messiah R2 SSPT Chap 16

Messiah Chapter 16 Stationary Perturbation Theory Notes, Round Two PhL 1.23.09 symbol grab: ΣkE H(m) H Ea0 (Ea0) I am not ready for "meta notes" yet. Here we take a second pass through Messiah's offering, and I try to impose my own structure on his information, adding details where I think useful. The Kato Topic III is not treated in these Round Two notes. Topic I. The Non-Degenerate Case (686) 2 2. Expansion in Powers of the Perturbation (686) 2 3. First Order Perturbations (688) 2 6. Higher Order Corrections (694) 2 a. Mapping Notations → and , and the Inverse of a Linear Operator 2 b. The General Projection operator R 3 c. Projection operators P0 and Q0 3 d. Projection-Sandwich Operators in H 4 e. The operator Q0a 5 f. Does Ω-1 = (Ea0 - H0)-1 exist? 6 g. The Perturbation Theory Equation Set 6 h. Energy Equations for the Non-degenerate Case. 8 i. The Utility of Qa0 and solving the non-degenerate problem to all orders. 8 j. State Corrections for the Non-degenerate Case. 9 k. Solution Algorithm for the Non-degenerate Case. 9 l. Examples and a possible graphical language. 10 Topic II. The Degenerate Case (698) 11 8. Elementary Theory (698) 11 Finding the ε1 eigenvalue problem. 12 Express the ε1 eigenvalue problem in matrix notation: 12 Express the ε1 eigenvalue problem as an operator equation. 13 Finding the ε2 eigenvalue problem. 14 Express the ε2 eigenvalue problem in matrix notation: 15 Express the ε2 eigenvalue problem as an operator equation. 16 Finding the ε3 eigenvalue problem. 16 Finding the εs eigenvalue problem. 18 P1. Finding the state corrections. 18 Computation of |1; n,ε1> when N > 1 and |n,ε1> is non degenerate at the ε1 level 18 Review of Computation of |1; n,ε1> when N > 1 22 Computation of |1; n,ε1> when N = 1 ( |n,ε1> is in this case non-degenerate at ε1 level) 24 Computation of |2; n,ε1> in the case N > 1 where |n,ε1> is non-degenerate at ε1 level 25 Computation of |2; n, ε1> in the case N = 1 ( |n,ε1> is non-degenerate at the ε1 level) 28 Computation of |1; m,ε2> in the case N > 1 and N1> 0 where |m,ε2> is non-degenerate at ε2 level 29 Computation of |2; m,ε2> in the case N > 1 and N1> 0 where |m,ε2> is non-degenerate at ε2 level 34 Comments 34 Topic I. The Non-Degenerate Case (686) Actually, I am going to try to stay as general as possible (ie, allow degeneracy) except where stated otherwise. 2. Expansion in Powers of the Perturbation (686) 3. First Order Perturbations (688) 6. Higher Order Corrections (694) Everything done here is analogous to Schiff (which I read first), but the notation is different, so here is a summary of the notation comparison: Schiff/PL Messiah m a // selected state (or space of N degenerate states) H0(m) Ea0 // space containing N deg unperturbed state(s) of interest H(m) Ea // space containing the N perturbed states H = H0 H = H0 // full Hilbert space H' V // perturbing Hamiltonian Wm = Σn=0 λn Wm,n E = Ea0 + Σn=1 λn εn // energy expansion Wm Ea Wm,0 Ea0 Wm,n εn n = 1,2,3... ψm = Σn=0 λn ψm,n |ψ> = Σn=0 λn |n> // state expansion ψm,n |n> n = 0,1,2.... // the tag "a" is implicit <um| ψm,n> = 0 n>0 <0|n> = 0 n>0 // special normalization The Main Equation set for perturbation theory is 31.4 in Schiff and appears as (7) page 687 in Messiah. Before we get to it, there are many "preliminaries" to be considered. There are several totally new ingredients in Messiah not present in Schiff. a. Mapping Notations → and , and the Inverse of a Linear Operator Consider an operator Ω0: H0→ H0 . For bound state problems, H0 could be either a finite or infinite dimensional Hilbert Space (we sort of ignore the continuous part of the spectrum, the collision part for H). When we "mess with operators", we are in general trying to find equations that are valid in all of H0 so we can use them freely. In the above notation, H0→ H0 shows the two spaces involved, it does not show what the domain and range actually might be for Ω0 , and you really need to know these two things about Ω0 to use