleftBig
1−r
6a/parenrightBig
re−r/3ar2e−r/3ar3dr
=24
38√
5a6/parenleftbigg3a
2/parenrightbigg7/integraldisplay∞
0/parenleftBig
1−x
4/parenrightBig
x6e−xdx=a
24√
5/parenleftbigg
6!−1
47!/parenrightbigg
=−9√
5
2a.
/integraldisplay
Y0
1Y0
2sinθcosθdθdφ =/radicalbigg
3
4π/radicalbigg
5
16π/integraldisplay
cosθ(3 cos2θ−1) cosθsinθdθdφ
=√
15
8π2π/integraldisplayπ
0(3 cos4θ−cos2θ) sinθdθ=√
15
4/bracketleftbigg
−3
5cos5θ+1
3cos3θ/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=2√
15.
/angbracketleft310|H/prime
s|320/angbracketright=eEext/parenleftBigg
−9√
5
2a/parenrightBigg/parenleftbigg2√
15/parenrightbigg
=−3√
3aeEext.
/angbracketleft31±1|H/prime
s|32±1/angbracketright=eEext/integraldisplay
R31R32r3dr/integraldisplay/parenleftbig
Y±1
1/parenrightbig∗Y±1
2cosθsinθdθdφ.
/integraldisplay/parenleftbig
Y±1
1/parenrightbig∗Y±1
2cosθsinθdθdφ =/parenleftBigg
∓/radicalbigg
3
8π/parenrightBigg/parenleftBigg
∓/radicalbigg
15
8π/parenrightBigg/integraldisplay
sinθe∓iφsinθcosθe±iφcosθsinθdθdφ
=3√
5
8π2π/integraldisplayπ
0cos2θ(1−cos2θ) sinθdθ=3
4√
5/parenleftbigg
−cos3θ
3+cos5θ
5/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0
=1√
5.
/angbracketleft31±1|H/prime
s|32±1/angbracketright=eEext/parenleftBigg
−9√
5
2a/parenrightBigg/parenleftbigg1√
5/parenrightbigg
=−9
2aeEext.
Thus the matrix representing H/prime
sis (all empty boxes are zero; all numbers multiplied by −aeEext):
(b)The perturbing matrix (below) breaks into a 3 ×3 block, two 2 ×2 blocks, and two 1 ×1 blocks, so we can
work out the eigenvalues in each block separately.
3×3:3√
3
0√
20√
201
01 0
;/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ√
20√
2−λ1
01−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λ
3+λ+2λ=−λ(λ2−3) = 0;
λ=0,±√
3⇒E1
1=0,E1
2=9aeEext,E1
3=−9aeEext.
2×2:9
2/parenleftbigg01
10/parenrightbigg
;/vextendsingle/vextendsingle/vextendsingle/vextendsingle−λ1
1−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle=λ
2−1=0⇒λ=±1.
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190 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
300 310 320 311 321 31-1 32-1 322 32-2
300
310
320
311
321
31-1
32-1
322
32-23√63√6
3√3
3√3
9/29/29/2
9/2
E1
4=9
2aeEext,E1
5=−9
2aeEext.From the other 2 ×2 we get E1
6=E1
4,E1
7=E1
5, and from the 1 ×1’s we
getE1
8=E1
9= 0. Thus the perturbations to the energy ( E3) are:0 (degeneracy 3)
(9/2)aeEext (degeneracy 2)
−(9/2)aeEext(degeneracy 2)
9aeEext (degeneracy 1)
−9aeEext (degeneracy 1)
Problem 6.38
Equation 6 .89⇒E1
hf=µ0gde2
3πmdmea3/angbracketleftSd·Se/angbracketright;E q.6.91⇒Sd·Se=1
2(S2−S2
e−S2
d).
Electron has spin1
2,s oS2
e=1
2/parenleftbig3
2/parenrightbig
/planckover2pi12=3
4/planckover2pi12; deuteron has spin 1, so S2
d= 1(2) /planckover2pi12=2/planckover2pi12.
Total spin could be3
2[in which case S2=3
2/parenleftbig5
2/parenrightbig
/planckover2pi12=15
4/planckover2pi12]o r1
2[in which case S2=3
4/planckover2pi12]. Thus
/angbracketleftSd·Se/angbracketright=
1
2/parenleftbig15
4/planckover2pi12−3
4/planckover2pi12−2/planckover2pi12/parenrightbig
=1
2/planckover2pi12
1
2/parenleftbig3
4/planckover2pi12−3
4/planckover2pi12−2/planckover2pi12/parenrightbig
=−/planckover2pi12
; the difference is3
2/planckover2pi12,so ∆E=µ0gde2/planckover2pi12
2πmdmea3.
Butµ0FepsilonC0=1
c2⇒µ0=1
FepsilonC0c2,so ∆E=2gde2/planckover2pi12
4πFepsilonC0mdmec2a3=2gd/planckover2pi14
mdm2ec2a4=3
2gd
gpmp
md∆Ehydrogen (Eq. 6.98) .
Now,λ=c
ν=ch
∆E,soλd=2
3gp
gdmd
mpλh, and since md=2mp,λd=4
3/parenleftbigg5.59
1.71/parenrightbigg
(21 cm) = 92 cm.
Problem 6.39
(a)The potential energy of the electron (charge −e)a t(x,y,z) due to q’s atx=±dalone is:
V=−eq
4πFepsilonC0/bracketleftBigg
1/radicalbig
(x+d)2+y2+z2+1/radicalbig
(x−d)2+y2+z2/bracketrightBigg
.Expanding (with d/greatermuchx,y,z):
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 191
1/radicalbig
(x±d)2+y2+z2=(x2±2dx+d2+y2+z2)−1/2=(d2±2dx+r2)−1/2=1
d/parenleftbigg
1±2x
d+r2
d2/parenrightbigg−1/2
≈1
d/parenleftbigg
1∓x
d−r2
2d2+3
84x2
d2/parenrightbigg
=1
d/bracketleftbigg
1∓x
d+1
2d2(3x2−r2)/bracketrightbigg
.
V=−eq
4πFepsilonC0d/bracketleftbigg
1−x
d+1
2d2(3x2−r2)+1+x
d+1
2d2(3x2−r2)/bracketrightbigg
=−2eq
4πFepsilonC0d−eq
4πFepsilonC0d3(3x2−r2)
=2βd2+3βx2−βr2,where β≡−e
4πFepsilonC0q
d3.
Thus with all six charges in place
H/prime=2 (β1d2
1+β2d2
2+β3d2
3)+3 (β1x2+β2y2+β3z2)−r2(β1+β2+β3).QED
(b)/angbracketleft100|H/prime|100/angbracketright=1
πa3/integraldisplay
e−2r/aH/primer2sinθdrdθdφ
=V0+3
πa3/integraldisplay
e−2r/a(β1x2+β2y2+β3z2)r2sinθdrdθdφ−(β1+β2+β3)
πa3/integraldisplay
r2e−2r/ar2sinθdrdθdφ.
I1≡/integraldisplay
r2e−2r/ar2sinθdrdθdφ =4π/integraldisplay∞
0r4e−2r/adr=4π4!(a
2)5=3πa5.
I2≡/integraldisplay
e−2r/a(β1x2+β2y2+β3z2)r2sinθdrdθdφ
=/integraldisplay
r4e−2r/a(β1sin2θcos2φ+β2sin2θsin2φ+β3cos2θ) sinθdrdθdφ.
But/integraldisplay2π
0cos2φdφ=/integraldisplay2π
0sin2φdφ=π,/integraldisplay2π
0dφ=2π.So
=/integraldisplay∞
0r4e−2r/adr/integraldisplayπ
0/bracketleftbig
π(β1+β2) sin2θ+2πβ3cos2θ/bracketrightbig
sinθdθ.
But/integraldisplayπ
0sin3θdθ=4
3,/integraldisplayπ
0cos2θsinθdθ=2
3.So
=4 !/parenleftBiga
2/parenrightBig5/bracketleftbigg4π
3(β1+β2)+4π
3β3/bracketrightbigg
=πa5(β1+β2+β3).
/angbracketleft100|H/prime|100/angbracketright=V0+3
πa3πa5(β1+β2+β3)−(β1+β2+β3)
πa33πa5=V0.
(c)The four states are
|200/angbracketright=R
20Y0
0
|211/angbracketright=R21Y1
1
|21−1/angbracketright=R21Y−1
1
|210/angbracketright=R21Y0
1
(functional forms in Problem 4.11).
Diagonal elements:/angbracketleftnlm|H/prime|nlm/angbracketright=V0+3/parenleftbig
β1/angbracketleftx2/angbracketright+β2/angbracketlefty2/angbracketright+β3/angbracketleftz2/angbracketright/parenrightbig
−(β1+β2+β3)/angbracketleftr2/angbracketright.
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192 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
For|200/angbracketright,/angbracketleftx2/angbracketright=/angbracketlefty2/angbracketright=/angbracketleftz2/angbracketright=1
3/angbracketleftr2/angbracketright(Y0
0does not depend on φ,θ; this state has spherical symmetry),
so/angbracketleft200|H/prime|200/angbracketright=V0.(I could have used the same argument in (b).)
From Problem 6.35(a), /angbracketleftr2/angbracketright=n2a2
2/bracketleftbig
5n2−3l(l+1 )+1/bracketrightbig
, so for n=2,l=1:/angbracketleftr2/angbracketright=3 0a2. Moreover,
since/angbracketleftx2/angbracketright={...}/integraldisplay2π
0cos2φdφ={...}/integraldisplay2π
0sin2φdφ=/angbracketlefty2/angbracketright, and/angbracketleftx2/angbracketright+/angbracketlefty2/angbracketright+/angbracketleftz2/angbracketright=/angbracketleftr2/angbracketright, it follows
that/angbracketleftx2/angbracketright=/angbracketlefty2/angbracketright=1
2(/angbracketleftr2/angbracketright−/angbracketleftz2/angbracketright)=1 5a2−1
2/angbracketleftz2/angbracketright. So all we need to calculate is /angbracketleftz2/angbracketright.
/angbracketleft210|z2|210/angbracketright=1
2πa1
16a4/integraldisplay
r2e−r/acos2θ(r2cos2θ)r2sinθdrdθdφ
=1
16a5/integraldisplay∞
0r6e−r/adr/integraldisplayπ
0cos4θsinθdθ=1
16a56!a72
5=1 8a2;/angbracketleftx2/angbracketright=/angbracketlefty2/angbracketright=1 5a2−9a2=6a2.
/angbracketleft210|H/prime|210/angbracketright=V0+ 3(6a2β1+6a2β2+1 8a2β3)−30a2(β1+β2+β3)
=V0−12a2(β1+β2+β3)+3 6a2β3.
/angbracketleft21±1|z2|21±1/angbracketright=1
πa1
64a4/integraldisplay
r2e−r/asin2θ(r2cos2θ)r2sinθdrdθdφ
=1
32a5/integraldisplay∞
0r6e−r/adr/integraldisplayπ
0(1−cos2θ) cos2θsinθdθ=1
32a56!a7/parenleftbigg2
3−2
5/parenrightbigg
=6a2;
/angbracketleftx2/angbracketright=/angbracketlefty2/angbracketright=1 5a2−3a2=1 2a2.
/angbracketleft21±1|H/prime|21±1/angbracketright=V0+ 3(12a2β1+1 2a2β2+6a2β3)−30a2(β1+β2+β3)
=V0+6a2(β1+β2+β3)−18a2β3.
Off-diagonal elements : We need/angbracketleft200|H/prime|210/angbracketright,/angbracketleft200|H/prime|21±1/angbracketright,/angbracketleft210|H/prime|21±1/angbracketright, and/angbracketleft21−1|H/prime|211/angbracketright.
Now/angbracketleftnlm|V0|n/primel/primem/prime/angbracketright= 0, by orthogonality, and /angbracketleftnlm|r2|n/primel/primem/prime/angbracketright= 0, by orthogonality of Ym
l,s o
all we need are the matrix elements of x2andy2(/angbracketleft|z2|/angbracketright=−/angbracketleft|x2|/angbracketright−/angbracketleft|y2/angbracketright). For/angbracketleft200|x2|21±1/angbracketrightand
/angbracketleft210|x2|21±1/angbracketrighttheφintegral is/integraltext2π
0cos2φe±iφdφ=/integraltext2π
0cos3φdφ±i/integraltext2π
0cos2φsinφdφ= 0, and the
same goes for y2.S o/angbracketleft200|H/prime|21±1/angbracketright=/angbracketleft210|H/prime|21±1/angbracketright=0 .
For/angbracketleft200|x2|210/angbracketrightand/angbracketleft200|y2|210/angbracketrighttheθintegral is/integraltextπ
0cosθ(sin2θ) sinθdθ= sin4θ/4/vextendsingle/vextendsingleπ
0=0 , s o
/angbracketleft200|H/prime|210/angbracketright=0.Finally:
/angbracketleft21−1|x2|211/angbracketright=−1
πa1
64a4/integraldisplay
r2e−r/asin2θe2iφ(r2sin2θcos2φ)r2sinθdrdθdφ
=−1
64πa5/integraldisplay∞
0r6e−r/adr
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
6!a7/integraldisplayπ
0sin5θdθ
/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
16/15/integraldisplay2π
0e2iφcos2φdφ
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
π/2
=−1
64πa56!a716
15π
2=−6a2.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 193
Fory2, theφintegral is/integraltext2π
0e2iφsin2φdφ=−π/2, so/angbracketleft21−1|y2|211/angbracketright=6a2, and/angbracketleft21−1|z2|211/angbracketright=0 .
/angbracketleft21−1|H/prime|211/angbracketright=3/bracketleftbig
β1(−6a2)+β2(6a2)/bracketrightbig
=−18a2(β1−β2).
The perturbation matrix is:
200 210 211 21 - 1
200 V0 0 0 0
210 0V0−12a2(β1+β2)+2 4a2β3 0 0
211 0 0 V0+6a2(β1+β2)−12a2β3−18a2(β1−β2)
21 - 1 0 0 −18a2(β1−β2) V0+6a2(β1+β2)−12a2β3
The 2×2 block has the form/parenleftbiggAB
BA/parenrightbigg
; its characteristic equation is ( A−λ)2−B2=0 ,s o A−λ=±B,
or
λ=A∓B=V0+6a2(β1+β2)−12a2β3±18a2(β1−β2)=/braceleftbiggV0+2 4a2β1−12a2β2−12a2β3,
V0−12a2β1+2 4a2β2−12a2β3.
The first-order corrections to the energy ( E2) are therefore:FepsilonC1=V0
FepsilonC2=V0−12a2(β1+β2−2β3)
FepsilonC3=V0−12a2(−2β1+β2+β3)
FepsilonC4=V0−12a2(β1−2β2+β3)
(i) Ifβ1=β2=β3,thenFepsilonC1=FepsilonC2=FepsilonC3=FepsilonC4=V0:one level (still 4-fold degenerate).
(ii) Ifβ1=β2/negationslash=β3,thenFepsilonC1=V0,FepsilonC2=V0−24a2(β1−β3),FepsilonC3=FepsilonC4=V0+1 2a2(β1−β3):three levels
(one remains doubly degenerate).
(iii) If all three β’s are different, there are four levels (no remaining degeneracy).
Problem 6.40
(a)(i) Equation 6.10: ( H0−E0
0)ψ1
0=−(H/prime−E1
0)ψ0
0.
H0=−/planckover2pi12
2m∇2−e2
4πFepsilonC01
r=−/planckover2pi12
2m/parenleftbigg
∇2+2
ar/parenrightbigg
,sincea=4πFepsilonC0/planckover2pi12
me2.
E0
0=−/planckover2pi12
2ma2.
H/prime=eEextrcosθ;E1
0= 0 (Problem6 .36(a)).
ψ0
0=1√
πa3e−r/a;ψ1
0=f(r)e−r/acosθ.
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194 CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Equation 4.13 ⇒
∇2ψ1
0=cosθ
r2d
dr/bracketleftbigg
r2d
dr/parenleftBig
fe−r/a/parenrightBig/bracketrightbigg
+fe−r/a
r2sinθd
dθ/bracketleftbigg
sinθd
dθ(cosθ)/bracketrightbigg
=cosθ
r2d
dr/bracketleftbigg
r2/parenleftbigg
f/prime−1
af/parenrightbigg
e−r/a/bracketrightbigg
+fe−r/a
r2sinθd
dθ/bracketleftbig
−sin2θ/bracketrightbig
=cosθ
r2/bracketleftbigg
2r/parenleftbigg
f/prime−1
af/parenrightbigg
e−r/a+r2/parenleftbigg
f/prime/prime−2
af/prime+1
a2f/parenrightbigg
e−r/a/bracketrightbigg
−2 cosθ
r2fe−r/a
= cosθe−r/a/bracketleftbigg/parenleftbigg
f/prime/prime−2
af/prime+1
a2f/parenrightbigg
+2/parenleftbigg
f/prime−1
af/parenrightbigg1
r−2f1
r2/bracketrightbigg
.
Plug this into Eq. 6.10:
−/planckover2pi12
2mcosθe−r/a/bracketleftbigg/parenleftbigg
f/prime/prime−2
af/prime+1
a2f/parenrightbigg
+2/parenleftbigg
f/prime−1
af/parenrightbigg1
r−2f1
r2+2f1
a1
r−f1
a2/bracketrightbigg
=−eEextrcosθ1√
πa3e−r/a,
/diamondsolid/parenleftbigg
f/prime/prime−2
af/prime/parenrightbigg
+2f/prime1
r−2f1
r2=/parenleftbigg2meE ext
/planckover2pi12√
πa3/parenrightbigg
r=4γ
ar,where γ≡meE ext
2/planckover2pi12√πa.
Now let f(r)=A+Br+Cr2,s of/prime=B+2Crandf/prime/prime=2C.Then
2C−2
a(B+2Cr)+2
r(B+2Cr)−2
r2(A+Br+Cr2)=4γ
ar.
Collecting like powers of r:
r−2:A=0.
r−1:2B−2B= 0 (automatic) .
r0:2C−2B/a+4C−2C=0⇒B=2aC.
r1:−4C/a=4γ/a⇒C=−γ.
Evidently the function suggested doessatisfy Eq. 6.10, with the coefficients A=0,B=−2aγ, C =−γ;
the second-order correction to the wavefunction is
ψ1
0=−γr(r+2a)e−r/acosθ.
(ii) Equation 6.11 says, in this case:
E2
0=/angbracketleftψ0
0|H/prime|ψ1
0/angbracketright=−1√
πa3meE ext
2/planckover2pi12√πaeEext/integraldisplay
e−r/a(rcosθ)r(r+2a)e−r/acosθr2sinθdrdθdφ
=−m(eEext)2
2πa2/planckover2pi122π/integraldisplay∞
0r4(r+2a)e−2r/adr/integraldisplayπ
0cos2θsinθdθ
=−m/parenleftbiggeEext
a/planckover2pi1/parenrightbigg2/bracketleftbigg
5!/parenleftBiga
2/parenrightBig6
+2a4!/parenleftBiga
2/parenrightBig5/bracketrightbigg/parenleftbigg
−cos3θ
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0
=−m/parenleftbiggeEext
a/planckover2pi1/parenrightbigg2/parenleftbigg27
8a6/parenrightbigg2
3=−m/parenleftbigg3eEexta2
2/planckover2pi1/parenrightbigg2
.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 195
(b)(i) This is the same as (a) [note that E1
0= 0, as before, since ψ0
0is spherically symmetric, so /angbracketleftcosθ/angbracketright=0 ]
except for the r-dependence of H/prime. So Eq. /diamondsolid⇒
f/prime/prime+2f/prime/parenleftbigg1
r−1
a/parenrightbigg
−2f1
r2=−/parenleftbigg2mep
4πFepsilonC0/planckover2pi12√
πa3/parenrightbigg1
r2=−2β
r2,where β≡mep
4πFepsilonC0/planckover2pi12√
πa3.
The solution this time it obvious: f(r)=β(constant). [For the general solution we would add the general
solution to the homogeneous equation (right side set equal to zero), but this would simply reproduce theunperturbed ground state, ψ
0
0, which we exclude—see p. 253.] So
ψ1
0=βe−r/acosθ.
(ii) The electric dipole moment of the electron is
/angbracketleftpe/angbracketright=/angbracketleft−ercosθ/angbracketright=−e/angbracketleftψ0
0+ψ1
0|rcosθ|ψ0
0+ψ1
0/angbracketright=−e/parenleftbig
/angbracketleftψ0
0|rcosθ|ψ0
0/angbracketright+2/angbracketleftψ0
0|rcosθ|ψ1
0/angbracketright+/angbracketleftψ1
0|rcosθ|ψ1
0/angbracketright/parenrightbig
.
But the first term is zero, and the third is higher order, so
/angbracketleftpe/angbracketright=−2e1√
πa3β/integraldisplay
e−r/a(rcosθ)e−r/acosθr2sinθdrdθdφ
=−2e/parenleftbiggmep
4πFepsilonC0/planckover2pi12πa3/parenrightbigg
2π/integraldisplay∞
0r3e−2r/adr/integraldisplayπ
0cos2θsinθdθ=−/parenleftbiggme2p
FepsilonC0/planckover2pi12πa3/parenrightbigg/bracketleftbigg
3!/parenleftBiga
2/parenrightBig4/bracketrightbigg/parenleftbigg2
3/parenrightbigg
=−/parenleftbiggme2p
FepsilonC0/planckover2pi12πa3/parenrightbigg/parenleftbigg3a4
8/parenrightbigg/parenleftbigg2
3/parenrightbigg
=−/parenleftbiggme2pa
4πFepsilonC0/planckover2pi12/parenrightbigg
=−p.
Evidently the dipole moment associated with the perturbation of the electron cloud cancels the dipole
moment of the nucleus, and the total dipole moment of the atom is zero.
(iii) The first-order correction is zero (as noted in (i)). The second-order correction is
E2
0=/angbracketleftψ0
0|H/prime|ψ1
0/angbracketright=1√
πa3/parenleftbigg
−ep
4πFepsilonC0/parenrightbigg/parenleftbiggmep
4πFepsilonC0/planckover2pi12√
πa3/parenrightbigg/integraldisplay
e−r/a/parenleftbiggcosθ
r2/parenrightbigg
e−r/acosθr2sinθdrdθdφ
=−m(ep)2
(4πFepsilonC0)2/planckover2pi12πa32π/integraldisplay∞
0e−2r/adr/integraldisplayπ
0cos2θsinθdθ=−2m(ep)2
(4πFepsilonC0)2/planckover2pi12a3/parenleftBiga
2/parenrightBig/parenleftbigg2
3/parenrightbigg
=4
3/parenleftbigg
−me4
2(4πFepsilonC0)2/planckover2pi12/parenrightbiggp2
e2a2=4
3/parenleftBigp
ea/parenrightBig2
E1.
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196 CHAPTER 7. THE VARIATIONAL PRINCIPLE
Chapter 7
The Variational Principle
Problem 7.1
(a)
/angbracketleftV/angbracketright=2αA2/integraldisplay∞
0xe−2bx2dx=2αA2/parenleftbigg
−1
4be−2bx2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0=αA2
2b=α
2b/radicalbigg
2b
π=α√
2bπ.
/angbracketleftH/angbracketright=/planckover2pi12b
2m+α√
2πb.∂/angbracketleftH/angbracketright
∂b=/planckover2pi12
2m−1
2α√
2πb−3/2=0=⇒b3/2=α√
2πm
/planckover2pi12;b=/parenleftbiggmα√
2π/planckover2pi12/parenrightbigg2/3
.
/angbracketleftH/angbracketrightmin=/planckover2pi12
2m/parenleftbiggmα√
2π/planckover2pi12/parenrightbigg2/3
+α√
2π/parenleftBigg√
2π/planckover2pi12
mα/parenrightBigg1/3
=α2/3/planckover2pi12/3
m1/3(2π)1/3/parenleftbigg1
2+1/parenrightbigg
=3
2/parenleftbiggα2/planckover2pi12
2πm/parenrightbigg1/3
.
(b)
/angbracketleftV/angbracketright=2αA2/integraldisplay∞
0x4e−2bx2dx=2αA23
8(2b)2/radicalbiggπ
2b=3α
16b2/radicalbiggπ
2b/radicalbigg
2b
π=3α
16b2.
/angbracketleftH/angbracketright=/planckover2pi12b
2m+3α
16b2.∂/angbracketleftH/angbracketright
∂b=/planckover2pi12
2m−3α
8b3=0=⇒b3=3αm
4/planckover2pi12;b=/parenleftbigg3αm
4/planckover2pi12/parenrightbigg1/3
.
/angbracketleftH/angbracketrightmin=/planckover2pi12
2m/parenleftbigg3αm
4/planckover2pi12/parenrightbigg1/3
+3α
16/parenleftbigg4/planckover2pi12
3αm/parenrightbigg2/3
=α1/3/planckover2pi14/3
m2/331/34−1/3/parenleftbigg1
2+1
4/parenrightbigg
=3
4/parenleftbigg3α/planckover2pi14
4m2/parenrightbigg1/3
.
Problem 7.2
Normalize: 1 = 2 |A|2/integraldisplay∞
01
(x2+b2)2dx=2|A|2π
4b3=π
2b3|A|2.A=/radicalbigg
2b3
π.
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CHAPTER 7. THE VARIATIONAL PRINCIPLE 197
Kinetic Energy: /angbracketleftT/angbracketright=−/planckover2pi12
2m|A|2/integraldisplay∞
−∞1
(x2+b2)d2
dx2/parenleftbigg1
(x2+b2)/parenrightbigg
dx.
Butd2
dx2/parenleftbigg1
(x2+b2)/parenrightbigg
=d
dx/parenleftbigg−2x
(x2+b2)2/parenrightbigg
=−2
(x2+b2)2+2x4x
(x2+b2)3=2(3x2−b2)
(x2+b2)3,so
/angbracketleftT/angbracketright=−/planckover2pi12
2m2b3
π/integraldisplay∞
0(3x2−b2)
(x2+b2)4dx=−4/planckover2pi12b3
πm/bracketleftbigg
3/integraldisplay∞
01
(x2+b2)3dx−4b2/integraldisplay∞
01
(x2+b2)4dx/bracketrightbigg
=−4/planckover2pi12b3
πm/bracketleftbigg
33π
16b5−4b25π
32b7/bracketrightbigg
=/planckover2pi12
4mb2.
Potential Energy: /angbracketleftV/angbracketright=1
2mω2|A|22/integraldisplay∞
0x2
(x2+b2)2dx=mω22b3
ππ
4b=1
2mω2b2.
/angbracketleftH/angbracketright=/planckover2pi12
4mb2+1
2mω2b2.∂/angbracketleftH/angbracketright
∂b=−/planckover2pi12
2mb3+mω2b=0=⇒b4=/planckover2pi12
2m2ω2=⇒b2=1√
2/planckover2pi1
mω.
/angbracketleftH/angbracketrightmin=/planckover2pi12
4m√
2mω
/planckover2pi1+1
2mω21√
2/planckover2pi1
mω=/planckover2pi1ω/parenleftBigg√
2
4+1
2√
2/parenrightBigg
=√
2
2/planckover2pi1ω=0.707/planckover2pi1ω>1
2/planckover2pi1ω./check
Problem 7.3
ψ(x)=
A(x+a/2),(−a/2<x< 0),
A(a/2−x),(0<x<a / 2),
0, (otherwise) .
1=|A|22/integraldisplaya/2
0/parenleftBiga
2−x/parenrightBig2
dx=−2|A|21
3/parenleftBiga
2−x/parenrightBig3/vextendsingle/vextendsingle/vextendsingle/vextendsinglea/2
0=2
3|A|2/parenleftBiga
3/parenrightBig3
=a3
12|A|2;A=/radicalbigg
12
a3(as before).
dψ
dx=
A, (−a/2<x< 0),
−A,(0<x<a / 2),
0,(otherwise) .d2ψ
dx2=Aδ/parenleftBig
x+a
2/parenrightBig
−2Aδ(x)+Aδ/parenleftBig
x−a
2/parenrightBig
.
/angbracketleftT/angbracketright=−/planckover2pi12
2m/integraldisplay
ψ/bracketleftBig
Aδ/parenleftBig
x+a
2/parenrightBig
−2Aδ(x)+Aδ/parenleftBig
x−a
2/parenrightBig/bracketrightBig
dx=/planckover2pi12
2m2Aψ(0) =/planckover2pi12
mA2a
2
=/planckover2pi12a
2m12
a3=6/planckover2pi12
ma2(as before).
/angbracketleftV/angbracketright=−α/integraldisplay
|ψ|2δ(x)dx=−α|ψ(0)|2=−αA2/parenleftBiga
2/parenrightBig2
=−3α
a./angbracketleftH/angbracketright=/angbracketleftT/angbracketright+/angbracketleftV/angbracketright=6/planckover2pi12
ma2−3α
a.
∂
∂a/angbracketleftH/angbracketright=−12/planckover2pi12
ma3+3α
a2=0⇒a=4/planckover2pi12
mα.
/angbracketleftH/angbracketrightmin=6/planckover2pi12
m/parenleftBigmα
4/planckover2pi12/parenrightBig2
−3α/parenleftBigmα
4/planckover2pi12/parenrightBig
=mα2
/planckover2pi12/parenleftbigg3
8−3
4/parenrightbigg
=−3mα2
8/planckover2pi12>−mα2
2/planckover2pi12./check
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198 CHAPTER 7. THE VARIATIONAL PRINCIPLE
Problem 7.4
(a)Follow the proof in §7.1:ψ=∞/summationdisplay
n=1cnψn, where ψ1is the ground state. Since /angbracketleftψ1|ψ/angbracketright= 0, we have:
∞/summationdisplay
n=1cn/angbracketleftψ1|ψ/angbracketright=c1= 0; the coefficient of the ground state is zero. So
/angbracketleftH/angbracketright=∞/summationdisplay
n=2En|cn|2≥Efe∞/summationdisplay
n=2|cn|2=Efe,sinceEn≥Efefor allnexcept 1 .
(b)
1=|A|2/integraldisplay∞
−∞x2e−2bx2dx=|A|221
8b/radicalbiggπ
2b=⇒|A|2=4b/radicalbigg
2b
π.
/angbracketleftT/angbracketright=−/planckover2pi12
2m|A|2/integraldisplay∞
−∞xe−bx2d2
dx2/parenleftBig
xe−bx2/parenrightBig
dx
d2
dx2/parenleftBig
xe−bx2/parenrightBig
=d
dx/parenleftBig
e−bx2−2bx2e−bx2/parenrightBig
=−2bxe−bx2−4bxe−bx2+4b2x3e−bx2
/angbracketleftT/angbracketright=−/planckover2pi12
2m4b/radicalbigg
2b
π2/integraldisplay∞
0/parenleftbig
−6bx2+4b2x4/parenrightbig
e−2bx2dx=−2/planckover2pi12b
m/radicalbigg
2b
π2/bracketleftbigg
−6b1
8b/radicalbiggπ
2b+4b23
32b2/radicalbiggπ
2b/bracketrightbigg
=−4/planckover2pi12b
m/parenleftbigg
−3
4+3
8/parenrightbigg
=3/planckover2pi12b
2m.
/angbracketleftV/angbracketright=1
2mω2|A|2/integraldisplay∞
−∞x2e−2bx2x2dx=1
2mω24b/radicalbigg
2b
π23
32b2/radicalbiggπ
2b=3mω2
8b.
/angbracketleftH/angbracketright=3/planckover2pi12b
2m+3mω
8b;∂/angbracketleftH/angbracketright
∂b=3/planckover2pi12
2m−3mω2
8b2=0=⇒b2=m2ω2
4/planckover2pi12=⇒b=mω
2/planckover2pi1.
/angbracketleftH/angbracketrightmin=3/planckover2pi12
2mmω
2/planckover2pi1+3mω2
82/planckover2pi1
mω=/planckover2pi1ω/parenleftbigg3
4+3
4/parenrightbigg
=3
2/planckover2pi1ω.
This is exact, since the trial wave function is in the form of the true first excited state.
Problem 7.5
(a)Use the unperturbed ground state ( ψ0
gs) as the trial wave function. The variational principle says
/angbracketleftψ0
gs|H|ψ0
gs/angbracketright≥E0
gs. ButH=H0+H/prime,s o/angbracketleftψ0
gs|H|ψ0
gs/angbracketright=/angbracketleftψ0
gs|H0|ψ0
gs/angbracketright+/angbracketleftψ0
gs|H/prime|ψ0
gs/angbracketright. But/angbracketleftψ0
gs|H0|ψ0
gs/angbracketright=
E0
gs(the unperturbed ground state energy), and /angbracketleftψ0
gs|H/prime|ψ0
gs/angbracketrightis precisely the first order correction to the
ground state energy (Eq. 6.9), so E0
gs+E1
gs≥Egs. QED
(b)The second order correction ( E2
gs)i sE2
gs=/summationdisplay
m/negationslash=gs|/angbracketleftψ0
m|H/prime|ψgs/angbracketright|2
E0gs−E0m. But the numerator is clearly positive ,
and the denominator is always negative (since E0
gs<E0
mfor allm), soE2
gsisnegative .
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CHAPTER 7. THE VARIATIONAL PRINCIPLE 199
Problem 7.6
He+is a hydrogenic ion (see Problem 4.16); its ground state energy is (2)2(−13.6 eV), or−54.4 eV. It takes
79.0−54.4=24.6 eV to remove one electron.
Problem 7.7
I’ll do the general case of a nucleus with Z0protons. Ignoring electron-electron repulsion altogether gives
ψ0=Z3
0
πa3e−Z0(r1+r2)/a,(generalizing Eq. 7.17)
and the energy is 2 Z2
0E1./angbracketleftVee/angbracketrightgoes like 1 /a(Eqs. 7.20 and 7.25), so the generalization of Eq. 7.25 is /angbracketleftVee/angbracketright=
−5
4Z0E1, and the generalization of Eq. 7.26 is /angbracketleftH/angbracketright=( 2Z2
0−5
4Z0)E1.
If we include shielding, the only change is that ( Z−2) in Eqs. 7.28, 7.29, and 7.32 is replaced by ( Z−Z0).
Thus Eq. 7.32 generalizes to
/angbracketleftH/angbracketright=/bracketleftbigg
2Z2−4Z(Z−Z0)−5
4Z/bracketrightbigg
E1=/bracketleftbigg
−2Z2+4ZZ0−5
4Z/bracketrightbigg
E1.
∂/angbracketleftH/angbracketright
∂Z=/bracketleftbigg
−4Z+4Z0−5
4/bracketrightbigg
E1=0=⇒Z=Z0−5
16.
/angbracketleftH/angbracketrightmin=/bracketleftBigg
−2/parenleftbigg
Z0−5
16/parenrightbigg2
+4/parenleftbigg
Z0−5
16/parenrightbigg
Z0−5
4/parenleftbigg
Z0−5
16/parenrightbigg/bracketrightBigg
E1
=/parenleftbigg
−2Z2
0+5
4Z0−25
128+4Z2
0−5
4Z0−5
4Z0+25
64/parenrightbigg
E1
=/parenleftbigg
2Z2
0−5
4Z0+25
128/parenrightbigg
E1=(16Z0−5)2
128E1,
generalizing Eq. 7.34. The first term is the naive estimate ignoring electron-electron repulsion altogether; the
second term is /angbracketleftVee/angbracketrightin the unscreened state, and the third term is the effect of screening.
Z0=1( H−):Z=1−5
16=11
16=0.688.The effective nuclear charge is less than 1, as expected.
/angbracketleftH/angbracketrightmin=112
128E1=121
128E1=−12.9 eV.
Z0= 2 (He) :Z=2−5
16=27
16=1.69 (as before); /angbracketleftH/angbracketrightmin=272
128E1=729
128E1=−77.5 eV.
Z0= 3 (Li+):Z=3−5
16=43
16=2.69(somewhat less than 3); /angbracketleftH/angbracketrightmin=432
128E1=1849
128E1=−196 eV.
Problem 7.8
D=a/angbracketleftψ0(r1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r2/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
0(r1)/angbracketright=a/angbracketleftψ0(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r1/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
0(r2)/angbracketright=a1
πa3/integraldisplay
e−2r2/a1
r1d3r
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
200 CHAPTER 7. THE VARIATIONAL PRINCIPLE
=1
πa3/integraldisplay
e−2
a√
r2+R2−2rRcosθ1
rr2sinθdrdθdφ =2π
πa3/integraldisplay∞
0r/bracketleftbigg/integraldisplayπ
0e−2
a√
r2+R2−2rRcosθsinθdθ/bracketrightbigg
dr.
[...]=1
rR/integraldisplayr+R
|r−R|e−2y/aydy=−a
2rR/bracketleftBig
e−2(r+R)/a/parenleftBig
r+R+a
2/parenrightBig
−e−2|r−R|/a/parenleftBig
|r−R|+a
2/parenrightBig/bracketrightBig
D=2
a2/parenleftBig
−a
2R/parenrightBig/bracketleftbigg
e−2R/a/integraldisplay∞
0e−2r/a/parenleftBig
r+R+a
2/parenrightBig
dr
−e−2R/a/integraldisplayR
0e2R/a/parenleftBig
R−r+a
2/parenrightBig
dr−e2R/a/integraldisplay∞
Re−2r/a/parenleftBig
r−R+a
2/parenrightBig
dr/bracketrightBigg
=−1
aR/braceleftbigg
e−2R/a/bracketleftbigg/parenleftBiga
2/parenrightBig2
+/parenleftBig
R+a
2/parenrightBig/parenleftBiga
2/parenrightBig/bracketrightbigg
−e−2R/a/parenleftBig
R+a
2/parenrightBig/parenleftBiga
2e2r/a/parenrightBig/vextendsingle/vextendsingle/vextendsingleR
0
+e−2R/a/parenleftBiga
2/parenrightBig2
e2r/a/parenleftbigg2r
a−1/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleR
0−e2R/a/parenleftBig
−R+a
2/parenrightBig/parenleftBig
−a
2e−2r/a/parenrightBig/vextendsingle/vextendsingle/vextendsingle∞
R−e2R/a/parenleftBiga
2/parenrightBig2
e−2r/a/parenleftbigg
−2r
a−1/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
R/bracerightbigg
=−1
aR/braceleftbigg
e−2R/a/bracketleftbigga2
4+aR
2+a2
4+aR
2+a2
4+a2
4/bracketrightbigg
+/bracketleftbigg
−aR
2−a2
4+a2
42R
a−a2
4+aR
2−a2
4−a2
42R
a−a2
4/bracketrightbigg/bracerightbigg
=−1
aR/bracketleftBig
e−2R/a/parenleftbig
a2+aR/parenrightbig
+/parenleftbig
−a2/parenrightbig/bracketrightBig
=⇒D=a
R−/parenleftBig
1+a
R/parenrightBig
e−2R/a(confirms Eq. 7.47).
X=a/angbracketleftψ0(r1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r1/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
0(r2)/angbracketright=a1
πa3/integraldisplay
e−r1/ae−r2/a1
r1d3r
=1
πa2/integraldisplay
e−r/ae−√
r2+R2−2rRcosθ/a1
rr2sinθdrdθdφ =2π
πa2/integraldisplay∞
0re−r/a/bracketleftbigg/integraldisplayπ
0e−√
r2+R2−2rRcosθ/asinθdθ/bracketrightbigg
dr.
[...]=−a
rR/bracketleftBig
e−(r+R)/a(r+R+a)−e−|r−R|/a(|r−R|+a)/bracketrightBig
X=2
a2/parenleftBig
−a
R/parenrightBig/bracketleftbigg
e−R/a/integraldisplay∞
0e−2r/a(r+R+a)dr
−e−R/a/integraldisplayR
0(R−r+a)dr−eR/a/integraldisplay∞
Re−2r/a(r−R+a)dr/bracketrightBigg
=−2
aR/braceleftbigg
e−R/a/bracketleftbigg/parenleftBiga
2/parenrightBig2
+(R+a)/parenleftBiga
2/parenrightBig/bracketrightbigg
−e−R/a/bracketleftbigg
(R+a)R−R2
2/bracketrightbigg
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 7. THE VARIATIONAL PRINCIPLE 201
−eR/a(−R+a)/parenleftBig
−a
2e−2r/a/parenrightBig/vextendsingle/vextendsingle/vextendsingle∞
R−eR/a/parenleftBiga
2/parenrightBig2
e−2r/a/parenleftbigg
−2r
a−1/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
R/bracerightbigg
=−2
aR/bracketleftbigg
e−R/a/parenleftbigga2
4+aR
2+a2
2−R2−aR+R2
2+aR
2−a2
2−a2
42R
a−a2
4/parenrightbigg/bracketrightbigg
=−2
aRe−R/a/parenleftbigg
−aR
2−R2
2/parenrightbigg
=⇒X=e−R/a/parenleftbigg
1+R
a/parenrightbigg
(confirms Eq. 7.48) .
Problem 7.9
There are two changes: (1) the 2 in Eq. 7.38 changes sign ...which amounts to changing the sign of Iin
Eq. 7.43; (2) the last term in Eq. 7.44 changes sign ...which amounts to reversing the sign of X. Thus Eq. 7.49
becomes
/angbracketleftH/angbracketright=/bracketleftbigg
1+2D−X
1−I/bracketrightbigg
E1,and hence Eq. 7.51 becomes
F(x)=Etot
−E1=2a
R−1−2D−X
1−I=−1+2
x−21/x−( 1+1/x)e−2x−(1 +x)e−x
1−(1 +x+x2/3)e−x
=−1+2
x/bracketleftbigg1−(1 +x+x2/3)e−x−1+(x+1 )e−2x+(x+x2)e−x
1−(1 +x+x2/3)e−x/bracketrightbigg
=−1+2
x/bracketleftBigg
(1 +x)e−2x+/parenleftbig2
3x2−1/parenrightbig
e−x
1−(1 +x+x2/3)e−x/bracketrightBigg
.
The graph (with plus sign for comparison) has no minimum, and remains above −1, indicating that the energy
is greater than for the proton and atom dissociated. Hence, no evidence of bonding here.
2 4 6 8
-1.1-0.9-0.8-0.7-0.6-0.5
xF(x)
(−)(+)
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202 CHAPTER 7. THE VARIATIONAL PRINCIPLE
Problem 7.10
According to Mathematica , the minimum occurs at x=2.493, and at this point F/prime/prime=0.1257.
mω2=V/prime/prime=−E1
a2F/prime/prime,soω=1
a/radicalbigg
−(0.1257)E1
m.
Heremis the reduced mass of the proton: m=mpmp
mp+mp=1
2mp.
ω=3×108m/s
(0.529×10−10m)/radicalBigg
(0.1257)(13 .6 eV)
(938×106eV)/2=3.42×1014/s.
1
2/planckover2pi1ω=1
2(6.58×10−16eV·s)(3.42×1014/s) =0.113 eV (ground state vibrational energy) .
Mathematica says that at the minimum F=−1.1297, so the binding energy is (0.1297)(13.6 eV) = 1.76 eV.
