rotation group generators and their powers
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Short note by Phil dated 1.22.08. It constructs the (2j+1)-dimensional matrices Jz, J+, J-, Jx and Jy from their matrix elements <m|J|m'>, citing Tinkham. It then examines powers: Jz powers are all independent for general j, while for j=1/2 Jz^2 = 1/4 and Ji^3 = Ji/4, and for j=1 Ji^3 = Ji, checked with Maple. The case j>1 for Jx and Jy is left open.
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Rotation Group Generator Matrices and their Powers PhL 1.22.08
1. The Generator Matrices. First of all, how do we build these matrices for an arbitrary representation j? The dimension of the matrices is going to be (2j+1), at least we know that much.
We always know that
<m|J3|m'> = mm'm so J3 = diag(j,j-1,j-2...,-j) // examples: diag(1/2,-1/2), diag(1,0,-1)
Then we can say [ where J = Jx iJy and the inverse Jx = (J+ + J-)/2 and Jy =( J+ - J- )/(2i) ]
<m|J|m'> = <m|m' 1> // Tinkham page 145
= m,m'1 = m,m'1
so these matrices have real numbers on a first off-diagonal. It follows that
<m|Jx|m'> = (1/2) m,m'+1 + (1/2) m,m'-1
= (1/2)[ m,m'+1 + m,m'-1]
<m|Jy|m'> = (1/2i)[ m,m'+1 - m,m'-1]
So we now have our answers:
Generator matrices for any value of j :
<m|Jz|m'> = mm'm = (Jz)mm'
<m|Jx|m'> =(1/2)[ m,m'+1 + m,m'-1] = (Jx)mm'
<m|Jy|m'> = (1/2i)[ m,m'+1 m,m'-1] = (Jy)mm'
<m|J|m'> = m,m'1 = (J)mm'
2. Powers of the generator matrices. Now we can look at powers of these matrices:
(Jz)2mm' = m2m,m' and more generally (Jz)Nmm' = mNm,m'
For general values of j, each power of (Jz) is a brand new matrix, and you cannot write (Jz)N as a linear combination of lower powers. For example:
Jz = diag[ 3/2, 1/2, -1/2, -3/2 ]
JzN = diag [ (3/2)N, (1/2)N, (-1/2)N,(-3/2)N ] j = 3/2
JzN = diag [ 2N, 1, 0, (-1)N ,(-2)N ] j = 2
However, in the cases of j=1/2 and j=1 we find some simplifications.
First, for j = 1/2 we get
(Jz)2mm' = m2m,m' = (1/4) mm' so Jz2 = (1/4) 1
(Jz)N = (1/2)N 1 N = even
(Jz)N+1 = (1/2)N Jz N = even
In general we can say Ji = (1/2) i and it turns out in that i2 = 1 for i = z and x and y.
Secondly, for j = 1 we get
Jz = diag[ 1, 0, -1 ]
Jz2 = diag[ 1, 0, 1 ]
JzN = diag[ 1, 0, 1 ] N even
JzN+1 = Jz N even so Jz3 = Jz
What about powers of Jx and Jy ? Consider:
<m|Jx|m'> =(1/2)[ m,m'+1 + m,m'-1]
Matrix elements vanish unless m,m' differ by 1. But if they so differ, then m*m' = 0 such as 1*0 = 0. Then we find that
<m|Jx|m'> = (1/)[ m,m'+1 + m,m'-1] = Jx = (1/)
If we tell Maple to square this we get
Jx2 = (1/2) and then Jx3 = (1/) = Jx
so we can use Jx3 = Jx to reduce powers. Similarly,
<m|Jy|m'> = <m|Jy|m'> = (1/2i)[ m,m'+1 - m,m'-1]
= (1/i) [ m,m'+1 - m,m'-1] Jy = (1/i)
The matrix in the brackets when cubed gives -2 times itself says Maple, so not too surprisingly we find that Jy3 = Jy.
Summary of the special cases of powers of generators: "powers"
j=1/2 Ji3 = Ji/4 for i = x,y or z and Ji2 = (1/4) *1
j=1 Ji3 = Ji for i = x, y or z
j > 1 powers of Jz are all different ( and not sure what happens for powers of Jx and Jy )