it. Usually in our application the domain will be all of H0. We know that if Ω0 has a non-trivial nullspace and if it is Hermitian or symmetric or self-adjoint of whatever, then the Alternative Theorem tells us that the range is less than all of H0. More simply put, if Ax0 = 0, then the problem Ax=f which has a solution (x,f) also has a solution A(x+x0) = f so that the image vector f maps back into lots of domain vectors, which means A is not one-to-one and this means A-1 does not exist. This fact is true regardless of the dimension of H0, finite or infinite. If we know that Ω0 has a certain range when the domain is all of H0, we will use this notation to show that fact: Ω0: H0 R0 . This is my own made-up notation just to clarify this precise point, Ω0:domain range. The various 0's here just remind us we are talking about the unperturbed state Hilbert Space H0 . b. The General Projection operator R Suppose also that R is a projection operator R: H0 S which projects onto some subspace S in H. Then as above, we have R representable as "a portion of unity" : R = ΣrS |r><r| 1 = Σn,all |n><n| where in general |k> form an orthonormal basis for H0. It is easy to show that R2 = R: just put in the sums shown and use <r|r'> = δr,r'. It is also easy to show that R = R†. The simplest proof is to realize that [|r>]† = <r| and similarly for the ket. Perhaps a more convincing proof is this: Rij = <i|R|j> = ΣrS <i | r><r | j> = ΣrS <r | j><i | r> = ΣrS <j | r>*<r |i>* = Rji* = (R†)ij where i and j can be any states, not just basis states. If we use basis states, then we have Rij = <i|R|j> = ΣrS <i | r><r | j> = δi,j if both in S, else = 0 and then we see that Rij is just a partially populated unit matrix (a portion of unity) which of course is Hermitian, and Hermiticity is in invariant under any unitary transformation to another basis, so R is Hermitian in any basis. So, our main ideas here are these: R: H0→ H0 R: H0 S R2 = R R = R† R|r> = |r>, rS R|k> =0, kS R-1 does not exist where the last follows since R has a non-trivial nullspace S as shown on the same line. c. Projection operators P0 and Q0 The first new ingredients appearing in Messiah are the P0 and Q0 operators. P0: H0 Ea0 is very clear as a projection operator. It's partner is Q0: H0 (Ea0) and of course P0 + Q0 = 1. Really these things should be called Pa0 and Qa0. Later on, he introduces an analogous operator P: H→ Ea , but we won't need that for a while. Here are some good equations: P0 = ΣjEoa |j><j| // portion of full unity accounted for by the Ea0 subspace Q0 = ΣkEoa |k><k| // portion of full unity accounted for by all other states in H0 1 = Σn |n><n| // sum over all states gives the full unity P0Q0 = Q0P0 = 0 // should be obvious P02 = P0 and P0†= P0 and Q02 = Q0 and Q0†= Q0 Warning: We know that P02 = 1. If we write 1 = P0P0-1 then we are tempted to say that P0-1 = P0. This is wrong! Because the range of P0 is less than H0 (or because P0 has a non-trivial nullspace), we know (as just discussed above) that the operator P0-1 does not exist, so don't try to use it in any calculations! This applies to any projection operator, and hence to Q0 as well. d. Projection-Sandwich Operators in H Suppose Ω is diagonal in our selected basis for space H with eigenvalues ωk. Suppose also that R is a projection operator which projects onto some subspace S in H. Then as above, we have R = ΣrS |r><r| Consider then this combination operator F = R f(Ω) R (1) where we sandwich an arbitrary function of Ω between two copies of our projection operator. It is trivial then to show that ( the eigenvalues ωr need not be real) F = ΣrS |r> f(ωr) <r| (2) Here we use the tradition of putting the number f(ωr) in the middle, but it could go anywhere: ΣrS |r>Cr <r| = ΣrS Cr |r> <r| = ΣrS |r> <r| Cr because this thing is just a linear combination of operators |r> <r| with coefficient Cr. As a special case, suppose f(Ω) = Ω-1 ( which we assume exists). Then we have F = R Ω-1 R = ΣrS |r> (1/ωr) <r| (3) where we assume ωr ≠ 0 at