Since this is substantially greater than the vibrational energy, it stays bound. The highest vibrational energy is
given by ( n+1
2)/planckover2pi1ω=1.76 eV, so n=1.76
0.226−1
2=7.29. I estimate eight bound vibrational states (including
n= 0).
Problem 7.11
(a)
1=/integraldisplay
|ψ|2dx=|A|2/integraldisplaya/2
−a/2cos2/parenleftBigπx
a/parenrightBig
dx=|A|2a
2⇒A=/radicalbigg
2
a.
/angbracketleftT/angbracketright=−/planckover2pi12
2m/integraldisplay
ψd2ψ
dx2dx=/planckover2pi12
2m/parenleftBigπ
a/parenrightBig2/integraldisplay
ψ2dx=π2/planckover2pi12
2ma2.
/angbracketleftV/angbracketright=1
2mω2/integraldisplay
x2ψ2dx=1
2mω22
a/integraldisplaya/2
−a/2x2cos2/parenleftBigπx
a/parenrightBig
dx=mω2
a/parenleftBiga
π/parenrightBig3/integraldisplayπ/2
−π/2y2cos2ydy
=mω2a2
π3/bracketleftbiggy3
6+/parenleftbiggy2
4−1
8/parenrightbigg
sin 2y+ycos 2y
4/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ/2
−π/2=mω2a2
4π2/parenleftbiggπ2
6−1/parenrightbigg
.
/angbracketleftH/angbracketright=π2/planckover2pi12
2ma2+mω2a2
4π2/parenleftbiggπ2
6−1/parenrightbigg
;∂/angbracketleftH/angbracketright
∂a=−π2/planckover2pi12
ma3+mω2a
2π2/parenleftbiggπ2
6−1/parenrightbigg
=0⇒
a=π/radicalbigg
/planckover2pi1
mω/parenleftbigg2
π2/6−1/parenrightbigg1/4
.
/angbracketleftH/angbracketrightmin=π2/planckover2pi12
2mπ2mω
/planckover2pi1/radicalbigg
π2/6−1
2+mω2
4π2/parenleftbiggπ2
6−1/parenrightbigg
π2/planckover2pi1
mω/radicalBigg
2
π2/6−1
=1
2/planckover2pi1ω/radicalbigg
π2
3−2=1
2/planckover2pi1ω(1.136)>1
2/planckover2pi1ω./check
[We do notneed to worry about the kink at ±a/2. It is true that d2ψ/dx2has delta functions there, but
sinceψ(±a/2) = 0 no “extra” contribution to Tcomes from these points.]
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CHAPTER 7. THE VARIATIONAL PRINCIPLE 203
(b)Because this trial function is odd, it is orthogonal to the ground state, so by Problem 7.4 /angbracketleftH/angbracketrightwill give
an upper bound to the first excited state.
1=/integraldisplay
|ψ|2dx=|B|2/integraldisplaya
−asin2/parenleftBigπx
a/parenrightBig
dx=|B|2a⇒B=1√a.
/angbracketleftT/angbracketright=−/planckover2pi12
2m/integraldisplay
ψd2ψ
dx2dx=/planckover2pi12
2m/parenleftBigπ
a/parenrightBig2/integraldisplay
ψ2dx=π2/planckover2pi12
2ma2.
/angbracketleftV/angbracketright=1
2mω2/integraldisplay
x2ψ2dx=1
2mω21
a/integraldisplaya
−ax2sin2/parenleftBigπx
a/parenrightBig
dx=mω2
2a/parenleftBiga
π/parenrightBig3/integraldisplayπ
−πy2sin2ydy
=mω2a2
2π3/bracketleftbiggy3
6−/parenleftbiggy2
4−1
8/parenrightbigg
sin 2y−ycos 2y
4/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
−π=mω2a2
4π2/parenleftbigg2π2
3−1/parenrightbigg
.
/angbracketleftH/angbracketright=π2/planckover2pi12
2ma2+mω2a2
4π2/parenleftbigg2π2
3−1/parenrightbigg
;∂/angbracketleftH/angbracketright
∂a=−π2/planckover2pi12
ma3+mω2a
2π2/parenleftbigg2π2
3−1/parenrightbigg
=0⇒
a=π/radicalbigg
/planckover2pi1
mω/parenleftbigg2
2π2/3−1/parenrightbigg1/4
.
/angbracketleftH/angbracketrightmin=π2/planckover2pi12
2mπ2mω
/planckover2pi1/radicalbigg
2π2/3−1
2+mω2
4π2/parenleftbigg2π2
3−1/parenrightbigg
π2/planckover2pi1
mω/radicalBigg
2
2π2/3−1
=1
2/planckover2pi1ω/radicalbigg
4π2
3−2=1
2/planckover2pi1ω(3.341)>3
2/planckover2pi1ω./check
Problem 7.12
We will need the following integral repeatedly:
/integraldisplay∞
0xk
(x2+b2)ldx=1
2b2l−k−1Γ/parenleftbigk+1
2/parenrightbig
Γ/parenleftbig2l−k−1
2/parenrightbig
Γ(l).
(a)
1=/integraldisplay∞
−∞|ψ|2dx=2|A|2/integraldisplay∞
01
(x2+b2)2ndx=|A|2
b4n−1Γ/parenleftbig1
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig
Γ(2n)⇒A=/radicalBigg
b4n−1Γ(2n)
Γ/parenleftbig1
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig.
/angbracketleftT/angbracketright=−/planckover2pi12
2m/integraldisplay∞
−∞ψd2ψ
dx2dx=−/planckover2pi12
2mA2/integraldisplay∞
−∞1
(x2+b2)nd
dx/bracketleftBigg
−2nx
(x2+b2)n+1/bracketrightBigg
dx
=n/planckover2pi12
mA2/integraldisplay∞
−∞1
(x2+b2)n/bracketleftBigg
1
(x2+b2)n+1−2(n+1 )x2
(x2+b2)n+2/bracketrightBigg
dx
=2n/planckover2pi12
mA2/bracketleftBigg/integraldisplay∞
01
(x2+b2)2n+1dx−2(n+1 )/integraldisplay∞
0x2
(x2+b2)2n+2dx/bracketrightBigg
=2n/planckover2pi12
mb4n−1Γ(2n)
Γ/parenleftbig1
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig/bracketleftBigg
1
2b4n−1Γ/parenleftbig1
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig
Γ(2n+1 )−2(n+1 )
2b4n−1Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n+1
2/parenrightbig
Γ(2n+2 )/bracketrightBigg
=/planckover2pi12
4mb2n(4n−1)
(2n+1 ).
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
204 CHAPTER 7. THE VARIATIONAL PRINCIPLE
/angbracketleftV/angbracketright=1
2mω2/integraldisplay∞
−∞ψ2x2dx=1
2mω22A2/integraldisplay∞
0x2
(x2+b2)2ndx
=mω2b4n−1Γ(2n)
Γ/parenleftbig1
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig1
2b4n−3Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n−3
2/parenrightbig
Γ(2n)=mω2b2
2(4n−3).
/angbracketleftH/angbracketright=/planckover2pi12
4mb2n(4n−1)
(2n+1 )+mω2b2
(4n−3);∂/angbracketleftH/angbracketright
∂b=−/planckover2pi12
2mb3n(4n−1)
(2n+1 )+mω2b
(4n−3)=0⇒
b=/radicalbigg
/planckover2pi1
mω/bracketleftbiggn(4n−1)(4n−3)
2(2n+1 )/bracketrightbigg1/4
.
/angbracketleftH/angbracketrightmin=/planckover2pi12
4mn(4n−1)
(2n+1 )mω
/planckover2pi1/radicalBigg
2(2n+1 )
n(4n−1)(4n−3)+mω2
2(4n−3)/planckover2pi1
mω/radicalBigg
n(4n−1)(4n−3)
2(2n+1 )
=1
2/planckover2pi1ω/radicalBigg
2n(4n−1)
(2n+ 1)(4n−3)=1
2/planckover2pi1ω/radicalbigg
8n2−2n
8n2−2n−3>1
2/planckover2pi1ω./check
(b)
1=2|B|2/integraldisplay∞
0x2
(x2+b2)2ndx=|B|2
b4n−3Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n−3
2/parenrightbig
Γ(2n)⇒B=/radicalBigg
b4n−3Γ(2n)
Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n−3
2/parenrightbig.
/angbracketleftT/angbracketright=−/planckover2pi12
2mB2/integraldisplay∞
−∞x
(x2+b2)nd
dx/bracketleftBigg
1
(x2+b2)n−2nx2
(x2+b2)n+1/bracketrightBigg
dx
=−/planckover2pi12B2
2m/integraldisplay∞
−∞x
(x2+b2)n/bracketleftBigg
−2nx
(x2+b2)n+1−4nx
(x2+b2)n+1+4n(n+1 )x3
(x2+b2)n+2/bracketrightBigg
dx
=4n/planckover2pi12B2
2m/bracketleftBigg
3/integraldisplay∞
0x2
(x2+b2)2n+1dx−2(n+1 )/integraldisplay∞
0x4
(x2+b2)2n+2dx/bracketrightBigg
=2n/planckover2pi12
mb4n−3Γ(2n)
Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n−3
2/parenrightbig/bracketleftBigg
3
2b4n−1Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig
Γ(2n+1 )−2(n+1 )
2b4n−1Γ/parenleftbig5
2/parenrightbig
Γ/parenleftbig4n−1
2/parenrightbig
Γ(2n+2 )/bracketrightBigg
=3/planckover2pi12
4mb2n(4n−3)
(2n+1 ).
/angbracketleftV/angbracketright=1
2mω22B2/integraldisplay∞
0x4
(x2+b2)2ndx=1
2mω2b4n−3Γ(2n)
Γ/parenleftbig3
2/parenrightbig
Γ/parenleftbig4n−3
2/parenrightbig2
2b4n−5Γ/parenleftbig5
2/parenrightbig
Γ/parenleftbig4n−5
2/parenrightbig
Γ(2n)=3
2mω2b2
(4n−5).
/angbracketleftH/angbracketright=3/planckover2pi12
4mb2n(4n−3)
(2n+1 )+3
2mω2b2
(4n−5);∂/angbracketleftH/angbracketright
∂b=−3/planckover2pi12
2mb3n(4n−3)
(2n+1 )+3mω2b
(4n−5)=0⇒
b=/radicalbigg
/planckover2pi1
mω/bracketleftbiggn(4n−3)(4n−5)
2(2n+1 )/bracketrightbigg1/4
.
/angbracketleftH/angbracketrightmin=3/planckover2pi12
4mn(4n−3)
(2n+1 )mω
/planckover2pi1/radicalBigg
2(2n+1 )
n(4n−3)(4n−5)+3
2mω2
(4n−5)/planckover2pi1
mω/radicalBigg
n(4n−3)(4n−5)
2(2n+1 )
=3
2/planckover2pi1ω/radicalBigg
2n(4n−3)
(2n+ 1)(4n−5)=3
2/planckover2pi1ω/radicalbigg
8n2−6n
8n2−6n−5>3
2/planckover2pi1ω./check
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 7. THE VARIATIONAL PRINCIPLE 205
(c)Asn→∞,ψbecomes more and more “gaussian”. In the figures I have plotted the trial wave functions
forn=2 ,n= 3, and n= 4, as well as the exact states (heavy line). Even for n= 2 the fit is pretty good,
so it is hard to see the improvement, but the successive curves do move perceptably toward the correctresult.
12340.10.20.30.40.50.60.70.8
12340.10.20.30.40.50.60.70.8
Analytically, for large n,b≈/radicalbigg
/planckover2pi1
mω/parenleftbiggn·4n·4n
2·2n/parenrightbigg1/4
=/radicalbigg
2n/planckover2pi1
mω,so
/parenleftbig
x2+b2/parenrightbign=b2n/parenleftbigg
1+x2
b2/parenrightbiggn
≈b2n/parenleftbigg
1+mωx2
2/planckover2pi1n/parenrightbiggn
→b2nemωx2/2/planckover2pi1.
Meanwhile, using Stirling’s approximation (Eq. 5.84), in the form Γ( z+1 )≈zze−z:
A2=b4n−1Γ(2n)
Γ/parenleftbig1
2/parenrightbig
Γ/parenleftbig
2n−1
2/parenrightbig≈b4n−1
√π(2n−1)2n−1e−(2n−1)
/parenleftbig
2n−3
2/parenrightbig2n−3/2e−(2n−3/2)≈b4n−1
√π1√e/parenleftbigg2n−1
2n−3
2/parenrightbigg2n−1/radicalbig
2n−3/2.
But/parenleftbigg1−1
2n
1−3
4n/parenrightbigg
≈/parenleftbigg
1−1
2n/parenrightbigg/parenleftbigg
1+3
4n/parenrightbigg
≈1+3
4n−1
2n=1+1
4n;
so/parenleftbigg2n−1
2n−3
2/parenrightbigg2n−1
≈/bracketleftbigg/parenleftbigg
1+1
4n/parenrightbiggn/bracketrightbigg21
1+1/4n→/parenleftBig
e1/4/parenrightBig2
=√e.
=b4n−1
√πe√e√
2n=/radicalbigg
2n
πb4n−1⇒A≈/parenleftbigg2n
π/parenrightbigg1/4
b2n−1/2.So
ψ≈/parenleftbigg2n
π/parenrightbigg1/4
b2n−1/21
b2ne−mωx2/2/planckover2pi1=/parenleftbigg2n
π/parenrightbigg1/4/parenleftBigmω
2n/planckover2pi1/parenrightBig1/4
e−mωx2/2/planckover2pi1=/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
e−mωx2/2/planckover2pi1,
which is precisely the ground state of the harmonic oscillator (Eq. 2.59). So it’s no accident that we get
the exact energies, in the limit n→∞.
Problem 7.13
1=|A|2/integraldisplay
e−2br2r2sinθdrdθdφ =4π|A|2/integraldisplay∞
0r2e−2br2dr=|A|2/parenleftBigπ
2b/parenrightBig3/2
⇒A=/parenleftbigg2b
π/parenrightbigg3/4
.
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206 CHAPTER 7. THE VARIATIONAL PRINCIPLE
/angbracketleftV/angbracketright=−e2
4πFepsilonC0|A|24π/integraldisplay∞
0e−2br21
rr2dr=−e2
4πFepsilonC0/parenleftbigg2b
π/parenrightbigg3/2
4π1
4b=−e2
4πFepsilonC02/radicalbigg
2b
π.
/angbracketleftT/angbracketright=−/planckover2pi12
2m|A|2/integraldisplay
e−br2(∇2e−br2)r2sinθdrdθdφ
But (∇2e−br2)=1
r2d
dr/parenleftbigg
r2d
dre−br2/parenrightbigg
=1
r2d
dr/parenleftBig
−2br3e−br2/parenrightBig
=−2b
r2/parenleftbig
3r2−2br4/parenrightbig
e−br2.
=−/planckover2pi12
2m/parenleftbigg2b
π/parenrightbigg3/2
(4π)(−2b)/integraldisplay∞
0(3r2−2br4)e−2br2dr=/planckover2pi12
mπb4/parenleftbigg2b
π/parenrightbigg3/2/bracketleftbigg
31
8b/radicalbiggπ
2b−2b3
32b2/radicalbiggπ
2b/bracketrightbigg
=/planckover2pi12
m4πb/parenleftbigg2b
π/parenrightbigg/parenleftbigg3
8b−3
16b/parenrightbigg
=3/planckover2pi12b
2m.
/angbracketleftH/angbracketright=3/planckover2pi12b
2m−e2
4πFepsilonC02/radicalbigg
2b
π;∂/angbracketleftH/angbracketright
∂b=3/planckover2pi12
2m−e2
4πFepsilonC0/radicalbigg
2
π1√
b=0⇒√
b=e2
4πFepsilonC0/radicalbigg
2
π2m
3/planckover2pi12.
/angbracketleftH/angbracketrightmin=3/planckover2pi12
2m/parenleftbigge2
4πFepsilonC0/parenrightbigg22
π4m2
9/planckover2pi14−e2
4πFepsilonC02/radicalbigg
2
π/parenleftbigge2
4πFepsilonC0/parenrightbigg/radicalbigg
2
π2m
3/planckover2pi12=/parenleftbigge2
4πFepsilonC0/parenrightbigg2m
/planckover2pi12/parenleftbigg4
3π−8
3π/parenrightbigg
=−m
2/planckover2pi12/parenleftbigge2
4πFepsilonC0/parenrightbigg28
3π=8
3πE1=−11.5 eV.
Problem 7.14
Letψ=1√
πb3e−r/b(same as hydrogen, but with a→badjustable). From Eq. 4.191, we have /angbracketleftT/angbracketright=−E1=
/planckover2pi12
2ma2for hydrogen, so in this case /angbracketleftT/angbracketright=/planckover2pi12
2mb2.
/angbracketleftV/angbracketright=−e2
4πFepsilonC04π
πb3/integraldisplay∞
0e−2r/be−µr
rr2dr=−e2
4πFepsilonC04
b3/integraldisplay∞
0e−(µ+2/b)rrdr=−e2
4πFepsilonC04
b31
(µ+2/b)2=−e2
4πFepsilonC01
b(1 +µb
2)2.
/angbracketleftH/angbracketright=/planckover2pi12
2mb2−e2
4πFepsilonC01
b(1 +µb
2)2.
∂/angbracketleftH/angbracketright
∂b=−/planckover2pi12
mb3+e2
4πFepsilonC0/bracketleftbigg1
b2(1 +µb/2)2+µ
b(1 +µb/2)3/bracketrightbigg
=−/planckover2pi12
mb3+e2
4πFepsilonC0( 1+3µb/2)
b2(1 +µb/2)3=0⇒
/planckover2pi12
m/parenleftbigg4πFepsilonC0
e2/parenrightbigg
=b( 1+3µb/2)
(1 +µb/2)3,orb( 1+3µb/2)
(1 +µb/2)3=a.
This determines b, but unfortunately it’s a cubic equation. So we use the fact that µis small to obtain a suitable
approximate solution. If µ= 0, then b=a(of course), so µa/lessmuch1=⇒µb/lessmuch1 too. We’ll expand in powers of
µb:
a≈b/parenleftbigg
1+3µb
2/parenrightbigg/bracketleftBigg
1−3µb
2+6/parenleftbiggµb
2/parenrightbigg2/bracketrightBigg
≈b/bracketleftbigg
1−9
4(µb)2+6
4(µb)2/bracketrightbigg
=b/bracketleftbigg
1−3
4(µb)2/bracketrightbigg
.
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CHAPTER 7. THE VARIATIONAL PRINCIPLE 207
Since the3
4(µb)2term is already a second-order correction, we can replace bbya:
b≈a/bracketleftbig
1−3
4(µb)2/bracketrightbig≈a/bracketleftbigg
1+3
4(µa)2/bracketrightbigg
.
/angbracketleftH/angbracketrightmin=/planckover2pi12
2ma2/bracketleftbig
1+3
4(µa)2/bracketrightbig2−e2
4πFepsilonC01
a/bracketleftbig
1+3
4(µa)2/bracketrightbig/bracketleftbig
1+1
2(µa)/bracketrightbig2
≈/planckover2pi12
2ma2/bracketleftbigg
1−23
4(µa)2/bracketrightbigg
−e2
4πFepsilonC01
a/bracketleftbigg
1−3
4(µa)2/bracketrightbigg/bracketleftbigg
1−2µa
2+3/parenleftBigµa
2/parenrightBig2/bracketrightbigg
=−E1/bracketleftbigg
1−3
2(µa)2/bracketrightbigg
+2E1/bracketleftbigg
1−µa+3
4(µa)2−3
4(µa)2/bracketrightbigg
=E1/bracketleftbigg
1−2(µa)+3
2(µa)2/bracketrightbigg
.
Problem 7.15
(a)
H=/parenleftbiggEah
hE b/parenrightbigg
; det( H−λ)=(Ea−λ)(Eb−λ)−h2=0=⇒λ2−λ(Ea+Eb)+EaEb−h2=0.
λ=1
2/parenleftbigg
Ea+Eb±/radicalBig
E2a+2EaEb+E2
b−4EaEb+4h2/parenrightbigg
⇒E±=1
2/bracketleftBig
Ea+Eb±/radicalbig
(Ea−Eb)2+4h2/bracketrightBig
.
(b)Zeroth order: E0
a=Ea,E0
b=Eb. First order: E1
a=/angbracketleftψa|H/prime|ψa/angbracketright=0,E1
b=/angbracketleftψb|H/prime|ψb/angbracketright= 0. Second
order:
E2
a=|/angbracketleftψb|H/prime|ψa/angbracketright|2
Ea−Eb=−h2
Eb−Ea;E2
b=|/angbracketleftψa|H/prime|ψb/angbracketright|2
Eb−Ea=h2
Eb−Ea;
E−≈Ea−h2
(Eb−Ea);E+≈Eb+h2
(Eb−Ea).
(c)
/angbracketleftH/angbracketright=/angbracketleftcosφψa+ sinφψb|(H0+H/prime)|cosφψa+ sinφψb/angbracketright
= cos2φ/angbracketleftψa|H0|ψa/angbracketright+ sin2φ/angbracketleftψb|H0|ψb/angbracketright+ sinφcosφ/angbracketleftψb|H/prime|ψa/angbracketright+ sinφcosφ/angbracketleftψa|H/prime|ψb/angbracketright
=Eacos2φ+Ebsin2φ+2hsinφcosφ.
∂/angbracketleftH/angbracketright
∂φ=−Ea2 cosφsinφ+Eb2 sinφcosφ+2h(cos2φ−sin2φ)=(Eb−Ea) sin 2φ+2hcos 2φ=0.
tan 2φ=−2h
Eb−Ea=−FepsilonCwhere FepsilonC≡2h
Eb−Ea.sin 2φ/radicalbig
1−sin22φ=−FepsilonC; sin22φ=FepsilonC2(1−sin22φ);
or sin22φ(1 +FepsilonC2)=FepsilonC2; sin 2 φ=±FepsilonC√
1+FepsilonC2; cos22φ=1−sin22φ=1−FepsilonC2
1+FepsilonC2=1
1+FepsilonC2;
cos 2φ=∓1√
1+FepsilonC2(sign dictated by tan 2 φ=sin 2φ
cos 2φ=−FepsilonC).
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208 CHAPTER 7. THE VARIATIONAL PRINCIPLE
cos2φ=1
2(1 + cos 2 φ)=1
2/parenleftbigg
1∓1√
1+FepsilonC2/parenrightbigg
; sin2φ=1
2(1−cos 2φ)=1
2/parenleftbigg
1±1√
1+FepsilonC2/parenrightbigg
.
/angbracketleftH/angbracketrightmin=1
2Ea/parenleftbigg
1∓1√
1+FepsilonC2/parenrightbigg
+1
2Eb/parenleftbigg
1±1√
1+FepsilonC2/parenrightbigg
±hFepsilonC√
1+FepsilonC2=1
2/bracketleftbigg
Ea+Eb±(Eb−Ea+2hFepsilonC)√
1+FepsilonC2/bracketrightbigg
But(Eb−Ea+2hFepsilonC)√
1+FepsilonC2=(Eb−Ea)+2h2h
(Eb−Ea)/radicalBig
1+4h2
(Eb−Ea)2=(Eb−Ea)2+4h2
/radicalbig
(Eb−Ea)2+4h2=/radicalbig
(Eb−Ea)2+4h2,So
/angbracketleftH/angbracketrightmin=1
2/bracketleftBig
Ea+Eb±/radicalbig
(Eb−Ea)2+4h2/bracketrightBig
we want the minus sign (+ is maximum)
=1
2/bracketleftBig
Ea+Eb−/radicalbig
(Eb−Ea)2+4h2/bracketrightBig
.
(d)Ifhis small, the exact result (a) can be expanded: E±=1
2/bracketleftBig
(Ea+Eb)±(Eb−Ea)/radicalBig
1+4h2
(Eb−Ea)2/bracketrightBig
.
=⇒E±≈1
2/braceleftbigg
Ea+Eb±(Eb−Ea)/bracketleftbigg
1+2h2
(Eb−Ea)2/bracketrightbigg/bracerightbigg
=1
2/bracketleftbigg
Ea+Eb±(Eb−Ea)±2h2
(Eb−Ea)/bracketrightbigg
,
soE+≈Eb+h2
(Eb−Ea),E −≈Ea−h2
(Eb−Ea),
confirming the perturbation theory results in (b). The variational principle (c) gets the ground state ( E−)
exactly right—not too surprising since the trial wave function Eq. 7.56 is almost the most general state
(there could be a relative phase factor eiθ).
Problem 7.16
For the electron, γ=−e/m,s oE±=±eBz/planckover2pi1/2m(Eq. 4.161). For consistency with Problem 7.15, Eb>E a,
soχb=χ+=/parenleftbigg1
0/parenrightbigg
,χ a=χ−=/parenleftbigg0
1/parenrightbigg
,E b=E+=eBz/planckover2pi1
2m,E a=E−=−eBz/planckover2pi1
2m.
(a)
/angbracketleftχa|H/prime|χa/angbracketright=eBx
m/planckover2pi1
2/parenleftbig01/parenrightbig/parenleftbigg01
10/parenrightbigg/parenleftbigg0
1/parenrightbigg
=eBx/planckover2pi1
2m/parenleftbig01/parenrightbig/parenleftbigg1
0/parenrightbigg
=0 ;
/angbracketleftχb|H/prime|χb/angbracketright=eBx/planckover2pi1
2m/parenleftbig10/parenrightbig/parenleftbigg01
10/parenrightbigg/parenleftbigg1
0/parenrightbigg
=0 ;/angbracketleftχb|H/prime|χa/angbracketright=eBx/planckover2pi1
2m/parenleftbig10/parenrightbig/parenleftbigg01
10/parenrightbigg/parenleftbigg0
1/parenrightbigg
=eBx/planckover2pi1
2m;
/angbracketleftχa|H/prime|χb/angbracketright=eBx/planckover2pi1
2m/parenleftbig01/parenrightbig/parenleftbigg01
10/parenrightbigg/parenleftbigg1
0/parenrightbigg
=eBx/planckover2pi1
2m/parenleftbig01/parenrightbig/parenleftbigg0
1/parenrightbigg
=eBx/planckover2pi1
2m.Soh=eBx/planckover2pi1
2m,
and the conditions of Problem 7.15 are met.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 7. THE VARIATIONAL PRINCIPLE 209
(b)From Problem 7.15(b),
Egs≈Ea−h2
(Eb−Ea)=−eBz/planckover2pi1
2m−(eBx/planckover2pi1/2m)2
(eBz/planckover2pi1/m)=−e/planckover2pi1
2m/parenleftbigg
Bz+B2
x
2Bz/parenrightbigg
.
(c)From Problem 7.15(c), Egs=1
2/bracketleftBig
Ea+Eb−/radicalbig
(Eb−Ea)2+4h2/bracketrightBig
(it’s actually the exact ground state).
Egs=−1
2/radicalBigg/parenleftbiggeBz/planckover2pi1
m/parenrightbigg2
+4/parenleftbiggeBx/planckover2pi1
2m/parenrightbigg2
=−e/planckover2pi1
2m/radicalbig
B2z+B2x
(which was obvious from the start, since the square root is simply the magnitude of the total field).
Problem 7.17
(a)
r1=1√
2(u+v);r2=1√
2(u−v);r2
1+r2
2=1
2(u2+2u·v+v2+u2−2u·v+v2)=u2+v2.
(∇2
1+∇2
2)f(r1,r2)=/parenleftbigg∂2f
∂x2
1+∂2f
∂y2
1+∂2f
∂z2
1+∂2f
∂x2
2+∂2f
∂y2
2+∂2f
∂z2
2/parenrightbigg
.
∂f
∂x1=∂f
∂ux∂ux
∂x1+∂f
∂vx∂vx
∂x1=1√
2/parenleftbigg∂f
∂ux+∂f
∂vx/parenrightbigg
;∂f
∂x2=∂f
∂ux∂ux
∂x2+∂f
∂vx∂vx
∂x2=1√
2/parenleftbigg∂f
∂ux−∂f
∂vx/parenrightbigg
.
∂2f
∂x2
1=1√
2∂
∂x1/parenleftbigg∂f
∂ux+∂f
∂vx/parenrightbigg
=1√
2/parenleftbigg∂2f
∂u2x∂ux
∂x1+∂2f
∂ux∂vx∂vx
∂x1+∂2f
∂vx∂ux∂ux
∂x1+∂2f
∂v2x∂vx
∂x1/parenrightbigg
=1
2/parenleftbigg∂2f
∂u2x+2∂2f
∂ux∂vx+∂2f
∂v2x/parenrightbigg
;
∂2f
∂x2
2=1√
2∂
∂x2/parenleftbigg∂f
∂ux−∂f
∂vx/parenrightbigg
=1√
2/parenleftbigg∂2f
∂u2x∂ux
∂x2+∂2f
∂ux∂vx∂vx
∂x2−∂2f
∂vx∂ux∂ux
∂x2−∂2f
∂v2x∂vx
∂x2/parenrightbigg
=1
2/parenleftbigg∂2f
∂u2x−2∂2f
∂ux∂vx+∂2f
∂v2x/parenrightbigg
.
So/parenleftbigg∂2f
∂x2
1+∂2f
∂x2
2/parenrightbigg
=/parenleftbigg∂2f
∂u2x+∂2f
∂v2x/parenrightbigg
,and likewise for yandz:∇2
1+∇2
2=∇2
u+∇2
v.
H=−/planckover2pi12
2m(∇2
u+∇2
v)+1
2mω2(u2+v2)−λ
4mω22v2
=/bracketleftbigg
−/planckover2pi12
2m∇2
u+1
2mω2u2/bracketrightbigg
+/bracketleftbigg
−/planckover2pi12
2m∇2
v+1
2mω2v2−1
2λmω2v2/bracketrightbigg
.QED
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210 CHAPTER 7. THE VARIATIONAL PRINCIPLE
(b)The energy is3
2/planckover2pi1ω(for the upart) and3
2/planckover2pi1ω√
1−λ(for the vpart): Egs=3
2/planckover2pi1ω/parenleftbig
1+√
1−λ/parenrightbig
.
(c)The ground state for a one-dimensional oscillator is
ψ0(x)=/parenleftBigmω
π/planckover2pi1/parenrightBig1/4
e−mωx2/2/planckover2pi1(Eq. 2.59).
So, for a 3-D oscillator, the ground state is ψ0(r)=/parenleftbigmω
π/planckover2pi1/parenrightbig3/4e−mωr2/2/planckover2pi1, and for two particles
ψ(r1,r2)=/parenleftBigmω
π/planckover2pi1/parenrightBig3/2
e−mω
2/planckover2pi1(r2
1+r2
2).(This is the analog to Eq. 7.17.)
/angbracketleftH/angbracketright=3
2/planckover2pi1ω+3
2/planckover2pi1ω+/angbracketleftVee/angbracketright=3/planckover2pi1ω+/angbracketleftVee/angbracketright(the analog to Eq. 7.19) .
/angbracketleftVee/angbracketright=−λ
4mω2/parenleftBigmω
π/planckover2pi1/parenrightBig3/integraldisplay
e−mω
/planckover2pi1(r2
1+r2
2)(r1−r2)2
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
r2
1−2r1·r2+r2
2d3r1d3r2(the analog to Eq. 7.20).
Ther1·r2term integrates to zero, by symmetry, and the r2
2term is the same as the r2
1term, so
/angbracketleftVee/angbracketright=−λ
4mω2/parenleftBigmω
π/planckover2pi1/parenrightBig3
2/integraldisplay
e−mω
/planckover2pi1(r2
1+r2
2)r2
1d3r1d3r2
=−λ
2mω2/parenleftBigmω
π/planckover2pi1/parenrightBig3
(4π)2/integraldisplay∞
0e−mωr2
2//planckover2pi1r2
2dr2/integraldisplay∞
0e−mωr2
1//planckover2pi1r4
1dr1
=−λ8m4ω5
π/planckover2pi13/bracketleftBigg
1
4/planckover2pi1
mω/radicalbigg
π/planckover2pi1
mω/bracketrightBigg/bracketleftBigg
3
8/parenleftbigg/planckover2pi1
mω/parenrightbigg2/radicalbigg
π/planckover2pi1
mω/bracketrightBigg
=−3
4λ/planckover2pi1ω.
/angbracketleftH/angbracketright=3/planckover2pi1ω−3
4λ/planckover2pi1ω=3/planckover2pi1ω/parenleftbigg
1−λ
4/parenrightbigg
.
The variational principle says this must exceed the exact ground-state energy (b); let’s check it:
3/planckover2pi1ω/parenleftbigg
1−λ
4/parenrightbigg
>3
2/planckover2pi1ω/parenleftBig
1+√
1−λ/parenrightBig
⇔2−λ
2>1+√
1−λ⇔1−λ
2>√
1−λ⇔1−λ+λ2
4>1−λ.
It checks. In fact, expanding the exact answer in powers of λ,Egs≈3
2/planckover2pi1ω(1 + 1−1
2λ)=3 /planckover2pi1ω/parenleftbig
1−λ
4/parenrightbig
,
we recover the variational result.
Problem 7.18
1==/integraldisplay
|ψ|2d3r1d3r2=|A|2/bracketleftbigg/integraldisplay
ψ2
1d3r1/integraldisplay
ψ2
2d3r2+2/integraldisplay
ψ1ψ2d3r1/integraldisplay
ψ1ψ2d3r2+/integraldisplay
ψ2
2d3r1/integraldisplay
ψ2
1d3r2/bracketrightbigg
=|A|2( 1+2S2+1 ),
where
S≡/integraldisplay
ψ1(r)ψ2(r)d3r=/radicalbig
(Z1Z2)3
πa3/integraldisplay
e−(Z1+Z2)r/a4πr2dr=4
a3/parenleftBigy
2/parenrightBig3/bracketleftbigg2a3
(Z1+Z2)3/bracketrightbigg
=/parenleftBigy
x/parenrightBig3
.
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CHAPTER 7. THE VARIATIONAL PRINCIPLE 211
A2=1
2/bracketleftBig
1+(y/x)6/bracketrightBig.
H=−/planckover2pi12
2m(∇2
1+∇2
2)−e2
4πFepsilonC0/parenleftbigg1
r1+1
r2/parenrightbigg
+e2
4πFepsilonC01
|r1−r2|,
Hψ=A/braceleftbigg/bracketleftbigg
−/planckover2pi12
2m(∇2
1+∇2
2)−e2
4πFepsilonC0/parenleftbiggZ1
r1+Z2
r2/parenrightbigg/bracketrightbigg
ψ1(r1)ψ2(r2)
+/bracketleftbigg
−/planckover2pi12
2m(∇2
1+∇2
2)−e2
4πFepsilonC0/parenleftbiggZ1
r1+Z2
r2/parenrightbigg/bracketrightbigg
ψ2(r1)ψ1(r2)/bracerightbigg
+Ae2
4πFepsilonC0/braceleftbigg/bracketleftbiggZ1−1
r1+Z2−1
r2/bracketrightbigg
ψ1(r1)ψ2(r2)+/bracketleftbiggZ2−1
r1+Z1−1
r2/bracketrightbigg
ψ2(r1)ψ1(r2)/bracerightbigg
+Veeψ,
whereVee≡e2
4πFepsilonC01
|r1−r2|.
The term in first curly brackets is ( Z2
1+Z2
2)E1ψ1(r1)ψ2(r2)+(Z2
2+Z2
1)ψ2(r1)ψ1(r2), so
Hψ=(Z2
1+Z2
2)E1ψ
+Ae2
4πFepsilonC0/braceleftbigg/bracketleftbiggZ1−1
r1+Z2−1
r2/bracketrightbigg
ψ1(r1)ψ2(r2)+/bracketleftbiggZ2−1
r1+Z1−1
r2/bracketrightbigg
ψ2(r1)ψ1(r2)/bracerightbigg
+Veeψ
/angbracketleftH/angbracketright=(Z2
1+Z2
2)E1+/angbracketleftVee/angbracketright+A2/parenleftbigge2
4πFepsilonC0/parenrightbigg
×/braceleftbigg
/angbracketleftψ1(r1)ψ2(r2)+ψ2(r1)ψ1(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg/bracketleftbiggZ
1−1
r1+Z2−1
r2/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r1)ψ2(r2)/angbracketright+/bracketleftbiggZ2−1
r1+Z1−1
r2/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r1)ψ1(r2)/angbracketright/parenrightbigg/bracerightbigg
.
/braceleftbigg/bracerightbigg
=(Z1−1)/angbracketleftψ1(r1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r1/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r1)/angbracketright+(Z2−1)/angbracketleftψ2(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r2/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r2)/angbracketright
+(Z2−1)/angbracketleftψ1(r1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r1/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r1)/angbracketright/angbracketleftψ2(r2)|ψ1(r2)/angbracketright
+(Z1−1)/angbracketleftψ1(r1)|ψ2(r1)/angbracketright/angbracketleftψ2(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r2/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r2)/angbracketright+(Z1−1)/angbracketleftψ2(r1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r1/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r1)/angbracketright/angbracketleftψ1(r2)|ψ2(r2)/angbracketright
+(Z2−1)/angbracketleftψ2(r1)|ψ1(r1)/angbracketright/angbracketleftψ1(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r2/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r2)/angbracketright+(Z2−1)/angbracketleftψ2(r1)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r1/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r1)/angbracketright
+(Z1−1)/angbracketleftψ1(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r2/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r2)/angbracketright
=2 (Z1−1)/angbracketleftbigg1
r/angbracketrightbigg
1+2 (Z1−1)/angbracketleftbigg1
r/angbracketrightbigg
2+2 (Z1−1)/angbracketleftψ1|ψ2/angbracketright/angbracketleftψ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2/angbracketright+2 (Z2−1)/angbracketleftψ1|ψ2/angbracketright/angbracketleftψ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2/angbracketright.
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212 CHAPTER 7. THE VARIATIONAL PRINCIPLE
But/angbracketleftbigg1
r/angbracketrightbigg
1=/angbracketleftψ1(r)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r)/angbracketright=Z1
a;/angbracketleftbigg1
r/angbracketrightbigg
2=Z2
a,so/angbracketleftH/angbracketright=(Z2
1+Z2
2)E1
+A2/parenleftbigge2
4πFepsilonC0/parenrightbigg
2/bracketleftbigg1
a(Z1−1)Z1+1
a(Z2−1)Z2+(Z1+Z2−2)/angbracketleftψ1|ψ2/angbracketright/angbracketleftψ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2/angbracketright/bracketrightbigg
+/angbracketleftVee/angbracketright.
And/angbracketleftψ1|ψ2/angbracketright=S=(y/x)3,so
/angbracketleftψ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
r/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2/angbracketright=/radicalbig
(Z1Z2)3
πa34π/integraldisplay
e−(Z1+Z2)r/ardr=y3
2a3/bracketleftbigga
Z1+Z2/bracketrightbigg2
=y3
2ax2.
/angbracketleftH/angbracketright=(x2−1
2y2)E1+A2/parenleftbigge2
4πFepsilonC0/parenrightbigg2
a/braceleftbigg/bracketleftbig
Z2
1+Z2
2−(Z1+Z2)/bracketrightbig
+(x−2)/parenleftBigy
x/parenrightBig3y3
2x2/bracerightbigg
+/angbracketleftVee/angbracketright
=(x2−1
2y2)E1+4E1A2/bracketleftbigg
x2−1
2y2−x+1
2(x−2)y6
x5/bracketrightbigg
+/angbracketleftVee/angbracketright.
/angbracketleftVee/angbracketright=e2
4πFepsilonC0/angbracketleftψ/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
|r1−r2|/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ/angbracketright
=/parenleftbigge
4πFepsilonC0/parenrightbigg
A2/angbracketleftψ1(r1)ψ2(r2)+ψ2(r1)+ψ1(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
|r1−r2|/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r1)ψ2(r2)+ψ2(r1)ψ1(r2)/angbracketright
=/parenleftbigge
4πFepsilonC0/parenrightbigg
A2/bracketleftbigg
2/angbracketleftψ1(r1)ψ2(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
|r1−r2|/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r1)ψ2(r2)/angbracketright+2/angbracketleftψ1(r1)ψ2(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
|r1−r2|/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r1)ψ1(r2)/angbracketright/bracketrightbigg
=2/parenleftbigge
4πFepsilonC0/parenrightbigg
A2(B+C),where
B≡/angbracketleftψ1(r1)ψ2(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
|r1−r2|/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
1(r1)ψ2(r2)/angbracketright;C≡/angbracketleftψ1(r1)ψ2(r2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
|r1−r2|/vextendsingle/vextendsingle/vextendsingle/vextendsingleψ
2(r1)ψ1(r2)/angbracketright.
B=Z3
1Z3
2
(πa3)2/integraldisplay
e−2Z1r1/ae−2Z2r2/a 1
|r1−r2|d3r1d3r2.As on pp 300-301, the r2integral is
/integraldisplay
e−2Z2r2/a 1/radicalbig
r2
1+r2
2−2r1r2cosθ2d3r2
=πa3
Z3
2r1/bracketleftbigg
1−/parenleftbigg
1+Z2r1
a/parenrightbigg
e−2Z2r1/a/bracketrightbigg
(Eq. 7.24, but with a→2
Z2a).
B=Z3
1Z3
2
(πa3)2(πa3)
Z3
24π/integraldisplay∞
0e−2Z1r1/a1
r1/bracketleftbigg
1−/parenleftbigg
1+Z2r1
a/parenrightbigg
e−2Z2r1/a/bracketrightbigg
r2
1dr1
=4Z3
1
a3/integraldisplay∞
0/bracketleftbigg
r1e−2Z1r1/a−r1e−2(Z1+Z2)r1/a−Z2
ar2
1e−2(Z1+Z2)r1/a/bracketrightbigg
dr1
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 7. THE VARIATIONAL PRINCIPLE 213
=4Z3
1
a3/bracketleftBigg/parenleftbigga
2Z1/parenrightbigg2
−/parenleftbigga
2(Z1+Z2)/parenrightbigg2
−Z2
a2/parenleftbigga
2(Z1+Z2)/parenrightbigg3/bracketrightBigg
=Z3
1
a/parenleftbigg1
Z2
1−1
(Z1+Z2)2−Z2
(Z1+Z2)3/parenrightbigg
=Z1Z2
a(Z1+Z2)/bracketleftbigg
1+Z1Z2
(Z1+Z2)2/bracketrightbigg
=y2
4ax/parenleftbigg
1+y2
4x2/parenrightbigg
.
C=Z3
1Z3
2
(πa3)2/integraldisplay
e−Z1r1/ae−Z2r2/ae−Z2r1/ae−Z1r2/a 1
|r1−r2|d3r1d3r2
=(Z1Z2)3
(πa3)2/integraldisplay
e−(Z1+Z2)(r1+r2)/a 1
|r1−r2|d3r1d3r2.