least in the set S. Then we find this interesting result: Ω F = R (4) Just insert the expansion and use Ω |r> = ωr |r> and then you get (1/ωr)ωr = 1 and you end up with R. The above implies that F† Ω† = R . If Ω is Hermitian, then so is F and we have F Ω = R // Ω = Ω† (5) In this case, we also have that F Ω|r1> = R|r1> = |r1> // Ω = Ω† (6) This says that for such a state, F acts in effect as the inverse of Ω, nothing very profound. Now here is another theorem that is easy to prove F1F2F3 = R f1(Ω) R R f2(Ω) R R f3(Ω) R = R f1(Ω) Rf2(Ω) Rf3(Ω) R (7) = ΣrS |r> f1(ωr) f2(ωr) f3(ωr)<r| In this proof, we took every fi(Ω) and acted to the right on |r> to get fi(ωr) |r> and then we used <r|r'> = δr,r' on three of the r sums. You can never get rid of the last sum! This result does not require fi to be a real function nor does it require that the ωr be real. As a special case of the above theorem we have the leftmost equality here Fn = ΣrS |r> f (ωr)n <r| = R f(Ω)n R (8) and the rightmost equality follows trivially by -- as usual -- inserting the expansion for R. e. The operator Q0a This operator is a special case of the above where R = Q0 and Ω = (Ea0 1 - H0) = (Ea0 - H0). Recall that Q0 projects onto the perp space of our space Ea0 "of interest". So Q0a ≡ Q0 (Ea0 - H0)-1Q0 = – Q0 G0 Q0 where both Q0 and H0 on the RHS are operators, and of course Ea0 is just a number. Later we might refer to this central operator (Ea0 - H0)-1 as (minus) the unperturbed resolvent G0, and I show that on the right above. Quoting result (d.8) above we have (Q0a)n = Q0 (Ea0 - H0)-nQ0 = Σ kEoa |k> (Ea0 - Ek0)-n < k| = " " Because the operator (Q0a)n has Q0 on each end, it becomes 0 if it is pressed against a P0 operator on either the left or right. For example, consider this where we have lots of operators: ABC....FP0(Q0a)nXY...Z = ABC....F (Q0a)n P0XY...Z = 0 Messiah uses the exceedingly unhelpful notation shown on the right to represent (Q0a)n . It is meant to remind the user that (Ea0 - Ek0)-n symbolized by a-n is sandwiched between the normal Q0 operators. I don't use this notation in any of my notes. Notice that Ω = (Ea0 - H0) is Hermitian, so as noted above before (5), Q0a is also Hermitian, so we get both results (4) and (5) , ΩF = R = FΩ F = R Ω-1 R (Ea0 - H0) Q0a = Q0 = Q0a (Ea0 - H0) Q0a ≡ Q0 (Ea0 - H0)-1Q0 f. Does Ω-1 = (Ea0 - H0)-1 exist? The SE tells us that Ho|ψ> = Ea0|ψ> for any ψ inside Ea0. Thus, we know that (Ea0 - H0) |ψ>= 0 so the operator (Ea0 - H0) clearly has some non-trivial nullspace. We therefore know that (Ea0 - H0)-1 does not exist if we think of (Ea0 - H0): H0→ H0. However, we can also think of a different operator (Ea0 - H0): (Ea0)→ (Ea0) where we have changed the mapping spaces. This is not domain and range necessarily, we use the → notation to refer to the spaces. Each space here is in fact a Hilbert Space, by the way. For any state |k> in (Ea0) , we know that the SE says Ho|k> = Ek0|k> where Ek0 ≠ Ea0 . So yes, (Ek0 - H0) |k> = 0. But then we have (Ea0 - H0) |k> = (Ea0 - H0 + Ek0 - Ek0) |k> = (Ea0 - Ek0) |k> ≠ 0 |k> (Ea0) Since we get ≠ 0 for all basis kets in (Ea0), it is clear that we get ≠ 0 for any ket in (Ea0) . Since we never get 0, that means (Ea0 - H0) has only the trivial nullspace in (Ea0) and thus (Ea0 - H0)-1 exists, where we have now carefully restricted what we mean by (Ea0 - H0). Now consider again Q0a ≡ Q0 (Ea0 - H0)-1Q0 . The central part here (Ea0 - H0)-1 can never act on a ket in Ea0 ! It has "protectors" on each side, in the form of Q0. Thus, in effect, (Ea0 - H0)-1 here can only act on |k> or <k| (Ea0) ! This is I think a key fact about this Q0a operator! You can think of this really as the product of three operator acting left to right as follows (we start on with the left Q0) Q0 : H0 (Ea0) (Ea0 - H0)-1: (Ea0) → (Ea0) Q0 : (Ea0) (Ea0) Q0a: H0 (Ea0) overall The last Q0 does not do anything when viewed this way, but we can alternatively think of the three operators as acting in the reverse direction on the bra space, and