The integral is the same as in Eq. 7.20, only with a→4
Z1+Z2a.Comparing Eqs. 7.20 and 7.25, we see that the
integral itself was
5
4a/parenleftbiggπa3
8/parenrightbigg2
=5
256π2a5.SoC=(Z1Z2)3
(πa3)25π2
25645a5
(Z1+Z2)5=20
a(Z1Z2)3
(Z1+Z2)5=5
16ay6
x5.
/angbracketleftVee/angbracketright=2/parenleftbigge
4πFepsilonC0/parenrightbigg
A2/bracketleftbiggy2
4ax/parenleftbigg
1+y2
4x2/parenrightbigg
+5
16ay6
x5/bracketrightbigg
=2A2(−2E1)y2
4x/parenleftbigg
1+y2
4x2+5y4
4x4/parenrightbigg
.
/angbracketleftH/angbracketright=E1/braceleftbigg
x2−1
2y2−2
[ 1+(y/x)6]/bracketleftbigg
x2−1
2y2−x+1
2(x−2)y6
x5/bracketrightbigg
−2
[ 1+(y/x)6]y2
4x/parenleftbigg
1+y2
4x2+5y4
4x4/parenrightbigg/bracerightbigg
=E1
(x6+y6)/braceleftbigg
(x2−1
2y2)(x6+y6)−2x6/bracketleftbigg
x2−1
2y2−x+1
2y6
x4−y6
x5+y2
4x+y4
16x3+5y6
16x5/bracketrightbigg/bracerightbigg
=E1
(x6+y6)/parenleftbigg
x8+x2y6−1
2x6y2−1
2y8−2x8+x6y2+2x7−x2y6+2xy6−1
2x5y2−1
8x3y4−5
8xy6/parenrightbigg
=E1
(x6+y6)/parenleftbigg
−x8+2x7+1
2x6y2−1
2x5y2−1
8x3y4+11
8xy6−1
2y8/parenrightbigg
.
Mathematica finds the minimum of /angbracketleftH/angbracketrightatx=1.32245,y=1.08505, corresponding to Z1=1.0392,Z2=
0.2832. At this point, /angbracketleftH/angbracketrightmin=1.0266E1=−13.962 eV, which isless than−13.6 eV—but not by much!
Problem 7.19
The calculation is the same as before, but with me→mµ(reduced), where
mµ(reduced) =mµmd
mµ+md=mµ2mp
mµ+2mp=mµ
1+mµ/2mp.From Problem 6.28, mµ= 207me,so
1+mµ
2mp=1+/parenleftbigg207
2/parenrightbigg(9.11×10−31)
(1.67×10−27)=1.056;mµ(reduced) =207me
1.056= 196me.
This shrinks the whole molecule down by a factor of almost 200, bringing the deuterons much closer together,
as desired. The equilibrium separation for the electron case was 2 .493a(Problem 7.10), so for muons, R=
2.493
196(0.529×10−10m) = 6.73×10−13m.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
214 CHAPTER 7. THE VARIATIONAL PRINCIPLE
Problem 7.20
(a)
−/planckover2pi12
2m/parenleftbigg∂2ψ
∂x2+∂2ψ
∂y2/parenrightbigg
=Eψ. Letψ(x,y)=X(x)Y(y).
Yd2X
dx2+Xd2Y
dy2=−2mE
/planckover2pi12XY;1
Xd2X
dx2+1
Yd2Y
dy2=−2mE
/planckover2pi12.
d2X
dx2=−k2
xX;d2Y
dy2=−k2
yY,withk2
x+k2
y=2mE
/planckover2pi12.The general solution to the yequation is
Y(y)=Acoskyy+Bsinkyy; the boundary conditions Y(±a) = 0 yield ky=nπ
2awith minimumπ
2a.
[Note that k2
yhas to be positive, or you cannot meet the boundary conditions at all.] So
E≥/planckover2pi12
2m/parenleftbigg
k2
x+π2
4a2/parenrightbigg
.For a traveling wavek2
xhas to be positive. Conclusion: Any solution with E<
π2/planckover2pi12
8ma2will be a bound state.
(b)
aa
xy
III
Integrate over regions I and II (in the figure), and multiply by 8.
III=A2/integraldisplay∞
x=a/integraldisplaya
y=0/parenleftbigg
1−y
a/parenrightbigg2
e−2αx/adxdy. Letu≡x
a,v≡y
a,d x=adu, dy =adv.
=A2a2/integraldisplay∞
1/integraldisplay1
0(1−v)2e−2αududv =A2a2/bracketleftbigg(1−v)3
3/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
0×e−2αu
2α/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
1/bracketrightbigg
=A2a2
6α(−1)/parenleftbig
−e−2α/parenrightbig
=A2a2
6αe−2α.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 7. THE VARIATIONAL PRINCIPLE 215
II=1
2A2/integraldisplaya
x=0/integraldisplaya
y=0/parenleftbigg
1−xy
a2/parenrightbigg2
e−2αdxdy
=1
2A2a2/integraldisplay1
0/integraldisplay1
0(1−uv)2e−2αdudv =1
2A2a2e−2α/integraldisplay1
0(1−uv)3
−3v/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
0dv
=−1
2A2a2e−2α1
3/integraldisplay1
0(1−v)3−1
vdv=1
6A2a2e−2α/integraldisplay1
0(v2−3v+3 )dv,
=1
6A2a2e−2α/parenleftbiggv3
3−3v2
2+3v/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
0=11
36A2a2e−2α.
Normalizing: 8/bracketleftbiggA2a2
6αe−2α+11
36A2a2e−2α/bracketrightbigg
=1⇒A2=9α
2a2e2α
( 6+1 1α).
/angbracketleftH/angbracketright=−/planckover2pi12
2m/angbracketleftψ|∂2
∂x2+∂2
∂y2|ψ/angbracketright=−8/planckover2pi12
2m(JI+JII).[Ignore roof-lines for the moment.]
JII=A2/integraldisplay∞
x=a/integraldisplaya
y=0/parenleftbigg
1−y
a/parenrightbigg
e−αx/a/parenleftbigg∂2
∂x2+
✼0
∂2
∂y2/parenrightbigg/bracketleftbigg/parenleftbigg
1−y
a/parenrightbigg
e−αx/a/bracketrightbigg
dxdy
=A2/integraldisplay∞
x=a/integraldisplaya
y=0/parenleftbigg
1−y
a/parenrightbigg2/parenleftbiggα
a/parenrightbigg2
e−2αx/adxdy =/parenleftbiggα
a/parenrightbigg2
III=/parenleftbiggα
✁a/parenrightbigg✄2A2 a2
6✚αe−2α=1
6A2αe−2α.
JI=1
2A2/integraldisplaya
0/integraldisplaya
0/parenleftbigg
1−xy
a2/parenrightbigg
e−α/parenleftbigg∂2
∂x2+∂2
∂y2/parenrightbigg/parenleftbigg
1−xy
a2/parenrightbigg
e−αdxdy =0.
[Note that∂2
∂x2/parenleftbigg
1−xy
a2/parenrightbigg
=∂
∂x/parenleftBig
−y
a2/parenrightBig
=0,and likewise for ∂2/∂y2.]
/angbracketleftH/angbracketrightso far=−2
3A2/planckover2pi12α
me−2α.
Now the roof-lines; label them as follows:
I. Right arm: aty=0:KI.
II. Central square: atx= 0 and at y=0:KII.
III. Boundaries: atx=±aand aty=±a:KIII.
KI=4/parenleftbigg
−/planckover2pi12
2m/parenrightbigg
A2/integraldisplay∞
x=a/integraldisplaya
y=−a/parenleftbigg
1−|y|
a/parenrightbigg
e−αx/a/parenleftbigg
∂2
∂x2+∂2
∂y2/parenrightbigg/parenleftbigg
1−|y|
a/parenrightbigg
e−αx/adxdy.
|y|=y/bracketleftbigg
θ(y)−θ(−y)/bracketrightbigg
,
∂
∂y/parenleftbigg
1−|y|
a/parenrightbigg
=−1
a/bracketleftbigg
θ(y)−θ(−y)+✟✟✟yδ(y)+✟✟✟yδ(y)/bracketrightbigg
,
∂2
∂y2/parenleftbigg
1−|y|
a/parenrightbigg
=−1
a/bracketleftbig
δ(y)−δ(−y)(−1)/bracketrightbig
=−2
aδ(y).
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216 CHAPTER 7. THE VARIATIONAL PRINCIPLE
KI=−2/planckover2pi12
mA2/integraldisplay∞
x=ae−2αx/adx
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
♣/integraldisplaya
y=−a/parenleftbigg
1−|y|
a/parenrightbigg/bracketleftbigg
−2
aδ(y)/bracketrightbigg
dy
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
♠
♣=e−2αx/a
(−2α/a)/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
a=−e−2α
(−2α/a)=a
2αe−2α;♠=−2
a,
= −2/planckover2pi12A2
m✁a
✁2αe−2α/parenleftbigg
−2
✁a/parenrightbigg
;KI=2/planckover2pi12
mαe−2αA2.
KII=4A2/parenleftbigg
−/planckover2pi12
2m/parenrightbigg/integraldisplaya
x=0/integraldisplaya
y=−a/parenleftbigg
1−x|y|
a2/parenrightbigg
e−α/parenleftbigg
∂2
∂x2+∂2
∂y2/parenrightbigg/parenleftbigg
1−x|y|
a2/parenrightbigg
e−αdxdy
=−2/planckover2pi12
mA2e−2α/integraldisplaya
x=0/integraldisplaya
y=−a/parenleftbigg
1−x|y|
a2/parenrightbigg/bracketleftbigg
−2x
a2δ(y)/bracketrightbigg
dxdy
= −2/planckover2pi12A2
me−2α/parenleftbigg
−✁✁✁2
a2/parenrightbigg/integraldisplaya
0xdx
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
a2
2;KII=2/planckover2pi12
me−2αA2.
KIII=8/parenleftbigg
−/planckover2pi12
2m/parenrightbigg/integraldisplaya
y=0/integraldisplaya+ρepsilono
x=a−ρepsilonoψ/parenleftbigg∂2
∂x2+∂2
∂y2/parenrightbigg
ψdxdy.
In this region ( x, yboth positive) ψ=A
/parenleftbigg
1−xy/a 2/parenrightbigg
e−α(x<a)
/parenleftbigg
1−y/a/parenrightbigg
e−αx/a(x>a)
,or
ψ=A/parenleftbigg/braceleftbigg
1−y
a/bracketleftbigg
θ(x−a)+x
aθ(a−x)/bracketrightbigg/bracerightbigg
e−α[θ(a−x)+x
aθ(x−a)]/parenrightbigg
.
∂ψ
∂x=A/parenleftbigg
−y
a/bracketleftbigg
✘✘✘✘δ(x−a)+1
aθ(a−x)−
✟✟✟✟✟ x
aδ(a−x)/bracketrightbigg
e−α[θ(a−x)+x
aθ(x−a)]
+/braceleftbigg
1−y
a/bracketleftbigg
θ(x−a)+x
aθ(a−x)/bracketrightbigg/bracerightbigg
e−α[θ(a−x)+x
aθ(x−a)]/bracketleftbigg
α✘✘✘✘δ(a−x)−α
aθ(x−a)−✘✘✘✘✘✘ αx
aδ(x−a)/bracketrightbigg/parenrightbigg
[Note:f(x)=xδ(x) should be zero—but perhaps we should check that this is still safe when we’re
planning to take it’s derivative: df/dx =δ(x)+xdδ/dx :
/integraldisplay
gdf
dxdx=/integraldisplay
g/bracketleftbigg
δ(x)+xdδ
dx/bracketrightbigg
dx=g(0) +/integraldisplay
gxdδ
dxdx
=g(0) +✘✘✘✘✘✘✿0
gxδ(x)|x=0−/integraldisplayd
dx(gx)δ(x)dx=g(0)−/integraldisplay/parenleftbigg
g+xdg
dx/parenrightbigg
δ(x)dx
=g(0)−g(0)−(xg/prime)|x=0=0.
This confirms that f(x) can be taken to be zero evenwhen differentiated.]
Soδ(x−a)−x
aδ(a−x)=1
a(a−x)δ(a−x)=0.Hence the cancellations above, leaving
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CHAPTER 7. THE VARIATIONAL PRINCIPLE 217
∂ψ
∂x=A/parenleftbigg
−y
a2θ(a−x)e−α[θ(a−x)+x
aθ(x−a)]
−α
aθ(x−a)/braceleftbigg
1−y
a/bracketleftbigg
θ(x−a)+x
aθ(a−x)/bracketrightbigg/bracerightbigg
e−α[θ(a−x)+x
aθ(x−a)]/parenrightbigg
=Ae−α[θ(a−x)+x
aθ(x−a)]/parenleftbigg
−y
a2θ(a−x)−α
aθ(x−a)/braceleftbigg
1−y
a/bracketleftbigg
θ(x−a)+x
aθ(a−x)/bracketrightbigg/bracerightbigg/parenrightbigg
=−A
ae−α[θ(a−x)+x
aθ(x−a)]/bracketleftbiggy
aθ(a−x)+αθ(x−a)/parenleftbigg
1−y
a/parenrightbigg/bracketrightbigg
.
∂2ψ
∂x2=−A
ae−α[θ(a−x)+x
aθ(x−a)]/braceleftbigg
−y
aδ(a−x)+αδ(x−a)/parenleftbigg
1−y
a/parenrightbigg
−α/bracketleftbigg
−✘✘✘✘δ(a−x)+1
aθ(x−a)
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
integral 0+
✟✟✟✟✟ x
aδ(x−a)/bracketrightbigg/bracerightbigg
=−A
ae−αδ(x−a)/bracketleftbigg
α−αy
a−y
a/bracketrightbigg
.
KIII=−4/planckover2pi12
m/integraldisplaya
y=0/integraldisplaya+ρepsilono
x=a−ρepsilonoψ(x,y)/bracketleftbigg
−A
ae−αδ(x−a)/parenleftbigg
α−αy
a−y
a/parenrightbigg/bracketrightbigg
dxdy
=4/planckover2pi12A
mae−α/integraldisplaya
y=0ψ(a,y)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
A(1−y/a)e−α/parenleftbigg
α−αy
a−y
a/parenrightbigg
dy=4/planckover2pi12A2
mae−2α/integraldisplaya
0/parenleftbigg
α−2αy
a−y
a+αy2
a2+y2
a2/parenrightbigg
dy
=4/planckover2pi12A2
mae−2α/parenleftbigg
αa−✁2α
✁aa✄2
✁2−1
aa2
2+α
a2a3
3+1
a2a3
3/parenrightbigg
=4/planckover2pi12A2
me−2α/parenleftbigg
✟✟αa−✟✟αa−✁a
2+α✁a
3+✁a
3/parenrightbigg
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
α
3−1
6=1
6(2α−1)
=4/planckover2pi12A2
6m(2α−1)e−2α;KIII=2/planckover2pi12A2
3m(2α−1)e−2α.
/angbracketleftH/angbracketright=−2
3A2/planckover2pi12α
me−2α+2/planckover2pi12
mαe−2αA2+2/planckover2pi12
me−2αA2+2/planckover2pi12
3mA2(2α−1)e−2α
=A2e−2α/planckover2pi12
m/bracketleftbigg
−2
3α+2
α+2+2
3(2α−1)/bracketrightbigg
=2A2e−2α/planckover2pi12
3m/parenleftbigg
−α+3
α+3+2α−1/parenrightbigg
=2A2e−2α/planckover2pi12
3m/parenleftbigg
α+2+3
α/parenrightbigg
=2A2e−2α/planckover2pi12
3mα/parenleftbig
α2+2α+3/parenrightbig
=✁2
3✘✘✘e−2α/planckover2pi12
m✚α/parenleftbig
α2+2α+3/parenrightbig9
✁2✚α
a2✟✟e2α
( 6+1 1α)
=3/planckover2pi12
ma2(α2+2α+3 )
( 6+1 1α).
d/angbracketleftH/angbracketright
dα=3/planckover2pi12
ma2( 6+1 1α)(2α+2 )−(α2+2α+ 3)(11)
( 6+1 1α)2=0⇒( 6+1 1α)(2α+ 2) = 11( α2+2α+3 ).
12α+1 2+2 2 α2+2 2α=1 1α2+2 2α+3 3⇒11α2+1 2α−21 = 0.
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218 CHAPTER 7. THE VARIATIONAL PRINCIPLE
α=−12±2/radicalbig
(12)2+4·11·21
22=−6±√36 + 231
11
=−6±16.34
11=10.34
11[αhas to be positive ]=0.940012239 .
/angbracketleftH/angbracketrightmin=3/planckover2pi12
ma22(α+1 )
11=6
11/planckover2pi12
ma2(α+1 ) =1.058/parenleftbigg/planckover2pi12
ma2/parenrightbigg
.ButEthreshold =π2
8/planckover2pi12
ma2=1.2337/planckover2pi12
ma2,
soE0is definitely lessthanEthreshold .
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CHAPTER 8. THE WKB APPROXIMATION 219
Chapter 8
The WKB Approximation
Problem 8.1
/integraldisplaya
0p(x)dx=nπ/planckover2pi1,withn=1,2,3,...andp(x)=/radicalbig
2m[E−V(x)] (Eq. 8.16) .
Here/integraldisplaya
0p(x)dx=√
2mE/parenleftBiga
2/parenrightBig
+/radicalbig
2m(E−V0)/parenleftBiga
2/parenrightBig
=√
2m/parenleftBiga
2/parenrightBig/parenleftBig√
E+/radicalbig
E−V0/parenrightBig
=nπ/planckover2pi1
⇒E+E−V0+2/radicalbig
E(E−V0)=4
2m/parenleftbiggnπ/planckover2pi1
a/parenrightbigg2
=4E0
n;2/radicalbig
E(E−V0)=( 4E0
n−2E+V0).
Square again: 4 E(E−V0)=4E2−4EV0=1 6E0
n2+4E2+V2
0−16EE0
n+8E0
nV0−4EV0
⇒16EE0
n=1 6E0
n2+8E0
nV0+V2
0⇒En=E0
n+V0
2+V2
0
16E0n.
Perturbation theory gave En=E0
n+V0
2; the extra term goes to zero for very small V0(or, since E0
n∼n2), for
largen.
Problem 8.2
(a)
dψ
dx=i
/planckover2pi1f/primeeif//planckover2pi1;d2ψ
dx2=i
/planckover2pi1/parenleftbigg
f/prime/primeeif//planckover2pi1+i
/planckover2pi1(f/prime)2eif//planckover2pi1/parenrightbigg
=/bracketleftbiggi
/planckover2pi1f/prime/prime−1
/planckover2pi12(f/prime)2/bracketrightbigg
eif//planckover2pi1.
d2ψ
dx2=−p2
/planckover2pi12ψ=⇒/bracketleftbiggi
/planckover2pi1f/prime/prime−1
/planckover2pi12(f/prime)2/bracketrightbigg
eif//planckover2pi1=−p2
/planckover2pi12eif//planckover2pi1=⇒i/planckover2pi1f/prime/prime−(f/prime)2+p2=0.QED
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220 CHAPTER 8. THE WKB APPROXIMATION
(b)f/prime=f/prime
0+/planckover2pi1f/prime
1+/planckover2pi12f/prime
2+···=⇒(f/prime)2=(f/prime
0+/planckover2pi1f/prime
1+/planckover2pi12f/prime
2+···)2=(f/prime
0)2+2/planckover2pi1f/prime
0f/prime
1+/planckover2pi12[2f/prime
0f/prime
2+(f/prime
1)2]+···
f/prime/prime=f/prime/prime
0+/planckover2pi1f/prime/prime
1+/planckover2pi12f/prime/prime
2+···.i/planckover2pi1(f/prime/prime
0+/planckover2pi1f/prime/prime
1+/planckover2pi12f/prime/prime
2)−(f/prime
0)2−2/planckover2pi1f/prime
0f/prime
1−/planckover2pi12[2f/prime
0f/prime
2+(f/prime
1)2]+p2+···=0.
/planckover2pi10:(f/prime
0)2=p2;/planckover2pi11:if/prime/prime
0=2f/prime
0f/prime
1;/planckover2pi12:if/prime/prime
1=2f/prime
0f/prime
2+(f/prime
1)2;...
(c)df0
dx=±p=⇒f0=±/integraldisplay
p(x)dx+ constant ;df1
dx=i
2f/prime/prime
0
f/prime
0=i
2/parenleftbigg±p/prime
±p/parenrightbigg
=i
2d
dxlnp=⇒f1=i
2lnp+ const .
ψ= exp/parenleftbiggif
/planckover2pi1/parenrightbigg
= exp/bracketleftbiggi
/planckover2pi1/parenleftbigg
±/integraldisplay
p(x)dx+/planckover2pi1i
2lnp+K/parenrightbigg/bracketrightbigg
= exp/parenleftbigg
±i
/planckover2pi1/integraldisplay
pdx/parenrightbigg
p−1/2eiK//planckover2pi1
=C√pexp/parenleftbigg
±i
/planckover2pi1/integraldisplay
pdx/parenrightbigg
.QED
Problem 8.3
γ=1
/planckover2pi1/integraldisplay
|p(x)|dx=1
/planckover2pi1/integraldisplay2a
0/radicalbig
2m(V0−E)dx=2a
/planckover2pi1/radicalbig
2m(V0−E).T≈e−4a√
2m(V0−E)//planckover2pi1.
From Problem 2.33, the exact answer is
T=1
1+V2
0
4E(V0−E)sinh2γ.
Now, the WKB approximation assumes the tunneling probability is small (p. 322)—which is to say that γis
large. In this case, sinh γ=1
2(eγ−e−γ)≈1
2eγ, and sinh2γ≈1
4e2γ, and the exact result reduces to
T≈1
1+V2
0
16E(V0−E)e2γ≈/braceleftbigg16E(V0−E)
V2
0/bracerightbigg
e−2γ.
The coefficient in {}is of order 1; the dominant dependence on Eis in the exponential factor. In this sense
T≈e−2γ(the WKB result).
Problem 8.4
I take the masses from Thornton and Rex, Modern Physics , Appendix 8. They are all atomic masses, but the
electron masses subtract out in the calculation of E. All masses are in atomic units (u): 1 u = 931 MeV/ c2.
The mass of He4is 4.002602 u, and that of the α-particle is 3727 MeV/ c2.
U238:Z=9 2,A= 238,m= 238.050784 u→Th234:m= 234.043593 u .
r1=( 1.07×10−15m)(238)1/3=6.63×10−15m.
E= (238.050784−234.043593−4.002602)(931) MeV = 4 .27 MeV .
V=/radicalbigg
2E
m=/radicalbigg
(2)(4.27)
3727×3×108m/s=1.44×107m/s.
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CHAPTER 8. THE WKB APPROXIMATION 221
γ=1.98090√
4.27−1.485/radicalbig
90(6.63) = 86 .19−36.28 = 49 .9.
τ=(2)(6.63×10−15)
1.44×107e98.8s=7.46×1021s=7.46×1021
3.15×107yr = 2.4×1014yrs.
Po212:Z=8 4,A= 212,m= 211.988842 u→Pb208:m= 207.976627 u .
r1=( 1.07×10−15m)(212)1/3=6.38×10−15m.
E= (211.988842−207.976627−4.002602)(931) MeV = 8 .95 MeV .
V=/radicalbigg
2E
m=/radicalbigg
(2)(8.95)
3727×3×108m/s=2.08×107m/s.
γ=1.98082√
8.95−1.485/radicalbig
82(6.38) = 54 .37−33.97 = 20 .4.
τ=(2)(6.38×10−15)
2.08×107e40.8s=3.2×10−4s.
These results are wayoff—but note the extraordinary sensitivity to nuclear masses: a tiny change in Eproduces
enormous changes in τ.
Much more impressive results are obtained when you plot the logarithm of lifetimes against 1 /√
E,a si n
Figure 8.6. Thanks to David Rubin for pointing this out. Some experimental values are listed below (all energiesin MeV):
Uranium
(Z= 92):AE τ
2384.1984.468×109yr
2364.4942.342×107yr
2344.7752.455×105yr
2325.320 68.9y r
2305.888 20.8d a y
2286.680 9.1 min
2267.570 0.35 sProtactinium (Z= 91):AEτ
2247.4880.79 s
2228.5402.9m s
2209.6500.78µs
2189.6140.12 ms
Thorium (Z= 90):AE τ
2324.0121.405×1010yr
2304.6877.538×104yr
2285.423 1.912 yr
2266.337 30.57 minRadium (Z= 88):AE τ
2264.7841600 yr
2245.6853.66 day
2226.559 38 s
2207.45518 ms
2188.38925.6µs
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222 CHAPTER 8. THE WKB APPROXIMATION
Problem 8.5
(a)V(x)=mgx.
(b)
−/planckover2pi12
2md2ψ
dx2+mgxψ =Eψ=⇒d2ψ
dx2=2m2g
/planckover2pi12/parenleftbigg
x−E
mg/parenrightbigg
.Lety≡x−E
mg,andα≡/parenleftbigg2m2g
/planckover2pi12/parenrightbigg1/3
.
Thend2ψ
dy2=α3yψ. Letz≡αy=α(x−E
mg), sod2ψ
dz2=zψ. This is the Airy equation (Eq. 8.36), and
the general solution is ψ=aAi(z)+bBi(z). However, Bi(z) blows up for large z,s ob= 0 (to make ψ
normalizable). Hence ψ(x)=aAi/bracketleftBig
α(x−E
mg)/bracketrightBig
.
(c)SinceV(x)=∞forx<0, we require ψ(0) = 0; hence Ai[α(−E/mg )] = 0. Now, the zeros of Aiare
an(n=1,2,3,...). Abramowitz and Stegun list a1=−2.338,a2=−4.088,a3=−5.521,a4=−6.787,
etc. Here−αEn
mg=an,o rEn=−mg
αan=−mg/parenleftbigg/planckover2pi12
2m2g/parenrightbigg1/3
an,o rEn=−(1
2mg2/planckover2pi12)1/3an.In this case
1
2mg2/planckover2pi12=1
2(0.1 kg)(9.8 m/s2)2(1.055×10−34J·s)2=5.34×10−68J3;(1
2mg2/planckover2pi12)1/3=3.77×10−23J.
E1=8.81×10−23J,E 2=1.54×10−22J,E 3=2.08×10−22J,E 4=2.56×10−22J.
(d)
2/angbracketleftT/angbracketright=/angbracketleftxdV
dX/angbracketright(Eq. 3.97); heredV
dx=mg, so/angbracketleftxdV
dx/angbracketright=/angbracketleftmgx/angbracketright=/angbracketleftV/angbracketright,so/angbracketleftT/angbracketright=1
2/angbracketleftV/angbracketright.
But/angbracketleftT/angbracketright+/angbracketleftV/angbracketright=/angbracketleftH/angbracketright=En,so3
2/angbracketleftV/angbracketright=En,or/angbracketleftV/angbracketright=2
3En.But/angbracketleftV/angbracketright=mg/angbracketleftx/angbracketright,so/angbracketleftx/angbracketright=2En
3mg.
For the electron ,/parenleftbigg1
2mg2/planckover2pi12/parenrightbigg1/3
=/bracketleftbigg1
2(9.11×10−31)(9.8)2(1.055×10−34)2/bracketrightbigg1/3
=7.87×10−33J.
E1=1.84×10−32J=1.15×10−13eV./angbracketleftx/angbracketright=2(1.84×10−32)
3(9.11×10−31)(9.8)=1.37×10−3=1.37 mm.
Problem 8.6
(a)
Eq.8.47 =⇒/integraldisplayx2
0p(x)dx=(n−1
4)π/planckover2pi1,wherep(x)=/radicalbig
2m(E−mgx) andE=mgx2=⇒x2=E/mg.
/integraldisplayx2
0p(x)dx=√
2m/integraldisplayx2
0/radicalbig
E−mgxdx =√
2m/bracketleftbigg
−2
3mg(E−mgx)3/2/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglex2
0
=−2
3/radicalbigg
2
m1
g/bracketleftBig
(E−mgx2)3/2−E3/2/bracketrightBig
=2
3/radicalbigg
2
m1
gE3/2.
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CHAPTER 8. THE WKB APPROXIMATION 223
EV(x)
xxmgx
2
1
3√mg(2E)3/2=(n−1
4)π/planckover2pi1,orEn=/bracketleftbig9
8π2mg2/planckover2pi12(n−1
4)2/bracketrightbig1/3.
(b)
/parenleftbigg9
8π2mg2/planckover2pi12/parenrightbigg1/3
=/bracketleftbigg9
8π2(0.1)(9.8)2(1.055×10−34)2/bracketrightbigg1/3
=1.0588×10−22J.
E1=( 1.0588×10−22)/parenleftbigg3
4/parenrightbigg2/3
=8.74×10−23J,
E2=( 1.0588×10−22)/parenleftbigg7
4/parenrightbigg2/3
=1.54×10−22J,
E3=( 1.0588×10−22)/parenleftbigg11
4/parenrightbigg2/3
=2.08×10−22J,
E4=( 1.0588×10−22)/parenleftbigg15
4/parenrightbigg2/3
=2.56×10−22J.
These are in very close agreement with the exact results (Problem 8.5(c)). In fact, they agree precisely
(to 3 significant digits), except for E1(for which the exact result was 8 .81×10−23J).
(c)From Problem 8.5(d),
/angbracketleftx/angbracketright=2En
3mg,so 1 =2
3(1.0588×10−22)
(0.1)(9.8)/parenleftbigg
n−1
4/parenrightbigg2/3
,or/parenleftbigg
n−1
4/parenrightbigg2/3
=1.388×1022.
n=1
4+( 1.388×1022)3/2=1.64×1033.
Problem 8.7
/integraldisplayx2
x1p(x)dx=/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1;p(x)=/radicalBigg
2m/parenleftbigg
E−1
2mω2x2/parenrightbigg
;x2=−x1=1
ω/radicalbigg
2E
m.
/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=mω/integraldisplayx2
−x2/radicalbigg
2E
mω2−x2dx=2mω/integraldisplayx2
0/radicalBig
x2
2−x2dx=mω/bracketleftbigg
x/radicalBig
x2
2−x2+x2
2sin−1(x/x2)/bracketrightbiggx2
0
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224 CHAPTER 8. THE WKB APPROXIMATION
=mωx2
2sin−1(1) =π
2mωx2
2=π
2mω2E
mω2=πE
ω.En=/parenleftbig
n−1
2/parenrightbig
/planckover2pi1ω(n=1,2,3,...)
Since the WKB numbering starts with n= 1, whereas for oscillator states we traditionally start with n=0 ,
lettingn→n+ 1 converts this to the usual formula En=(n+1
2)/planckover2pi1ω. In this case the WKB approximation
yields the exact results.
Problem 8.8
(a)
1
2mω2x2
2=En=/parenleftbigg
n+1
2/parenrightbigg
/planckover2pi1ω(counting n=0,1,2,...);x2=/radicalbigg
(2n+1 )/planckover2pi1
mω.
(b)
Vlin(x)=1
2mω2x2
2+(mω2x2)(x−x2)=⇒Vlin(x2+d)=1
2mω2x2
2+mω2x2d.
V(x2+d)−Vlin(x2+d)
V(x2)=1
2mω2(x2+d)2−1
2mω2x2
2−mω2x2d
1
2mω2x2
2
=x2
2+2x2d+d2−x2
2−2x2d
x2
2=/parenleftbiggd
x2/parenrightbigg2
=0.01.d=0.1x2.
(c)
α=/bracketleftbigg2m
/planckover2pi12mω2x2/bracketrightbigg1/3
(Eq.8.34),so 0.1x2/bracketleftbigg2m2ω2
/planckover2pi12x2/bracketrightbigg1/3
≥5=⇒/bracketleftbigg2m2ω2
/planckover2pi12x4
2/bracketrightbigg1/3
≥50.
2m2ω2
/planckover2pi12(2n+1 )2/planckover2pi12
m2ω2≥(50)3;o r ( 2 n+1 )2≥(50)3
2= 62500; 2 n+1≥250;n≥249
2= 124.5.
nmin= 125.However, as we saw in Problems 8.6 and 8.7, WKB may be valid at much smaller n.
Problem 8.9
Shift origin to the turning point.
ψWKB=
1
/radicalbig
|p(x)|De−1
/planckover2pi1/integraltext0
x|p(x/prime)|dx/prime(x<0)
1/radicalbig
|p(x)|/bracketleftBig
Bei
/planckover2pi1/integraltextx
0p(x/prime)dx/prime+Ce−i
/planckover2pi1/integraltextx
0p(x/prime)dx/prime/bracketrightBig
(x>0)
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CHAPTER 8. THE WKB APPROXIMATION 225
x0Nonclassical Classicaloverlap 2 overlap 1
WKBP
ψψ
WKBψ
patching regionE
Linearized potential in the patching region:
V(x)≈E+V/prime(0)x.Note :V/prime(0) isnegative .d2ψp
dx2=2mV/prime(0)
/planckover2pi12xψp=−α3xψp,whereα≡/parenleftbigg2m|V/prime(0)|
/planckover2pi12/parenrightbigg1/3
.
ψp(x)=aAi(−αx)+bBi(−αx).(Note change of sign, as compared with Eq. 8.37).
p(x)=/radicalbig
2m[E−E−V/prime(0)x]=/radicalbig
−2mV/prime(0)x=/radicalbig
2m|V/prime(0)|x=√
α3/planckover2pi12x=/planckover2pi1α3/2√x.
Overlap region 1 ( x<0):
/integraldisplay0
x|p(x/prime)|dx/prime=/planckover2pi1α3/2/integraldisplay0
x√
−x/primedx/prime=/planckover2pi1α3/2/parenleftbigg
−2
3(−x/prime)3/2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle0
x=2
3/planckover2pi1α3/2(−x)3/2=2
3/planckover2pi1(−αx)3/2.
ψWKB≈1
/planckover2pi11/2α3/2(−x)1/4De−2
3(−αx)3/2.For large positive argument ( −αx/greatermuch1) :
ψp≈a1
2√π(−αx)1/4e−2
3(−αx)3/2+b1√π(−αx)1/4e2
3(−αx)3/2.Comparing⇒a=2D/radicalbiggπ
α/planckover2pi1;b=0.
Overlap region 2 ( x>0):
/integraldisplayx
0|p(x/prime)|dx/prime=/planckover2pi1α3/2/integraldisplayx
0√
x/primedx/prime=/planckover2pi1α3/2/bracketleftbigg2
3(x/prime)3/2/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglex
0=2
3/planckover2pi1(αx)3/2.
ψWKB≈1
/planckover2pi11/2α3/4x1/4/bracketleftBig
Bei2
3(αx)3/2+Ce−i2
3(αx)3/2/bracketrightBig
.For large negative argument ( −αx/lessmuch−1) :
ψp(x)≈a1√π(αx)1/4sin/bracketleftbigg2
3(αx)3/2+π
4/bracketrightbigg
=a√π(αx)1/41
2i/bracketleftBig
eiπ/4ei2
3(αx)3/2−e−iπ/4e−i2
3(αx)3/2/bracketrightBig
(remember : b=0 ).
Comparing the two: B=a
2i/radicalbigg
α/planckover2pi1
πeiπ/4,C=−a
2i/radicalbigg
α/planckover2pi1
πe−iπ/4.
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226 CHAPTER 8. THE WKB APPROXIMATION
Inserting the expression for afrom overlap region 1 : B=−ieiπ/4D;C=ie−iπ/4D.Forx>0,then,
ψWKB=−iD/radicalbig
p(x)/bracketleftBig
ei
/planckover2pi1/integraltextx
0p(x/prime)dx/prime+iπ
4−e−i
/planckover2pi1/integraltextx
0p(x/prime)dx/prime−iπ
4/bracketrightBig
=2D/radicalbig
p(x)sin/bracketleftbigg1
/planckover2pi1/integraldisplayx
0p(x/prime)dx/prime+π
4/bracketrightbigg
.
Finally, switching the origin back to x1:
ψWKB(x)=
D
/radicalbig
|p(x)|e−1
/planckover2pi1/integraltextx1
x|p(x/prime)|dx/prime, (x<x 1);
2D/radicalbig
p(x)sin/bracketleftbigg1
/planckover2pi1/integraldisplayx
x1p(x/prime)dx/prime+π
4/bracketrightbigg
,(x>x 1).
QED
Problem 8.10
Atx1, we have an upward-sloping turning point. Follow the method in the book. Shifting origin to x1:
ψWKB(x)=
1
/radicalbig
p(x)/bracketleftBig
Aei
/planckover2pi1/integraltext0
xp(x/prime)dx/prime+B−i
/planckover2pi1/integraltext0
xp(x/prime)dx/prime/bracketrightBig
(x<0)
1/radicalbig
p(x)/bracketleftBig
Ce1
/planckover2pi1/integraltextx
0|p(x/prime)|dx/prime+D−1
/planckover2pi1/integraltextx
0|p(x/prime)|dx/prime/bracketrightBig
(x>0)
In overlap region 2 , Eq. 8.39 becomes ψWKB≈1
/planckover2pi11/2α3/4x1/4/bracketleftBig
Ce2
3(αx)3/2+De−2
3(αx)3/2/bracketrightBig
,
whereas Eq. 8.40 is unchanged. Comparing them = ⇒a=2D/radicalbiggπ
α/planckover2pi1,b=C/radicalbiggπ
α/planckover2pi1.
In overlap region 1 , Eq. 8.43 becomes ψWKB≈1
/planckover2pi11/2α3/4(−x)1/4/bracketleftBig
Aei2
3(−αx)3/2+Be−i2
3(−αx)3/2/bracketrightBig
,
and Eq. 8.44 (with b/negationslash= 0) generalizes to
ψp(x)≈a√π(−αx)1/4sin/bracketleftbigg2
3(−αx)3/2+π
4/bracketrightbigg
+b√π(−αx)1/4cos/bracketleftbigg2
3(−αx)3/2+π
4/bracketrightbigg
=1
2√π(−αx)1/4)/bracketleftBig
(−ia+b)ei2
3(−αx)3/2eiπ/4+(ia+b)e−i2
3(−αx)3/2e−iπ/4/bracketrightBig
.Comparing them = ⇒
A=/radicalbigg
/planckover2pi1α
π/parenleftbigg−ia+b
2/parenrightbigg
eiπ/4;B=/radicalbigg
/planckover2pi1α
π/parenleftbiggia+b
2/parenrightbigg
e−iπ/4.Putting in the expressions above for aandb:
A=/parenleftbiggC
2−iD/parenrightbigg
eiπ/4;B=/parenleftbiggC
2+iD/parenrightbigg
e−iπ/4.
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CHAPTER 8. THE WKB APPROXIMATION 227
These are the connection formulas relating A,B,C , andD,a tx1.
Atx2, we have a downward-sloping turning point, and follow the method of Problem 8.9. First rewrite the
middle expression in Eq. 8.52:
ψWKB=1/radicalbig
|p(x)|/bracketleftBig
Ce1
/planckover2pi1/integraltextx2
x1|p(x/prime)|dx/prime+1
/planckover2pi1/integraltextx
x2|p(x/prime)|dx/prime
+De−1
/planckover2pi1/integraltextx2
x1|p(x/prime)|dx/prime−1
/planckover2pi1/integraltextx
x2|p(x/prime)|dx/prime/bracketrightBig
.
Letγ≡/integraltextx2
x1|p(x)|dx, as before (Eq. 8.22), and let C/prime≡De−γ,D/prime≡Ceγ. Then (shifting the origin to x2):
ψWKB=
1
/radicalbig
|p(x)|/bracketleftBig
C/primee1
/planckover2pi1/integraltext0
x|p(x/prime)|dx/prime+D/primee−1
/planckover2pi1/integraltext0
x|p(x/prime)|dx/prime/bracketrightBig
,(x<0);
1/radicalbig
p(x)Fei
/planckover2pi1/integraltextx
0p(x/prime)dx/prime, (x>0).
In the patching region ψp(x)=aAi(−αx)+bBi(−αx),whereα≡/parenleftbigg2m|V/prime(0)|
/planckover2pi12/parenrightbigg1/3
;p(x)=/planckover2pi1α3/2√x.
In overlap region 1 (x<0):/integraldisplay0
x|p(x/prime)|dx/prime=2
3/planckover2pi1(−αx)3/2,so
ψWKB≈1
/planckover2pi11/2α3/4(−x)1/4/bracketleftBig
C/primee2
3(−αx)3/2+D/primee−2
3(−αx)3/2/bracketrightBig
ψp≈a
2√π(−αx)1/4e−2
3(−αx)3/2+b√π(−αx)1/4e2
3(−αx)3/2
Comparing =⇒
a=2/radicalbigg
π
/planckover2pi1αD/prime
b=/radicalbiggπ
/planckover2pi1αC/prime
In overlap region 2 (x>0):/integraldisplayx
0p(x/prime)dx/prime=2
3/planckover2pi1(αx)3/2=⇒ψWKB≈1
/planckover2pi11/2α3/4x1/4Fei2
3(αx)3/2.
ψp≈a√π(αx)1/4sin/bracketleftbigg2
3(αx)3/2+π
4/bracketrightbigg
+b√π(αx)1/4cos/bracketleftbigg2
3(αx)3/2+π
4/bracketrightbigg
=1
2√π(αx)1/4/bracketleftBig
(−ia+b)eiπ
4ei2
3(αx)3/2+(ia+b)e−iπ
4e−i2
3(αx)3/2/bracketrightBig
.Comparing =⇒(ia+b)=0 ;
F=/radicalbigg
/planckover2pi1α
π/parenleftbigg−ia+b
2/parenrightbigg
eiπ/4=b/radicalbigg
/planckover2pi1α
πeiπ/4.b=/radicalbiggπ
/planckover2pi1αe−iπ/4F;a=i/radicalbiggπ
/planckover2pi1αe−iπ/4F.
C/prime=/radicalbigg
/planckover2pi1α
πb=e−iπ/4F, D/prime=1
2/radicalbigg
/planckover2pi1α
πa=i
2e−iπ/4F. D =eγe−iπ/4F;C=i
2e−γe−iπ/4F.
These are the connection formulas at x2. Putting them into the equation for A:
A=/parenleftbiggC
2−iD/parenrightbigg
eiπ/4=/parenleftbiggi
4e−γe−iπ/4F−ieγe−iπ/4F/parenrightbigg
eiπ/4=i/parenleftbigge−γ
4−eγ/parenrightbigg
F.
T=/vextendsingle/vextendsingle/vextendsingle/vextendsingleF
A/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=1
(eγ−e−γ
4)2=e−2γ
/bracketleftbig
1−(e−γ/2)2/bracketrightbig2.
Ifγ/greatermuch1,the denominator is essentially 1, and we recover T=e−2γ(Eq. 8.22) .