then you need the left side Q0 to be present since it is then the first to act! So basically our two "protectors" allow us to have the function (Ea0 - H0)-1: (Ea0) → (Ea0) exist in the middle, and this inverse does exist, as we just showed ! It is amazing how much is going on here that Messiah glosses over in the usual writer's need to minimize pages spent on obscure topics in a textbook. g. The Perturbation Theory Equation Set We must now look at the main perturbation theory equation set which is Schiff 31.4 or Messiah 16.7 page 687. In my Schiff notes I give an exact derivation of the equation set and I write it in this way (H0 - Em) ψ(m)s = ( W(m)1 - H' ) ψ(m)s-1 + Σj=2,3..s W(m)j ψ(m)s-j s = 1,2,3... so (H0 - Em) ψ(m)0 = 0 s = 0 (H0 - Em) ψ(m)1 = ( W(m)1 - H' ) ψ(m)0 s = 1 (H0 - Em) ψ(m)2 = ( W(m)1 - H' ) ψ(m)1 + W(m)2 ψ(m)0 s = 2 (H0 - Em) ψ(m)3 = ( W(m)1 - H' ) ψ(m)2 + W(m)2 ψ(m)1 + W(m)3 ψ(m)0 s = 3 where each value of s is an equality for the power λs in the equation (H+λH')ψ = Eψ where we expand both ψ and E in λ. Here is a translation of the above into Messiah notation: (H0 - Ea0) |s> = ( ε1 - V ) |s-1> + Σj=2,3..s εj |s-j> s = 1,2,3... so (H0 - Ea0) |0> = 0 s = 0 (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (*) (H0 - Ea0) |3> = (ε1 - V) |2> + ε2 |1> + ε3 |0> s = 3 (H0 - Ea0) |4> = (ε1 - V) |3> + ε2 |2> + ε3 |1> + ε4 |0> s = 4 (H0 - Ea0) |5> = (ε1 - V) |4> + ε2 |3> + ε3 |2> + ε4 |1> + ε5 |0> s = 5 where I show lots of equations to make the pattern clear. You can see the benefit of Messiah's notation in that things are compacted down very nicely. State |0> is some unperturbed eigenstate of H0 that lies inside Ea0 (see s=0 equation above). WARNING #1: In our write-up here, we now have two distinct meanings for the notation |s>. In our discussion earlier on projection operators, |s> was an arbitrary eigenstate of H0 |s> = En0 |s>. But in the above perturbation expansion, |s> refers to the λs order correction of some H eigenstate state un. We might expand Messiah's notation to |s; n> to mean the sth correction to the state |n>. Reader beware! WARNING #2. This state |0> = |0; n> we have said is an eigenstate of H0 with energy Ea0, that is what the s=0 equation above says. BUT, it is not just any eigenstate of H0 !! This is a potential pitfall if you don't understand why. The state |0; n> is one of the special eigenstates of H0 within Ea0 which is the λ = 0 component of a state we might just call |n_full> which solves H |n> = En |n>, where now H includes the perturbation and En is the exact energy to all orders, and |n_full> is the exact state to all orders. So in the above set of equations, if you start off with some arbitrary spanning set of basis states |r> that span Ea0, it is very unlikely that the state |r> will be a legal state |0;n> for the perturbation equations shown above! You have to somehow discover what the legal states |0; n> really are ! . While we are here, suppose we close from the left with a basis bra <r| from the degenerate subspace Ea0. We then get, 0 = <r| ( ε1 - V ) |s-1;n> + Σj=2,3..s εj <r| s-j;n> s = 1,2,3... 0 = <r| ( ε1 - V ) |s-1;n> + Σj=2,3..s-1 εj <r| s-j;n> + εs<0;r| 0;n> s = 2,3... so 0 = 0 s = 0 0 = <r| (ε1 - V ) |0;n > s = 1 0 = <r| (ε1 - V ) |1;n > + ε2<r|0;n> s = 2 0 = <r| (ε1 - V) |2;n > + ε2<r|1;n > + ε3<r| 0;n> s = 3 0 = <r| (ε1 - V) |3;n > + ε2<r|2;n > + ε3 <r |1;n > + ε4<r| 0;n> s = 4 0 = <r| (ε1 - V) |4;n > + ε2<r|3;n > + ε3 <r |2;n > + ε4 <r |1;n > + ε5<r| 0;n> s = 5 and as just noted, we cannot replace <r| 0;n> with δr,n since | 0;n> is in general a lincomb of the basis states |r> in Ea0. In the general degenerate case, I don't think there is much one can do with the above set of equations. h. Energy Equations for the Non-degenerate Case. Now, in the special case that Ea0 is non-degenerate, we can identify | 0;n> = |r> because there is only one state in Ea0. Then we can replace <r|0;n> = 1. Moreover, factors like <r|1;n > = 0 due to the tricky method of normalization, so things