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228 CHAPTER 8. THE WKB APPROXIMATION
Problem 8.11
Equation 8.51 ⇒/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=2/integraldisplayx2
0/radicalbig
2m(E−αxν)dx=2√
2mE/integraldisplayx2
0/radicalbigg
1−α
Exνdx;E=αxν
2.Let
z≡α
Exν,sox=/parenleftbiggzE
α/parenrightbigg1/ν
;dx=/parenleftbiggE
α/parenrightbigg1/ν1
νz1
ν−1dz.Then
/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=2√
2mE/parenleftbiggE
α/parenrightbigg1/ν1
ν/integraldisplay1
0z1
ν−1√
1−zdz=2√
2mE/parenleftbiggE
α/parenrightbigg1/ν1
νΓ(1/ν)Γ(3/2)
Γ(1
ν+3
2)
=2√
2mE/parenleftbiggE
α/parenrightbigg1/νΓ(1
ν+1 )1
2√π
Γ(1
ν+3
2)=√
2πmE/parenleftbiggE
α/parenrightbigg1/νΓ(1
ν+1 )
Γ(1
ν+3
2).
E1
ν+1
2=(n−1
2)π/planckover2pi1√
2πmα1/νΓ(1
ν+3
2)
Γ(1
ν+1 );En=/bracketleftbigg/parenleftbigg
n−1
2/parenrightbigg
/planckover2pi1/radicalbiggπ
2mαΓ(1
ν+3
2)
Γ(1
ν+1 )/bracketrightbigg(2ν
ν+2)
α.
Forν=2 :En=/bracketleftbigg/parenleftbigg
n−1
2/parenrightbigg
/planckover2pi1/radicalbiggπ
2mαΓ(2)
Γ(3/2)/bracketrightbigg
α=(n−1
2)/planckover2pi1/radicalbigg
2α
m.
For a harmonic oscillator, with α=1
2mω2,E n=/parenleftbig
n−1
2/parenrightbig
/planckover2pi1ω(n=1,2,3,...)./check
Problem 8.12
V(x)=−/planckover2pi12a2
msech2(ax).Eq. 8.51 =⇒/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=2/integraldisplayx2
0/radicalBigg
2m/bracketleftbigg
E+/planckover2pi12a2
msech2(ax)/bracketrightbigg
dx
=2√
2/planckover2pi1a/integraldisplayx2
0/radicalbigg
sech2(ax)+mE
/planckover2pi12a2dx.
E=−/planckover2pi12a2
msech2(ax2) defines x2.Letb≡−mE
/planckover2pi12a2,z≡sech2(ax),so that x=1
asech−1√z,and hence
dx=1
a/parenleftbigg−1√z√1−z/parenrightbigg1
2√zdz=−1
2a1
z√1−zdz.Then/parenleftbigg
n−1
2/parenrightbigg
π=2√
2a/parenleftbigg
−1
2a/parenrightbigg/integraldisplayz2
z1√
z−b
z√1−zdz.
Limits :/braceleftBiggx=0 =⇒z= sech2(0) = 1
x=x2=⇒z= sech2(ax2)=−mE
/planckover2pi12a2=b/bracerightBigg
./parenleftbigg
n−1
2/parenrightbigg
π=√
2/integraldisplay1
b1
z/radicalbigg
z−b
1−zdz.
1
z/radicalbigg
z−b
1−z=1
z(z−b)/radicalbig
(1−z)(z−b)=1/radicalbig
(1−z)(z−b)−b
z/radicalbig
(1−z)(z−b).
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CHAPTER 8. THE WKB APPROXIMATION 229
/parenleftbigg
n−1
2/parenrightbigg
π=√
2/bracketleftBigg/integraldisplay1
b1/radicalbig
(1−z)(z−b)dz−b/integraldisplay1
b1
z/radicalbig
−b+( 1+b)z−z2dz/bracketrightBigg
=√
2/braceleftBigg
−2 tan−1/radicalbigg
1−z
z−b−√
bsin−1/bracketleftbigg(1 +b)z−2b
z(1−b)/bracketrightbigg/bracerightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
b
=√
2/bracketleftBig
−2 tan−1(0) + 2 tan−1(∞)−√
bsin−1(1) +√
bsin−1(−1)/bracketrightBig
=√
2/parenleftBig
0+2π
2−√
bπ
2−√
bπ
2/parenrightBig
=√
2π(1−√
b);(n−1
2)√
2=1−√
b;√
b=1−1√
2/parenleftbigg
n−1
2/parenrightbigg
.
Since the left side is positive, the right side must also be: ( n−1
2)<√
2,n <1
2+√
2=0.5+1.4 1 4=1 .914.
So the only possible nis 1; there is only onebound state (which is correct—see Problem 2.51).
Forn=1,√
b=1−1
2√
2;b=1−1√
2+1
8=9
8−1√
2;E1=−/planckover2pi12a2
m/parenleftbigg9
8−1√
2/parenrightbigg
=−0.418/planckover2pi12a2
m.
Theexact answer (Problem 2.51(c)) is −0.5/planckover2pi12a2
m. Not bad.
Problem 8.13
/parenleftbigg
n−1
4/parenrightbigg
π/planckover2pi1=/integraldisplayr0
0/radicalbig
2m[E−V0ln(r/a)]dr;E=V0ln(r0/a) defines r0.
=√
2m/integraldisplayr0
0/radicalbig
V0ln(r0/a)−V0ln(r/a)dr=/radicalbig
2mV0/integraldisplayr0
0/radicalbig
ln(r0/r)dr.
Letx≡ln(r0/r),soex=r0/r,orr=r0e−x=⇒dr=−r0e−xdx.
/parenleftbigg
n−1
4/parenrightbigg
π/planckover2pi1=/radicalbig
2mV0(−r0)/integraldisplayx2
x1√xe−edx.Limits :/braceleftbiggr=0=⇒x1=∞
r=r0=⇒x2=0/bracerightbigg
.
/parenleftbigg
n−1
4/parenrightbigg
π/planckover2pi1=/radicalbig
2mV0r0/integraldisplay∞
0√xe−xdx=/radicalbig
2mV0r0Γ(3/2) =/radicalbig
2mV0r0√π
2.
r0=/radicalbigg
2π
mV0/planckover2pi1/parenleftbigg
n−1
4/parenrightbigg
⇒En=V0ln/bracketleftbigg/planckover2pi1
a/radicalbigg
2π
mV0/parenleftbigg
n−1
4/parenrightbigg/bracketrightbigg
=V0ln/parenleftbigg
n−1
4/parenrightbigg
+V0ln/bracketleftbigg/planckover2pi1
a/radicalbigg
2π
mV0/bracketrightbigg
.
En+1−En=V0ln/parenleftbigg
n+3
4/parenrightbigg
−V0ln/parenleftbigg
n−1
4/parenrightbigg
=V0ln/parenleftbiggn+3/4
n−1/4/parenrightbigg
,which is indeed independent of m(anda).
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
230 CHAPTER 8. THE WKB APPROXIMATION
Problem 8.14
/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=/integraldisplayr2
r1/radicalBigg
2m/parenleftbigg
E+e2
4πFepsilonC01
r−/planckover2pi12
2ml(l+1 )
r2/parenrightbigg
dr=√
−2mE/integraldisplayr2
r1/radicalbigg
−1+A
r−B
r2dr,
whereA≡−e2
4πFepsilonC01
EandB≡−/planckover2pi12
2ml(l+1 )
Eare positive constants, since Eis negative.
/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=√
−2mE/integraldisplayr2
r1√
−r2+Ar−B
rdr.
Letr1andr2be the roots of the polynomial in the numerator: −r2+Ar−B=(r−r1)(r2−r).
/parenleftbigg
n−1
2/parenrightbigg
π/planckover2pi1=√
−2mE/integraldisplayr2
r1/radicalbig
(r−r1)(r2−r)
rdr=√
−2mEπ
2(√r2−√r1)2.
2/parenleftbigg
n−1
2/parenrightbigg
/planckover2pi1=√
−2mE(r2+r1−2√r1r2).But−r2+Ar−B=−r2+(r1+r2)r−r1r2
=⇒r1+r2=A;r1r2=B.Therefore
2/parenleftbigg
n−1
2/parenrightbigg
/planckover2pi1=√
−2mE/parenleftBig
A−2√
B/parenrightBig
=√
−2mE/parenleftBigg
−e2
4πFepsilonC01
E−2/radicalbigg
−/planckover2pi12
2ml(l+1 )
E/parenrightBigg
=e2
4πFepsilonC0/radicalbigg
−2m
E−2/planckover2pi1/radicalbig
l(l+1 ).
e2
4πFepsilonC0/radicalbigg
−2m
E=2/planckover2pi1/bracketleftbigg
n−1
2+/radicalbig
l(l+1 )/bracketrightbigg
;−E
2m=(e2/4πFepsilonC0)2
4/planckover2pi12/bracketleftBig
n−1
2+/radicalbig
l(l+1 )/bracketrightBig2.
E=−(m/2/planckover2pi12)(e2/4πFepsilonC0)2
/bracketleftBig
n−1
2+/radicalbig
l(l+1 )/bracketrightBig2=−13.6e V
/parenleftBig
n−1
2+/radicalbig
l(l+1 )/parenrightBig2.
Problem 8.15
(a) (i)ψWKB(x)=D/radicalbig
|p(x)|e−1
/planckover2pi1/integraltextx
x2|p(x/prime)|dx/prime
(x>x 2);
(ii)ψWKB(x)=1/radicalbig
p(x)/bracketleftBig
Bei
/planckover2pi1/integraltextx2
xp(x/prime)dx/prime+Ce−i
/planckover2pi1/integraltextx2
xp(x/prime)dx/prime/bracketrightBig
(x1<x<x 2);
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 8. THE WKB APPROXIMATION 231
(iii)ψWKB(x)=1/radicalbig
|p(x)|/bracketleftBig
Fe1
/planckover2pi1/integraltextx1
x|p(x/prime)|dx/prime+Ge−1
/planckover2pi1/integraltextx1
x|p(x/prime)|dx/prime/bracketrightBig
(0<x<x 1).
Equation 8.46 = ⇒(ii)ψWKB=2D/radicalbig
p(x)sin/bracketleftbigg1
/planckover2pi1/integraldisplayx2
xp(x/prime)dx/prime+π
4/bracketrightbigg
(x1<x<x 2).
To effect the join at x1, first rewrite (ii):
(ii)ψWKB=2D/radicalbig
p(x)sin/bracketleftbigg1
/planckover2pi1/integraldisplayx2
x1p(x/prime)dx/prime−1
/planckover2pi1/integraldisplayx
x1p(x/prime)dx/prime+π
4/bracketrightbigg
=−2D/radicalbig
p(x)sin/bracketleftbigg1
/planckover2pi1/integraldisplayx
x1p(x/prime)dx/prime−θ−π
4/bracketrightbigg
,
whereθis defined in Eq. 8.58. Now shift the origin to x1:
ψWKB=
1
/radicalbig
|p(x)|/bracketleftBig
Fe1
/planckover2pi1/integraltext0
x|p(x/prime)|dx/prime+Ge−1
/planckover2pi1/integraltext0
x|p(x/prime)|dx/prime/bracketrightBig
(x<0)
−2D/radicalbig
p(x)sin/bracketleftbigg1
/planckover2pi1/integraldisplayx
0p(x/prime)dx/prime−θ−π
4/bracketrightbigg
(x>0)
.
Following Problem 8.9: ψ
p(x)=aAi(−αx)+bBi(−αx),withα≡/parenleftbigg2m|V/prime(0)|
/planckover2pi12/parenrightbigg1/3
;p(x)=/planckover2pi1α3/2√x.
Overlap region 1 ( x<0):/integraldisplay0
x|p(x/prime)|dx/prime=2
3/planckover2pi1(−αx)3/2.
ψWKB≈1
/planckover2pi11/2α3/4(−x)1/4/bracketleftBig
Fe2
3(−αx)3/2+Ge−2
3(−αx)3/2/bracketrightBig
ψp≈a
2√π(−αx)1/4e−2
3(−αx)3/2+b√π(−αx)1/4e2
3(−αx)3/2
=⇒a=2G/radicalbiggπ
/planckover2pi1α;b=F/radicalbiggπ
/planckover2pi1α.
Overlap region 2 ( x>0):/integraldisplayx
0p(x/prime)dx/prime=2
3/planckover2pi1(αx)3/2.
=⇒ψWKB≈−2D
/planckover2pi11/2α3/4x1/4sin/bracketleftbigg2
3(αx)3/2−θ−π
4/bracketrightbigg
,
ψp≈a√π(αx)1/4sin/bracketleftbigg2
3(αx)3/2+π
4/bracketrightbigg
+b√π(αx)1/4cos/bracketleftbigg2
3(αx)3/2+π
4/bracketrightbigg
.
Equating the two expressions:−2D
/planckover2pi11/2α3/41
2i/bracketleftBig
ei2
3(αx)3/2e−iθe−iπ/4−e−i2
3(αx)3/2eiθeiπ/4/bracketrightBig
=1√πα1/4/braceleftbigga
2i/bracketleftBig
ei2
3(αx)3/2eiπ/4−e−i2
3(αx)3/2e−iπ/4/bracketrightBig
+b
2/bracketleftBig
ei2
3(αx)3/2eiπ/4+e−i2
3(αx)3/2e−iπ/4/bracketrightBig/bracerightbigg
=⇒
−2D/radicalbigg
π
α/planckover2pi1e−iθe−iπ/4=(a+ib)eiπ/4,or (a+ib)=2D/radicalbiggπ
α/planckover2pi1ie−iθ
2D/radicalbiggπ
α/planckover2pi1eiθeiπ/4=(−a+ib)e−iπ/4,or (a−ib)=−2D/radicalbiggπ
α/planckover2pi1ieiθ
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232 CHAPTER 8. THE WKB APPROXIMATION
=⇒
2a=2D/radicalbigg
π
α/planckover2pi1i(e−iθ−eiθ)⇒a=2D/radicalbiggπ
α/planckover2pi1sinθ,
2ib=2D/radicalbiggπ
α/planckover2pi1i(e−iθ+eiθ)⇒b=2D/radicalbiggπ
α/planckover2pi1cosθ.
Combining these with the results from overlap region 1 = ⇒
2G/radicalbigg
π
α/planckover2pi1=2D/radicalbiggπ
α/planckover2pi1sinθ,orG=Dsinθ;F/radicalbiggπ
α/planckover2pi1=2D/radicalbiggπ
α/planckover2pi1cosθ,orF=2Dcosθ.
Putting these into (iii) : ψWKB(x)=D/radicalbig
|p(x)|/bracketleftBig
2 cosθe1
/planckover2pi1/integraltextx1
x|p(x/prime)|dx/prime+ sinθe−1
/planckover2pi1/integraltextx1
x|p(x/prime)|dx/prime/bracketrightBig
(0<x<x 1).
(b)
Odd(−) case: (iii) =⇒ψ(0) = 0⇒2 cosθe1
/planckover2pi1/integraltextx1
0|p(x/prime)|dx/prime+ sinθe−1
/planckover2pi1/integraltextx1
0|p(x/prime)|dx/prime=0.
1
/planckover2pi1/integraldisplayx1
0|p(x/prime)|dx/prime=1
2φ,withφdefined by Eq. 8.60. So sin θe−φ/2=−2 cosθeφ/2,or tanθ=−2eφ.
Even(+) case: (iii) = ⇒ψ/prime(0) = 0⇒−1
2D
(|p(x)|)3/2d|p(x)|
dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle
0/bracketleftBig
2 cosθeφ/2+ sinθe−φ/2/bracketrightBig
+D/radicalbig
|p(x)|/bracketleftbigg
2 cosθe1
/planckover2pi1/integraltextx1
0|p(x/prime)|dx/prime/parenleftbigg
−1
/planckover2pi1|p(0)|/parenrightbigg
+ sinθe−1
/planckover2pi1/integraltextx1
0|p(x/prime)|dx/prime/parenleftbigg1
/planckover2pi1|p(0)|/parenrightbigg/bracketrightbigg
=0.
Nowd|p(x)|
dx=d
dx/radicalbig
2m[V(x)−E]=√
2m1
21√
V−EdV
dx,anddV
dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle
0=0,sod|p(x)|
dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle
0=0.
2 cosθeφ/2= sinθe−φ/2,or tanθ=2eφ.Combining the two results: tan θ=±2eφ.QED
(c)
tanθ= tan/bracketleftbigg/parenleftbigg
n+1
2/parenrightbigg
π+FepsilonC/bracketrightbigg
=sin/bracketleftbig/parenleftbig
n+1
2/parenrightbig
π+FepsilonC/bracketrightbig
cos/bracketleftbig/parenleftbig
n+1
2/parenrightbig
π+FepsilonC/bracketrightbig=(−1)ncosFepsilonC
(−1)n+1sinFepsilonC=−cosFepsilonC
sinFepsilonC≈−1
FepsilonC.
So−1
FepsilonC≈±2eφ,orFepsilonC≈∓1
2e−φ,orθ−/parenleftbigg
n+1
2/parenrightbigg
π≈∓1
2e−φ,soθ≈/parenleftbigg
n+1
2/parenrightbigg
π∓1
2e−φ.QED
[Note: Since θ(Eq. 8.58) is positive, nmust be a non-negative integer: n=0,1,2,.... This is like
harmonic oscillator (conventional) numbering, since it starts with n= 0.]
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CHAPTER 8. THE WKB APPROXIMATION 233
a -axxxV(x)
1 2
(d)
θ=1
/planckover2pi1/integraldisplayx2
x1/radicalBigg
2m/bracketleftbigg
E−1
2mω2(x−a)2/bracketrightbigg
dx.Letz=x−a(shifts the origin to a).
=2
/planckover2pi1/integraldisplayz2
0/radicalBigg
2m/bracketleftbigg
E−1
2mω2z2/bracketrightbigg
dz,where E=1
2mω2z2
2.
=2
/planckover2pi1mω/integraldisplayz2
0/radicalBig
z2
2−z2dz=mω
/planckover2pi1/bracketleftbigg
z/radicalBig
z2
2−z2+z2
2sin−1(z/z2)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglez2
0=mω
/planckover2pi1z2
2sin−1(1) =π
2mω
/planckover2pi1z2
2,
=π
2mω
/planckover2pi12E
mω2=πE
/planckover2pi1ω.
Putting this into Eq. 8.61 yieldsπE
/planckover2pi1ω≈/parenleftbigg
n+1
2/parenrightbigg
π∓1
2e−φ,orE±
n≈/parenleftbigg
n+1
2/parenrightbigg
/planckover2pi1ω∓/planckover2pi1ω
2πe−φ.QED
(e)
Ψ(x,t)=1√
2/parenleftBig
ψ+
ne−iE+
nt//planckover2pi1+ψ−
ne−iE−
nt//planckover2pi1/parenrightBig
=⇒
|Ψ(x,t)|2=1
2/bracketleftBig
|ψ+
n|2+|ψ−
n|2+ψ+
nψ−
n/parenleftBig
ei(E−
n−E+
n)t//planckover2pi1+e−i(E−
n−E+
n)t//planckover2pi1/parenrightBig/bracketrightBig
.
(Note that the wavefunctions (i), (ii), (iii) are real). ButE−
n−E+
n
/planckover2pi1≈1
/planckover2pi12/planckover2pi1ω
2πe−φ=ω
πe−φ,s o
|Ψ(x,t)|2=1
2/bracketleftbig
ψ+
n(x)2+ψ−
n(x)2/bracketrightbig
+ψ+
n(x)ψ−
n(x) cos/parenleftBigω
πe−φt/parenrightBig
.
It oscillates back and forth, with period τ=2π
(ω/π)e−φ=2π2
ωeφ.QED
(f)
φ=21
/planckover2pi1/integraldisplayx1
0/radicalBigg
2m/bracketleftbigg1
2mω2(x−a)2−E/bracketrightbigg
dx=2
/planckover2pi1√
2mE/integraldisplayx1
0/radicalbigg
mω2
2E(x−a)2−1dx.
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234 CHAPTER 8. THE WKB APPROXIMATION
Letz≡/radicalbiggm
2Eω(a−x),sodx=−/radicalbigg
2E
m1
ωdz.Limits:
x=0 =⇒z=/radicalbiggm
2Eωa≡z0
x=x1=⇒radicand = 0 = ⇒z=1
.
φ=2
/planckover2pi1√
2mE/radicalbigg
2E
m1
ω/integraldisplayz0
1/radicalbig
z2−1dz=4E
/planckover2pi1ω/integraldisplayz0
1/radicalbig
z2−1dz=4E
/planckover2pi1ω1
2/bracketleftBig
z/radicalbig
z2−1−ln(z+/radicalbig
z2−1)/bracketrightBig/vextendsingle/vextendsingle/vextendsinglez0
1
=2E
/planckover2pi1ω/bracketleftbigg
z0/radicalBig
z2
0−1−ln/parenleftbigg
z0+/radicalBig
z2
0−1/parenrightbigg/bracketrightbigg
,
wherez0=aω/radicalbiggm
2E.V(0) =1
2mω2a2,s oV(0)/greatermuchE⇒m
2ω2a2/greatermuchE⇒aω/radicalbiggm
2E/greatermuch1, orz0/greatermuch1.
In that case
φ≈2E
/planckover2pi1ω/bracketleftbig
z2
0−ln(2z0)/bracketrightbig
≈2E
/planckover2pi1ωz2
0=2E
/planckover2pi1ωa2ω2m
2E=mωa2
/planckover2pi1.
This, together with Eq. 8.64, gives us the period of oscillation in a double well.
Problem 8.16
(a)En≈n2π2/planckover2pi12
2m(2a)2.Withn=1 , E1=π2/planckover2pi12
8ma2.
(b)
V(x)
x
V0
E1a -a
V(x)
x
E1a -a x0
Etunneling
(c)
γ=1
/planckover2pi1/integraldisplayx0
a|p(x)|dx. αx 0=V0−E1⇒x0=V0−E1
α.
p(x)=/radicalbig
2m[E−V(x)];V(x)=−αx, E =E1−V0.
=/radicalbig
2m(E1−V0+αx)=√
2mα√x−x0;|p(x)|=√
2mα√x0−x.
γ=1
/planckover2pi1√
2mα/integraldisplayx0
a√x0−xdx=√
2mα
/planckover2pi1/bracketleftbigg
−2
3(x0−x)3/2/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglex0
a=2
3√
2mα
/planckover2pi1(x0−a)3/2.
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CHAPTER 8. THE WKB APPROXIMATION 235
Nowx0−a=(V0−E1−aα)/α, andαa/lessmuch/planckover2pi12/ma2≈E1/lessmuchV0, so we can drop E1andαa. Then
γ≈2
3√
2mα
/planckover2pi1/parenleftbiggV0
α/parenrightbigg3/2
=/radicalbig
8mV3
0
3α/planckover2pi1.
Equation 8.28 ⇒τ=4a
ve2γ,where1
2mv2≈π2/planckover2pi12
8ma2⇒v2=π2/planckover2pi12
4m2a2,o r v=π/planckover2pi1
2ma.So
τ=4a
π/planckover2pi12mae2γ=8ma2
π/planckover2pi1e2γ.
(d)
τ=(8)/parenleftbig
9.1×10−31/parenrightbig/parenleftbig
10−10/parenrightbig2
π(1.05×10−34)e2γ=/parenleftbig
2×10−19/parenrightbig
e2γ;
γ=/radicalBig
(8) (9.1×10−31) (20×1.6×10−19)3
(3) (1.6×10−19)( 7×106)( 1.05×10−34)=4.4×104;e2γ=e8.8×104=/parenleftbig
10loge/parenrightbig8.8×104
=1 038,000.
τ=/parenleftbig
2×10−19/parenrightbig
×1038,000s=1038,000yr.
Seconds, years ...it hardly matters; nor is the factor out front significant. This is a huge number—the
age of the universe is about 1010years. In any event, this is clearly notsomething to worry about.
Problem 8.17
Equation 8.22 ⇒the tunneling probability: T=e−2γ, where
γ=1
/planckover2pi1/integraldisplayx0
0/radicalbig
2m(V−E)dx.HereV(x)=mgx, E =0,x0=/radicalbig
R2+(h/2)2−h/2 (half the diagonal) .
=√
2m
/planckover2pi1√mg/integraldisplayx0
0x1/2dx=m
/planckover2pi1/radicalbig
2g2
3x3/2/vextendsingle/vextendsingle/vextendsingle/vextendsinglex0
0=2m
3/planckover2pi1/radicalbig
2gx3/2
0.
I estimate: h= 10 cm, R= 3 cm, m= 300 gm; let g=9.8 m/s2. Then x0=√9+2 5−5=0.83 cm, and
γ=(2)(0.3)
(3)(1.05×10−34)/radicalbig
(2)(9.8) (0.0083)3/2=6.4×1030.
Frequency of “attempts”: say f=v/2R. We want the product of the number of attempts ( ft) and the
probability of toppling at each attempt ( T), to be 1:
tv
2Re−2γ=1⇒t=2R
ve2γ.
Estimating the thermal velocity:1
2mv2=1
2kBT(I’m done with the tunneling probability; from now on T
is the temperature, 300 K) ⇒v=/radicalbig
kBT/m.
t=2R/radicalbiggm
kBTe2γ= 2(0.03)/radicalBigg
0.3
(1.4×10−23)(300)e12.8×1030=5×108/parenleftbig
10loge/parenrightbig13×1030
=( 5×108)×105.6×1030s
=16×105.6×1030yr.
Don’t hold your breath.
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236 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Chapter 9
Time-Dependent Perturbation Theory
Problem 9.1
ψnlm=RnlYm
l.From Tables 4.3 and 4.7:
ψ100=1√
πa3e−r/a;ψ200=1√
8πa3/parenleftBig
1−r
2a/parenrightBig
e−r/2a;
ψ210=1√
32πa3r
ae−r/2acosθ;ψ21±1=∓1√
64πa3r
aer/2asinθe±iφ.
Butrcosθ=zandrsinθe±iφ=rsinθ(cosφ±isinφ)=rsinθcosφ±irsinθsinφ=x±iy. So|ψ|2is an
evenfunction of zin all cases, and hence/integraltext
z|ψ|2dxdydz =0 ,s o H/prime
ii=0 . Moreover, ψ100is even in z, and
so areψ200,ψ211, andψ21−1,s oH/prime
ij=0for all except
H/prime
100,210=−eE1√
πa31√
32πa31
a/integraldisplay
e−r/ae−r/2az2d3r=−eE
4√
2πa4/integraldisplay
e−3r/2ar2cos2θr2sinθdrdθdφ
=−eE
4√
2πa4/integraldisplay∞
0r4e−3r/2adr/integraldisplayπ
0cos2θsinθdθ/integraldisplay2π
0dφ=−eE
4√
2πa44!/parenleftbigg2a
3/parenrightbigg52
32π=−/parenleftbigg28
35√
2/parenrightbigg
eEa,
or−0.7449eEa.
Problem 9.2
˙ca=−i
/planckover2pi1H/prime
abe−iω0tcb;˙cb=−i
/planckover2pi1H/prime
baeiω0tca.Differentiating with respect to t:
¨cb=−i
/planckover2pi1H/prime
ba/bracketleftbig
iω0eiω0tca+eiω0t˙ca/bracketrightbig
=iω0/bracketleftbigg
−i
/planckover2pi1H/prime
baeiω0tca/bracketrightbigg
−i
/planckover2pi1H/prime
baeiωot/bracketleftbigg
−i
/planckover2pi1H/prime
abe−iω0tcb/bracketrightbigg
,or
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 237
¨cb=iω0˙cb−1
/planckover2pi12|H/prime
ab|2cb.Letα2≡1
/planckover2pi12|H/prime
ab|2.Then ¨cb−iω0˙cb+α2cb=0.
This is a linear differential equation with constant coefficients, so it can be solved by a function of the form
cb=eλt:
λ2−iω0λ+α2=0=⇒λ=1
2/bracketleftbigg
iω0±/radicalBig
−ω2
0−4α2/bracketrightbigg
=i
2(ω0±ω),whereω≡/radicalBig
ω2
0+4α2.
The general solution is therefore
cb(t)=Aei(ω0+ω)/2+Bei(ω0−ω)/2=eiω0t/2/parenleftBig
Aeiωt/2+Be−iωt/2/parenrightBig
,or
cb(t)=eiω0t/2[Ccos (ωt/2) +Dsin (ωt/2)].Butcb(0) = 0,soC=0,and hence
cb(t)=Deiω0t/2sin (ωt/2).Then
˙cb=D/bracketleftbiggiω0
2eiω0t/2sin (ωt/2) +ω
2eiω0t/2cos (ωt/2)/bracketrightbigg
=ω
2Deiω0t/2/bracketleftBig
cos (ωt/2) +iω0
ωsin (ωt/2)/bracketrightBig
=−i
/planckover2pi1H/prime
baeiω0tca.
ca=i/planckover2pi1
H/prime
baω
2e−iω0t/2D/bracketleftBig
cos (ωt/2) +iω0
ωsin (ωt/2)/bracketrightBig
.Butca(0) = 1,soi/planckover2pi1
H/prime
baω
2D=1.Conclusion:
ca(t)=e−iω0t/2/bracketleftBig
cos (ωt/2) +iω0
ωsin (ωt/2)/bracketrightBig
,
cb(t)=2H/prime
ba
i/planckover2pi1ωeiω0t/2sin (ωt/2),whereω≡/radicalBig
ω2
0+4|H/prime
ab|2
/planckover2pi12.
|ca|2+|cb|2= cos2(ωt/2) +ω2
0
ω2sin2(ωt/2) +4|H/prime
ab|2
/planckover2pi12ω2sin2(ωt/2)
= cos2(ωt/2) +1
ω2/parenleftbigg
ω2
0+4|H/prime
ab|2
/planckover2pi12/parenrightbigg
sin2(ωt/2) = cos2(ωt/2) + sin2(ωt/2) = 1./check
Problem 9.3
This is a tricky problem, and I thank Prof. Onuttom Narayan for showing me the correct solution. The safest
approach is to represent the delta function as a sequence of rectangles:
δρepsilono(t)=/braceleftbigg(1/2FepsilonC),−FepsilonC<t<FepsilonC ,
0, otherwise ./bracerightbigg
Then Eq. 9.13 ⇒
t<−FepsilonC:c
a(t)=1,cb(t)=0,
t>FepsilonC:ca(t)=a, cb(t)=b,
−FepsilonC<t<FepsilonC :
˙ca=−iα
2ρepsilono/planckover2pi1e−iω0tcb,
˙cb=−iα∗
2ρepsilono/planckover2pi1eiω0tca.
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238 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
In the interval −FepsilonC<t<FepsilonC ,
d2cb
dt2=−iα∗
2FepsilonC/planckover2pi1/bracketleftbigg
iω0eiω0tca+eiω0t/parenleftbigg−iα
2FepsilonC/planckover2pi1e−iω0tcb/parenrightbigg/bracketrightbigg
=−iα∗
2FepsilonC/planckover2pi1/bracketleftbigg
iω0i2FepsilonC/planckover2pi1
α∗dcb
dt−iα
2FepsilonC/planckover2pi1cb/bracketrightbigg
=iω0dcb
dt−|α|2
(2FepsilonC/planckover2pi1)2cb.
Thuscbsatisfies a homogeneous linear differential equation with constant coefficients:
d2cb
dt2−iω0dcb
dt+|α|2
(2FepsilonC/planckover2pi1)2cb=0.
Try a solution of the form cb(t)=eλt:
λ2−iω0λ+|α|2
(2FepsilonC/planckover2pi1)2=0⇒λ=iω0±/radicalbig
−ω2
0−|α|2/(FepsilonC/planckover2pi1)2
2,
or
λ=iω0
2±iω
2,whereω≡/radicalBig
ω2
0+|α|2/(FepsilonC/planckover2pi1)2.
The general solution is
cb(t)=eiω0t/2/parenleftBig
Aeiωt/2+Be−iωt/2/parenrightBig
.
But
cb(−FepsilonC)=0⇒Ae−iωρepsilono/2+Beiωρepsilono/2=0⇒B=−Ae−iωρepsilono,
so
cb(t)=Aeiω0t/2/parenleftBig
eiωt/2−e−iω(ρepsilono+t/2)/parenrightBig
.
Meanwhile
ca(t)=2iFepsilonC/planckover2pi1
α∗e−iω0t˙cb=2iFepsilonC/planckover2pi1
α∗e−iω0t/2A/bracketleftbiggiω0
2/parenleftBig
eiωt/2−e−iω(ρepsilono+t/2)/parenrightBig
+iω
2/parenleftBig
eiωt/2+e−iω(ρepsilono+t/2)/parenrightBig/bracketrightbigg
=−FepsilonC/planckover2pi1
α∗e−iω0t/2A/bracketleftBig
(ω+ω0)eiωt/2+(ω−ω0)e−iω(ρepsilono+t/2)/bracketrightBig
.
Butca(−FepsilonC)=1=−FepsilonC/planckover2pi1
α∗ei(ω0−ω)ρepsilono/2A[(ω+ω0)+(ω−ω0)] =−2FepsilonC/planckover2pi1ω
α∗ei(ω0−ω)ρepsilono/2A,soA=−α∗
2FepsilonC/planckover2pi1ωei(ω−ω0)ρepsilono/2.
ca(t)=1
2ωe−iω0(t+ρepsilono)/2/bracketleftBig
(ω+ω0)eiω(t+ρepsilono)/2+(ω−ω0)e−iω(t+ρepsilono)/2/bracketrightBig
=e−iω0(t+ρepsilono)/2/braceleftbigg
cos/bracketleftbiggω(t+FepsilonC)
2/bracketrightbigg
+iω0
ωsin/bracketleftbiggω(t+FepsilonC)
2/bracketrightbigg/bracerightbigg
;
cb(t)=−iα∗
2FepsilonC/planckover2pi1ωeiω0(t−ρepsilono)/2/bracketleftBig
eiω(t+ρepsilono)/2−e−iω(t+ρepsilono)/2/bracketrightBig
=−iα∗
FepsilonC/planckover2pi1ωeiω0(t−ρepsilono)/2sin/bracketleftbiggω(t+FepsilonC)
2/bracketrightbigg
.
Thus
a=ca(FepsilonC)=e−iω0ρepsilono/bracketleftBig
cos(ωFepsilonC)+iω0
ωsin(ωFepsilonC)/bracketrightBig
,b=cb(FepsilonC)=−iα∗
FepsilonC/planckover2pi1ωsin(ωFepsilonC).
This is for the rectangular pulse; it remains to take the limit FepsilonC→0:ω→|α|/FepsilonC/planckover2pi1,s o
a→cos/parenleftbigg|α|
/planckover2pi1/parenrightbigg
+iω0FepsilonC/planckover2pi1
|α|sin/parenleftbigg|α|
/planckover2pi1/parenrightbigg
→cos/parenleftbigg|α|
/planckover2pi1/parenrightbigg
,b→−iα∗
|α|sin/parenleftbigg|α|
/planckover2pi1/parenrightbigg
,
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 239
and we conclude that for the delta function
ca(t)=/braceleftbigg
1,t < 0,
cos(|α|//planckover2pi1),t>0;
cb(t)=
0,t < 0,
−i/radicalbigg
α∗
αsin(|α|//planckover2pi1),t>0.
Obviously,|ca(t)|2+|cb(t)|2= 1 in both time periods. Finally,
Pa→b=|b|2= sin2(|α|//planckover2pi1).
Problem 9.4
(a)
Eq. 9.10 =⇒˙ca=−i
/planckover2pi1/bracketleftbig
caH/prime
aa+cbH/prime
abe−iω0t/bracketrightbig
Eq. 9.11 =⇒˙cb=−i
/planckover2pi1/bracketleftbig
cbH/prime
bb+caH/prime
baeiω0t/bracketrightbig
(these are exact, and replace Eq. 9.13) .
Initial conditions :ca(0) = 1,c b(0) = 0.
Zeroth order :ca(t)=1,c b(t)=0.
First order :
˙c
a=−i
/planckover2pi1H/prime
aa =⇒
˙cb=−i
/planckover2pi1H/prime
baeiω0t=⇒ca(t)=1−i
/planckover2pi1/integraldisplayt
0H/prime
aa(t/prime)dt/prime
cb(t)=−i
/planckover2pi1/integraldisplayt
0H/prime
ba(t/prime)eiω0t/primedt/prime
|ca|2=/bracketleftbigg
1−i
/planckover2pi1/integraldisplayt
0H/prime
aa(t/prime)dt/prime/bracketrightbigg/bracketleftbigg
1+i
/planckover2pi1/integraldisplayt
0H/prime
aa(t/prime)dt/prime/bracketrightbigg
=1+/bracketleftbigg1
/planckover2pi1/integraldisplayt
0H/prime
aa(t/prime)dt/prime/bracketrightbigg2
= 1 (to first order in H/prime).
|cb|2=/bracketleftbigg
−i
/planckover2pi1/integraldisplayt
0H/prime
ba(t/prime)eiω0t/primedt/prime/bracketrightbigg/bracketleftbiggi
/planckover2pi1/integraldisplayt
0H/prime
ab(t/prime)e−iω0t/primedt/prime/bracketrightbigg
= 0 (to first order in H/prime).
So|ca|2+|cb|2= 1 (to first order).
(b)
˙da=ei
/planckover2pi1/integraltextt
0H/prime
aa(t/prime)dt/prime/parenleftbiggi
/planckover2pi1H/prime
aa/parenrightbigg
ca+ei
/planckover2pi1/integraltextt
0H/prime
aa(t/prime)dt/prime˙ca.But ˙ca=−i
/planckover2pi1/bracketleftbig
caH/prime
aa+cbH/prime
abe−iω0t/bracketrightbig
Two terms cancel, leaving
˙da=−i
/planckover2pi1ei
/planckover2pi1/integraltextt
0H/prime
aa(t/prime)dt/primecbH/prime
abe−iω0t.Butcb=e−i
/planckover2pi1/integraltextt
0H/prime
bb(t/prime)dt/primedb.
=−i
/planckover2pi1ei
/planckover2pi1/integraltextt
0[H/prime
aa(t/prime)−H/prime
bb(t/prime)]dt/primeH/prime
abe−iω0tdb,or˙da=−i
/planckover2pi1eiφH/prime
abe−iω0tdb.
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240 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Similarly,
˙db=ei
/planckover2pi1/integraltextt
0H/prime
bb(t/prime)dt/prime/parenleftbiggi
/planckover2pi1H/prime
bb/parenrightbigg
cb+ei
/planckover2pi1/integraltextt
0H/prime
bb(t/prime)dt/prime˙cb.But ˙cb=−i
/planckover2pi1/bracketleftbig
cbH/prime
bb+caH/prime
baeiω0t/bracketrightbig
.
=−i
/planckover2pi1ei
/planckover2pi1/integraltextt
0H/prime
bb(t/prime)dt/primecaH/prime
baeiω0t.Butca=e−i
/planckover2pi1/integraltextt
0H/prime
aa(t/prime)dt/primeda.
=−i
/planckover2pi1ei
/planckover2pi1/integraltextt
0[H/prime
bb(t/prime)−H/prime
aa(t/prime)]dt/primeH/prime
baeiω0tda=−i
/planckover2pi1e−iφH/prime
baeiω0tda.QED
(c)
Initial conditions :ca(0) = 1 =⇒da(0) = 1; cb(0) = 0 =⇒db(0) = 0.
Zeroth order :da(t)=1,d b(t)=0.
First order :˙da=0=⇒da(t)=1=⇒ca(t)=e−i
/planckover2pi1/integraltextt
0H/prime
aa(t/prime)dt/prime.
˙db=−i
/planckover2pi1e−iφH/prime
baeiω0t=⇒db=−i
/planckover2pi1/integraldisplayt
0e−iφ(t/prime)H/prime
ba(t/prime)eiω0t/primedt/prime=⇒
cb(t)=−i
/planckover2pi1e−i
/planckover2pi1/integraltextt
0H/prime
bb(t/prime)dt/prime/integraldisplayt
0e−iφ(t/prime)H/prime
ba(t/prime)eiω0t/primedt/prime.
These don’t lookmuch like the results in (a), but remember, we’re only working to first order inH/prime,
soca(t)≈1−i
/planckover2pi1/integraltextt
0H/prime
aa(t/prime)dt/prime(to this order), while for cb, the factor Hbain the integral means it is
already first order and hence both the exponential factor in front and e−iφshould be replaced by 1. Then
cb(t)≈−i
/planckover2pi1/integraltextt
0H/prime
ba(t/prime)eiω0t/primedt/prime, and we recover the results in (a).
Problem 9.5
Zeroth order :c(0)
a(t)=a, c(0)
b(t)=b.
First order :
˙ca=−i
/planckover2pi1H/prime
abe−iω0tb=⇒c(1)
a(t)=a−ib
/planckover2pi1/integraldisplayt
0H/prime
ab(t/prime)e−iω0t/primedt/prime.
˙cb=−i
/planckover2pi1H/prime
baeiω0ta=⇒c(1)
b(t)=b−ia
/planckover2pi1/integraldisplayt
0H/prime
ba(t/prime)eiω0t/primedt/prime.
Second order :˙ca=−i
/planckover2pi1H/prime
abe−iω0t/bracketleftbigg
b−ia
/planckover2pi1/integraldisplayt
0H/prime
ba(t/prime)eiω0t/primedt/prime/bracketrightbigg
=⇒
c(2)
a(t)=a−ib
/planckover2pi1/integraldisplayt
0H/prime
ab(t/prime)e−iω0t/primedt/prime−a
/planckover2pi12/integraldisplayt
0H/prime
ab(t/prime)e−iω0t/prime/bracketleftBigg/integraldisplayt/prime
0H/prime
ba(t/prime/prime)eiω0t/prime/primedt/prime/prime/bracketrightBigg
dt/prime.
To getcb, just switch a↔b(which entails also changing the sign of ω0):
c(2)
b(t)=b−ia
/planckover2pi1/integraldisplayt
0H/prime
ba(t/prime)eiω0t/primedt/prime−b
/planckover2pi12/integraldisplayt
0H/prime
ba(t/prime)eiω0t/prime/bracketleftBigg/integraldisplayt/prime
0H/prime
ab(t/prime/prime)e−iω0t/prime/primedt/prime/prime/bracketrightBigg
dt/prime.
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 241
Problem 9.6
ForH/primeindependent of t, Eq. 9.17 =⇒c(2)
b(t)=c(1)
b(t)=−i
/planckover2pi1H/prime
ba/integraldisplayt
0eiω0t/primedt/prime=⇒
c(2)
b(t)=−i
/planckover2pi1H/prime
baeiω0t/prime
iω0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglet
0=−H/prime
ba
/planckover2pi1ω0/parenleftbig
eiω0t−1/parenrightbig
.Meanwhile Eq. 9.18 = ⇒
c(2)
a(t)=1−1
/planckover2pi12|H/prime
ab|2/integraldisplayt
0e−iω0t/prime/bracketleftBigg/integraldisplayt/prime
0eiω0t/prime/primedt/prime/prime/bracketrightBigg
dt/prime=1−1
/planckover2pi12|H/prime
ab|21
iω0/integraldisplayt
0/parenleftBig
1−e−iω0t/prime/parenrightBig
dt/prime
=1 +i
ω0/planckover2pi12|H/prime
ab|2/parenleftBigg
t/prime+e−iω0t/prime
iω0/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglet
0=1+i
ω0/planckover2pi12|H/prime
ab|2/bracketleftbigg
t+1
iω0/parenleftbig
e−iω0t−1/parenrightbig/bracketrightbigg
.