simplify a lot: ( do the s=1 case manually from above): 0 = <0;n| ( ε1 - V ) |s-1;n> + εs s = 2,3... so 0 = <0;n| (- V ) |0;n > + ε1 s = 1 0 = <0;n| (- V ) |1;n > + ε2 s = 2 0 = <0;n| (- V) |2;n > + ε3 s = 3 0 = <0;n| (- V) |3;n > + ε4 s = 4 0 = <0;n| (- V) |4;n > + ε5 s = 5 All the equations have exactly the same form and we conclude that εs = <0;n| V |s-1;n> including ε1 = <0;n| V |0;n > = Vnn We see the result that each energy correction is determined by the state correction of the preceding order. But, going down just this path, we don't have any formulas for the state corrections, so we cannot really go on to compute the higher energy corrections. We now resume talking about the general degenerate case. i. The Utility of Qa0 and solving the non-degenerate problem to all orders. Now, back to the utility of Qa0. So far, Q0a has been just an amusing mathematical toy with some odd properties. What happens if we apply Qa0 to all these equations in set (*) in the previous section. Recall from just above our theorem that Q0a (Ea0 - H0) = Q0 Thus, on the LHS we can replace the fairly complicated operator (H0 - Ea0) by - Q0. And on the RHS we know that Q0 on the right end of Q0a kills off the |0> state which we know lies in Ea0. So the above equations become -Q0 |s> = Q0a ( ε1 - V ) |s-1> + Σj=2,3..s εj Q0a |s-j> s = 1,2,3... Q0 |s> = Q0a V |s-1> – ε1 Q0a |s-1> – Σj=2,3..s εj Q0a |s-j> s = 1,2,3... Q0 |s> = Q0a V |s-1> – Σj=1,3..s εj Q0a |s-j> s = 1,2,3... so Q0 |1> = Q0aV |0> s = 1 (**) Q0 |2> = Q0aV |1> – ε1 Q0a |1> s = 2 Q0 |3> = Q0aV |2> – ε1 Q0a |2> – ε2 Q0a |1> s = 3 Q0 |4> = Q0aV |3> – ε1 Q0a |3> – ε2 Q0a |2> – ε3 Q0a |1> s = 4 Q0 |5> = Q0aV |4> – ε1 Q0a |4> – ε2 Q0a |3> – ε3 Q0a |2> – ε4 Q0a |1> s = 5 In general, each of these equations tells us something about the portion of the state correction |n> which lies in the space (Ea0). The above equations are valid in both the degenerate and non-degenerate case. j. State Corrections for the Non-degenerate Case. Now just for the moment, suppose we were doing non-degenerate with the usual normalization. In this case, Qo|n> = |n> for n > 1. Then we get |s> = Q0a V |s-1> – Σj=1,3..s εj Q0a |s-j> s = 1,2,3... |1> = Q0aV |0> // agrees with Messiah p 689 (14) s = 1 |2> = Q0aV |1> – ε1 Q0a |1> s = 2 |3> = Q0aV |2> – ε1 Q0a |2> – ε2 Q0a |1> s = 3 |4> = Q0aV |3> – ε1 Q0a |3> – ε2 Q0a |2> – ε3 Q0a |1> s = 4 |5> = Q0aV |4> – ε1 Q0a |4> – ε2 Q0a |3> – ε3 Q0a |2> – ε4 Q0a |1> s = 5 and now we have explicit formulas for each state correction. We can then jam these states into our energy correction equation we found above which said εs = <0;r|V|s-1;r> for our non-degenerate (now) state r. k. Solution Algorithm for the Non-degenerate Case. So here is a viable algorithm for solving to all orders in this case: [ each pair of computations can be done in either order ] for s = 1 to ∞ do begin compute |s> = Q0a V |s-1> – Σj=1,3..s εj Q0a |s-j> compute εs = <0;n| V |s-1;n> end Here is the thing linearized for a bit: # terms |1> = Q0aV |0> 1 s = 1 ε1 = <0| V |0 > = Vnn // Messiah p 689 (12) 1 |2> = Q0aV |1> – ε1 Q0a |1> 2 s = 2 ε2 = <0| V |1> = <0|V Q0aV |0> // Messiah p 694 (26) 1 |3> = Q0aV |2> – ε1 Q0a |2> – ε2 Q0a |1> 5 s = 3 ε3 = <0| V |2> 2 |4> = Q0aV |3> – ε1 Q0a |3> – ε2 Q0a |2> – ε3 Q0a |1> 13 s = 4 ε4 = <0| V |3> 5 |5> = Q0aV |4> – ε1 Q0a |4> – ε2 Q0a |3> – ε3 Q0a |2> – ε4 Q0a |1> 34 s = 5 ε5 = <0| V |4> 13 You can see that expressions are going to get very messy. I will write out some of the simpler ones: |2> = Q0aV Q0aV |0> – ε1 Q0a2 V |0> |3> = Q0aV{ Q0aV Q0aV |0> – ε1 Q0a Q0aV |0>} – ε1 Q0a { Q0aV Q0aV |0> – ε1 Q0a Q0aV |0>} – ε2 Q0a Q0aV |0> = Q0aV Q0aV Q0aV |0> – ε1 Q0a2V |0> - ε1 Q0a 2V Q0aV |0> + ε12 Q0a3 V |0> – ε2 Q0a2V |0> so here are the 5 terms alluded to in the table above. We know that dim(Q0a) = 1/energy, so the operator in each term above (and the term overall) is dimensionless, a way perhaps to check things. Each time there is a power of Q0a, we are going to have a sum as shown in the examples