For comparison with the exact answers (Problem 9.2), note first that cb(t) is already first order (because of
theH/prime
bain front), whereas ωdiffers from ω0only in second order, so it suffices to replace ω→ω0in the exact
formula to get the second-order result:
cb(t)≈2H/prime
ba
i/planckover2pi1ω0eiω0t/2sin (ω0t/2) =2H/prime
ba
i/planckover2pi1ω0eiω0t/21
2i/parenleftBig
eiω0t/2−e−iω0t/2/parenrightBig
=−H/prime
ba
/planckover2pi1ω0/parenleftbig
eiω0t−1/parenrightbig
,
in agreement with the result above. Checking cais more difficult. Note that
ω=ω0/radicalBigg
1+4|H/prime
ab|2
ω2
0/planckover2pi12≈ω0/parenleftbigg
1+2|H/prime
ab|2
ω2
0/planckover2pi12/parenrightbigg
=ω0+2|H/prime
ab|2
ω0/planckover2pi12;ω0
ω≈1−2|H/prime
ab|2
ω2
0/planckover2pi12.
Taylor expansion:
cos(x+FepsilonC) = cosx−FepsilonCsinx=⇒cos (ωt/2) = cos/parenleftbiggω
0t
2+|H/prime
ab|2t
ω0/planckover2pi12/parenrightbigg
≈cos (ω0t/2)−|H/prime
ab|2t
ω0/planckover2pi12sin (ω0t/2)
sin(x+FepsilonC) = sinx+FepsilonCcosx=⇒sin (ωt/2) = sin/parenleftbiggω0t
2+|H/prime
ab|2t
ω0/planckover2pi12/parenrightbigg
≈sin (ω0t/2) +|H/prime
ab|2t
ω0/planckover2pi12cos (ω0t/2)
ca(t)≈e−iω0t/2/braceleftbigg
cos/parenleftbiggω0t
2/parenrightbigg
−|H/prime
ab|2t
ω0/planckover2pi12sin/parenleftbiggω0t
2/parenrightbigg
+i/parenleftbigg
1−2|H/prime
ab|2
ω2
0/planckover2pi12/parenrightbigg/bracketleftbigg
sin/parenleftbiggω0t
2/parenrightbigg
+|H/prime
ab|2t
ω0/planckover2pi12cos/parenleftbiggω0t
2/parenrightbigg/bracketrightbigg/bracerightbigg
=e−iω0t/2/braceleftbigg/bracketleftbigg
cos/parenleftbiggω0t
2/parenrightbigg
+isin/parenleftbiggω0t
2/parenrightbigg/bracketrightbigg
−|H/prime
ab|2
ω0/planckover2pi12/bracketleftbigg
t/parenleftbigg
sin/parenleftbiggω0t
2/parenrightbigg
−icos/parenleftbiggω0t
2/parenrightbigg/parenrightbigg
+2i
ω0sin/parenleftbiggω0t
2/parenrightbigg/bracketrightbigg/bracerightbigg
=e−iω0t/2/braceleftbigg
eiω0t/2−|H/prime
ab|2
ω0/planckover2pi12/bracketleftbigg
−iteiω0t/2+2i
ω1
2i/parenleftBig
eiω0t/2−e−iω0t/2/parenrightBig/bracketrightbigg/bracerightbigg
=1−|H/prime
ab|2
ω0/planckover2pi12/bracketleftbigg
−it+1
ω0/parenleftbig
1−e−iω0t/parenrightbig/bracketrightbigg
=1+i
ω0/planckover2pi12|H/prime
ab|2/bracketleftbigg
t+1
iω0/parenleftbig
e−iω0t−1/parenrightbig/bracketrightbigg
,as above ./check
Problem 9.7
(a)
˙ca=−i
2/planckover2pi1Vabeiωte−iω0tcb;˙cb=−i
2/planckover2pi1Vbae−iωteiω0tca.
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242 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Differentiate the latter, and substitute in the former:
¨cb=−iVba
2/planckover2pi1/bracketleftBig
i(ω0−ω)ei(ω0−ω)tca+ei(ω0−ω)t˙ca/bracketrightBig
=i(ω0−ω)/bracketleftbigg
−iVba
2/planckover2pi1ei(ω0−ω)tca/bracketrightbigg
−iVba
2/planckover2pi1ei(ω0−ω)t/bracketleftbigg
−iVab
2/planckover2pi1e−i(ω0−ω)tcb/bracketrightbigg
=i(ω0−ω)˙cb−|Vab|2
(2/planckover2pi1)2cb.
d2cb
dt2+i(ω−ω0)dcb
dt+|Vab|2
4/planckover2pi12cb=0.Solution is of the form cb=eλt:λ2+i(ω−ω0)λ+|Vab|2
4/planckover2pi12=0.
λ=1
2/bracketleftBigg
−i(ω−ω0)±/radicalbigg
−(ω−ω0)2−|Vab|2
/planckover2pi12/bracketrightBigg
=i/bracketleftbigg
−(ω−ω0)
2±ωr/bracketrightbigg
,withωrdefined in Eq. 9.30.
General solution: cb(t)=Aei/bracketleftBig
−(ω−ω0)
2+ωr/bracketrightBig
t+Bei/bracketleftBig
−(ω−ω0)
2+ωr/bracketrightBig
t=e−i(ω−ω0)t/2/bracketleftbig
Aeiωrt+Be−iωrt/bracketrightbig
,
or, more conveniently: cb(t)=e−i(ω−ω0)t/2[Ccos(ωrt)+Dsin(ωrt)].Butcb(0) = 0,soC=0:
cb(t)=Dei(ω0−ω)t/2sin(ωrt).˙cb=D/bracketleftbigg
i/parenleftbiggω0−ω
2/parenrightbigg
ei(ω0−ω)t/2sin(ωrt)+ωrei(ω0−ω)t/2cos(ωrt)/bracketrightbigg
;
ca(t)=i2/planckover2pi1
Vbaei(ω−ω0)t˙cb=i2/planckover2pi1
Vbaei(ω−ω0)t/2D/bracketleftbigg
i/parenleftbiggω0−ω
2/parenrightbigg
sin(ωrt)+ωrcos(ωrt)/bracketrightbigg
.Butca(0) = 1 :
1=i2/planckover2pi1
VbaDωr,orD=−iVba
2/planckover2pi1ωr.
cb(t)=−i
2/planckover2pi1ωrVbaei(ω0−ω)t/2sin(ωrt),c a(t)=ei(ω−ω0)t/2/bracketleftbigg
cos(ωrt)+i/parenleftbiggω0−ω
2ωr/parenrightbigg
sin(ωrt)/bracketrightbigg
.
(b)
Pa→b(t)=|cb(t)|2=/parenleftbigg|Vab|
2/planckover2pi1ωr/parenrightbigg2
sin2(ωrt).The largest this gets (when sin2=1 )i s|Vab|2//planckover2pi12
4ω2r,
and the denominator, 4 ω2
r=(ω−ω0)2+|Vab|2//planckover2pi12,exceeds the numerator, so P≤1 (and 1 only if ω=ω0).
|ca|2+|cb|2= cos2(ωrt)+/parenleftbiggω0−ω
2ωr/parenrightbigg2
sin2(ωrt)+/parenleftbigg|Vab|
2/planckover2pi1ωr/parenrightbigg2
sin2(ωrt)
= cos2(ωrt)+(ω−ω0)2+(|Vab|//planckover2pi1)2
4ω2rsin2(ωrt) = cos2(ωrt) + sin2(ωrt)=1./check
(c)If|Vab|2/lessmuch/planckover2pi12(ω−ω0)2,thenωr≈1
2|ω−ω0|,andPa→b≈|Vab|2
/planckover2pi12sin2/parenleftbigω−ω0
2t/parenrightbig
(ω−ω0)2, confirming
Eq. 9.28.
(d)ωrt=π=⇒t=π/ωr.
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 243
Problem 9.8
Spontaneous emission rate (Eq. 9.56): A=ω3|℘|2
3πFepsilonC0/planckover2pi1c3.Thermally stimulated emission rate (Eq. 9.47):
R=π
3FepsilonC0/planckover2pi12|℘|2ρ(ω),with ρ(ω)=/planckover2pi1
π2c3ω3
(e/planckover2pi1ω/kBT−1)(Eq. 9.52).
So the ratio is
A
R=ω3|℘|2
3πFepsilonC0/planckover2pi1c3·3FepsilonC0/planckover2pi12
π|℘|2·π2c3/parenleftbig
e/planckover2pi1ω/kBT−1/parenrightbig
/planckover2pi1ω3=e/planckover2pi1ω/kBT−1.
The ratio is a monotonically increasing function of ω, and is 1 when
e/planckover2pi1ω/kbt=2,or/planckover2pi1ω
kBT=l n2,ω=kBT
/planckover2pi1ln 2,orν=ω
2π=kBT
hln 2.ForT= 300 K ,
ν=(1.38×10−23J/K)(300 K)
(6.63×10−34J·s)ln 2 = 4 .35×1012Hz.
For higher frequencies, (including light, at 1014Hz), spontaneous emission dominates.
Problem 9.9
(a)Simply remove the factor/parenleftBig
e/planckover2pi1ω/kBT−1/parenrightBig
in the denominator of Eq. 5.113: ρ0(ω)=/planckover2pi1ω3
π2c3.
(b)Plug this into Eq. 9.47:
Rb→a=π
3FepsilonC0/planckover2pi12|℘|2/planckover2pi1ω3
π2c3=ω3|℘|2
3πFepsilonC0/planckover2pi1c3,
reproducing Eq. 9.56.
Problem 9.10
N(t)=e−t/τN(0) (Eqs. 9.58 and 9.59). After one half-life, N(t)=1
2N(0),so1
2=e−t/τ,o r2 = et/τ,
sot/τ=l n2 , o r t1/2=τln 2.
Problem 9.11
In Problem 9.1 we calculated the matrix elements of z; all of them are zero except /angbracketleft100|z|210/angbracketright=28
35√
2a.A s
forxandy, we noted that |100/angbracketright,|200/angbracketright, and|210/angbracketrightareeven(inx,y), whereas|21±1/angbracketrightis odd. So the only
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244 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
non-zero matrix elements are /angbracketleft100|x|21±1/angbracketrightand/angbracketleft100|y|21±1/angbracketright. Using the wave functions in Problem 9.1:
/angbracketleft100|x|21±1/angbracketright=1√
πa3/parenleftbigg∓1
8√
πa3/parenrightbigg1
a/integraldisplay
e−r/are−r/2asinθe±iφ(rsinθcosφ)r2sinθdrdθdφ
=∓1
8πa4/integraldisplay∞
0r4e−3r/2adr/integraldisplayπ
0sin3θdθ/integraldisplay2π
0(cosφ±isinφ) cosφdφ
=∓1
8πa4/bracketleftBigg
4!/parenleftbigg2a
3/parenrightbigg5/bracketrightBigg/parenleftbigg4
3/parenrightbigg
(π)=∓27
35a.
/angbracketleft100|y|21±1/angbracketright=∓1
8πa4/bracketleftBigg
4!/bracketleftbigg2a
3/parenrightbigg5/bracketrightBigg/parenleftbigg4
3/parenrightbigg/integraldisplay2π
0(cosφ±isinφ) sinφdφ
=∓1
8πa4/bracketleftBigg
4!/parenleftbigg2a
3/parenrightbigg5/bracketrightBigg/parenleftbigg4
3/parenrightbigg
(±iπ)=−i27
35a.
/angbracketleft100|r|200/angbracketright=0 ;/angbracketleft100|r|210/angbracketright=27√
2
35aˆk;/angbracketleft100|r|21±1/angbracketright=27
35a/parenleftBig
∓ˆi−iˆj/parenrightBig
,and hence
℘2= 0 (for|200/angbracketright→|100/angbracketright),and|℘|2=(qa)2215
310(for|210/angbracketright→100/angbracketrightand|21±1/angbracketright→|100/angbracketright).
Meanwhile, ω=E2−E1
/planckover2pi1=1
/planckover2pi1/parenleftbiggE1
4−E1/parenrightbigg
=−3E1
4/planckover2pi1,so for the three l= 1 states:
A=−33E3
1
26/planckover2pi13(ea)2215
3101
3πFepsilonC0/planckover2pi1c3=−29
38πE3
1e2a2
FepsilonC0/planckover2pi14c3=210
38/parenleftbiggE1
mc2/parenrightbigg2c
a
=210
38/parenleftbigg13.6
0.511×106/parenrightbigg2(3.00×108m/s)
(0.529×10−10m)=6.27×108/s;τ=1
A=1.60×10−9s
for the three l= 1 states (all have the same lifetime); τ=∞for thel= 0 state.
Problem 9.12
[L2,z]=[L2
x,z]+[L2
y,z]+[L2
z,z]=Lx[Lx,z]+[Lx,z]Lx+Ly[Ly,z]+[Ly,z]Ly+Lz[Lz,z]+[Lz,z]Lz
But
[Lx,z]=[ypz−zpy,z]= [ypz,z]−[zpy,z]=y[pz,z]=−i/planckover2pi1y,
[Ly,z]=[zpx−xpz,z]=[zpx,z]−[xpz,z]=−x[pz,z]=i/planckover2pi1x,
[Lz,z]=[xpy−ypx,z]=[xpy,z]−[ypx,z]=0.
So: [L2,z]=Lx(−i/planckover2pi1y)+(−i/planckover2pi1y)Lx+Ly(i/planckover2pi1x)+(i/planckover2pi1x)Ly=i/planckover2pi1(−Lxy−yLx+Lyx+xLy).
But/braceleftbiggLxy=Lxy−yLx+yLx=[Lx,y]+yLx=i/planckover2pi1z+yLx,
Lyx=Lyx−xLy+xLy=[Ly,x]+xLy=−i/planckover2pi1z+xLy.
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 245
So: [L2,z]=i/planckover2pi1(2xLy−i/planckover2pi1z−2yLx−i/planckover2pi1z)=⇒[L2,z]=2i/planckover2pi1(xLy−yLx−i/planckover2pi1z).
/bracketleftbig
L2,[L2,z]/bracketrightbig
=2i/planckover2pi1/braceleftbig
[L2,xLy]−[L2,yLx]−i/planckover2pi1[L2,z]/bracerightbig
=2i/planckover2pi1/braceleftbig
[L2,x]Ly+x[L2,Ly]−[L2,y]Lx−y[L2,Lx]−i/planckover2pi1(L2z−zL2)/bracerightbig
.
But [L2,Ly]=[L2,Lx] = 0 (Eq. 4.102), so
/bracketleftbig
L2,[L2,z]/bracketrightbig
=2i/planckover2pi1/braceleftbig
(yLz−zLy−i/planckover2pi1x)Ly−2i/planckover2pi1(zLx−xLz−i/planckover2pi1y)Lx−i/planckover2pi1/parenleftbig
L2z−zL2/parenrightbig/bracerightbig
,or
/bracketleftbig
L2,[L2,z]/bracketrightbig
=−2/planckover2pi12/parenleftbigg
2yLzLy−2zL2
y−2zL2
x/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
−2z(L2
x+L2
y+L2
z)+2zL2
z−2i/planckover2pi1xLy+2xLzLx+2i/planckover2pi1yLx−L2z+zL2/parenrightbigg
=−2/planckover2pi12/parenleftbig
2yLzLy−2i/planckover2pi1xLy+2xLzLx+2i/planckover2pi1yLx+2zL2
z−2zL2−L2z+zL2/parenrightbig
=−2/planckover2pi12/parenleftbig
zL2+L2z/parenrightbig
−4/planckover2pi12/bracketleftbigg
(yLz−i/planckover2pi1x)/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
LzyLy+(xLz+i/planckover2pi1y)/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright
LzxLx+zLzLz/bracketrightbigg
=2/planckover2pi12/parenleftbig
zL2+L2z/parenrightbig
−4/planckover2pi12(LzyLy+LzxLx+LzzLz)/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
Lz(r·L)=0=2/planckover2pi12(zL2+L2z).QED
Problem 9.13
|n00/angbracketright=Rn0(r)Y0
0(θ,φ)=1√
4πRn0(r),so/angbracketleftn/prime00|r|n00/angbracketright=1
4π/integraldisplay
Rn/prime0(r)Rn0(r)(xˆi+yˆj+zˆk)dxdydz.
But the integrand is odd in x,y,o rz, so the integral is zero.
Problem 9.14
(a)
|300/angbracketright→
|210/angbracketright
|211/angbracketright
|21−1/angbracketright
→|100/angbracketright.(|300/angbracketright→|200/angbracketrightand|300/angbracketright→|100/angbracketrightviolate ∆ l=±1 rule.)
(b)
From Eq. 9.72: /angbracketleft210|r|300/angbracketright=/angbracketleft210|z|300/angbracketrightˆk.
From Eq. 9.69: /angbracketleft21±1|r|300/angbracketright=/angbracketleft21±1|x|300/angbracketrightˆi+/angbracketleft21±1|y|300/angbracketrightˆj.
From Eq. 9.70: ±/angbracketleft21±1|x|300/angbracketright=i/angbracketleft21±1|y|300/angbracketright.
Thus|/angbracketleft210|r|300/angbracketright|2=|/angbracketleft210|z|300/angbracketright|2and|/angbracketleft21±1|r|300/angbracketright|2=2|/angbracketleft21±1|x|300/angbracketright|2,
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246 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
so there are really just two matrix elements to calculate.
ψ21m=R21Ym
1,ψ 300=R30Y0
0.From Table 4.3:
/integraldisplay
Y0
1Y0
0cosθsinθdθdφ =/radicalbigg
3
4π/radicalbigg
1
4π/integraldisplayπ
0cos2θsinθdθ/integraldisplay2π
0dφ=√
3
4π/parenleftbigg
−cos3θ
3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0(2π)=√
3
2/parenleftbigg2
3/parenrightbigg
=1√
3.
/integraldisplay/parenleftbig
Y±1
1/parenrightbig∗Y0
0sin2θcosφdθdφ =∓/radicalbigg
3
8π/radicalbigg
1
4π/integraldisplayπ
0sin3θdθ/integraldisplay2π
0cosφe∓iφdφ
=∓1
4π/radicalbigg
3
2/parenleftbigg4
3/parenrightbigg/bracketleftbigg/integraldisplay2π
0cos2φdφ∓i/integraldisplay2π
0cosφsinφdφ/bracketrightbigg
=∓1
π√
6(π∓0) =∓1√
6.
From Table 4.7:
K≡/integraldisplay∞
0R21R30r3dr=1√
24a3/22√
27a3/2/integraldisplay∞
0r
ae−r/2a/bracketleftbigg
1−2
3r
a+2
27/parenleftBigr
a/parenrightBig2/bracketrightbigg
e−r/3ar3dr
=1
9√
2a3a4/integraldisplay∞
0/parenleftbigg
1−2
3u+2
27u2/parenrightbigg
u4e−5u/6du=a
9√
2/bracketleftBigg
4!/parenleftbigg6
5/parenrightbigg5
−2
35!/parenleftbigg6
5/parenrightbigg6
+2
276!/parenleftbigg6
5/parenrightbigg7/bracketrightBigg
=a
9√
24! 65
56/parenleftbigg
5−2
36·5+2
2763/parenrightbigg
=a
9√
24! 65
56=2734
56√
2a.
So:
/angbracketleft21±1|x|300/angbracketright=/integraldisplay
R21(Y±1
1)∗(rsinθcosφ)R30Y0
0r2sinθdrdθdφ =K/parenleftbigg
∓1√
6/parenrightbigg
.
/angbracketleft210|z|300/angbracketright=/integraldisplay
R21Y0
1(rcosθ)R30Y0
0r2sinθdrdθdφ =K/parenleftbigg1√
3/parenrightbigg
.
|/angbracketleft210|r|300/angbracketright|2=|/angbracketleft210|z|300/angbracketright|2=K2/3;
|/angbracketleft21±1|r|300/angbracketright|2=2|/angbracketleft21±1|x|300/angbracketright|2=K2/3.
Evidently the three transition rates are equal, and hence 1/3go by each route.
(c)For each mode, A=ω3e2|/angbracketleftr/angbracketright|2
3πFepsilonC0/planckover2pi1c3; here ω=E3−E2
/planckover2pi1=1
/planckover2pi1/parenleftbiggE1
9−E1
4/parenrightbigg
=−5
36E1
/planckover2pi1,so the total
decay rate is
R=3/parenleftbigg
−5
36E1
/planckover2pi1/parenrightbigg3e2
3πFepsilonC0/planckover2pi1c31
3/parenleftbigg2734
56√
2a/parenrightbigg2
=6/parenleftbigg2
5/parenrightbigg9/parenleftbiggE1
mc2/parenrightbigg2/parenleftBigc
a/parenrightBig
=6/parenleftbigg2
5/parenrightbigg9/parenleftbigg13.6
0.511×106/parenrightbigg2/parenleftbigg3×108
0.529×10−10/parenrightbigg
/s=6.32×106/s.τ=1
R=1.58×10−7s.
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 247
Problem 9.15
(a)
Ψ(t)=/summationdisplay
cn(t)e−iEnt//planckover2pi1ψn.HΨ=i/planckover2pi1∂Ψ
∂t;H=H0+H/prime(t);H0ψn=Enψn.So
/summationdisplay
cne−iEnt//planckover2pi1Enψn+/summationdisplay
cne−iEnt//planckover2pi1H/primeψn=i/planckover2pi1/summationdisplay
˙cne−iEnt//planckover2pi1ψn+i/planckover2pi1/parenleftbigg
−i
/planckover2pi1/parenrightbigg/summationdisplay
cnEne−iEnt//planckover2pi1ψn.
The first and last terms cancel, so
/summationdisplay
cne−iEnt//planckover2pi1H/primeψn=i/planckover2pi1/summationdisplay
˙cne−iEnt//planckover2pi1ψn.Take the inner product with ψm:
/summationdisplay
cne−iEnt//planckover2pi1/angbracketleftψm|H/prime|ψn/angbracketright=i/planckover2pi1/summationdisplay
˙cne−iEnt//planckover2pi1/angbracketleftψm|ψn/angbracketright.
Assume orthonormality of the unperturbed states, /angbracketleftψm|ψn/angbracketright=δmn,and define H/prime
mn≡/angbracketleftψm|H/prime|ψn/angbracketright.
/summationdisplay
cne−iEnt//planckover2pi1H/prime
mn=i/planckover2pi1˙cme−iEmt//planckover2pi1,or ˙cm=−i
/planckover2pi1/summationdisplay
ncnH/prime
mnei(Em−En)t//planckover2pi1.
(b)Zeroth order: cN(t)=1,c m(t) = 0 for m/negationslash=N. Then in first order:
˙cN=−i
/planckover2pi1H/prime
NN,o r cN(t)=1−i
/planckover2pi1/integraldisplayt
0H/prime
NN(t/prime)dt/prime,whereas for m/negationslash=N:
˙cm=−i
/planckover2pi1H/prime
mNei(Em−EN)t//planckover2pi1,o r cm(t)=−i
/planckover2pi1/integraldisplayt
0H/prime
mN(t/prime)ei(Em−EN)t/prime//planckover2pi1dt/prime.
(c)
cM(t)=−i
/planckover2pi1H/prime
MN/integraldisplayt
0ei(EM−EN)t/prime//planckover2pi1dt/prime=−i
/planckover2pi1H/prime
MN/bracketleftBigg
ei(EM−EN)t/prime//planckover2pi1
i(EM−EN)//planckover2pi1/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglet
0=−H/prime
MN/bracketleftbiggei(EM−EN)t//planckover2pi1−1
EM−EN/bracketrightbigg
=−H/prime
MN
(EM−EN)ei(EM−EN)t/2/planckover2pi12isin/parenleftbiggEM−EN
2/planckover2pi1t/parenrightbigg
.
PN→M=|cM|2=4|H/prime
MN|2
(EM−EN)2sin2/parenleftbiggEM−EN
2/planckover2pi1t/parenrightbigg
.
(d)
cM(t)=−i
/planckover2pi1VMN1
2/integraldisplayt
0/parenleftBig
eiωt/prime+e−iωt/prime/parenrightBig
ei(EM−EN)t/prime//planckover2pi1dt/prime
=−iVMN
2/planckover2pi1/bracketleftBigg
ei(/planckover2pi1ω+EM−EN)t/prime//planckover2pi1
i(/planckover2pi1ω+EM−EN)//planckover2pi1+ei(−/planckover2pi1ω+EM−EN)t/prime//planckover2pi1
i(−/planckover2pi1ω+EM−EN)//planckover2pi1/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglet
0.
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248 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
IfEM>E N, the second term dominates, and transitions occur only for ω≈(EM−EN)//planckover2pi1:
cM(t)≈−iVMN
2/planckover2pi11
(i//planckover2pi1)(EM−EN−/planckover2pi1ω)ei(EM−EN−/planckover2pi1ω)t/2/planckover2pi12isin/parenleftbiggEM−EN−/planckover2pi1ω
2/planckover2pi1t/parenrightbigg
,so
PN→M=|cM|2=|VMN|2
(EM−EN−/planckover2pi1ω)2sin2/parenleftbiggEM−EN−/planckover2pi1ω
2/planckover2pi1t/parenrightbigg
.
IfEM<E Nthe first term dominates, and transitions occur only for ω≈(EN−EM)//planckover2pi1:
cM(t)≈−iVMN
2/planckover2pi11
(i//planckover2pi1)(EM−EN+/planckover2pi1ω)ei(EM−EN+/planckover2pi1ω)t/2/planckover2pi12isin/parenleftbiggEM−EN+/planckover2pi1ω
2/planckover2pi1t/parenrightbigg
,and hence
PN→M=|VMN|2
(EM−EN+/planckover2pi1ω)2sin2/parenleftbiggEM−EN+/planckover2pi1ω
2/planckover2pi1t/parenrightbigg
.
Combining the two results, we conclude that transitions occur to states with energy EM≈EN±/planckover2pi1ω, and
PN→M=|VMN|2
(EM−EN±/planckover2pi1ω)2sin2/parenleftbiggEM−EN±/planckover2pi1ω
2/planckover2pi1t/parenrightbigg
.
(e)For light, Vba=−℘E0(Eq. 9.34). The rest is as before (Section 9.2.3), leading to Eq. 9.47:
RN→M=π
3FepsilonC0/planckover2pi12|℘|2ρ(ω),withω=±(EM−EN)//planckover2pi1(+ sign⇒absorption,−sign⇒stimulated emission) .
Problem 9.16
For example (c):
cN(t)=1−i
/planckover2pi1H/prime
NNt;cm(t)=−2iH/prime
mN
(Em−EN)ei(Em−EN)t/2/planckover2pi1sin/parenleftbiggEm−EN
2/planckover2pi1t/parenrightbigg
(m/negationslash=N).
|cN|2=1+1
/planckover2pi12|H/prime
NN|2t2,|cm|2=4|H/prime
mN|2
(Em−EN)2sin2/parenleftbiggEm−EN
2/planckover2pi1t/parenrightbigg
,so
/summationdisplay
m|cm|2=1+t2
/planckover2pi12|H/prime
NN|2+4/summationdisplay
m/negationslash=N|H/prime
mN|2
(Em−EN)2sin2/parenleftbiggEm−EN
2/planckover2pi1t/parenrightbigg
.
This is plainly greater than 1! But remember: The c’s are accurate only to firstorder in H/prime; to this order the
|H/prime|2terms do not belong. Only if terms of firstorder appeared in the sum would there be a genuine problem
with normalization.
For example (d):
cN=1−i
/planckover2pi1VNN/integraldisplayt
0cos(ωt/prime)dt/prime=1−i
/planckover2pi1VNNsin(ωt/prime)
ω/vextendsingle/vextendsingle/vextendsingle/vextendsinglet
0=⇒cN(t)=1−i
/planckover2pi1ωVNNsin(ωt).
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 249
cm(t)=−VmN
2/bracketleftbiggei(Em−EN+/planckover2pi1ω)t//planckover2pi1−1
(Em−EN+/planckover2pi1ω)+ei(Em−EN−/planckover2pi1ω)t//planckover2pi1−1
(Em−EN−/planckover2pi1ω)/bracketrightbigg
(m/negationslash=N).So
|cN|2=1+|VNN|2
(/planckover2pi1ω)2sin2(ωt); and in the rotating waveapproximation
|cm|2=|VmN|2
(Em−EN±/planckover2pi1ω)2sin2/parenleftbiggEm−EN±/planckover2pi1ω
2/planckover2pi1t/parenrightbigg
(m/negationslash=N).
Again, ostensibly/summationtext|cm|2>1, but the “extra” terms are of second order in H/prime, and hence do not belong (to
first order).
You would do better to use 1 −/summationtext
m/negationslash=N|cm|2.Schematically: cm=a1H+a2H2+···,s o|cm|2=
a2
1H2+2a1a2H3+···, whereas cN=1+b1H+b2H2+···,s o|cN|2=1+2 b1H+( 2b2+b2
1)H2+···.
Thus knowing cmtofirstorder (i.e., knowing a1) gets you|cm|2tosecond order, but knowing cNto first order
(i.e.,b1)d o e s notget you|cN|2to second order (you’d also need b2). It is precisely this b2term that would
cancel the “extra” (second-order) terms in the calculations of/summationtext|cm|2above.
Problem 9.17
(a)
Equation 9.82 ⇒˙cm=−i
/planckover2pi1/summationdisplay
ncnH/prime
mnei(Em−En)t//planckover2pi1.HereH/prime
mn=/angbracketleftψm|V0(t)|ψn/angbracketright=δmnV0(t).
˙cm=−i
/planckover2pi1cmV0(t);dcm
cm=−i
/planckover2pi1V0(t)dt⇒lncm=−i
/planckover2pi1/integraldisplay
V0(t/prime)dt/prime+constant.
cm(t)=cm(0)e−i
/planckover2pi1/integraltextt
0V0(t/prime)dt/prime.Let Φ(t)≡−1
/planckover2pi1/integraldisplayt
0V0(t/prime)dt/prime;cm(t)=eiΦcm(0).Hence
|cm(t)|2=|cm(0)|2,and there are notransitions. Φ(T)=−1
/planckover2pi1/integraldisplayT
0V0(t)dt.
(b)
Eq. 9.84⇒cN(t)≈1−i
/planckover2pi1/integraldisplayt
0V0(t/prime)dt=1+iΦ.
Eq. 9.85⇒cm(t)=−i
/planckover2pi1/integraldisplayt
0δmNV0(t/prime)ei(Em−EN)t/prime//planckover2pi1dt/prime=0(m/negationslash=N).
cN(t)=1 + iΦ(t),
cm(t)=0(m/negationslash=N).
Theexact answer is cN(t)=eiΦ(t),cm(t) = 0, and they areconsistent, since eiΦ≈1+iΦ, to first order.
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250 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Problem 9.18
Use result of Problem 9.15(c). Here En=n2π2/planckover2pi12
2ma2,soE2−E1=3π2/planckover2pi12
2ma2.
H/prime
12=2
a/integraldisplaya/2
0sin/parenleftBigπ
ax/parenrightBig
V0sin/parenleftbigg2π
ax/parenrightbigg
dx
=2V0
a/bracketleftBigg
sin/parenleftbigπ
ax/parenrightbig
2(π/a)−sin/parenleftbig3π
ax/parenrightbig
2(3π/a)/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea/2
0=V0
π/bracketleftbigg
sin/parenleftBigπ
2/parenrightBig
−1
3sin/parenleftbigg3π
2/parenrightbigg/bracketrightbigg
=4V0
3π.
Eq. 9.86 =⇒P1→2=4/parenleftbigg4V0
3π/parenrightbigg/parenleftbigg2ma2
3π2/planckover2pi12/parenrightbigg2
sin2/parenleftbigg3π2/planckover2pi1
4ma2t/parenrightbigg
=/bracketleftbigg16ma2V0
9π3/planckover2pi12sin/parenleftbigg3π2/planckover2pi1T
4ma2/parenrightbigg/bracketrightbigg2
.
[Actually, in this case H/prime
11andH/prime
22are nonzero:
H/prime
11=/angbracketleftψ1|H/prime|ψ1/angbracketright=2
aV0/integraldisplaya/2
0sin2/parenleftBigπ
ax/parenrightBig
dx=V0
2,H/prime
22=/angbracketleftψ2|H/prime|ψ2/angbracketright=2
aV0/integraldisplaya/2
0sin2/parenleftbigg2π
ax/parenrightbigg
dx=V0
2.
However, this does not affect the answer, for according to Problem 9.4, c1(t) picks up an innocuous phase factor,
whilec2(t) is not affected at all, in first order (formally, this is because H/prime
bbis multiplied by cb, in Eq. 9.11, and
in zeroth order cb(t) = 0).]
Problem 9.19
Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic
field. But you can’t dothat, because the gound state already has the lowest allowed energy.
Problem 9.20
(a)
H=−γB·S=−γ(BxSx+BySy+BzSz);
H=−γ/planckover2pi1
2(Bxσx+Byσy+Bzσz)=−γ/planckover2pi1
2/bracketleftbigg
Bx/parenleftbigg01
10/parenrightbigg
+By/parenleftbigg0−i
i0/parenrightbigg
+Bz/parenleftbigg10
0−1/parenrightbigg/bracketrightbigg
=−γ/planckover2pi1
2/parenleftbiggBzBx−iBy
Bx+iBy−Bz/parenrightbigg
=−γ/planckover2pi1
2/parenleftbiggB0 Brf(cosωt+isinωt)
Brf(cosωt−isinωt)−B0/parenrightbigg
=−γ/planckover2pi1
2/parenleftbiggB0Brfeiωt
Brfe−iωt−B0/parenrightbigg
.
(b)i/planckover2pi1˙χ=Hχ⇒
i/planckover2pi1/parenleftbigg˙a
˙b/parenrightbigg
=−γ/planckover2pi1
2/parenleftbiggB0Brfeiωt
Brfe−iωt−B0/parenrightbigg/parenleftbigga
b/parenrightbigg
=−γ/planckover2pi1
2/parenleftbiggB0aB rfeiωtb
Brfe−iωta−B0b/parenrightbigg
⇒
˙a=iγ
2/parenleftbig
B0a+Brfeiωtb/parenrightbig
=i
2/parenleftbig
Ωeiωtb+ω0a/parenrightbig
,
˙b=−iγ
2/parenleftbig
B0b−Brfe−iωta/parenrightbig
=i
2/parenleftbig
Ωe−iωta−ω0b/parenrightbig
.
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 251
(c)You can decouple the equations by differentiating with respect to t, but it is simpler just to check the quoted
results. First of all, they clearly satisfy the initial conditions: a(0) =a0andb(0) =b0. Differentiating a:
˙a=iω
2a+/braceleftbigg
−a0ω/prime
2sin(ω/primet/2) +i
ω/prime[a0(ω0−ω)+b0Ω]ω/prime
2cos(ω/primet/2)/bracerightbigg
eiωt/2
=i
2eiωt/2/braceleftbigg
ωa0cos(ω/primet/2) +iω
ω/prime[a0(ω0−ω)+b0Ω] sin(ω/primet/2)
+iω/primea0sin(ω/primet/2 )+[a0(ω0−ω)+b0Ω] cos(ω/primet/2)/bracerightbigg
Equation 9.90 says this should be equal to
i
2/parenleftbig
Ωeiωtb+ω0a/parenrightbig
=i
2eiωt/2/braceleftbigg
Ωb0cos(ω/primet/2) +iΩ
ω/prime[b0(ω−ω0)+a0Ω] sin(ω/primet/2)
+ω0a0cos(ω/primet/2) +iω0
ω/prime[a0(ω0−ω)+b0Ω] sin(ω/primet/2)/bracerightbigg
.
By inspection the cos( ω/primet/2) terms in the two expressions are equal; it remains to check that
iω
ω/prime[a0(ω0−ω)+b0Ω] +iω/primea0=iΩ
ω/prime[b0(ω−ω0)+a0Ω] +iω0
ω/prime[a0(ω0−ω)+b0Ω],
which is to say
a0ω(ω0−ω)+b0ωΩ+a0(ω/prime)2=b0Ω(ω−ω0)+a0Ω2+a0ω0(ω0−ω)+b0ω0Ω,
or
a0/bracketleftbig
ωω0−ω2+(ω/prime)2−Ω2−ω2
0+ω0ω/bracketrightbig
=b0[Ωω−ω0Ω+ω0Ω−ωΩ] = 0.
Substituting Eq. 9.91 for ω/prime, the coefficient of a0on the left becomes
2ωω0−ω2+(ω−ω0)2+Ω2−Ω2−ω2
0=0./check
The check of b(t) is identical, with a↔b,ω0→−ω0, andω→−ω.
(d)
b(t)=iΩ
ω/primesin(ω/primet/2)e−iωt/2;P(t)=|b(t)|2=/parenleftbiggΩ
ω/prime/parenrightbigg2
sin2(ω/primet/2).
(e)
P(ω)
ω1
1/2
ω0∆ω
The maximum ( Pmax= 1) occurs (obviously) at ω=ω0.
P=1
2⇒(ω−ω0)2=Ω2⇒ω=ω0±Ω,so ∆ω=ω+−ω−=2Ω.
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252 CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
(f)B0=1 0,000 gauss = 1 T; Brf=0.01 gauss = 1 ×10−6T.ω0=γB0.Comparing Eqs. 4.156 and
6.85,γ=gpe
2mp, where gp=5.59. So
νres=ω0
2π=gpe
4πmpB0=(5.59)(1.6×10−19)
4π(1.67×10−27)(1) = 4.26×107Hz.
∆ν=∆ω
2π=Ω
π=γ
2π2Brf=νres2Brf
B0=( 4.26×107)(2×10−6)=85.2H z.
Problem 9.21
(a)
H/prime=−qE·r=−q(E0·r)(k·r) sin(ωt).WriteE0=E0ˆn,k=ω
cˆk.Then
H/prime=−qE0ω
c(ˆn·r)(ˆk·r) sin(ωt).H/prime
ba=−qE0ω
c/angbracketleftb|(ˆn·r)(ˆk·r)|a/angbracketrightsin(ωt).
This is the analog to Eq. 9.33: H/prime
ba=−qE0/angbracketleftb|ˆn·r|a/angbracketrightcosωt.The rest of the analysis is identical to the
dipole case (except that it is sin( ωt) instead of cos( ωt), but this amounts to resetting the clock, and clearly
has no effect on the transition rate). We can skip therefore to Eq. 9.56, except for the factor of 1/3, whichcame from the averaging in Eq. 9.46:
A=ω
3
πFepsilonC0/planckover2pi1c3q2ω2
c2|/angbracketleftb|(ˆn·r)(ˆk·r)|a/angbracketright|2=q2ω5
πFepsilonC0/planckover2pi1c5|/angbracketleftb|(ˆn·r)(ˆk·r)|a/angbracketright|2.
(b)Let the oscillator lie along the xdirection, so (ˆ n·r)=ˆnxxandˆk·r=ˆkxx. For a transition from nton/prime,
we have
A=q2ω5
πFepsilonC0/planckover2pi1c5/parenleftBig
ˆkxˆnx/parenrightBig2
|/angbracketleftn/prime|x2|n/angbracketright|2.From Example 2.5, /angbracketleftn/prime|x2|n/angbracketright=/planckover2pi1
2m¯ω/angbracketleftn/prime|(a2
++a+a−+a−a++a2
−)|n/angbracketright,
where ¯ωis the frequency of the oscillator , not to be confused with ω, the frequency of the electromagnetic
wave. Now, for spontaneous emission the final state must be lower in energy, so n/prime<n, and hence the
only surviving term is a2
−. Using Eq. 2.66:
/angbracketleftn/prime|x2|n/angbracketright=/planckover2pi1
2m¯ω/angbracketleftn/prime|/radicalbig
n(n−1)|n−2/angbracketright=/planckover2pi1
2m¯ω/radicalbig
n(n−1)δn/prime,n−2.
Evidently transitions only go from |n/angbracketrightto|n−2/angbracketright, and hence
ω=En−En−2
/planckover2pi1=1
/planckover2pi1/bracketleftbig
(n+1
2)/planckover2pi1¯ω−(n−2+1
2)/planckover2pi1¯ω/bracketrightbig
=2 ¯ω.
/angbracketleftn/prime|x2|n/angbracketright=/planckover2pi1
mω/radicalbig
n(n−1)δn/prime,n−2;Rn→n−2=q2ω5
πFepsilonC0/planckover2pi1c5(ˆkxˆnx)2/planckover2pi12
m2ω2n(n−1).
It remains to calculate the average of ( ˆkxˆnx)2. It’s easiest to reorient the oscillator along a direction ˆ r,
making angle θwith the zaxis, and let the radiation be incident from the zdirection (so ˆkx→ˆkr= cosθ).
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY 253
Averaging over the two polarizations ( ˆiandˆj):/angbracketleftˆn2
r/angbracketright=1
2/parenleftBig
ˆi2
r+ˆj2
r/parenrightBig
=1
2/parenleftbig
sin2θcos2φ+ sin2θsin2φ/parenrightbig
=
1
2sin2θ. Now average overall directions:
/angbracketleftˆk2
rˆn2
r/angbracketright=1
4π/integraldisplay1
2sin2θcos2θsinθdθdφ =1
8π2π/integraldisplayπ
0(1−cos2θ) cos2θsinθdθ
=1
4/bracketleftbigg
−cos3θ
3+cos5θ
5/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=1
4/parenleftbigg2
3−2
5/parenrightbigg
=1
15.
R=1
15q2/planckover2pi1ω3
πFepsilonC0m2c5n(n−1).Comparing Eq. 9.63:R(forbidden)
R(allowed)=2
5(n−1)/planckover2pi1ω
mc2.
For a nonrelativistic system, /planckover2pi1ω/lessmuchmc2; hence the term “forbidden”.
(c)If both the initial state and the final state have l= 0, the wave function is independent of angle ( Y0
0=
1/√
4π), and the angular part of the integral is:
/angbracketlefta|(ˆn·r)(ˆk·r)|b/angbracketright=···/integraldisplay
(ˆn·r)(ˆk·r) sinθdθdφ =···4π
3(ˆn·ˆk) (Eq. 6.95).
But ˆn·ˆk= 0, since electromagnetic waves are transverse. So R= 0 in this case, both for allowed and
for forbidden transitions.
Problem 9.22
[This is done in Fermi’s Notes on Quantum Mechanics (Chicago, 1995), Section 24, but I am looking for a more
accessible treatment.]
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254 CHAPTER 10. THE ADIABATIC APPROXIMATION
Chapter 10
The Adiabatic Approximation
Problem 10.1
(a)
Let (mvx2−2Ei
nat)/2/planckover2pi1w=φ(x,t).Φn=/radicalbigg
2
wsin/parenleftBignπ
wx/parenrightBig
eiφ,so
∂Φn
∂t=√
2/parenleftbigg
−1
21
w3/2v/parenrightbigg
sin/parenleftBignπ
wx/parenrightBig
eiφ+/radicalbigg
2
w/bracketleftBig
−nπx
w2vcos/parenleftBignπ
wx/parenrightBig/bracketrightBig
eiφ+/radicalbigg
2
wsin/parenleftBignπ
wx/parenrightBig/parenleftbigg
i∂φ
∂t/parenrightbigg
eiφ
=/bracketleftbigg
−v
2w−nπxv
w2cot/parenleftBignπ
wx/parenrightBig
+i∂φ
∂t/bracketrightbigg
Φn.∂φ
∂t=1
2/planckover2pi1/bracketleftbigg
−2Ei
na
w−v
w2/parenleftbig
mvx2−2Ei
nat/parenrightbig/bracketrightbigg
=−Ei
na
/planckover2pi1w−v
wφ.
i/planckover2pi1∂Φn
∂t=−i/planckover2pi1/bracketleftbiggv
2w+nπxv
w2cot/parenleftBignπ
wx/parenrightBig
+iEi
na
/planckover2pi1w+iv
wφ/bracketrightbigg
Φn.
HΦn=−/planckover2pi12
2m∂2Φn
∂x2.∂Φn
∂x=/radicalbigg
2
w/bracketleftBignπ
wcos/parenleftBignπ
wx/parenrightBig/bracketrightBig
eiφ+/radicalbigg
2
wsin/parenleftBignπ
wx/parenrightBig
eiφ/parenleftbigg
i∂φ
∂x/parenrightbigg
.
∂φ
∂x=mvx
/planckover2pi1w.∂Φn
∂x=/bracketleftBignπ
wcot/parenleftBignπ
wx/parenrightBig
+imvx
/planckover2pi1w/bracketrightBig
Φn.
∂2Φn
∂x2=/bracketleftbigg
−/parenleftBignπ
w/parenrightBig2
csc2/parenleftBignπ
wx/parenrightBig
+imb
/planckover2pi1w/bracketrightbigg
Φn+/bracketleftBignπ
wcot/parenleftBignπ
wx/parenrightBig
+imvx
/planckover2pi1w/bracketrightBig2
Φn.
So the Schr¨ odinger equation ( i/planckover2pi1∂Φn/∂t=HΦn) is satisfied⇔
−i/planckover2pi1/bracketleftbiggv
2w+nπxv
w2cot/parenleftBignπ
wx/parenrightBig
+iEi
na
/planckover2pi1w+iv
wφ/bracketrightbigg
=−/planckover2pi12
2m/braceleftbigg
−/parenleftBignπ
w/parenrightBig2
csc2/parenleftBignπ
wx/parenrightBig
+imv
/planckover2pi1w+/bracketleftBignπ
wcot/parenleftBignπ
wx/parenrightBig
+imvx
/planckover2pi1w/bracketrightBig2/bracerightbigg
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CHAPTER 10. THE ADIABATIC APPROXIMATION 255
Cotangent terms :−i/planckover2pi1/parenleftBignπxv
w2/parenrightBig?=−/planckover2pi12
2m/parenleftBig
2nπ
wimvx
/planckover2pi1w/parenrightBig
=−i/planckover2pi1nπvx
w2./check
Remaining trig terms on right :
−/parenleftBignπ
w/parenrightBig2
csc2/parenleftBignπ
wx/parenrightBig
+/parenleftBignπ
w/parenrightBig2
cot2/parenleftBignπ
wx/parenrightBig
=−/parenleftBignπ
w/parenrightBig2/bracketleftbigg1−cos2(nπx/w )
sin2(nπx/w )/bracketrightbigg
=−/parenleftBignπ
w/parenrightBig2
.
This leaves:
i/bracketleftbiggv
2w+iEi
na
/planckover2pi1w+iv
w/parenleftbiggmvx2−2Ei
nat
2/planckover2pi1w/parenrightbigg/bracketrightbigg
?=/planckover2pi1
2m/bracketleftbigg
−/parenleftBignπ
w/parenrightBig2
+imv
/planckover2pi1w−m2v2x2
/planckover2pi12w2/bracketrightbigg
✁✁✁iv
2−Ei
na
/planckover2pi1−
✚✚✚✚mv2x2
2/planckover2pi1w+vEi
nat
/planckover2pi1w?=−/planckover2pi1n2π2
2mw+✁✁✁iv
2−
✚✚✚✚mv2x2
2/planckover2pi1w
−Ei
na
/planckover2pi1w(w−vt)=−Ei
na2
/planckover2pi1w?=−/planckover2pi1n2π2
2mw⇔−n2π2/planckover2pi12
2ma2a2
/planckover2pi1w=−/planckover2pi1n2π2
2mw= r.h.s. /check
So Φ ndoessatisfy the Schr¨ odinger equation, and since Φ n(x,t)=(···) sin (nπx/w ), it fits the boundary
conditions: Φ n(0,t)=Φ n(w,t)=0 .
(b)
Equation 10.4 = ⇒Ψ(x,0) =/summationdisplay
cnΦn(x,0) =/summationdisplay
cn/radicalbigg
2
asin/parenleftBignπ
ax/parenrightBig
eimvx2/2/planckover2pi1a.
Multiply by/radicalbigg
2
asin/parenleftbiggn/primeπ
ax/parenrightbigg
e−imvx2/2/planckover2pi1aand integrate:
/radicalbigg
2
a/integraldisplaya
0Ψ(x,0) sin/parenleftbiggn/primeπ
ax/parenrightbigg
e−imvx2/2/planckover2pi1adx=/summationdisplay
cn/bracketleftbigg2
a/integraldisplayπ
0sin/parenleftBignπ
ax/parenrightBig
sin/parenleftbiggn/primeπ
ax/parenrightbigg
dx
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
δnn/prime/bracketrightbigg
=c/prime
n.
So, in general: cn=/radicalbigg
2
a/integraldisplaya
0e−imvx2/2/planckover2pi1asin/parenleftBignπ
ax/parenrightBig
Ψ(x,0)dx.In this particular case,
cn=2
a/integraldisplaya
0e−imvx2/2/planckover2pi1asin/parenleftBignπ
a/parenrightBig
sin/parenleftBigπ
ax/parenrightBig
dx.Letπ
ax≡z;dx=a
πdz;mvx2
2/planckover2pi1a=mvz2
2/planckover2pi1aa2
π2=mva
2π2/planckover2pi1z2.
cn=2
π/integraldisplayπ
0e−iαz2sin(nz) sin(z)dz.QED
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256 CHAPTER 10. THE ADIABATIC APPROXIMATION
(c)
w(Te)=2a⇒a+vTe=2a⇒vTe=a⇒Tea/v;e−iE1t//planckover2pi1⇒ω=E1
/planckover2pi1⇒Ti=2π
ω=2π/planckover2pi1
E1,or
Ti=2π/planckover2pi1
π2/planckover2pi122ma2=4
πma2
/planckover2pi1.Ti=4ma2
π/planckover2pi1.Adiabatic⇒Te/greatermuchTi⇒a
v/greatermuch4ma2
π/planckover2pi1⇒4
πmav
/planckover2pi1/lessmuch1,or
8π/parenleftBigmav
2π2/planckover2pi1/parenrightBig
=8πα/lessmuch1,soα/lessmuch1.Thencn=2
π/integraldisplayπ
0sin(nz) sin(z)dz=δn1.Therefore
Ψ(x,t)=/radicalbigg
2
wsin/parenleftBigπx
w/parenrightBig
ei(mvx2−2Ei
1at)/2/planckover2pi1w,
which (apart from a phase factor) is the ground state of the instantaneous well, of width w, as required
by the adiabatic theorem. (Actually, the first term in the exponent, which is at mostmva2
2/planckover2pi1a=mva
2/planckover2pi1/lessmuch1
and could be dropped, in the adiabatic regime.)
(d)
θ(t)=−1
/planckover2pi1/parenleftbiggπ2/planckover2pi12
2m/parenrightbigg/integraldisplayt
01
(a+vt/prime)2dt/prime=−π2/planckover2pi1
2m/bracketleftbigg
−1
v/parenleftbigg1
a+vt/prime/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglet
0
=−π2/planckover2pi1
2mv/parenleftbigg1
a−1
w/parenrightbigg
=−π2/planckover2pi1
2mv/parenleftbiggvt
aw/parenrightbigg
=−π2/planckover2pi1t
2maw.
So (dropping themvx2
2/planckover2pi1wterm, as explained in (c)) Ψ( x,t)=/radicalbigg
2
wsin/parenleftBigπx
w/parenrightBig
e−iEi
1at//planckover2pi1wcan be written
(since−Ei
1at
/planckover2pi1w=−π2/planckover2pi12
2ma2at
/planckover2pi1w=−π2/planckover2pi1t
2maw=θ):Ψ(x,t)=/radicalbigg
2
wsin/parenleftBigπx
w/parenrightBig
eiθ.
This is exactly what one would naively expect: For a fixed well (of width a) we’d have Ψ( x,t)=
Ψ1(x)e−iE1t//planckover2pi1; for the (adiabatically) expanding well, simply replace aby the (time-dependent) width
w, andintegrate to get the accumulated phase factor, noting that E1is now a function of t.
Problem 10.2
To show: i/planckover2pi1∂χ
∂t=Hχ, whereχis given by Eq. 10.31 and His given by Eq. 10.25 .
∂χ
∂t=
λ
2/bracketleftBig
−sin/parenleftbigλt
2/parenrightbig
−i(ω1−ω)
λcos/parenleftbigλt
2/parenrightbig/bracketrightBig
cos/parenleftbigα
2/parenrightbig
e−iωt/2−iω
2/bracketleftBig
cos/parenleftbigλt
2/parenrightbig
−i(ω1−ω)
λsin/parenleftbigλt
2/parenrightbig/bracketrightBig
cos/parenleftbigα
2/parenrightbig
e−iωt/2
λ
2/bracketleftBig
−sin/parenleftbigλt
2/parenrightbig
−i(ω1+ω)
λcos/parenleftbigλt
2/parenrightbig/bracketrightBig
sin/parenleftbigα
2/parenrightbig
eiωt/2+iω
2/bracketleftBig
cos/parenleftbigλt
2/parenrightbig
−i(ω1+ω)
λsin/parenleftbigλt
2/parenrightbig/bracketrightBig
sin/parenleftbigα
2/parenrightbig
eiωt/2
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CHAPTER 10. THE ADIABATIC APPROXIMATION 257
Hχ=
/planckover2pi1ω1
2
cosα/bracketleftBig
cos(λt
2)−i(ω1−ω)
λsin(λt
2)/bracketrightBig
cosα
2e−iωt/2+e−iωtsinα/bracketleftBig
cos(λt
2)−i(ω1+ω)
λsin(λt
2)/bracketrightBig
sinα
2eiωt/2
eiωtcosα/bracketleftBig
cos(λt
2)−i(ω1−ω)
λsin(λt
2)/bracketrightBig
cosα
2e−iωt/2−cosα/bracketleftBig
cos(λt
2)−i(ω1+ω)
λsin(λt
2)/bracketrightBig
sinα
2eiωt/2
(1) Upper elements:
i✁/planckover2pi1/braceleftbiggλ
✁2/bracketleftbigg
−sin/parenleftbiggλt
2/parenrightbigg
−i(ω1−ω)
λcos/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
✚✚✚cosα
2−iω
✁2/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1−ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
✚✚✚cosα
2/bracerightbigg
?=✁/planckover2pi1ω1
✁2/braceleftbigg/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1−ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
cosα✚✚✚cosα
2+/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1+ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
sinα/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
⋆sinα
2/bracerightbigg
,
where⋆= 2 sinα
2✚✚✚cosα
2
The sine terms:
sin/parenleftbiggλt
2/parenrightbigg/bracketleftbigg
−iλ−iω(ω1−ω)
λ+ω1(ω1−ω)
λcosα+iω1(ω1+ω)
λ2 sin2α
2/bracketrightbigg
?=0.
i
λsin/parenleftbiggλt
2/parenrightbigg/bracketleftBig
✟✟✟−ω2−ω2
1+2ωω1cosα−ωω1+✚✚ω2+(ω2
1−ωω1) cosα+(ω2
1+ωω1)(1−cosα)/bracketrightBig
=−i
λsin/parenleftbiggλt
2/parenrightbigg/bracketleftBig
✟✟✟−ω2
1+2ωω1cosα−✘✘ωω1+✘✘✘✘ω2
1cosα−ωω1cosα+ ω2
1+✘✘ωω1−✘✘✘✘ω2
1cosα−ωω1cosα/bracketrightBig
=0./check
The cosine terms:
cos/parenleftbiggλt
2/parenrightbigg/bracketleftBig
(ω1−✚ω)+✚ω−ω1cosα−ω12 sin2α
2/bracketrightBig
=−ω1cos/parenleftbiggλt
2/parenrightbigg
[−1 + cosα+( 1−cosα)] = 0./check
(2) Lower elements:
i✁/planckover2pi1/braceleftbiggλ
✁2/bracketleftbigg
−sin/parenleftbiggλt
2/parenrightbigg
−i(ω1+ω)
λcos/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
✟✟✟✟sin/parenleftBigα
2/parenrightBig
+iω
✁2/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1+ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
✟✟✟✟sin/parenleftBigα
2/parenrightBig/bracerightbigg
?=✁/planckover2pi1ω1
✁2/braceleftbigg/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1−ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
2✚✚✚sinα
2cos2α
2−/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1+ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
cosα✚✚✚sinα
2/bracerightbigg
.
The sine terms:
sin/parenleftbiggλt
2/parenrightbigg/bracketleftbigg
−iλ+iω(ω1+ω)
λ+iω1(ω1−ω)
λ2 cos2/parenleftBigα
2/parenrightBig
−iω1(ω1+ω)
λcosα/bracketrightbigg
?=0.
i
λsin/parenleftbiggλt
2/parenrightbigg/bracketleftBig
−✚✚ω2−ω2
1+2ωω1cosα+ωω1+✚✚ω2+(ω2
1−ωω1)(1 + cos α)−(ω2
1+ωω1) cosα/bracketrightBig
=i
λsin/parenleftbiggλt
2/parenrightbigg/bracketleftBig
− ω2
1+2ωω1cosα+✘✘ωω1+ ω2
1−✘✘ωω1+✘✘✘✘ω2
1cosα−ωω1cosα−✘✘✘✘ω2
1cosα−ωω1cosα/bracketrightBig
=0./check
The cosine terms:
cos/parenleftbiggλt
2/parenrightbigg/bracketleftBig
(ω1+✚ω)−✚ω−ω12 cos2α
2+ω1cosα/bracketrightBig
= cos/parenleftbiggλt
2/parenrightbigg
[ω1−ω1(1 + cos α)+ω1cosα]=0./check
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258 CHAPTER 10. THE ADIABATIC APPROXIMATION
As for Eq. 10.33:
/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1−ωcosα)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
e−iωt/2/parenleftbigg
cosα
2
eiωtsinα
2/parenrightbigg
+i/bracketleftbiggω
λsinαsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
e−iωt/2/parenleftbigg
sinα
2
−eiωtcosα
2/parenrightbigg
=/parenleftbigg
α
β/parenrightbigg
,with
α=/braceleftbigg/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−iω1
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
cosα
2+iω
λ/bracketleftbigg
cosαcosα
2+ sinαsinα
2/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
cos(α−α
2)=cosα
2/bracketrightbigg
sin/parenleftbiggλt
2/parenrightbigg/bracerightbigg
e−iωt/2
=/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1−ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
cosα
2e−iωt/2(confirming the top entry).
β=/braceleftbigg/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−iω1
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
sinα
2+iω
λ/bracketleftbigg
cosαsinα
2−sinαcosα
2/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
sin(α
2−α)=−sinα
2/bracketrightbigg
sin/parenleftbiggλt
2/parenrightbigg/bracerightbigg
eiωt/2
=/bracketleftbigg
cos/parenleftbiggλt
2/parenrightbigg
−i(ω1+ω)
λsin/parenleftbiggλt
2/parenrightbigg/bracketrightbigg
sinα
2eiωt/2(confirming the bottom entry).
|c+|2+|c−|2= cos2/parenleftbiggλt
2/parenrightbigg
+(ω1−ωcosα)2
λ2sin2/parenleftbiggλt
2/parenrightbigg
+ω2
λ2sin2αsin2/parenleftbiggλt
2/parenrightbigg
= cos2/parenleftbiggλt
2/parenrightbigg
+1
λ2/parenleftbigg
ω2
1−2ωω1cosα+ω2cos2α+ω2sin2α/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
ω2+ω2
1−2ωω1cosα=λ2/parenrightbigg
sin2/parenleftbiggλt
2/parenrightbigg
= cos2/parenleftbiggλt
2/parenrightbigg
+ sin2/parenleftbiggλt
2/parenrightbigg
=1./check
Problem 10.3
(a)
ψn(x)=/radicalbigg
2
wsin/parenleftBignπ
wx/parenrightBig
.In this case R=w.
∂ψn
∂R=√
2/parenleftbigg
−1
21
w3/2/parenrightbigg
sin/parenleftBignπ
wx/parenrightBig
+/radicalbigg
2
w/parenleftBig
−nπ
w2x/parenrightBig
cos/parenleftBignπ
wx/parenrightBig
;
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CHAPTER 10. THE ADIABATIC APPROXIMATION 259
/angbracketleftbigg
ψn/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂ψ
n
∂R/angbracketrightbigg
=/integraldisplayw
0ψn∂ψn
∂Rdx
=−1
w2/integraldisplayw
0sin2/parenleftBignπ
wx/parenrightBig
dx−2nπ
w3/integraldisplayw
0xsin/parenleftBignπ
wx/parenrightBig
cos/parenleftBignπ
wx/parenrightBig
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
1
2sin(2nπ
wx)dx
=−1
w2/parenleftBigw
2/parenrightBig
−nπ
w3/integraldisplayw
0xsin/parenleftbigg2nπ
wx/parenrightbigg
dx
=−1
2w−nπ
w3/bracketleftbigg/parenleftBigw
2nπ/parenrightBig2
sin/parenleftbigg2nπ
wx/parenrightbigg
−wx
2nπcos/parenleftbigg2nπ
wx/parenrightbigg/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglew
0
=−1
2w−nπ
w3/bracketleftbigg
−w2
2nπcos(2nπ)/bracketrightbigg
=−1
2w+1
2w=0.
So Eq. 10.42 = ⇒γn(t)=0.(If the eigenfunctions are real, the geometric phase vanishes.)
(b)
Equation 10.39 = ⇒θn(t)=1
/planckover2pi1/integraldisplayt
0n2π2/planckover2pi12
2mw2dt/prime=−n2π2/planckover2pi1
2m/integraldisplay1
w2dt/prime
dwdw;
θn=−n2π2/planckover2pi1
2mv/integraldisplayw2
w11
w2dw=n2π2/planckover2pi1
2mv/parenleftbigg1
w/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglew2
w1=n2π2/planckover2pi1
2mv/parenleftbigg1
w2−1
w1/parenrightbigg
.
(c)Zero.
Problem 10.4
ψ=√mα
/planckover2pi1e−mα|x|//planckover2pi12.HereR=α,so
∂ψ
∂R=√m
/planckover2pi1/parenleftbigg1
21√α/parenrightbigg
e−mα|x|//planckover2pi12+√mα
/planckover2pi1/parenleftbigg
−m|x|
/planckover2pi12/parenrightbigg
e−mα|x|//planckover2pi12.
ψ∂ψ
∂R=√mα
/planckover2pi1/bracketleftbigg1
2/planckover2pi1/radicalbiggm
α−m√mα
/planckover2pi13|x|/bracketrightbigg
e−2mα|x|//planckover2pi12=/parenleftbiggm
2/planckover2pi12−m2α
/planckover2pi14|x|/parenrightbigg
e−2mα|x|//planckover2pi12.
/angbracketleftbigg
ψ/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂ψ
∂R/angbracketrightbigg
=2/bracketleftbiggm
2/planckover2pi12/integraldisplay∞
0e−2mαx/ /planckover2pi12dx−m2α
/planckover2pi14/integraldisplay∞
0xe−2mαx/ /planckover2pi12dx/bracketrightbigg
=m
/planckover2pi12/parenleftbigg/planckover2pi12
2mα/parenrightbigg
−2m2α
/planckover2pi14/parenleftbigg/planckover2pi12
2mα/parenrightbigg2
=1
2α−1
2α=0.So Eq. 10.42 = ⇒γ(t)=0.
E=−mα2
2/planckover2pi12,soθ(t)=−1
/planckover2pi1/integraldisplayT
0/parenleftbigg
−mα2
2/planckover2pi12/parenrightbigg
dt/prime=m
2/planckover2pi13/integraldisplayα2
α1α2dt/prime
dαdα=m
2/planckover2pi13c/integraldisplayα2
α1α2dα=m
6/planckover2pi12c/parenleftbig
α3
2−α3
1/parenrightbig
.
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260 CHAPTER 10. THE ADIABATIC APPROXIMATION
Problem 10.5
According to Eq. 10.44 the geometric phase is
γn(t)=i/integraldisplayRf
Ri/angbracketleftψn|∇Rψn/angbracketright·dR.
Now/angbracketleftψn|ψn/angbracketright=1 , s o
∇R/angbracketleftψn|ψn/angbracketright=/angbracketleft∇Rψn|ψn/angbracketright+/angbracketleftψn|∇Rψn/angbracketright=/angbracketleftψn|∇Rψn/angbracketright∗+/angbracketleftψn|∇Rψn/angbracketright=0,
and hence/angbracketleftψn|∇Rψn/angbracketrightispure imaginary .I fψnis real, then,/angbracketleftψn|∇Rψn/angbracketrightmust in fact be zero.
Suppose we introduce a phase factor to make the (originally real) wavefunction complex:
ψ/prime
n=eiφn(R)ψn,whereψnis real. Then ∇Rψ/prime
n=eiφn∇Rψn+i(∇Rφn)eiφnψn.So
/angbracketleftψ/prime
n|∇Rψ/prime
n/angbracketright=e−iφneiφn/angbracketleftψn|∇Rψn/angbracketright+ie−iφn(∇Rφn)eiφn/angbracketleftψn|ψn/angbracketright.But/angbracketleftψn|ψn/angbracketright=1,and
/angbracketleftψn|∇Rψn/angbracketright= 0 (as we just found), so /angbracketleftψ/prime
n|∇Rψ/prime
n/angbracketright=i∇Rφn,and Eq. 10.44 = ⇒
γ/prime
n(t)=i/integraldisplayRf
Rii∇R(φn)·dR=−[φn(Rf)−φn(Ri)],so Eq. 10.38 gives:
Ψ/prime
n(x,t)=ψ/prime
n(x,t)e−i
/planckover2pi1/integraltextt
0En(t/prime)dt/primee−i[φn(Rf)−φn(Ri)].
The wave function picks up a (trivial) phase factor, whose only function is precisely to kill the phase factor we
put in “by hand”:
Ψ/prime
n(x,t)=/bracketleftBig
ψn(x,t)e−i
/planckover2pi1/integraltextt
0En(t/prime)dt/prime/bracketrightBig
eiφn(Ri)=Ψ n(x,t)eiφn(Ri).
In particular, for a closed loopφn(Rf)=φn(Ri), soγ/prime
n(T)=0 .
Problem 10.6
H=e
mB·S.Here B=B0/bracketleftBig
sinθcosφˆi+ sinθsinφˆj+ cosθˆk/bracketrightBig
; take spin matrices from Problem 4.31.
H=eB0
m/planckover2pi1√
2
sinθcosφ
010
101010
+ sinθsinφ
0−i0
i0−i
0i0
+ cosθ
√
20 0
00 0
00−√
2
=eB0/planckover2pi1√
2m
√
2 cosθe−iφsinθ 0
eiφsinθ 0e−iφsinθ
0eiφsinθ−√
2 cosθ
.
We need the “spin up” eigenvector: Hχ+=eB0
m/planckover2pi1χ+.
√
2 cosθe−iφsinθ 0
eiφsinθ 0e−iφsinθ
0eiφsinθ−√
2 cosθ
a
b
c
=√
2
a
b
c
=⇒
(i)√
2 cosθa+e−iφsinθb=√
2a.
(ii)eiφsinθa+e−iφsinθc=√
2b.
(iii)eiφsinθb−√
2 cosθc=√
2c.
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CHAPTER 10. THE ADIABATIC APPROXIMATION 261
(i)⇒b=√
2eiφ/parenleftbigg1−cosθ
sinθ/parenrightbigg
a=√
2eiφtan (θ/2)a; (iii)⇒b=√
2e−iφ/parenleftbigg1 + cosθ
sinθ/parenrightbigg
c=√
2e−iφcot (θ/2)c.
Thusc=e2iφtan2(θ/2)a; (ii) is redundant. Normalize: |a|2+ 2 tan2(θ/2)|a|2+ tan4(θ/2)|a|2=1⇒
|a|2/bracketleftbig
1 + tan2(θ/2)/bracketrightbig2=|a|2/bracketleftbigg1
cos(θ/2)/bracketrightbigg4
=1⇒|a|2= cos4(θ/2).
Picka=e−iφcos2(θ/2); then b=√
2 sin(θ/2) cos(θ/2) and c=eiφsin2(θ/2),and
χ+=
e−iφcos2(θ/2)√
2 sin (θ/2) cos (θ/2)
eiφsin2(θ/2)
.This is the spin-1 analog to Eq. 10.57.
∇χ+=∂χ+
∂rˆr+1
r∂χ+
∂θˆθ+1
rsinθ∂χ+
∂φˆφ
=1
r
−e−iφcos (θ/2) sin (θ/2)√
2/bracketleftbig
cos2(θ/2)−sin2(θ/2)/bracketrightbig
/2
eiφsin (θ/2) cos (θ/2)
ˆθ+1
rsinθ
−ie−iφcos2(θ/2)
0
ieiφsin2(θ/2)
ˆφ.
/angbracketleftχ+|∇χ+/angbracketright=1
r/braceleftbig
−cos2(θ/2) [cos (θ/2) sin (θ/2)] + sin ( θ/2) cos (θ/2)/bracketleftbig
cos2(θ/2)−sin2(θ/2)/bracketrightbig
+ sin2(θ/2) [sin (θ/2) cos (θ/2)]/bracerightbigˆθ
+1
rsinθ/braceleftbig
cos2(θ/2)/bracketleftbig
−icos2(θ/2)/bracketrightbig
+ sin2(θ/2)/bracketleftbig
isin2(θ/2)/bracketrightbig/bracerightbigˆφ
=i
rsinθ/bracketleftbig
sin4(θ/2)−cos4(θ/2)/bracketrightbigˆφ
=i
rsinθ/bracketleftbig
sin2(θ/2) + cos2(θ/2)/bracketrightbig/bracketleftbig
sin2(θ/2)−cos2(θ/2)/bracketrightbigˆφ
=i
rsinθ(1)(−cosθ)ˆφ=−i
rcotθˆφ.
∇×/angbracketleftχ+|∇χ+/angbracketright=1
rsinθ∂
∂θ/bracketleftbigg
sinθ/parenleftbigg
−i
rcotθ/parenrightbigg/bracketrightbigg
ˆr=−i
r2sinθ∂
∂θ(cosθ)ˆr=isinθ
r2sinθˆr=i
r2ˆr.
Equation 10.51 = ⇒γ+(T)=i/integraldisplayi
r2r2dΩ=−Ω.
Problem 10.7
(a)GivingHa test function fto act upon:
Hf=1
2m/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg
·/parenleftbigg/planckover2pi1
i∇f−qAf/parenrightbigg
+qϕf
=1
2m/bracketleftbigg
−/planckover2pi12∇·(∇f)−q/planckover2pi1
i∇·(Af)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
(∇·A)f+A·(∇f)−q/planckover2pi1
iA·(∇f)+q2A·Af/bracketrightbigg
+qϕf.
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262 CHAPTER 10. THE ADIABATIC APPROXIMATION
But∇·A= 0 and ϕ= 0 (see comments after Eq. 10.66), so
Hf=1
2m/bracketleftbig
−/planckover2pi12∇2f+2iq/planckover2pi1A·∇f+q2A2f/bracketrightbig
,orH=1
2m/bracketleftbig
−/planckover2pi12∇2+q2A2+2iq/planckover2pi1A·∇/bracketrightbig
.QED
(b)Apply/parenleftbig/planckover2pi1
i∇−qA/parenrightbig
·to both sides of Eq. 10.78:
/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg2
Ψ=/parenleftbigg/planckover2pi1
i∇−qA/parenrightbigg
·/parenleftbigg/planckover2pi1
ieig∇Ψ/prime/parenrightbigg
=−/planckover2pi12∇·(eig∇Ψ/prime)−q/planckover2pi1
ieigA·∇Ψ/prime.
But∇·(eig∇Ψ/prime)=ieig(∇g)·(∇Ψ/prime)+eig∇·(∇Ψ/prime) and∇g=q
/planckover2pi1A, so the right side is
−i/planckover2pi12q
/planckover2pi1eigA·∇Ψ/prime−/planckover2pi12eig∇2Ψ/prime+iq/planckover2pi1eigA·∇Ψ/prime=−/planckover2pi12eig∇2Ψ/prime.QED
Problem 10.8
(a)Schr¨odinger equation:
−/planckover2pi12
2md2ψ
dx2=Eψ, ord2ψ
dx2=−k2ψ(k≡√
2mE/ /planckover2pi1)/braceleftBigg
0<x<1
2a+FepsilonC,
1
2a+FepsilonC<x<a .
Boundary conditions: ψ(0) =ψ(1
2a+FepsilonC)=ψ(a)=0.
Solution:(1) 0<x<
1
2a+FepsilonC:ψ(x)=Asinkx+Bcoskx.Butψ(0) = 0⇒B=0,and
ψ(1
2a+FepsilonC)=0⇒/braceleftBigg
k(1
2a+FepsilonC)=nπ(n=1,2,3,...)⇒En=n2π2/planckover2pi12/2m(a/2+FepsilonC)2,
or else A=0.
(2)1
2a+FepsilonC<x<a :ψ(x)=Fsink(a−x)+Gcosk(a−x).Butψ(a)=0⇒G=0,and
ψ(1
2a+FepsilonC)=0⇒/braceleftBigg
k(1
2a−FepsilonC)=n/primeπ(n/prime=1,2,3,...)⇒En/prime=(n/prime)2π2/planckover2pi12/2m(a/2−FepsilonC)2,
or else F=0.
The ground state energy is
either E
1=π2/planckover2pi12
2m(1
2a+FepsilonC)2(n=1 ),withF=0,
or else E1/prime=π2/planckover2pi12
2m(1
2a−FepsilonC)2(n/prime=1 ),withA=0.
Both are allowed energies, but E1is (slightly) lower (assuming FepsilonCis positive), so the ground state is
ψ(x)=
/radicalBig
2
1
2a+ρepsilonosin/parenleftBig
πx
1
2a+ρepsilono/parenrightBig
,0≤x≤1
2a+FepsilonC;
0,1
2a+FepsilonC≤x≤a.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
CHAPTER 10. THE ADIABATIC APPROXIMATION 263
xψ(x)
a a_
2_
2a+ε
(b)
−/planckover2pi12
2md2ψ
dx2+f(t)δ(x−1
2a−FepsilonC)ψ=Eψ⇒ψ(x)=/braceleftbiggAsinkx, 0≤x<1
2a+FepsilonC,
Fsink(a−x),1
2a+FepsilonC<x≤a,/bracerightbigg
where k≡√
2mE
/planckover2pi1.
Continuity in ψatx=1
2a+FepsilonC:
Asink/parenleftbig1
2a+FepsilonC/parenrightbig
=Fsink/parenleftbig
a−1
2a−FepsilonC/parenrightbig
=Fsink/parenleftbig1
2a−FepsilonC/parenrightbig
⇒F=Asink/parenleftbig1
2a+FepsilonC/parenrightbig
sink/parenleftbig1
2a−FepsilonC/parenrightbig.
Discontinuity in ψ/primeatx=1
2a+FepsilonC(Eq. 2.125):
−Fkcosk(a−x)−Akcoskx=2mf
/planckover2pi12Asinkx⇒Fcosk/parenleftbig1
2a−FepsilonC/parenrightbig
+Acosk/parenleftbig1
2a+FepsilonC/parenrightbig
=−/parenleftbigg2mf
/planckover2pi12k/parenrightbigg
Asink/parenleftbig1
2a+FepsilonC/parenrightbig
.
Asink/parenleftbig1
2a+FepsilonC/parenrightbig
sink/parenleftbig1
2a−FepsilonC/parenrightbigcosk/parenleftbig1
2a−FepsilonC/parenrightbig
+Acosk/parenleftbig1
2a+FepsilonC/parenrightbig
=−/parenleftbigg2T
z/parenrightbigg
Asink/parenleftbig1
2a+FepsilonC/parenrightbig
.
sink/parenleftbig1
2a+FepsilonC/parenrightbig
cosk/parenleftbig1
2a−FepsilonC/parenrightbig
+ cosk/parenleftbig1
2a+FepsilonC/parenrightbig
sink/parenleftbig1
2a−FepsilonC/parenrightbig
=−/parenleftbigg2T
z/parenrightbigg
sink/parenleftbig1
2a+FepsilonC/parenrightbig
sink/parenleftbig1
2a−FepsilonC/parenrightbig
.
sink/parenleftbig1
2a+FepsilonC+1
2a−FepsilonC/parenrightbig
=−/parenleftbigg2T
z/parenrightbigg1
2/bracketleftbig
cosk/parenleftbig1
2a+FepsilonC−1
2a+FepsilonC/parenrightbig
−cosk/parenleftbig1
2a+FepsilonC+1
2a−FepsilonC/parenrightbig/bracketrightbig
.
sinka=−T
z(cos 2kFepsilonC−coska)⇒zsinz=T[cosz−cos(zδ)].
(c)
sinz=T
z(cosz−1)⇒z
T=cosz−1
sinz=−tan(z/2)⇒ tan(z/2) =−z
T.
Plot tan( z/2) and−z/Ton the same graph, and look for intersections:
tan(z/2)
πz2π 3π
-z/T
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264 CHAPTER 10. THE ADIABATIC APPROXIMATION
Ast:0→∞,T:0→∞, and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock,
so the smallest zgoes from πto 2π, and the ground state energy goes from ka=π⇒E(0) =/planckover2pi12π2
2ma2
(appropriate to a well of width a)t oka=2π⇒E(∞)=/planckover2pi12π2
2m(a/2)2(appropriate for a well of width a/2.
(d)Mathematica yields the following table:T 0 1 5 20 100 1000
z3.14159 3.67303 4.76031 5.72036 6.13523 6.21452
(e)Pr=Ir
Ir+Il=1
1+(Il/Ir), where
Il=/integraldisplaya/2+ρepsilono
0A2sin2kxdx =A2/bracketleftbigg1
2x−1
4ksin(2kx)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea/2+ρepsilono
0
=A2/braceleftbigg1
2/parenleftBiga
2+FepsilonC/parenrightBig
−1
4ksin/bracketleftBig
2k/parenleftBiga
2+FepsilonC/parenrightBig/bracketrightBig/bracerightbigg
=a
4A2/bracketleftbigg
1+2FepsilonC
a−1
kasin/parenleftbigg
ka+2FepsilonC
aka/parenrightbigg/bracketrightbigg
=a
4A2/bracketleftbigg
1+δ−1
zsin(z+zδ)/bracketrightbigg
.
Ir=/integraldisplaya
a/2+ρepsilonoF2sin2k(a−x)dx.Letu≡a−x, du =−dx.
=−F2/integraldisplay0
a/2−ρepsilonosin2kudu =F2/integraldisplaya/2−ρepsilono
0sin2kudu =a
4F2/bracketleftbigg
1−δ−1
zsin(z−zδ)/bracketrightbigg
.
Il
Ir=A2[1 +δ−(1/z) sin(z+zδ)]
F2[1−δ−(1/z) sin(z−zδ)].But (from (b))A2
F2=sin2k(a/2−FepsilonC)
sin2k(a/2+FepsilonC)=sin2[z(1−δ)/2]
sin2[z(1 +δ)/2].
=I+
I−,where I±≡/bracketleftbigg
1±δ−1
zsinz(1±δ)/bracketrightbigg
sin2[z(1∓δ)/2].Pr=1
1+(I+/I−).
Usingδ=0.01 and the z’s from (d), Mathematica gives
T 0 1 5 20 100 1000
Pr0.490001 0.486822 0.471116 0.401313 0.146529 0.00248443
Ast:0→∞ (soT:0→∞), the probability of being in the right half drops from almost 1/2 to zero—the
particle gets sucked out of the slightly smaller side, as it heads for the ground state in (a).
(f)
T=0 T=1 T=5
T=20 T=100 T=1000
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CHAPTER 10. THE ADIABATIC APPROXIMATION 265
Problem 10.9
(a)Check the answer given: xc=ω/integraltextt
0f(t/prime) sin [ω(t−t/prime)]dt/prime=⇒xc(0) = 0./check
˙xc=ωf(t) sin [ω(t−t)] +ω2/integraldisplayt
0f(t/prime) cos [ω(t−t/prime)]dt/prime=ω2/integraldisplayt
0f(t/prime) cos [ω(t−t/prime)]dt/prime⇒˙xc(0) = 0./check
¨xc=ω2f(t) cos [ω(t−t)]−ω3/integraldisplayt
0f(t/prime) sin [ω(t−t/prime)]dt/prime=ω2f(t)−ω2xc.
Now the classical equation of motion is m(d2x/dt2)=−mω2x+mω2f. For the proposed solution,
m(d2xc/dt2)=mω2f−mω2xc,s o i t doessatisfy the equation of motion, with the appropriate boundary
conditions.
(b)Letz≡x−xc(soψn(x−xc)=ψn(z), and zdepends on tas well as x).
∂Ψ
∂t=dψn
dz(−˙xc)ei{}+ψnei{}i
/planckover2pi1/bracketleftbigg
−(n+1
2)/planckover2pi1ω+m¨xc(x−xc
2)−m
2˙x2
c+mω2
2fxc/bracketrightbigg
[]=−(n+1
2)/planckover2pi1ω+mω2
2/bracketleftbigg
2x(f−xc)+x2
c−˙x2
c
ω2/bracketrightbigg
.
∂Ψ
∂t=−˙xcdψn
dzei{}+iΨ/braceleftbigg
−(n+1
2)/planckover2pi1ω+mω2
2/planckover2pi1/bracketleftbigg
2x(f−xc)+x2
c−˙x2
c
ω2/bracketrightbigg/bracerightbigg
.
∂Ψ
∂x=dψn
dzei{}+ψnei{}i
/planckover2pi1(m˙xc);∂2Ψ
∂x2=d2ψn
dz2ei{}+2dψn
dzei{}i
/planckover2pi1(m˙xc)−/parenleftbiggm˙xc
/planckover2pi1/parenrightbigg2
ψnei{}.
HΨ=−/planckover2pi12
2m∂2Ψ
∂x2+1
2mω2x2Ψ−mω2fxΨ
=−/planckover2pi12
2md2ψn
dz2ei{}−/planckover2pi12
2m2dψn
dzei{}im˙xc
/planckover2pi1+/planckover2pi12
2m/parenleftbiggm˙xc
/planckover2pi1/parenrightbigg2
Ψ+1
2mω2x2Ψ−mω2fxΨ.
But−/planckover2pi12
2md2ψn
dz2+1
2mω2z2ψn=(n+1
2)/planckover2pi1ωψn,s o
HΨ=
✟✟✟✟✟
(n+1
2)/planckover2pi1ωΨ−1
2mω2z2Ψ−✟✟✟✟✟
i/planckover2pi1˙xcdΨn
dzei{}+m
2˙x2
cΨ+1
2mω2x2Ψ−mω2fxΨ
?=i/planckover2pi1∂Ψ
∂t=
✟✟✟✟✟
−i/planckover2pi1˙xcdψn
dzei{}−/planckover2pi1Ψ/bracketleftbigg
✟✟✟✟✟
−(n+1
2)ω+mω2
2/planckover2pi1(2xf−2xxc+x2
c−1
ω2˙x2
c)/bracketrightbigg
−1
2mω2z2+
m
2˙x2
c+1
2mω2x2−✘✘✘✘mω2fx?=−mω2
2/parenleftBigg
✟✟2xf−2xxc+x2
c−
1
ω2˙x2
c/parenrightBigg
z2−x2?=−2xxc+x2
c;z2?=(x2−2xxc+x2
c)=(x−xc)2./check
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266 CHAPTER 10. THE ADIABATIC APPROXIMATION
(c)
Eq. 10.90⇒H=−/planckover2pi12
2m∂2
∂x2+1
2mω2/parenleftbig
x2−2xf+f2/parenrightbig
−1
2mω2f2.Shift origin: u≡x−f.
H=/bracketleftbigg
−/planckover2pi12
2m∂2
∂u2+1
2mω2u2/bracketrightbigg
−/bracketleftbigg1
2mω2f2/bracketrightbigg
.
The first term is a simple harmonic oscillator in the variable u; the second is a constant (with respect
to position). So the eigenfunctions are ψn(u), and the eigenvalues are harmonic oscillator ones, ( n+
1
2)/planckover2pi1ω, less the constant: En=(n+1
2)/planckover2pi1ω−1
2mω2f2.
(d)Note that sin [ ω(t−t/prime)] =1
ωd
dt/primecos [ω(t−t/prime)], so xc(t)=/integraldisplayt
0f(t/prime)d
dt/primecos [ω(t−t/prime)]dt/prime,o r
xc(t)=f(t/prime) cos [ω(t−t/prime)]/vextendsingle/vextendsingle/vextendsinglet
0−/integraldisplayt
0/parenleftbiggdf
dt/prime/parenrightbigg
cos [ω(t−t/prime)]dt/prime=f(t)−/integraldisplayt
0/parenleftbiggdf
dt/prime/parenrightbigg
cos [ω(t−t/prime)]dt/prime
(sincef(0) = 0). Now, for an adiabatic process we want df/dt very small; specifically:df
dt/prime/lessmuchωf(t)
(0<t/prime≤t). Then the integral is negligible compared to f(t), and we have xc(t)≈f(t).(Physically,
this says that if you pull on the spring very gently, no fancy oscillations will occur; the mass just moves
along as though attached to a string of fixed length.)
(e)Putxc≈finto Eq. 10.92, using Eq. 10.93:
Ψ(x,t)=ψn(x,t)ei
/planckover2pi1/bracketleftBig
−(n+1
2)/planckover2pi1ωt+m˙f(x−f/2)+mω2
2/integraltextt
0f2(t/prime)dt/prime/bracketrightBig
.