below, so the first term on the last time implies three sums. So, the reader perhaps now appreciates "the utility" of the operator Qa0 and sees why it was introduced. At least we have seen its use in the non-degenerate case. l. Examples and a possible graphical language. Here are some examples showing the sums implied by the Qa0 factors: |1;n> = Q0aV |0;n> = Σ k≠n |0;k> (Ea0 - Ek0)-1 <0;k| V |0;n> = Σ k'≠n |0;k'> { (Ea0 - Ek'0)-1Vk'n } Note: This appears as Messiah p 689 (16) where he uses <E0α | 1> for my <k|1;n> |2;n> = Q0a(V - ε1) |1;n> = Σ k'≠n |0;k'> (Ea0 - Ek'0)-1<0;k'| (V - ε1) |1;n> = Σ k'≠n |0;k'> (Ea0 - Ek'0)-1<0;k' | (V - ε1) { Σ k≠n |0;k> (Ea0 - Ek0)-1Vkn } = Σ k',k≠n |0;k'> (Ea0 - Ek'0)-1<0;k' | (V - ε1) |0;k> (Ea0 - Ek0)-1Vkn = Σ k',k≠n |0;k'> (Ea0 - Ek'0)-1{ Vk'k - Vnn δk',k} (Ea0 - Ek0)-1Vkn ε1 = Vnn = Σ k'≠n |0;k'> { Σ k≠n (Ea0 - Ek'0)-1(Ea0 - Ek0)-1Vk'k Vkn - (Ea0 - Ek'0)-2 Vk'n Vnn } where we can think of the thing in {..} as a coefficient of |0;k'> , namely, <0;k'| 2;n>. Thus we have computed that <0;k'| 1;n> = (Ea0 - Ek'0)-1Vk'n <0;k'| 2;n> = - (Ea0 - Ek'0)-2 Vk'n Vnn + Σ k≠n (Ea0 - Ek'0)-1(Ea0 - Ek0)-1Vk'k Vkn We can compare these with results from Schiff (from my raw notes on Chap 8) ψ(m)1 = Σk≠m a(m)k,1 uk = Σk≠m { <k| H' |m> /(Em - Ek) } uk a (m)k,2 = – <k| H' |m><m|H'|m> / (Em - Ek)2 + Σn≠m < k| H' |n><n| H' |m>/ [(Em - En) (Em - Ek)] // which is 31.13 so I think both my results for |2;n> above is correct. You can get rid of some minus signs by negating the terms in the energy difference denominators. One is very tempted to try to make some kind of Feynman like graph for this amplitude <0;k'| 2;n>, something like this: (remember that each V is of order λ ) In the degenerate case, we cannot carry out the above program because of the Q0 factors on the LHS of all our perturbation equations (**) shown above, so we need a different plan. Topic II. The Degenerate Case (698) 8. Elementary Theory (698) Continuing the degenerate case, let's go back to our original equations (H0 - Ea0) |s> = ( ε1 - V ) |s-1> + Σj=2,3..s εj |s-j> s = 1,2,3... so (H0 - Ea0) |0> = 0 s = 0 (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (*) (H0 - Ea0) |3> = (ε1 - V) |2> + ε2 |1> + ε3 |0> s = 3 (H0 - Ea0) |4> = (ε1 - V) |3> + ε2 |2> + ε3 |1> + ε4 |0> s = 4 (H0 - Ea0) |5> = (ε1 - V) |4> + ε2 |3> + ε3 |2> + ε4 |1> + ε5 |0> s = 5 Finding the ε1 eigenvalue problem. Apply P0 on the left (this is really Pa0). P0 kills off the LHS's. Also, since we know |0> is in Ea0 (though we don't know the right linear combinations yet for such states |0>), we have P0|0>= |0>. So we get this: (0 = ( ε1 - V ) |s-1> + Σj=2,3..s εj |s-j> s = 1,2,3... so 0 = 0 s = 0 0 = P0 (ε1 - V ) |0> = ε1 |0> – P0V ) |0> s = 1 0 = P0 (ε1 - V ) |1> + ε2 |0> s = 2 0 = P0 (ε1 - V) |2> + ε2 P0 |1> + ε3 |0> s = 3 0 = P0 (ε1 - V) |3> + ε2 P0 |2> + ε3 P0|1> + ε4 |0> s = 4 0 = P0 (ε1 - V) |4> + ε2 P0 |3> + ε3 P02> + ε4 P0 |1> + ε5 |0> s = 5 We now write out the s=1 equation which we see is the ε1 eigenvalue problem: P0V |0; n, ε1> = ε1,n |0; n,ε1> Our states |0; n,ε1> are thus eigenstates of the operator [P0V]. Since P0 and V are each Hermitian, so is the product, and we expect to have real eigenvalues and N eigenvectors for n = 1,2,...N where N is the dimension of our space Ea0. Note: we could write the above as P0VP0 |0> = ε1|0> and we would then be in agreement with the first full sentence of Messiah's on page 699. Express the ε1 eigenvalue problem in matrix notation: Let's close on the left with our initial set <r| of Ea0 basis states to get <r| P0V |0; n,ε1> = ε1,n <r| 0; n,ε1> <r| V P0 |0; n,ε1> = ε1,n <r |0; n,ε1> since P0 does nothing to Ea0 Now expand the projection operator Σr'<r| V |r'><r'| 0; n,ε1> = ε1,n <r | 0; n,ε1> // see Messiah p 699 A, he uses <Ea0α| for <r| Σr' Vrr' ψ(n)r' = ε1,n ψ(n)r ψ(n)r = <r | 0; n,ε1> V ψ(n) = ε1,n ψ(n) | 0; n,ε1> = ΣrEa0 |r><r | 0; n,ε1> So this is a well-defined NxN matrix diagonalization problem which we can solve both for the eigenvalues and the eigenvectors ψ(n). The secular equation says det(V- ε1,n) = 0 and tells us the eigenvalues. So we have our problem in full control to the point where we can determine the | 0; n,ε1> and the first order energy corrections ε1,n . Compare this to our computer program above for solving the whole problem. This is the first two steps. It could happen that some or even all of the ε1,n are degenerate and we will deal with that below. We can regard the new eigenstates | 0; n,ε1> as a unitary transformation of the original bases states |r> that spanned our subspace Ea0. Let's define U(1)r,n = <r | 0; n,ε1> U(1)*r,n = < 0; n,ε1| r> = U(1)†n,r = U(1)-1n,r U(1) will be a unitary matrix of all states are properly orthonormalized which we shall assume. Then we can say | 0; n,ε1> = ΣrEa0 |r><r | 0; n,ε1> = ΣrEa0 U(1)r,n |r> <0; n,ε1| = ΣrEa0 <0; n,ε1| r><r| = ΣrEa0 U(1)-1n,r <r| Thus unitary transformation is the matrix that diagonalizes V: (U(1)-1 V U(1))nn' = εnδn,n' Note 1: It may be that some values of ε1 have multiplicity > 1 so groups of states might still be degenerate at the ε1 level. Within each group, we will assume we have carried out the GSO process so that we have a well defined set of N orthonormal states which we call | 0; n,ε1>. Each of these is then a certain linear combination of our original base states |r>. Note 2: Remember that the level 0 states k in (Ea0) are orthogonal to all the |r> in Ea0 and are thus orthogonal to the diagonalizing states | 0; n,ε1> as well. This NxN diagonalization problem does not involve those other states k in (Ea0) . However, as we shall see later, the state corrections do involve these other states. Express the ε1 eigenvalue problem as an operator equation. Above we had (in simplified notation) P0V|0> = ε1 |0> [P0 V] |0; n, ε1> = ε1,n |0; n,ε1> where the eigenvectors |0> all exist in Ea0 . If we replace |0> with P0 |0> on both sides, the resulting equation will then be true not just for the |0> states which span Ea0, but also for all other states, and thus we end up with an operator equation version of our eigenvalue problem P0V P0 = ε1 P0 This is a wonderful trick. In effect, both sides of this equation have "protectors" (filters) facing both left and right. Thus, this equation will be valid in both the bra space and the ket space when acting on any state! This is the beauty of using projectors. Finding the ε2 eigenvalue problem. In the ε1 EV problem we found a set of N states |0; n> with eigenvalues ε1,n. These states are the level 0 components of eigenstates of the full H through first order (which means V is taken into account). Let's assume some number of these states N1 are still degenerate with some ε1 and let's focus on this group of states and ignore all the other states. The number N1 could range anywhere from 1 to N. The space spanned by these vectors |0; n,ε1> we could call Ea0(1). All N1 of these states have the same value for ε1 so they are degenerate both at the 0 level and at the 1 level of correction. But only certain linear combinations of the states |0; n,ε1> will be the level 0 components of actual solutions of the full SE. It is very unlikely that the Ea0(1) "base states" |0; n,ε1> we obtain from the ε1 EV problem are the right states! This is similar to how we said that <r| states which spanned Ea0 were unlikely to be the right states |0;n> which were level 0 components of full SE solutions. So we are going to discover an ε2 eigenvalue problem which is going to tell us the proper states |0;n,ε2> that are the right linear combinations of the |0; n,ε1> states which will have well-defined values of ε2. Obviously we might find that some of these |0; n,ε2> states are still degenerate. We would then sort of "recur" our algorithm here in a space Ea0(1)(1), say, and look for an ε3 eigenvalue problem to determine the right linear combinations of the |0; n,ε2> states which are then eigenstates of H up through third order. And we could continue this process to any level. For now, let's