The dynamic phase (Eq. 10.39) is
θn(t)=−1
/planckover2pi1/integraldisplayt
0En(t/prime)dt/prime=−(n+1
2)/planckover2pi1ωt+mω2
2/planckover2pi1/integraldisplayt
0f2(t/prime)dt/prime,so Ψ( x,t)=ψn(x,t)eiθn(t)eiγn(t),
confirming Eq. 10.94, with the geometric phase given (ostensibly) by γn(t)=m
/planckover2pi1˙f(x−f/2). But the
eigenfunctions here are real, and hence(Problem 10.5) the geometric phase should be zero. The point is that
(in the adiabatic approximation) ˙fis extremely small (see above), and hence in this limitm
/planckover2pi1˙f(x−f/2)≈0
(at least, in the only region of xwhereψn(x,t) is nonzero).
Problem 10.10
(a)
˙cm=−/summationdisplay
jδjneiγn/angbracketleftψm|˙ψj/angbracketrightei(θj−θm)=−/angbracketleftψm|∂ψn
∂t/angbracketrighteiγnei(θn−θm)⇒
cm(t)=cm(0)−/integraldisplayt
0/angbracketleftψm|∂ψn
∂t/prime/angbracketrighteiγnei(θn−θm)dt/prime.
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CHAPTER 10. THE ADIABATIC APPROXIMATION 267
(b)From Problem 10.9:
ψn(x,t)=ψn(x−f)=ψn(u),where u≡x−f,
andψn(u) is the nth state of the ordinary harmonic oscillator;∂ψn
∂t=∂ψn
∂u∂u
∂t=−˙f∂ψn
∂u.
But ˆp=/planckover2pi1
i∂
∂u,so/angbracketleftψm|∂ψn
∂t/angbracketright=−i
/planckover2pi1˙f/angbracketleftm|p|n/angbracketright,where (from Problem 3.33):
/angbracketleftm|p|n/angbracketright=i/radicalbigg
m/planckover2pi1ω
2/parenleftbig√mδn,m−1−√nδm,n−1/parenrightbig
.Thus:
/angbracketleftψm|∂ψn
∂t/angbracketright=˙f/radicalbiggmω
2/planckover2pi1/parenleftbig√mδn,m−1−√nδm,n−1/parenrightbig
.
Evidently transitions occur only to the immediately adjacent states, n±1, and
(1)m=n+1:
cn+1=−/integraldisplayt
0/parenleftbigg
˙f/radicalbiggmω
2/planckover2pi1√
n+1/parenrightbigg
eiγnei(θn−θn+1)dt/prime.
Butγn=0,because the eigenfunctions are real (Problem 10.5), and (Eq. 10.39)
θn=−1
/planckover2pi1(n+1
2)/planckover2pi1ωt=⇒θn−θn+1=/bracketleftbigg
−(n+1
2)+(n+1+1
2)/bracketrightbigg
ωt=ωt.
Socn+1=−/radicalbiggmω
2/planckover2pi1√
n+1/integraldisplayt
0˙feiωt/primedt/prime.
(2)m=n−1:
cn−1=−/integraldisplayt
0/parenleftbigg
−˙f/radicalbiggmω
2/planckover2pi1√n/parenrightbigg
eiγnei(θn−θn−1)dt/prime;
θn−θn−1=/bracketleftbigg
−(n+1
2)+(n−1+1
2)/bracketrightbigg
ωt=−ωt.cn−1=/radicalbiggmω
2/planckover2pi1√n/integraldisplayt
0˙fe−iωt/primedt/prime.
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268 CHAPTER 11. SCATTERING
Chapter 11
Scattering
Problem 11.1
(a)
qq
b r1
2φθ
Conservation of energy: E=1
2m(˙r+r2˙φ2)+V(r),where V(r)=q1q2
4πFepsilonC1
r.
Conservation of angular momentum: J=mr2˙φ.So ˙φ=J
mr2.
˙r2+J2
m2r2=2
m(E−V).We want ras a function of φ(nott). Also, let u≡1/r. Then
˙r=dr
dt=dr
dudu
dφdφ
dt=/parenleftbigg
−1
u2/parenrightbiggdu
dφJ
mu2=−J
mdu
dφ.Then:/parenleftbigg
−J
mdu
dφ/parenrightbigg2
+J2
m2u2=2
m(E−V),or
/parenleftbiggdu
dφ/parenrightbigg2
=2m
J2(E−V)−u2;du
dφ=/radicalbigg
2m
J2(E−V)−u2;dφ=du/radicalBig
2m
J2(E−V)−u2=du/radicalbig
I(u),where
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CHAPTER 11. SCATTERING 269
I(u)≡2m
J2(E−V)−u2.Now, the particle q1starts out at r=∞(u= 0),φ= 0, and the point
of closest approach is rmin(umax),Φ: Φ=/integraldisplayumax
0du√
I.It now swings through an equal angle Φ
on the way out,s o Φ+Φ+ θ=π,orθ=π−2Φ.θ=π−2/integraldisplayumax
0du/radicalbig
I(u).
So far this is general ; now we put in the specific potential:
I(u)=2mE
J2−2m
J2q1q2
4πFepsilonC0u−u2=(u2−u)(u−u1),whereu1andu2are the two roots .
(Since du/dφ =/radicalbig
I(u),umaxis one of the roots; setting u2>u1,umax=u2.)
θ=π−2/integraldisplayu2
0du/radicalbig
(u2−u)(u−u1)=π+ 2 sin−1/parenleftbigg−2u+u1+u2
u2−u1/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleu2
0
=π+2/bracketleftbigg
sin−1(−1)−sin−1/parenleftbiggu1+u2
u2−u1/parenrightbigg/bracketrightbigg
=π+2/bracketleftbigg
−π
2−sin−1/parenleftbiggu1+u2
u2−u1/parenrightbigg/bracketrightbigg
=−2 sin−1/parenleftbiggu1+u2
u2−u1/parenrightbigg
.
NowJ=mvb,E=1
2mv2, where vis the incoming velocity, so J2=m2b2(2E/m)=2mb2E, and hence
2m/J2=1/b2E.S o
I(u)=1
b2−1
b2/parenleftbigg1
Eq1q2
4πFepsilonC0/parenrightbigg
u−u2.LetA≡q1q2
4πFepsilonC0E,so−I(u)=u2+A
b2u−1
b2.
To get the roots: u2+A
b2u−1
b2=0=⇒u=1
2/bracketleftBigg
−A
b2±/radicalbigg
A2
b4+4
b2/bracketrightBigg
=A
2b2
−1±/radicalBigg
1+/parenleftbigg2b
A/parenrightbigg2
.
Thus u2=A
2b2
−1+/radicalBigg
1+/parenleftbigg2b
A/parenrightbigg2
,u1=A
2b2
−1−/radicalBigg
1+/parenleftbigg2b
A/parenrightbigg2
;u1+u2
u2−u1=−1/radicalBig
1+( 2b/A)2.
θ= 2 sin−1
1/radicalBig
1+( 2b/A)2
,or1/radicalBig
1+( 2b/A)2= sin/parenleftbiggθ
2/parenrightbigg
;1 +/parenleftbigg2b
A/parenrightbigg2
=1
sin2(θ/2);
/parenleftbigg2b
A/parenrightbigg2
=1−sin2(θ/2)
sin2(θ/2)=cos2(θ/2)
sin2(θ/2);2b
A= cot(θ/2),orb=q1q2
8πFepsilonC0Ecot(θ/2).
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270 CHAPTER 11. SCATTERING
(b)
D(θ)=b
sinθ/vextendsingle/vextendsingle/vextendsingle/vextendsingledb
dθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle.Heredb
dθ=q1q2
8πFepsilonC0E/parenleftbigg
−1
2 sin2(θ/2)/parenrightbigg
.
=1
2 sin(θ/2) cos(θ/2)q1q2
8πFepsilonC0Ecos(θ/2)
sin(θ/2)q1q2
8πFepsilonC0E1
2 sin2(θ/2)=/bracketleftbiggq1q2
16πFepsilonC0Esin2(θ/2)/bracketrightbigg2
.
(c)
σ=/integraldisplay
D(θ) sinθdθdφ =2π/parenleftbiggq1q2
8πFepsilonC0E/parenrightbigg2/integraldisplayπ
0sinθ
sin4(θ/2)dθ.
This integral does not converge, for near θ= 0 (and again near π) we have sin θ≈θ, sin(θ/2)≈θ/2, so
the integral goes like 16/integraltextρepsilono
0θ−3dθ=−8θ−2/vextendsingle/vextendsingleρepsilono
0→∞.
Problem 11.2
xr
θ
Two dimensions: ψ(r,θ)≈A/bracketleftbigg
eikx+f(θ)eikr
√r/bracketrightbigg
.
One dimension: ψ(x)≈A/bracketleftbig
eikx+f(x/|x|)e−ikx/bracketrightbig
.
Problem 11.3
Multiply Eq. 11.32 by Pl/prime(cosθ) sinθdθ and integrate from 0 to π, exploiting the orthogonality of the Leg-
endre polynomials (Eq. 4.34)—which, with the change of variables x≡cosθ,s a y s
/integraldisplayπ
0Pl(cosθ)Pl/prime(cosθ) sinθdθ=/parenleftbigg2
2l+1/parenrightbigg
δll/prime.
The delta function collapses the sum, and we get
2il/prime/bracketleftBig
jl/prime(ka)+ikal/primeh(1)
l/prime(ka)/bracketrightBig
=0,
and hence (dropping the primes)
al=−jl(ka)
ikh(1)
l(ka).QED
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CHAPTER 11. SCATTERING 271
Problem 11.4
Keeping only the l= 0 terms, Eq. 11.29 says that in the exterior region:
ψ≈A/bracketleftBig
j0(kr)+ika0h(1)
0(kr)/bracketrightBig
P0(cosθ)=A/bracketleftbiggsin(kr)
kr+ika0/parenleftbigg
−ieikr
kr/parenrightbigg/bracketrightbigg
=A/bracketleftbiggsin(kr)
kr+a0eikr
r/bracketrightbigg
(r>a).
In the internal region Eq. 11.18 (with nleliminated because it blows up at the origin) yields
ψ(r)≈bj0(kr)=bsin(kr)
kr(r<a).
The boundary conditions hold independently for each l, as you can check by keeping the summation over land
exploiting the orthogonality of the Legendre polynomials:
(1)ψcontinuous at r=a:A/bracketleftbiggsinka
ka+a0eika
a/bracketrightbigg
=bsinka
ka.
(2)ψ/primediscontinuous at r=a: Integrating the radial equation across the delta function gives
−/planckover2pi12
2m/integraldisplayd2u
dr2dr+/integraldisplay/bracketleftbigg
αδ(r−a)+/planckover2pi12
2ml(l+1 )
r2/bracketrightbigg
udr⇒−/planckover2pi12
2m∆u/prime+αu(a)=0,or ∆u/prime=2mα
/planckover2pi12u(a).
Nowu=rR, sou/prime=R+rR/prime;∆u/prime=∆R+a∆R/prime=a∆R/prime=2mα
/planckover2pi12aR(a),or ∆ψ/prime=2mα
/planckover2pi12ψ(a)=β
aψ(a).
A
ka/bracketleftbig
kcos(ka)+a0ik2eika/bracketrightbig
−A
ka2✭✭✭✭✭✭✭✭✭ /bracketleftbig
sin(ka)+a0keika/bracketrightbig
−b
kakcos(ka)+
✟✟✟✟✟b
ka2sinka=β
absin(ka)
ka.
The indicated terms cancel (by (1)), leaving A/bracketleftbig
cos(ka)+ia0keika/bracketrightbig
=b/bracketleftbigg
cos(ka)+β
kasin(ka)/bracketrightbigg
.
Using(1)to eliminate b:A/bracketleftbig
cos(ka)+ia0keika/bracketrightbig
=/bracketleftbigg
cot(ka)+β
ka/bracketrightbigg/bracketleftbig
sin(ka)+a0keika/bracketrightbig
A.
✘✘✘✘cos(ka)+ia0keika=✘✘✘✘cos(ka)+β
kasin(ka)+a0kcot(ka)eika+βa0
aeika.
ia0keika/bracketleftbigg
1+icot(ka)+iβ
ka/bracketrightbigg
=β
kasin(ka).Butka/lessmuch1,so sin(ka)≈ka,and cot( ka)=cos(ka)
sin(ka)≈1
ka.
ia0k(1 +ika)/bracketleftbigg
1+i
ka(1 +β)/bracketrightbigg
=β;ia0k/bracketleftbigg
1+i
ka(1 +β)+ika−1−β/bracketrightbigg
≈ia0k/bracketleftbiggi
ka(1 +β)/bracketrightbigg
=β.
a0=−aβ
1+β.Equation 11.25 ⇒f(θ)≈a0=−aβ
1+β.Equation 11.14 ⇒D=|f|2=/parenleftbiggaβ
1+β/parenrightbigg2
.
Equation 11.27 ⇒σ=4πD=4π/parenleftbiggaβ
1+β/parenrightbigg2
.
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272 CHAPTER 11. SCATTERING
Problem 11.5
(a)In the region to the left
ψ(x)=Aeikx+B−ikx(x≤−a).
In the region −a<x< 0, the Schr¨ odinger equation gives
−h2
2md2ψ
dx2−V0ψ=Eψ⇒d2ψ
dx2=−k/primeψ
wherek/prime=/radicalbig
2m(E+V0)//planckover2pi1. The general solution is
ψ=Csin(k/primex)+Dcos(k/primex)
Butψ(0) = 0 implies D=0 ,s o
ψ(x)=Csin(k/primex)(−a≤x≤0).
The continuity of ψ(x) andψ/prime(x)a tx=−asays
Ae−ika+Beika=−Csin(k/primea),i k A e−ika−ikBika=k/primeCcos(k/primea).
Divide and solve for B:
ikAe−ika−ikBeika
Ae−ika+Beika=−k/primecot(k/primea),
ikAe−ika−ikBeika=−Ae−ikak/primecot(k/primea)−Beikak/primecot(k/primea),
Beika[−ik+k/primecot(k/primea)] =Ae−ika[−ik−k/primecot(k/primea)].
B=Ae−2ika/bracketleftbiggk−ik/primecot(k/primea)
k+ik/primecot(k/primea)/bracketrightbigg
.
(b)
|B|2=|A|2/bracketleftbiggk−ik/primecot(k/primea)
k+ik/primecot(k/primea)/bracketrightbigg
·/bracketleftbiggk+ik/primecot(k/primea)
k−ik/primecot(k/primea)/bracketrightbigg
=|A|2./check
(c)From part (a) the wavefunction for x<−ais
ψ(x)=Aeikx+Ae−2ika/bracketleftbiggk−ik/primecot(k/primea)
k+ik/primecot(k/primea)/bracketrightbigg
e−ikx.
But by definition of the phase shift (Eq. 11.40)
ψ(x)=A/bracketleftBig
eikx−ei(2δ−kx)/bracketrightBig
.
so
e−2ika/bracketleftbiggk−ik/primecot(k/primea)
k+ik/primecot(k/primea)/bracketrightbigg
=−e2iδ.
This is exact. For a very deep well, E/lessmuchV0,k=√
2mE/ /planckover2pi1/lessmuch/radicalbig
2m(E+V0)//planckover2pi1=k/prime,s o
e−2ika/bracketleftbigg−ik/primecot(k/primea)
ik/primecot(k/primea)/bracketrightbigg
=−e2iδ;e−2ika=e2iδ;δ=−ka.
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CHAPTER 11. SCATTERING 273
Problem 11.6
From Eq. 11.46, al=1
keiδlsinδl, and Eq. 11.33, al=ijl(ka)
kh(1)
l(ka), it follows that eiδlsinδl=ijl(ka)
h(1)
l(ka).
But (Eq. 11.19) h(1)
l(x)=jl(x)+inl(x), so
eiδlsinδl=ijl(ka)
jl(x)+inl(x)=i1
1+i(n/j)=i1−i(n/j)
1+(n/j)2=(n/j)+i
1+(n/j)2,
(writing ( n/j) as shorthand for nl(ka)/jl(ka)). Equating the real and imaginary parts:
cosδlsinδl=(n/j)
1+(n/j)2; sin2δl=1
1+(n/j)2.
Dividing the second by the first, I conclude that
tanδl=1
(n/j),orδl= tan−1/bracketleftbiggjl(ka)
nl(ka)/bracketrightbigg
.
Problem 11.7
r>a :u(r)=Asin(kr+δ);
r<a :u(r)=Bsinkr+Dcoskr=Bsinkr,because u(0) = 0 =⇒D=0.
Continuity at r=a=⇒Bsin(ka)=Asin(ka+δ)=⇒B=Asin(ka+δ)
sin(ka).Sou(r)=Asin(ka+δ)
sin(ka)sinkr.
From Problem 11.4,
∆/parenleftbiggdu
dr/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle
r=a=β
au(a)⇒Akcos(ka+δ)−Asin(ka+δ)
sin(ka)kcos(ka)=β
aAsin(ka+δ).
cos(ka+δ)−sin(ka+δ)
sin(ka)cos(ka)=β
kasin(ka+δ),
sin(ka) cos(ka+δ)−sin(ka+δ) cos(ka)=β
kasin(ka+δ) sin(ka),
sin(ka−ka−δ)=β
kasin(ka) [sin(ka) cosδ+ cos(ka) sinδ],
−sinδ=βsin2(ka)
ka[cosδ+ cot(ka) sinδ];−1=βsin2(ka)
ka[cotδ+ cot(ka)].
cotδ=−cot(ka)−ka
βsin2(ka);cotδ=−/bracketleftbigg
cot(ka)+ka
βsin2(ka)/bracketrightbigg
.
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274 CHAPTER 11. SCATTERING
Problem 11.8
G=−eikr
4πr=⇒∇G=−1
4π/parenleftbigg1
r∇eikr+eikr∇1
r/parenrightbigg
=⇒
∇2G=∇·(∇G)=−1
4π/bracketleftbigg
2/parenleftbigg
∇1
r/parenrightbigg
·(∇eikr)+1
r∇2(eikr)+eikr∇2/parenleftbigg1
r/parenrightbigg/bracketrightbigg
.
But∇1
r=−1
r2ˆr;∇(eikr)=ikeikrˆr;∇2eikr=ik∇·(eikrˆr)=ik1
r2d
dr(r2eikr)
(see reference in footnote 12) = ⇒∇2eikr=ik
r2(2reikr+ikr2eikr)=ikeikr/parenleftbigg2
r+ik/parenrightbigg
;
∇2/parenleftbigg1
r/parenrightbigg
=−4πδ3(r).So∇2G=−1
4π/bracketleftbigg
2/parenleftbigg
−1
r2ˆr/parenrightbigg
·/parenleftbig
ikeikrˆr/parenrightbig
+1
rikeikr/parenleftbigg2
r+ik/parenrightbigg
−4πeikrδ3(r)/bracketrightbigg
.
Buteikrδ3(r)=δ3(r),so
∇2G=δ3(r)−1
4πeikr/bracketleftbigg
−2ik
r2+2ik
r2−k2
r/bracketrightbigg
=δ3(r)+k2eikr
4πr=δ3(r)−k2G.
Therefore (∇2+k2)G=δ3(r).QED
Problem 11.9
ψ=1√
πa3e−r/a;V=−e2
4πFepsilonC0r=−/planckover2pi12
ma1
r(Eq. 4.72); k=i√
−2mE
/planckover2pi1=i
a.
In this case there is no “incoming” wave, and ψ0(r) = 0. Our problem is to show that
−m
2π/planckover2pi12/integraldisplayeik|r−r0|
|r−r0|V(r0)ψ(r0)d3r0=ψ(r).
We proceed to evaluate the left side (call it I):
I=/parenleftBig
−m
2π/planckover2pi12/parenrightBig/parenleftbigg
−/planckover2pi12
ma/parenrightbigg1√
πa3/integraldisplaye−|r−r0|/a
|r−r0|1
r0e−r0/ad3r0
=1
2πa1√
πa3/integraldisplaye−√
r2+r2
0−2rr0cosθ/ae−r0/a
/radicalbig
r2+r2
0−2rr0cosθr0r2
0sinθdr0dθdφ.
(I have set the z0axis along the—fixed—direction r, for convenience.) Doing the φintegral (2 π):
I=1
a√
πa3/integraldisplay∞
0r0e−r0/a/bracketleftBigg/integraldisplayπ
0e−√
r2+r2
0−2rr0cosθ/a
/radicalbig
r2+r2
0−2rr0cosθsinθdθ/bracketrightBigg
dr0.Theθintegral is
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CHAPTER 11. SCATTERING 275
/integraldisplayπ
0e−√
r2+r2
0−2rr0cosθ/a
/radicalbig
r2+r2
0−2rr0cosθsinθdθ=−a
rr0e−√
r2+r2
0−2rr0cosθ/a/vextendsingle/vextendsingle/vextendsingleπ
0=−a
rr0/bracketleftBig
e−(r+r0)/a−e−|r−r0|/a/bracketrightBig
.
I=−1
r√
πa3/integraldisplay∞
0e−r0/a/bracketleftBig
e−(r0+r)/a−e−|r0−r|/a/bracketrightBig
dr0
=−1
r√
πa3/bracketleftbigg
e−r/a/integraldisplay∞
0e−2r0/adr0−e−r/a/integraldisplayr
0dr−er/a/integraldisplay∞
re−2r0/adr0/bracketrightbigg
=−1
r√
πa3/bracketleftBig
e−r/a/parenleftBiga
2/parenrightBig
−e−r/a(r)−er/a/parenleftBig
−a
2e−2r0/a/parenrightBig/vextendsingle/vextendsingle/vextendsingle∞
r/bracketrightBig
=−1
r√
πa3/bracketleftBiga
2e−r/a−re−r/a−a
2er/ae−2r/a/bracketrightBig
=1√
πa3e−r/a=ψ(r).QED
Problem 11.10
For the potential in Eq. 11.81, Eq. 11.88 = ⇒
f(θ)=−2m
/planckover2pi12κV0/integraldisplaya
0rsin(κr)dr=−2mV0
/planckover2pi12κ/bracketleftbigg1
κ2sin(κr)−r
κcos(κr)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
0=−2mV0
/planckover2pi12κ3[sin(κa)−κacos(κa)],
where (Eq. 11.89) κ=2ksin(θ/2). For low-energy scattering ( ka/lessmuch1):
sin(κa)≈κa−1
3!(κa)3; cos(κa)=1−1
2(κa)2;s o
f(θ)≈−2mV0
/planckover2pi12κ3/bracketleftbigg
κa−1
6(κa)3−κa+1
2(κa)3/bracketrightbigg
=−2
3mV0a3
/planckover2pi12,in agreement with Eq. 11.82.
Problem 11.11
sin(κr)=1
2i/parenleftbig
eiκr−e−iκr/parenrightbig
,so/integraldisplay∞
0e−µrsin(κr)dr=1
2i/integraldisplay∞
0/bracketleftBig
e−(µ−iκ)r−e−(µ+iκ)r/bracketrightBig
dr
=1
2i/bracketleftbigge−(µ−iκ)r
−(µ−iκ)−e−(µ+iκ)r
−(µ+iκ)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0=1
2i/bracketleftbigg1
µ−iκ−1
µ+iκ/bracketrightbigg
=1
2i/parenleftbiggµ+iκ−µ+iκ
µ2+κ2/parenrightbigg
=κ
µ2+κ2.
Sof(θ)=−2mβ
/planckover2pi12κκ
µ2+κ2=−2mβ
/planckover2pi12(µ2+κ2).QED
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276 CHAPTER 11. SCATTERING
Problem 11.12
Equation 11.91 = ⇒D(θ)=|f(θ)|2=/parenleftbigg2mβ
/planckover2pi12/parenrightbigg21
(µ2+κ2)2,where Eq. 11.89 ⇒κ=2ksin(θ/2).
σ=/integraldisplay
D(θ) sinθdθdφ =2π/parenleftbigg2mβ
/planckover2pi12/parenrightbigg21
µ4/integraldisplayπ
01
/bracketleftBig
1+( 2k/µ)2sin2(θ/2)/bracketrightBig22 sin(θ/2) cos(θ/2)dθ.
Let2k
µsin(θ/2)≡x,so 2 sin( θ/2) =µ
kx,and cos( θ/2)dθ=µ
kdx.Then
σ=2π/parenleftbigg2mβ
/planckover2pi12/parenrightbigg21
µ4/parenleftBigµ
k/parenrightBig2/integraldisplayx1
x0x
(1 +x2)2dx.The limits are/braceleftbiggθ=0=⇒x=x0=0,
θ=π=⇒x=x1=2k/µ./bracerightbigg
So
σ=2π/parenleftbigg2mβ
/planckover2pi12/parenrightbigg21
(µk)2/bracketleftbigg
−1
21
(1 +x2)/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle2k/µ
0=π/parenleftbigg2mβ
/planckover2pi12/parenrightbigg21
(µk)2/bracketleftbigg
1−1
1+( 2k/µ)2/bracketrightbigg
=π/parenleftbigg2mβ
/planckover2pi12/parenrightbigg21
(µk)2/bracketleftbigg4(k/µ)2
1+4k2/µ2/bracketrightbigg
=π/parenleftbigg4mβ
/planckover2pi12/parenrightbigg21
µ21
µ2+4k2.Butk2=2mE
/planckover2pi12,so
σ=π/parenleftbigg4mβ
µ/planckover2pi1/parenrightbigg21
(µk)2+8mE.
Problem 11.13
(a)
V(r)=αδ(r−a).Eq. 11.80 =⇒f=−m
2π/planckover2pi12/integraldisplay
V(r)d3r=−m
2π/planckover2pi12α4π/integraldisplay∞
0δ(r−a)r2dr.
f=−2mα
/planckover2pi12a2;D=|f|2=/parenleftbigg2mα
/planckover2pi12a2/parenrightbigg2
;σ=4πD=π/parenleftbigg4mα
/planckover2pi12a2/parenrightbigg2
.
(b)
Eq. 11.88 =⇒f=−2m
/planckover2pi12κα/integraldisplay∞
0rδ(r−a) sin(κr)dr=−2mα
/planckover2pi12κasin(κa)(κ=2ksin(θ/2)).
(c)Note first that (b)reduces to (a)in the low-energy regime ( ka/lessmuch1=⇒κa/lessmuch1). Since Problem 11.4
was also for low energy, what we must confirm is that Problem 11.4 reproduces (a)in the regime for
which the Born approximation holds. Inspection shows that the answer to Problem 11.4 does reduce tof=−2mαa
2//planckover2pi12whenβ/lessmuch1, which is to say when f/a/lessmuch1. This is the appropriate condition, since
(Eq. 11.12) f/ais a measure of the relative size of the scattered wave, in the interaction region.
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CHAPTER 11. SCATTERING 277
Problem 11.14
F=1
4πFepsilonC0q1q2
r2ˆr;F⊥=1
4πFepsilonC0q1q2
r2cosφ; cosφ=b
r,soF⊥=1
4πFepsilonC0q1q2b
r3;dt=dx
v.
b
qqF
r
r
xφ
1
2
I⊥=/integraldisplay
F⊥dt=1
4πFepsilonC0q1q2b
v/integraldisplay∞
−∞dx
(x2+b2)3/2.But
/integraldisplay∞
−∞dx
(x2+b2)3/2=2/integraldisplay∞
0dx
(x2+b2)3/2=2x
b2√
x2+b2/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0=2
b2,soI⊥=1
4πFepsilonC02q1q2
bv.
tanθ=I⊥
mv=q1q2
4πFepsilonC01
b(1
2mv2)=q1q2
4πFepsilonC01
bE.θ= tan−1/bracketleftbiggq1q2
4πFepsilonC0bE/bracketrightbigg
.
b=q1q2
4πFepsilonC01
Etanθ=/parenleftbiggq1q2
8πFepsilonC0E/parenrightbigg
(2 cotθ).
The exact answer is the same, only with cot( θ/2) in place of 2 cot θ. So I must show that cot( θ/2)≈2 cotθ,
for small θ(that’s the regime in which the impulse approximation should work). Well:
cot(θ/2) =cos(θ/2)
sin(θ/2)≈1
θ/2=2
θ,for small θ, while 2 cot θ=2cosθ
sinθ≈21
θ.So it works.
Problem 11.15
First let’s set up the general formalism. From Eq. 11.101:
ψ(r)=ψ0(r)+/integraldisplay
g(r−r0)V(r0)ψ0(r0)d3r0+/integraldisplay
g(r−r0)V(r0)/bracketleftbigg/integraldisplay
g(r0−r1)V(r1)ψ0(r1)d3r1/bracketrightbigg
d3r0+···
Put inψ0(r)=Aeikz,g(r)=−m
2π/planckover2pi12eikr
r:
ψ(r)=Aeikz−mA
2π/planckover2pi12/integraldisplayeik|r−r0|
|r−r0|V(r0)eikz0d3r0
+/parenleftBigm
2π/planckover2pi12/parenrightBig2
A/integraldisplayeik|r−r0|
|r−r0|V(r0)/bracketleftbigg/integraldisplayeik|r0−r1|
|r0−r1|V(r1)eikz1d3r1/bracketrightbigg
d3r0.
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278 CHAPTER 11. SCATTERING
In the scattering region r/greatermuchr0,Eq. 11.73 =⇒eik|r−r0|
|r−r0|≈eikr
re−ik·r0,with k≡kˆr,so
ψ(r)=A/braceleftbigg
eikz−m
2π/planckover2pi12eikr
r/integraldisplay
e−ik·r0V(r0)eikz0d3r0
/parenleftBigm
2π/planckover2pi12/parenrightBig2eikr
r/integraldisplay
e−ik·r0V(r0)/bracketleftbigg/integraldisplayeik|r0−r1|
|r0−r1|V(r1)eikz1d3r1/bracketrightbigg
d3r0/bracerightbigg
f(θ,φ)=−m
2π/planckover2pi12/integraldisplay
ei(k/prime−k)·rV(r)d3r+/parenleftBigm
2π/planckover2pi12/parenrightBig2/integraldisplay
e−ik·rV(r)/bracketleftbigg/integraldisplayeik|r−r0|
|r−r0|V(r0)eikz0d3r0/bracketrightbigg
d3r.
I simplified the subscripts, since there is no longer any possible ambiguity. For low-energy scattering we drop
the exponentials (see p. 414):
f(θ,φ)≈−m
2π/planckover2pi12/integraldisplay
V(r)d3r+/parenleftBigm
2π/planckover2pi12/parenrightBig2/integraldisplay
V(r)/bracketleftbigg/integraldisplay1
|r−r0|V(r0)d3r0/bracketrightbigg
d3r.
Now apply this to the potential in Eq. 11.81:
/integraldisplay1
|r−r0|V(r0)d3r0=V0/integraldisplaya
01
|r−r0|r2
0sinθ0dr0dθ0dφ0.
Orient the z0axis along r,s o|r−r0|=r2+r2
0−2rr0cosθ0.
/integraldisplay1
|r−r0|V(r0)d3r0=V02π/integraldisplaya
0r2
0/bracketleftbigg/integraldisplayπ
01/radicalbig
r2+r2
0−2rr0cosθ0sinθ0dθ0/bracketrightbigg
dr0.But
/integraldisplayπ
01/radicalbig
r2+r2
0−2rr0cosθ0sinθ0dθ0=1
rr0/radicalBig
r2+r2
0−2rr0cosθ0/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ
0=1
rr0[(r0+r)−|r0−r|]=/braceleftbigg
2/r, r 0<r;
2/r0,r0>r .
Herer<a (from the “outer” integral), so
/integraldisplay1
|r−r0|V(r0)d3r0=4πV0/bracketleftbigg1
r/integraldisplayr
0r2
0dr0+/integraldisplaya
rr0dr0/bracketrightbigg
=4πV0/bracketleftbigg1
rr3
3+1
2(a2−r2)/bracketrightbigg
=2πV0/parenleftbigg
a2−1
3r2/parenrightbigg
.
/integraldisplay
V(r)/bracketleftbigg/integraldisplay1
|r−r0|V(r0)d3r0/bracketrightbigg
d3r=V0(2πV0)4π/integraldisplaya
0/parenleftbigg
a2−1
3r2/parenrightbigg
r2dr=8π2V2
0/bracketleftbigg
a2a3
3−1
3a5
5/bracketrightbigg
=32
15π2V2
0a5.
f(θ)=−m
2π/planckover2pi12V04
3πa3+/parenleftBigm
2π/planckover2pi12/parenrightBig232
15π2V2
0a5=−/parenleftbigg2mV0a3
3/planckover2pi12/parenrightbigg/bracketleftbigg
1−4
5/parenleftbiggmV0a2
/planckover2pi12/parenrightbigg/bracketrightbigg
.
Problem 11.16
/parenleftbiggd2
dx2+k2/parenrightbigg
G(x)=δ(x) (analog to Eq. 11.52) .G(x)=1√
2π/integraldisplay
eisxg(s)ds(analog to Eq. 11.54) .
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CHAPTER 11. SCATTERING 279
/parenleftbiggd2
dx2+k2/parenrightbigg
G=1√
2π/integraldisplay
(−s2+k2)g(s)eisxds=δ(x)=1
2π/integraldisplay
eisxds=⇒g(s)=1√
2π(k2−s2).
G(x)=1
2π/integraldisplay∞
−∞eisx
k2−s2ds.Skirt the poles as in Fig. 11.10. For x>0,close above:
G(x)=−1
2π/contintegraldisplay/parenleftbiggeisx
s+k/parenrightbigg1
s−kds=−1
2π2πi/parenleftbiggeisx
s+k/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle
s=k=−ieikx
2k.Forx<0,close below:
G(x)=+1
2π/contintegraldisplay/parenleftbiggeisx
s−k/parenrightbigg1
s+kds=1
2π2πi/parenleftbiggeisx
s−k/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle
s=−k=−ie−ikx
2k.
In either case, then, G(x)=−i
2keik|x|.(Analog to Eq. 11.65.)
ψ(x)=G(x−x0)2m
/planckover2pi12V(x0)ψ(x0)dx0=−i
2k2m
/planckover2pi12/integraldisplay
eik|x−x0|V(x0)ψ(x0)dx0,
plus any solution ψ0(x) to the homogeneous Schr¨ odinger equation:
/parenleftbiggd2
dx2+k2/parenrightbigg
ψ0(x)=0.So:
ψ(x)=ψ0(x)−im
/planckover2pi12k/integraldisplay∞
−∞eik|x−x0|V(x0)ψ(x0)dx0.
Problem 11.17
For the Born approximation let ψ0(x)=Aeikx, andψ(x)≈Aeikx.
ψ(x)≈A/bracketleftbigg
eikx−im
/planckover2pi12k/integraldisplay∞
−∞eik|x−x0|V(x0)eikx0dx0/bracketrightbigg
=A/bracketleftbigg
eikx−im
/planckover2pi12k/integraldisplayx
−∞eik(x−x0)V(x0)eikx0dx0−im
/planckover2pi12k/integraldisplay∞
xeik(x0−x)V(x0)eikx0dx0/bracketrightbigg
.
ψ(x)=A/bracketleftbigg
eikx−im
/planckover2pi12keikx/integraldisplayx
−∞V(x0)dx0−im
/planckover2pi12ke−ikx/integraldisplay∞
xe2ikx0V(x0)dx0/bracketrightbigg
.
Now assume V(x) is localized; for large positive x, the third term is zero, and
ψ(x)=Aeikx/bracketleftbigg
1−im
/planckover2pi12k/integraldisplay∞
−∞V(x0)dx0/bracketrightbigg
.This is the transmitted wave .
For large negative xthe middle term is zero:
ψ(x)=A/bracketleftbigg
eikx−im
/planckover2pi12ke−ikx/integraldisplay∞
−∞e2ikx0V(x0)dx0/bracketrightbigg
.
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280 CHAPTER 11. SCATTERING
Evidently the first term is the incident wave and the second the reflected wave:
R=/parenleftBigm
/planckover2pi12k/parenrightBig2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay
∞
−∞e2ikxV(x)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
.
If you try in the same spirit to calculate the transmission coefficient, you get
T=/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−im
/planckover2pi12k/integraldisplay∞
−∞V(x)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle2
=1+/parenleftBigm
/planckover2pi12k/parenrightBig2/bracketleftbigg/integraldisplay∞
−∞V(x)dx/bracketrightbigg2
,
which is nonsense (greater than 1). The first Born approximation gets Rright, but all you can say to this order
isT≈1 (you would do better using T=1−R).
Problem 11.18
Delta function :V(x)=−αδ(x)./integraldisplay∞
−∞e2ikxV(x)dx=−α,soR=/parenleftBigmα
/planckover2pi12k/parenrightBig2
,
or, in terms of energy ( k2=2mE/ /planckover2pi12):
R=m2α2
2mE/planckover2pi12=mα2
2/planckover2pi12E;T=1−R=1−mα2
2/planckover2pi12E.
The exact answer (Eq. 2.141) is1
1+mα2
2/planckover2pi12E≈1−mα2
2/planckover2pi12E, so they agree provided E/greatermuchmα
2/planckover2pi12.
Finite square well :V(x)=/braceleftbigg−V0(−a<x<a )
0 (otherwise)/bracerightbigg
.
/integraldisplay∞
−∞e2ikxV(x)dx=−V0/integraldisplaya
−ae2ikxdx=−V0e2ikx
2ik/vextendsingle/vextendsingle/vextendsingle/vextendsinglea
−a=−V0
k/parenleftbigge2ika−e−2ika
2i/parenrightbigg
=−V0
ksin(2ka).
SoR=/bracketleftBigm
/planckover2pi12k/parenrightBig2/parenleftbiggV0
ksin(2ka)/bracketrightbigg2
.T=1−/bracketleftbiggV0
2Esin/parenleftbigg2a
/planckover2pi1√
2mE/parenrightbigg/bracketrightbigg2
.
IfE/greatermuchV0, the exact answer (Eq. 2.169) becomes
T−1≈1+/bracketleftbiggV0
2Esin/parenleftbigg2a
/planckover2pi1√
2mE/parenrightbigg/bracketrightbigg2
=⇒T≈1−/parenleftbiggV0
2Esin/bracketleftbigg2a
/planckover2pi1√
2mE/parenrightbigg/bracketrightbigg2
,
so they agree provided E/greatermuchV0.
Problem 11.19
The Legendre polynomials satisfy Pl(1) = 1 (see footnote 30, p. 124), so Eq. 11.47 ⇒
f(0) =1
k∞/summationdisplay
l=0(2l+1 )eiδlsinδl.Therefore Im[ f(0)] =1
k∞/summationdisplay
l=0(2l+ 1) sin2δl,
and hence (Eq. 11.48):
σ=4π
kIm[f(0)].QED
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CHAPTER 11. SCATTERING 281
Problem 11.20
Using Eq. 11.88 and integration by parts:
f(θ)=−2m
/planckover2pi12κ/integraldisplay∞
0rAe−µr2sin(κr)dr=−2mA
/planckover2pi12κ/integraldisplay∞
0d
dr/parenleftbigg
−1
2µe−µr2/parenrightbigg
sin(κr)dr
=2mA
2µ/planckover2pi12κ/braceleftbigg
e−µr2sin(κr)/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞
0−/integraldisplay∞
0e−µr2d
dr[sin(κr)]dr/bracerightbigg
=mA
µ/planckover2pi12κ/braceleftbigg
0−κ/integraldisplay∞
0e−µr2cos(κr)dr/bracerightbigg
=−mA
µ/planckover2pi12/parenleftbigg√π
2√µe−κ2/4µ/parenrightbigg
=−mA√π
2/planckover2pi12µ3/2e−κ2/4µ,where κ=2ksin(θ/2) (Eq .11.89).
From Eq. 11.14, then,
dσ
dΩ=πm2A2
4/planckover2pi14µ3e−κ2/2µ,
and hence
σ=/integraldisplaydσ
dΩdΩ=πm2A2
4/planckover2pi14µ3/integraldisplay
e−4k2sin2(θ/2)/2µsinθdθdφ
=π2m2A2
2/planckover2pi14µ3/integraldisplayπ
0e−2k2sin2(θ/2)/µsinθdθ; write sin θ= 2 sin( θ/2) cos(θ/2) and let x≡sin(θ/2)
=π2m2A2
2/planckover2pi14µ3/integraldisplay1
0e−2k2x2/µ2x2dx=2π2m2A2
/planckover2pi14µ3/integraldisplay1
0xe−2k2x2/µdx
=2π2m2A2
/planckover2pi14µ3/bracketleftBig
−µ
4k2e−2k2x2/µ/bracketrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsingle1
0=−π2m2A2
2/planckover2pi14µ2k2/parenleftBig
e−2k2/µ−1/parenrightBig
=π2m2A2
2/planckover2pi14µ2k2/parenleftBig
1−e−2k2/µ/parenrightBig
.
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282 CHAPTER 12. AFTERWORD
Chapter 12
Afterword
Problem 12.1
Suppose, on the contrary, that
α|φa(1)/angbracketright|φb(2)/angbracketright+β|φb(1)/angbracketright|φa(2)/angbracketright=|ψr(1)/angbracketright|ψs(2)/angbracketright,
for some one-particle states |ψr/angbracketrightand|ψs/angbracketright. Because|φa/angbracketrightand|φb/angbracketrightconstitute a complete set of one-particle states
(this is a two-level system), any other one-particle state can be expressed as a linear combination of them. Inparticular,
|ψ
r/angbracketright=A|φa/angbracketright+B|φb/angbracketright,and|ψs/angbracketright=C|φa/angbracketright+D|φb/angbracketright,
for some complex numbers A,B,C, andD.T h u s
α|φa(1)/angbracketright|φb(2)/angbracketright+β|φb(1)/angbracketright|φa(2)/angbracketright=/bracketleftbig
A|φa(1)/angbracketright+B|φb(1)/angbracketright/bracketrightbig/bracketleftbig
C|φa(2)/angbracketright+D|φb(2)/angbracketright/bracketrightbig
=AC|φa(1)/angbracketright|φa(2)/angbracketright+AD|φa(1)/angbracketright|φb(2)/angbracketright+BC|φb(1)/angbracketright|φa(2)/angbracketright+BD|φb(1)/angbracketright|φb(2)/angbracketright.
(i) Take the inner product with /angbracketleftφa(1)|/angbracketleftφb(2)|:α=AD.
(ii) Take the inner product with /angbracketleftφa(1)|/angbracketleftφa(2)|:0 = AC.
(iii) Take the inner product with /angbracketleftφb(1)|/angbracketleftφa(2)|:β=BC.
(iv) Take the inner product with /angbracketleftφb(1)|/angbracketleftφb(2)|:0 = BD.
(ii)⇒eitherA=0o rC= 0. But if A= 0, then (i) ⇒α= 0, which is excluded by assumption, whereas if
C= 0, then (iii) ⇒β= 0, which is likewise excluded. Conclusion: It is impossible to express this state as a
product of one-particle states. QED
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APPENDIX. LINEAR ALGEBRA 283
Appendix A
Linear Algebra
Problem A.1
(a)Yes; two-dimensional.
(b)No;the sum of two such vectors has az= 2, and is not in the subset. Also, the null vector (0,0,0) is not
in the subset.
(c)Yes; one-dimensional.
Problem A.2
(a)Yes; 1 ,x,x2,...,xN−1is a convenient basis. Dimension: N.
(b)Yes; 1 ,x2,x4,....Dimension N/2(ifNis even) or (N+1 )/2(ifNis odd).
(c)No. The sum of two such “vectors” is not in the space.
(d)Yes; (x−1),(x−1)2,(x−1)3,...,(x−1)N−1.Dimension: N−1.
(e)No. The sum of two such “vectors” would have value 2 at x=0 .
Problem A.3
Suppose|α/angbracketright=a1|e1/angbracketright+a2|e2/angbracketright+···an|en/angbracketrightand|α/angbracketright=b1|e1/angbracketright+b2|e2/angbracketright+···+bn|en/angbracketright.Subtract: 0 = ( a1−b1)|e1/angbracketright+
(a2−b2)|e2/angbracketright+···+(an−bn)|en/angbracketright.Suppose aj/negationslash=bjfor some j; then we can divide by ( aj−bj) to get:
|ej/angbracketright=−(a1−b1)
(aj−bj)|e1/angbracketright−(a2−b2)
(aj−bj)|e2/angbracketright−···− 0|ej/angbracketright−···−(an−bn)
(aj−bj)|en/angbracketright,
so|ej/angbracketrightis linearly dependent on the others, and hence {|ej/angbracketright}is not a basis. If {|ej/angbracketright}isa basis, therefore, the
components must all be equal ( a1=b1,a2=b2,...,a n=bn). QED
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284 APPENDIX. LINEAR ALGEBRA
Problem A.4
(i)
/angbracketlefte1|e1/angbracketright=|1+i|2+1+|i|2=( 1+i)(1−i)+1+( i)(−i)=1+1+1+1=4 ./bardble1/bardbl=2.
|e/prime
1/angbracketright=1
2(1 +i)ˆi+1
2ˆj+i
2ˆk.
(ii)
/angbracketlefte/prime
1|e2/angbracketright=1
2(1−i)(i)+1
2(3) +/parenleftbigg−i
2/parenrightbigg
1=1
2(i+1+3−i)=2.
|e/prime/prime
2/angbracketright≡|e2/angbracketright−/angbracketlefte/prime
1|e2/angbracketright|e/prime
1/angbracketright=(i−1−i)ˆi+( 3−1)ˆj+( 1−i)ˆk=(−1)ˆi+ (2)ˆj+( 1−i)ˆk.
/angbracketlefte/prime/prime
2|e/prime/prime
2/angbracketright=1+4+2=7 .|e/prime
2/angbracketright=1√
7[−ˆi+2ˆj+( 1−i)ˆk].
(iii)
/angbracketlefte/prime
1|e3/angbracketright=1
228 = 14;/angbracketlefte/prime
2|e3/angbracketright=2√
72 8=8√
7.
|e/prime/prime
3/angbracketright=|e3/angbracketright−/angbracketlefte/prime
1|e3/angbracketright|e/prime
1/angbracketright−/angbracketlefte/prime
2|e3/angbracketright|e/prime
2/angbracketright=|e3/angbracketright−7|e1/angbracketright−8|e/prime/prime
2/angbracketright
=( 0−7−7i+8 )ˆi+ (28−7−16)ˆj+( 0−7i−8+8i)ˆk=( 1−7i)ˆi+5ˆj+(−8+i)ˆk.
/bardble/prime/prime
3/bardbl2=1+4 9+2 5+6 4+1=1 4 0 .|e/prime
3/angbracketright=1
2√
35[(1−7i)ˆi+5ˆj+(−8+i)ˆk].
Problem A.5
From Eq. A.21: /angbracketleftγ|γ/angbracketright=/angbracketleftγ|/parenleftbigg
|β/angbracketright−/angbracketleftα|β/angbracketright
/angbracketleftα|α/angbracketright|α/angbracketright/parenrightbigg
=/angbracketleftγ|β/angbracketright−/angbracketleftα|β/angbracketright
/angbracketleftα|α/angbracketright/angbracketleftγ|α/angbracketright.From Eq. A.19:
/angbracketleftγ|β/angbracketright∗=/angbracketleftβ|γ/angbracketright=/angbracketleftβ|/parenleftbigg
|β/angbracketright−/angbracketleftα|β/angbracketright
/angbracketleftα|α/angbracketright|α/angbracketright/parenrightbigg
=/angbracketleftβ|β/angbracketright−/angbracketleftα|β/angbracketright
/angbracketleftα|α/angbracketright/angbracketleftβ|α/angbracketright=/angbracketleftβ|β/angbracketright−|/angbracketleftα|β/angbracketright|2
/angbracketleftα|α/angbracketright,which is real.
/angbracketleftγ|α/angbracketright∗=/angbracketleftα|γ/angbracketright=/angbracketleftα|/parenleftbigg
|β/angbracketright−/angbracketleftα|β/angbracketright
/angbracketleftα|α/angbracketright|α/angbracketright/parenrightbigg
=/angbracketleftα|β/angbracketright−/angbracketleftα|β/angbracketright
/angbracketleftα|α/angbracketright/angbracketleftα|α/angbracketright=0./angbracketleftγ|α/angbracketright=0.So (Eq.A.20) :
/angbracketleftγ|γ/angbracketright=/angbracketleftβ|β/angbracketright−|/angbracketleftα|β/angbracketright|2
/angbracketleftα|α/angbracketright≥0,and hence|/angbracketleftα|β/angbracketright|2≤/angbracketleftα|α/angbracketright/angbracketleftβ|β/angbracketright.QED
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APPENDIX. LINEAR ALGEBRA 285
Problem A.6
/angbracketleftα|β/angbracketright=( 1−i)(4−i) + (1)(0) + (−i)(2−2i)=4−5i−1−2i−2=1−7i;/angbracketleftβ|α/angbracketright=1+7 i;
/angbracketleftα|α/angbracketright=1+1+1+1=4 ; /angbracketleftβ|β/angbracketright=1 6+1+4+4=2 5 ; c o s θ=/radicalbigg
1+4 9
4·25=1√
2;θ=4 5◦.
Problem A.7
Let|γ/angbracketright≡|α/angbracketright+|β/angbracketright;/angbracketleftγ|γ/angbracketright=/angbracketleftγ|α/angbracketright+/angbracketleftγ|β/angbracketright.
/angbracketleftγ|α/angbracketright∗=/angbracketleftα|γ/angbracketright=/angbracketleftα|α/angbracketright+/angbracketleftα|β/angbracketright=⇒/angbracketleftγ|α/angbracketright=/angbracketleftα|α/angbracketright+/angbracketleftβ|α/angbracketright.
/angbracketleftγ|β/angbracketright∗=/angbracketleftβ|γ/angbracketright=/angbracketleftβ|α/angbracketright+/angbracketleftβ|β/angbracketright=⇒/angbracketleftγ|β/angbracketright=/angbracketleftα|β/angbracketright+/angbracketleftβ|β/angbracketright.
/bardbl(|α/angbracketright+|β/angbracketright)/bardbl2=/angbracketleftγ|γ/angbracketright=/angbracketleftα|α/angbracketright+/angbracketleftβ|β/angbracketright+/angbracketleftα|β/angbracketright+/angbracketleftβ|α/angbracketright.
But/angbracketleftα|β/angbracketright+/angbracketleftβ|α/angbracketright= 2Re(/angbracketleftα|β/angbracketright)≤2|/angbracketleftα|β/angbracketright|≤2/radicalbig
/angbracketleftα|α/angbracketright/angbracketleftβ|β/angbracketright(by Schwarz inequality) ,so
/bardbl(|α/angbracketright+|β/angbracketright)/bardbl2≤/bardblα/bardbl2+/bardblβ/bardbl2+2/bardblα/bardbl/bardblβ/bardbl=(/bardblα/bardbl+/bardblβ/bardbl)2,and hence/bardbl(|α/angbracketright+|β/angbracketright)/bardbl≤/bardblα/bardbl+/bardblβ/bardbl.QED
Problem A.8
(a)
11 0
21 3
3i(3−2i)4
.
(b)
(−2+0−1 ) ( 0+1+3 i)(i+0+2i)
( 4+0+3 i) ( 0+0+9 ) ( −2i+0+6 )
(4i+0+2i)( 0−2i+6 ) ( 2+0+4 )
=
−3 ( 1+3 i)3i
( 4+3i)9( 6−2i)
6i(6−2i)6
.
(c) BA=
(−2+0+2 ) ( 2+0 −2) (2i+0−2i)
( 0+2+0 ) ( 0+0+0 ) ( 0+3+0 )
(−i+6+4i)(i+0−4i)(−1+9+4 )
=
00 0
20 3
( 6+3i)−3i12
.
[A,B]=AB−BA=
−3 ( 1+3 i)3i
( 2+3i)9 ( 3 −2i)
(−6+3i)( 6 +i)−6
.
(d)
−12 2i
10−2i
i32
.
(e)
−11−i
20 3
−2i2i2
.
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286 APPENDIX. LINEAR ALGEBRA
(f)
−12−2i
102i
−i32
.
(g)4+0+0−1−0−0=3.
(h)
B−1=1
3˜C;C=
|10
32|− |00
i2||01
i3|
−/vextendsingle/vextendsingle0−i
32/vextendsingle/vextendsingle/vextendsingle/vextendsingle2−i
i2/vextendsingle/vextendsingle−|20
i3|/vextendsingle/vextendsingle0−i
10/vextendsingle/vextendsingle−/vextendsingle/vextendsingle2−i
00/vextendsingle/vextendsingle|20
01|
=
20−i
−3i3−6
i02
.B−1=1
3
2−3ii
03 0
−i−62
.
BB−1=1
3
( 4+0−1) (−6i+0+6i)( 2i+0−2i)
( 0+0+0 ) ( 0+3+0 ) ( 0+0+0 )
(2i+0−2i) ( 3+9−12) (−1+0+4 )
=1
3
300
030003
=
100
010001
./check
detA=0+6 i+4−0−6i−4=0.
No; Adoesnothave an inverse.
Problem A.9
(a)
−i+2i+2i
2i+0+6
−2+4+4
=
3i
6+2i
6
.
(b)
/parenleftbig−i−2i2/parenrightbig
2
1−i
0
=−2i−2i(1−i)+0=−2−4i.
(c)
/parenleftbigi2i2/parenrightbig
20−i
01 0
i32
2
1−i
0
=/parenleftbigi2i2/parenrightbig
4
1−i
3−i
=4i+2i(1−i) + 2(3−i)=8+4i.
(d)
i
2i
2
/parenleftbig2( 1 +i)0/parenrightbig
=
2i(−1+i)0
4i(−2+2i)0
4 ( 2+2 i)0
.
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APPENDIX. LINEAR ALGEBRA 287
Problem A.10
(a)S=1
2(T+˜T); A=1
2(T−˜T).
(b)R=1
2(T+T∗);M=1
2(T−T∗).
(c)H=1
2(T+T†);K=1
2(T−T†).
Problem A.11
(/tildewiderST)ki=(ST)ik=n/summationdisplay
j=1SijTjk=n/summationdisplay
j=1˜Tkj˜Sji=(˜T˜S)ki⇒/tildewiderST=˜T˜S.QED
(ST)†=(/tildewiderST)∗=(˜T˜S)∗=˜T∗˜S∗=T†S†.QED
(T−1S−1)(ST)=T−1(S−1S)T=T−1T=I⇒(ST)−1=T−1S−1.QED
U†=U−1,W†=W−1⇒(WU)†=U†W†=U−1W−1=(WU)−1⇒WUis unitary .
H=H†,J=J†⇒(HJ)†=J†H†=JH;
the product is hermitian ⇔this is HJ,i.e.⇔[H,J]=0 (theycommute ).
(U+W)†=U†+W†=U−1+W−1?=(U+W)−1.No;the sum of two unitary matrices is notunitary.
(H+J)†=H†+J†=H+J.Yes;the sum of two hermitian matrices ishermitian.
Problem A.12
U†U=I=⇒(U†U)ik=δik=⇒n/summationdisplay
j=1U†
ijUjk=n/summationdisplay
j=1U∗
jiUjk=δik.
Construct the set of nvectors a(j)i≡Uij(a(j)is thej-th column of U; itsi-th component is Uij). Then
a(i)†a(k)=n/summationdisplay
j=1a(i)∗
ja(k)
j=n/summationdisplay
j=1U∗
jiUjk=δik,
so these vectors are orthonormal. Similarly,
UU†=I=⇒(UU†)ik=δik=⇒n/summationdisplay
j=1UijU†
jk=n/summationdisplay
j=1U∗
kjUij=δki.
This time let the vectors b(j)be the rowsofU:b(j)i≡Uji. Then
b(k)†b(i)=n/summationdisplay
j=1b(k)∗
jb(i)
j=n/summationdisplay
j=1U∗
kjUij=δki,
so the rows are also orthonormal.
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288 APPENDIX. LINEAR ALGEBRA
Problem A.13
H†=H(hermitian)⇒detH= det( H†) = det( ˜H∗) = (det ˜H)∗= (det H)∗⇒detHis real. /check
U†=U−1(unitary)⇒det(UU†) = (det U)(detU†) = (det U)(det˜U)∗=|detU|2= det I=1,so det U=1./check
˜S=S−1(orthogonal)⇒det(S˜S) = (det S)(det˜S) = (det S)2=1,so det S=±1./check
Problem A.14
(a)
ˆi/prime= cosθˆi+ sinθˆj;ˆj/prime=−sinθˆi+ cosθˆj;ˆk/prime=ˆk.Ta=
cosθ−sinθ0
sinθcosθ0
00 1
.
x
z, z'x'y y'
θθ
(b)
ˆi/prime=ˆj;ˆj/prime=ˆk;ˆk/prime=ˆi.Tb=
001
100010
.
x, z'
z, y'y, x'
(c)
ˆi/prime=ˆi;ˆj/prime=ˆj;ˆk/prime=−ˆk.Tc=
10 0
01 000−1
.
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APPENDIX. LINEAR ALGEBRA 289
(d)
˜TaTa=
cosθsinθ0
−sinθcosθ0
00 1
cosθ−sinθ0
sinθcosθ0
00 1
=
100
010001
./check
˜T
bTb=
010
001100
001
100010
=
100
010001
./check˜T
cTc=
10 0
01 000−1
10 0
01 000−1
=
100
010001
./check
detT
a= cos2θ+ sin2θ=1.detTb=1.detTc=-1.
Problem A.15
x, x'
zy
θ
θ
z'y'
ˆi/prime=ˆi;ˆj/prime= cosθˆj+ sinθˆk;ˆk/prime= cosθˆk−sinθˆj.Tx(θ)=
10 0
0 cosθ−sinθ
0 sinθcosθ
.
x
zx'y, y'
θ
θ
z'
ˆi/prime= cosθˆi−sinθˆk;ˆj/prime=ˆj;ˆk/prime= cosθˆk+ sinθˆi.Ty(θ)=
cosθ0 sinθ
01 0
−sinθ0 cosθ
.
ˆi/prime=ˆj;ˆj/prime=−ˆi;ˆk/prime=ˆk.S=
0−10
100001
.
S−1=
01 0
−100
00 1
.
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290 APPENDIX. LINEAR ALGEBRA
STxS−1=
0−10
100001
10 0
0 cosθ−sinθ
0 sinθcosθ
01 0
−100
00 1
=
0−10
100001
010
−cosθ0−sinθ
−sinθ0 cosθ
=
cosθ0 sinθ
01 0
−sinθ0 cosθ
=T
y(θ).
STyS−1=
0−10
100001
cosθ0 sinθ
01 0
−sinθ0 cosθ
01 0
−100
00 1
=
0−10
100001
0 cosθsinθ
−10 0
0−sinθcosθ
=
10 0
0 cosθsinθ
0−sinθcosθ
=T
x(−θ).
Is this what we would expect? Yes, for rotation about the xaxis now means rotation about the yaxis, and
rotation about the yaxis has become rotation about the −xaxis—which is to say, rotation in the opposite
direction about the + xaxis.
Problem A.16
From Eq. A.64 we have
AfBf=SAeS−1SBeS−1=S(AeBe)S−1=SCeS−1=Cf./check
Suppose S†=S−1andHe=He†(Sunitary, Hehermitian). Then
Hf†=(SHeS−1)†=(S−1)†He†S†=SHeS−1=Hf,soHfis hermitian ./check
In an orthonormal basis, /angbracketleftα|β/angbracketright=a†b(Eq. A.50). So if {|fi/angbracketright}is orthonormal, /angbracketleftα|β/angbracketright=af†bf.Butbf=Sbe
(Eq. A.63), and also af†=ae†S†.S o/angbracketleftα|β/angbracketright=ae†S†Sbe.This is equal to ae†be(and hence{|ei/angbracketright}is also
orthonormal), for all vectors |α/angbracketrightand|β/angbracketright⇔S†S=I, i.e.Sis unitary.
Problem A.17
Tr(T1T2)=n/summationdisplay
i=1(T1T2)ii=n/summationdisplay
i=1n/summationdisplay
j=1(T1)ij(T2)ji=n/summationdisplay
j=1n/summationdisplay
i=1(T2)ji(T1)ij=n/summationdisplay
j=1(T2T1)jj=T r (T2T1).
Is Tr( T1T2T3)=T r ( T2T1T3)? No. Counterexample:
T1=/parenleftbigg01
00/parenrightbigg
,T2=/parenleftbigg00
10/parenrightbigg
,T3=/parenleftbigg10
00/parenrightbigg
.
T1T2T3=/parenleftbigg
01
00/parenrightbigg/parenleftbigg
00
10/parenrightbigg/parenleftbigg
10
00/parenrightbigg
=/parenleftbigg
01
00/parenrightbigg/parenleftbigg
00
10/parenrightbigg
=/parenleftbigg
10
00/parenrightbigg
=⇒Tr(T1T2T3)=1.
T2T1T3=/parenleftbigg00
10/parenrightbigg/parenleftbigg01
00/parenrightbigg/parenleftbigg10
00/parenrightbigg
=/parenleftbigg00
10/parenrightbigg/parenleftbigg00
00/parenrightbigg
=/parenleftbigg00
00/parenrightbigg
=⇒Tr(T2T1T3)=0.
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APPENDIX. LINEAR ALGEBRA 291
Problem A.18
Eigenvalues:
/vextendsingle/vextendsingle/vextendsingle/vextendsingle(cosθ−λ)−sinθ
sinθ(cosθ−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle= (cosθ−λ)
2+ sin2θ= cos2θ−2λcosθ+λ2+ sin2θ=0,orλ2−2λcosθ+1=0 .
λ=2 cosθ±√
4 cos2θ−4
2= cosθ±/radicalbig
−sin2θ= cosθ±isinθ=e±iθ.
So there are two eigenvalues, both of them complex. Only if sin θ= 0 does this matrix possess realeigenvalues,
i.e., only if θ=0o rπ.
Eigenvectors:
/parenleftbiggcosθ−sinθ
sinθcosθ/parenrightbigg/parenleftbiggα
β/parenrightbigg
=e±iθ/parenleftbiggα
β/parenrightbigg
=⇒cosθα−sinθβ= (cosθ±isinθ)α⇒β=∓iα.Normalizing:
a(1)=1√
2/parenleftbigg1
−i/parenrightbigg
;a(2)=1√
2/parenleftbigg1
i/parenrightbigg
.
Diagonalization:
(S−1)11=a(1)
1=1√
2;(S−1)12=a(2)
1=1√
2;(S−1)21=a(1)
2=−i√
2;(S−1)22=a(2)
2=i√
2.
S−1=1√
2/parenleftbigg11
−ii/parenrightbigg
; inverting: S=1√
2/parenleftbigg1i
1−i/parenrightbigg
.
STS−1=1
2/parenleftbigg1i
1−i/parenrightbigg/parenleftbiggcosθ−sinθ
sinθcosθ/parenrightbigg/parenleftbigg11
−ii/parenrightbigg
=1
2/parenleftbigg1i
1−i/parenrightbigg/parenleftbigg(cosθ+isinθ) (cosθ−isinθ)
(sinθ−icosθ) (sinθ+icosθ)/parenrightbigg
=1
2/parenleftbigg1i
1−i/parenrightbigg/parenleftbiggeiθe−iθ
−ieiθie−iθ/parenrightbigg
=1
2/parenleftbigg2eiθ0
02e−iθ/parenrightbigg
=/parenleftbiggeiθ0
0e−iθ/parenrightbigg
./check
Problem A.19
/vextendsingle/vextendsingle/vextendsingle/vextendsingle(1−λ)1
0( 1−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=( 1−λ)
2=0=⇒λ=1 (only one eigenvalue) .
/parenleftbigg11
01/parenrightbigg/parenleftbiggα
β/parenrightbigg
=/parenleftbiggα
β/parenrightbigg
=⇒α+β=α=⇒β=0 ; a=/parenleftbigg1
0/parenrightbigg
(only one eigenvector—up to an arbitrary constant factor). Since the eigenvectors do not span the space, this
matrix cannot be diagonalized. [If itcould be diagonalized, the diagonal form would have to be/parenleftbigg10
01/parenrightbigg
, since
the only eigenvalue is 1. But in that case I=SMS−1. Multiplying from the left by S−1and on the right by
S:S−1IS=S−1SMS−1S=M.ButS−1IS=S−1S=I.SoM=I, which is false.]
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292 APPENDIX. LINEAR ALGEBRA
Problem A.20
Expand the determinant (Eq. A.72) by minors, using the first column:
det(T−λ1)=(T11−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(T
22−λ)... ...
......
...( T
nn−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+
n/summationdisplay
j=2Tj1cofactor( Tj1).
But the cofactor of Tj1(forj>1) is missing twoof the original diagonal elements: ( T11−λ) (from the first
column), and ( Tjj−λ) (from the j-th row). So its highest power of λwill be ( n−2). Thus terms in λnand
λn−1come exclusively from the first term above. Indeed, the same argument applied now to the cofactor of
(T11−λ) – and repeated as we expand thatdeterminant – shows that only the product of the diagonal elements
contributes to λnandλn−1:
(T11−λ)(T22−λ)···(Tnn−λ)=(−λ)n+(−λ)n−1(T11+T22+···+Tnn)+···
Evidently then, Cn=(−1)n, andCn−1=(−1)n−1Tr(T). To get C0– the term with nofactors of λ– we simply
setλ=0 . T h u s C0= det( T) .F o ra3×3 matrix:
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(T
11−λ)T12 T13
T21 (T22−λ)T23
T31 T32 (T33−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle
=(T
11−λ)(T22−λ)(T33−λ)+T12T23T31+T13T21T32
−T31T13(T22−λ)−T32T23(T11−λ)−T12T21(T33−λ)
=−λ3+λ2(T11+T22+T33)−λ(T11T22+T11T33+T22T33)+λ(T13T31+T23T32+T12T21)
+T11T22T33+T12T23T31+T13T21T32−T31T13T22−T32T23T11−T12T21T33
=−λ3+λ2Tr(T)+λC1+ det( T),with
C1=(T13T31+T23T32+T12T21)−(T11T22+T11T33+T22T33).
Problem A.21
The characteristic equation is an n-th order polynomial, which can be factored in terms of its n(complex) roots:
(λ1−λ)(λ2−λ)···(λn−λ)=(−λ)n+(−λ)n−1(λ1+λ2+···+λn)+···+(λ1λ2···λn)=0.
Comparing Eq. A.84, it follows that Tr( T)=λ1+λ2+···λnand det( T)=λ1λ2···λn. QED
Problem A.22
(a)
[Tf
1,Tf
2]=Tf
1Tf2−Tf
2Tf1=STe
1S−1STe
2S−1−STe
2S−1STe
1S−1=STe
1Te
2S−1−STe
2Te
1S−1=S[Te
1,Te
2]S−1=0./check
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APPENDIX. LINEAR ALGEBRA 293
(b)Suppose SAS−1=DandSBS−1=E, where DandEarediagonal :
D=
d
10···0
0d2···0
.........
00···d
n
,E=
e
10···0
0e2···0
.........
00···e
n
.
Then
[A,B]=AB−BA=(S−1DS)(S−1ES)−(S−1ES)(S−1DS)=S−1DES−S−1EDS=S−1[D,E]S.
Butdiagonal matrices always commute:
DE=
d
1e10··· 0
0d2e2··· 0
.........
00···d
nen
=ED,
so [A,B]=0.QED
Problem A.23
(a)
M†=/parenleftbigg11
1−i/parenrightbigg
;MM†=/parenleftbigg2( 1−i)
(1 +i)2/parenrightbigg
,M†M=/parenleftbigg2( 1 + i)
(1−i)2/parenrightbigg
;[M,M†]=/parenleftbigg0−2i
2i0/parenrightbigg
/negationslash=0.No.
(b)Find the eigenvalues:
/vextendsingle/vextendsingle/vextendsingle/vextendsingle(1−λ)1
1(i−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=( 1−λ)(i−λ)−1=i−λ(1 +i)+λ
2−1=0 ;
λ=(1 +i)±/radicalbig
(1 +i)2−4(i−1)
2=(1 +i)±√4−2i
2.
Since there are two distinct eigenvalues, there must be two linearly independent eigenvectors, and that’s
enough to span the space. So this matrix isdiagonalizable, even though it is not normal.
Problem A.24
Let|γ/angbracketright=|α/angbracketright+c|β/angbracketright, for some complex number c. Then
/angbracketleftγ|ˆTγ/angbracketright=/angbracketleftα|ˆTα/angbracketright+c/angbracketleftα|ˆTβ/angbracketright+c∗/angbracketleftβ|ˆTα/angbracketright+|c|2/angbracketleftβ|ˆTβ/angbracketright,and
/angbracketleftˆTγ|γ/angbracketright=/angbracketleftˆTα|α/angbracketright+c∗/angbracketleftˆTβ|α/angbracketright+c/angbracketleftˆTα|β/angbracketright+|c|2/angbracketleftˆTβ|β/angbracketright.
Suppose/angbracketleftˆTγ|γ/angbracketright=/angbracketleftγ|ˆTγ/angbracketrightforallvectors. For instance, /angbracketleftˆTα|α/angbracketright=/angbracketleftα|ˆTα/angbracketrightand/angbracketleftˆTβ|β/angbracketright=/angbracketleftβ|ˆTβ/angbracketright), so
c/angbracketleftα|ˆTβ/angbracketright+c∗/angbracketleftβ|ˆTα/angbracketright=c/angbracketleftˆTα|β/angbracketright+c∗/angbracketleftˆTβ|α/angbracketright,and this holds for anycomplex number c.
In particular, for c=1 :/angbracketleftα|ˆTβ/angbracketright+/angbracketleftβ|ˆTα/angbracketright=/angbracketleftˆTα|β/angbracketright+/angbracketleftˆTβ|α/angbracketright, while for c=i:/angbracketleftα|ˆTβ/angbracketright−/angbracketleftβ|ˆTα/angbracketright=/angbracketleftˆTα|β/angbracketright−/angbracketleftˆTβ|α/angbracketright.
(I canceled the i’s). Adding:/angbracketleftα|ˆTβ/angbracketright=/angbracketleftˆTα|β/angbracketright.QED
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294 APPENDIX. LINEAR ALGEBRA
Problem A.25
(a)
T†=˜T∗=/parenleftbigg
11−i
1+i0/parenrightbigg
=T./check
(b)
/vextendsingle/vextendsingle/vextendsingle/vextendsingle(1−λ)( 1−i)
(1 +i)( 0−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−(1−λ)λ−1−1=0 ;λ
2−λ−2=0 ;λ=1±√1+8
2=1±3
2.λ1=2,λ2=−1.
(c)
/parenleftbigg
1( 1−i)
(1 +i)0/parenrightbigg/parenleftbigg
α
β/parenrightbigg
=2/parenleftbigg
α
β/parenrightbigg
=⇒α+( 1−i)β=2α=⇒α=( 1−i)β.
|α|2+|β|2=1=⇒2|β|2+|β|2=1=⇒β=1√
3.a(1)=1√
3/parenleftbigg1−i
1/parenrightbigg
.
/parenleftbigg1( 1−i)
(1 +i)0/parenrightbigg/parenleftbiggα
β/parenrightbigg
=−/parenleftbiggα
β/parenrightbigg
=⇒α+( 1−i)β=−α;α=−1
2(1−i)β.
1
42|β|2+|β|2=1=⇒3
2|β|2=1 ;β=/radicalbigg
2
3.a(2)=1√
6/parenleftbiggi−1
2/parenrightbigg
.
a(1)†a(2)=1
3√
2/parenleftbig(1 +i)1/parenrightbig/parenleftbigg(i−1)
2/parenrightbigg
=1
3√
2(i−1−1−i+2 )=0 ./check
(d)
Eq. A.81 =⇒(S−1)11=a(1)
1=1√
3(1−i); (S−1)12=a(2)
1=1√
6(i−1);
(S−1)21=a(1)
2=1√
3;(S−1)22=a(2)
2=2√
6.
S−1=1√
3/parenleftbigg(1−i)(i−1)/√
2
1√
2/parenrightbigg
;S=(S−1)†=1√
3/parenleftbigg(1 +i)1
(−i−1)/√
2√
2/parenrightbigg
.
STS−1=1
3/parenleftbigg(1 +i)1
−(1 +i)/√
2√
2/parenrightbigg/parenleftbigg1( 1−i)
(1 +i)0/parenrightbigg/parenleftbigg(1−i)(i−1)/√
2
1√
2/parenrightbigg
=1
3/parenleftbigg(1 +i)1
−(1 +i)/√
2√
2/parenrightbigg/parenleftbigg2(1−i)( 1−i)/√
2
2−√
2/parenrightbigg
=1
3/parenleftbigg60
0−3/parenrightbigg
=/parenleftbigg20
0−1/parenrightbigg
./check
(e)
Tr(T)=1 ; det(T)=0−(1 +i)(1−i)=−2.Tr(STS−1)=2−1=1./checkdet(STS−1)=−2./check
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
APPENDIX. LINEAR ALGEBRA 295
Problem A.26
(a)
det(T)=8−1−1−2−2−2=0.Tr(T)=2+2+2= 6.
(b)
/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(2−λ)i 1
−i(2−λ)i
1−i(2−λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=( 2−λ)
3−1−1−(2−λ)−(2−λ)−(2−λ)=8−12λ+6λ2−λ3−8+3λ=0.
−λ3+6λ2−9λ=−λ(λ2−6λ+9 )=−λ(λ−3)2=0.λ1=0,λ2=λ3=3.
λ1+λ2+λ3=6=T r ( T)./checkλ1λ2λ3= 0 = det( T)./checkDiagonal form:
000
030003
.
(c)
2i1
−i2i
1−i2
α
β
γ
=0=⇒/braceleftbigg2α+iβ+γ=0
−iα+2β+iγ=0=⇒α+2iβ−γ=0/bracerightbigg
.
Add the two equations: 3 α+3iβ=0=⇒β=iα;2α−α+γ=0=⇒γ=−α.
a(1)=
α
iα
−α
.Normalizing: |α|2+|α|2+|α|2=1=⇒α=1√
3.a(1)=1√
3
1
i
−1
.
2i1
−i2i
1−i2
α
β
γ
=3
α
β
γ
=⇒
2α+iβ+γ=3α=⇒−α+iβ+γ=0,
−iα+2β+iγ=3β=⇒α−iβ−γ=0,
α−iβ+2γ=3γ=⇒α−iβ−γ=0.
The three equations are redundant – there is only onecondition here: α−iβ−γ=0.We could pick
γ=0,β=−iα,orβ=0,γ=α.Then
a(2)
0=
α
−iα
0
;a(3)
0=
α
0
α
.
But these are not orthogonal, so we use the Gram-Schmidt procedure (Problem A.4); first normalize a(2)
0:
a(2)=1√
2
1
−i
0
.
a(2)†a(3)
0=α√
2/parenleftbig1i0/parenrightbig
1
01
=α
√
2.Soa(3)
0−(a(2)†a(3)
0)a(2)=α
1
01
−α
2
1
−i
0
=α
1/2
i/2
1
.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
296 APPENDIX. LINEAR ALGEBRA
Normalize: |α|2/parenleftbigg1
4+1
4+1/parenrightbigg
=3
2|α|2=1 =⇒α=/radicalbigg
2
3.a(3)=1√
6
1
i
2
.
Check orthogonality:
a(1)†a(2)=1√
6/parenleftbig
1−i−1/parenrightbig
1
−i
0
=1√
6(1−1+0 )=0 ./check
a(1)†a(3)=1
3√
2/parenleftbig
1−i−1/parenrightbig
1
i
2
=1
3√
2( 1+1−2 )=0./check
(d)S−1is the matrix whose columns are the eigenvectors of T(Eq. A.81):
S−1=1√
6
√
2√
31√
2i−√
3ii
−√
202
;S=(S−1)†=1√
6
√
2−√
2i−√
2√
3√
3i0
1−i2
.
STS−1=1
6
√
2−√
2i−√
2√
3√
3i0
1−i2
2i1
−i2i
1−i2
√
2√
31√
2i−√
3ii
−√
202
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
03√
33
0−3√
3i3i
00 6
=1
6
00 0
01 8 0001 8
=
000
030003
./check
Problem A.27
(a)/angbracketleftˆUα|ˆUβ/angbracketright=/angbracketleftˆU†ˆUα|β/angbracketright=/angbracketleftα|β/angbracketright./check
(b)ˆU|α/angbracketright=λ|α/angbracketright=⇒/angbracketleftˆUα|ˆUα/angbracketright=|λ|2/angbracketleftα|α/angbracketright.But from (a) this is also /angbracketleftα|α/angbracketright.S o|λ|=1./check
(c)ˆU|α/angbracketright=λ|α/angbracketright,ˆU|β/angbracketright=µ|β/angbracketright=⇒|β/angbracketright=µˆU−1|β/angbracketright,s oˆU†|β/angbracketright=1
µ|β/angbracketright=µ∗|β/angbracketright(from (b)).
/angbracketleftβ|ˆUα/angbracketright=λ/angbracketleftβ|α/angbracketright=/angbracketleftˆU†β|α/angbracketright=µ/angbracketleftβ|α/angbracketright,o r(λ−µ)/angbracketleftβ|α/angbracketright=0.So ifλ/negationslash=µ, then/angbracketleftβ|α/angbracketright=0.QED
Problem A.28
(a) (i)
M2=
004
000000
;M
3=
000
000000
,so
c/circlecopyrt2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
APPENDIX. LINEAR ALGEBRA 297
eM=
100
010001
+
013
004000
+1
2
004
000000
=
115
014001
.
(ii)
M2=/parenleftbigg
−θ20
0−θ2/parenrightbigg
=−θ2I;M3=−θ3M;M4=θ4I; etc.
eM=I+θ/parenleftbigg
01
−10/parenrightbigg
−1
2θ2I−θ3
3!/parenleftbigg
01
−10/parenrightbigg
+θ4
4!I+···
=/parenleftbigg
1−θ2
2+θ4
4!−···/parenrightbigg
I+/parenleftbigg
θ−θ3
3!+θ5
5!−···/parenrightbigg/parenleftbigg01
−10/parenrightbigg
= cosθ/parenleftbigg10
01/parenrightbigg
+ sinθ/parenleftbigg01
−10/parenrightbigg
=/parenleftbiggcosθsinθ
−sinθcosθ/parenrightbigg
.
(b)
SMS−1=D=
d1 0
...
0dn
for some S.
SeMS−1=S/parenleftbigg
I+M+1
2M2+1
3!M3+···/parenrightbigg
S−1.Insert SS−1=I:
SeMS−1=I+SMS−1+1
2SMS−1SMS−1+1
3!SMS−1SMS−1SMS−1+···
=I+D+1
2D2+1
3!D3+···=eD.Evidently
det(eD) = det( SeMS−1) = det( S) det(eM) det(S−1) = det( eM).But
D2=
d2
1 0
...
0d2
n
,D3=
d3
1 0
...
0d3
n
,Dk=
dk
1 0
...
0dk
n
,so
eD=I+
d1 0
...
0dn
+1
2
d2
1 0
...
0d2
n
+1
3!
d3
1 0
...
0d3
n
+···=
ed1 0
...
0edn
.
det(eD)=ed1ed2···edn=e(d1+d2+···dn)=eTrD=eTrM(Eq. A.68), so det( eM)=eTrM.QED
(c)Matrices that commute obey the same algebraic rules as ordinary numbers , so the standard proofs of
ex+y=exeywill do the job. Here are two:
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
298 APPENDIX. LINEAR ALGEBRA
nm
(i)Combinatorial Method: Use the binomial theorem (valid if multiplication is commutative):
eM+N=∞/summationdisplay
n=01
n!(M+N)n=∞/summationdisplay
n=01
n!n/summationdisplay
m=0/parenleftbiggn
m/parenrightbigg
MmNn−m=∞/summationdisplay
n=0n/summationdisplay
m=01
m!(n−m)!MmNn−m.
Instead of summing vertically first, for fixed n(m:0→n), sum horizontally first, for fixed m(n:
m→∞,o rk≡n−m:0→∞)—see diagram (each dot represents a term in the double sum).
eM+N=∞/summationdisplay
m=01
m!Mm∞/summationdisplay
k=01
k!Nk=eMeN.QED
(ii)Analytic Method: Let
S(λ)≡eλMeλN;dS
dλ=MeλMeλN+eλMNeλN=(M+N)eλMeλN=(M+N)S.
(The second equality, in which we pull Nthrough eλM, would not hold if MandNdid not commute.)
Solving the differential equation: S(λ)=Ae(M+N)λ, for some constant A. But S(0) = I,s oA=1 ,
and hence eλMeλN=eλ(M+N), and (setting λ= 1) we conclude that eMeN=e(M+N).[This method
generalizes most easily when MandNdonotcommute—leading to the famous Baker-Campbell-
Hausdorf lemma.]
As a counterexample when [ M,N]/negationslash=0, letM=/parenleftbigg01
00/parenrightbigg
,N=/parenleftbigg00
−10/parenrightbigg
.Then M2=N2=0,s o
eM=I+M=/parenleftbigg11
01/parenrightbigg
,eN=I+N=/parenleftbigg10
−11/parenrightbigg
;eMeN=/parenleftbigg11
01/parenrightbigg/parenleftbigg10
−11/parenrightbigg
=/parenleftbigg01
−11/parenrightbigg
.
But (M+N)=/parenleftbigg01
−10/parenrightbigg
,so (from a(ii)): eM+N=/parenleftbiggcos(1) sin(1)
−sin(1) cos(1)/parenrightbigg
.
The two are clearly not equal.
(d)
eiH=∞/summationdisplay
n=01
n!inHn=⇒(eiH)†=∞/summationdisplay
n=01
n!(−i)n(H†)n=∞/summationdisplay
n=01
n!(−i)nHn=e−iH(forHhermitian) .
(eiH)†(eiH)=e−iHeiH=ei(H−H)=I,using (c). So eiHis unitary ./check
c/circlecopyrt2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
299
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 1
2/e 1/e
1 1
2N
3 6
4 7
5 8
6 11
7 12
8 13
9 14
10 2
11 3
12 4
13 5
14 9M
15 10
16N
17N
18N
Chapter 2
2/e 1/e
1 1
2 2
3 3
4 5
5 6M
6 7
7N
8N
9N
10 13M
11 14
12 37
13 17M
14N
15 15
16 16
17 18
18 19M
19N
20 20
21N
22 22
23 23
24 24
25N
26 25
27 26
28 27
29 28
30 29
31 30
32 31
33 32
34 33
35 41M
36 4M
37 36
38 3.48
39N
40N
41N
42 38
43 40
44 39 Chapter 2 (cont.)
2/e 1/e
45 42
46 43
47 44
48N
49 45
50 47
51 48M
52 34M, 35M
53 49
54N
55N
56N
300
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
(M) = 1/e problem number (distant model for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 3
2/e 1/e
1N
2N (33M)
3N (21M)
4N (12M)
5N
6N
7N
8N
9N
10N
11 38
12 51
13 41M
14 39
15N
16 42
17 43
18 44
19 45
20 46
21 57M
22N
23N
24 57M
25 25M
26N
27N
28 52M
29N
30N
31 53
32 56
33 50
34 49M
35N
36N
37N
38N
39 55
40N
Chapter 4
2/e 1/e
1 1
2 2
3 3
4 4
5 5
6 6
7 7M
8 8
9 9M
10 10
11 11
12 12
13 13
14N
15N
16 17
17 16
18 19
19 20
20 21
21N
22 22
23 23
24 25
25 26
26 27
27 28
28 29
29 30
30 31M
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41N
42 42 Chapter 4 (cont.)
2/e 1/e
43 43
44N
45 14
46 15
47N
48N
49N
50 44
51 45M
52 46
53N
54 47
55 48
56 49
57 50
58N
59 51
60 52M
61 53
301
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 5
2/e 1/e
1 1
2 2
3N
4 3
5 4
6 5
7 6
8 7
9 8
10 9
11 10
12 11M
13 11M
14 12
15N
16 13
17 14
18 15M
19 16M
20 17M
21 18
22 19M
23 20
24 21M
25 22
26 23
27 24
28 25
29 26
30 27M
31 28
32N
33 29
34 30
35 31
36 32
37 33
Chapter 6
2/e 1/e
1 1M
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10N
11 10
12 11
13 12
14 13
15N
16 14
17 15
18 16
19 17
20 18
21 19
22 20
23 21
24 22
25 23
26 24
27 25
28 26
29N
30N
31N
32 27
33 28
34 29
35 30
36 31
37 32
38 33
39 34
40N
Chapter 7
2/e 1/e
1 1
2 2M
3 3M
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11N
12N
13 11
14 12
15 13
16 14
17 15
18 16
19 17
20N
302
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 8
2/e 1/e
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16N
17N
Chapter 9
2/e 1/e
1 1
2 2
3 3M
4 4
5 5
6 6
7 7
8 8
9N
10 9
11 10
12 11
13 12
14 13
15 14
16 15
17 16
18 17
19 21
20 19M
21 20
22N
Chapter 10
2/e 1/e
1 1
2 3M
3 4
4 5
5 6
6 8
7 9
8N
9 10
10 11M
303
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 11
2/e 1/e
1 1
2 2
3 3
4 4
5N
6N
7N
8 5
9 6
10 7
11 8
12 9
13 10
14 11
15 12
16 13
17 14
18 15
19N
20N
Chapter 12
2/e 1/e
1N
Appendix
2/e 1/e
1 3.1
2 3.2
3 3.3
4 3.4
5 3.5
6 3.6
7 3.7
8 3.9
9 3.10.
10 3.11
11 3.12
12 3.16
13N
14N
15 3.13
16 3.14
17 3.15
18 3.17
19 3.18
20 3.19
21 3.20.
22 3.40M
23N
24 3.21M
25 3.22
26 3.23
27 3.24
28 3.47