just try to find the ε2 level EV problem. So, we define P(1) P0 as the sub-projector for our space Ea0(1) which does with the bundle of states labeled N1 in the above picture. We can then write our s=2 perturbation equation after we apply P(1) to both sides, the result is this: [ see ( ) above ] 0 = P(1)(ε1-V ) |1> + ε2 |0> (1) We also have our s=1 "Qa0 projected equation" [ see ( ) above ] Q0 |1> = Q0aV |0> (2) In order for (1) to become an eigenvalue equation, we need to have the |1> be gone and somehow get a |0> in its place. For higher order, we will have the same type of problem. We want to use (2) to "reduce" (1), but as things stand, we cannot do it because we need a Q0 in front of |1> which we don't have. This difficulty gets resolved by the discovery of the following operator equation which I will derive below: P(1) (ε1-V ) = – P(1)V Q0 D' which you see manages to make Q0 appear on the right just where we need it to be! Then if we insert D' into (1) and then insert (2) into that result, we get 0 = P(1)(ε1 - V ) |1> + ε2 |0> = – P(1)V Q0 |1> + ε2 |0> = – P(1)V Q0aV |0> + ε2 |0> => P(1)(V Q0aV) |0; m,ε2> = ε2 |0; m,ε2> (3) // = Messiah p 700 (39) and this is our ε2 eigenvalue equation! Below we will show how to express it in matrix form, and how to also express it as an operator equation, but first we want to derive D' . Recall from above our first order eigenvalue equation expressed as an operator equation: P0VP0 = ε1P0 (4) Of we apply P(1)from the left on both sides, we are in effect restricting this equation when it acts on the bra space to the subspace Ea0(1) -- we are certainly allowed to do that. Then P(1)P0 = P(1) from the obvious filtering property, and we end up with this restriction of the above operator equation: P(1)V P0 = ε1 P(1) (5) We now replace P0 = (1 - Q0) to get our result D' P(1)V (1 - Q0) = ε1 P(1) => P(1) (ε1-V ) = – P(1)V Q0 D' Express the ε2 eigenvalue problem in matrix notation: We start with (3) above P(1)(V Q0aV) |0; m,ε2> = ε2,m |0; m,ε2> Close from the left with our ε1 level "basis states" <0; m',ε1| just from the N1 group in the figure, <0; m',ε1| P(1)(V Q0aV) |0; m,ε2> = ε2,m <0; m',ε1|0; m,ε2> <0; m',ε1| (V Q0aV) P(1) |0; m,ε2> = ε2,m <0; m',ε1|0; m,ε2> In the second equation, we repeated what we did in the ε1 case, moving now P(1)to the right. Now expand the projector P(1): Σm"=1,N1 <0; m',ε1| (V Q0aV) | 0;m",ε1><0;m",ε1 |0; m,ε2> = ε2,m <0; m',ε1|0; m,ε2> We could regard this directly as a matrix equation Σm"=1,N1 (V Q0aV)m'm" ψ(m,ε2)m" = ε2,m ψ(m,ε2)m' but we have to go off and compute the matrix V Q0aV in the basis of | 0;m",ε1> states. But we are "comfortable" doing this from our calculation of the states |1;n,ε1>, so no need to reduce to |r>-basis matrix elements. So we have thus arrived at our matrix equation (V Q0aV) ψ(m,ε2) = ε2 ψ(m,ε2) Express the ε2 eigenvalue problem as an operator equation. We start with the eigenvector equation from above P(1)(V Q0aV) |0; m,ε2> = ε2,m |0; m,ε2> where the eigenvectors |0> all exist in Ea0(1) . If we replace |0; m,ε2> with P(1) |0; m,ε2> on both sides, the resulting equation will then be true not just for the |0; m,ε2> states which span Ea0(1), but also for all other states, and thus we end up with an operator equation version of our eigenvalue problem P(1)(V Q0aV) P(1) = ε1 P(1) This is a wonderful trick. In effect, both sides of this equation have "protectors" (filters) facing both left and right. Thus, this equation will be valid in both the bra space and the ket space when acting on any state! This is the beauty of using projectors. Finding the ε3 eigenvalue problem. Flushed with a feeling of success at lower levels, let's try to go one more level and see what happens. So, we define P(1,1) P(1) P0 as the sub-projector for our space Ea0(1,1) of some dimension N1,1. We can then write our s=3 perturbation equation after we apply P(1,1) to both sides, the result is this: ( I have also copied down the modified s = 2 equation) 0 = P(1,1)(ε1-V ) |1> + ε2 |0> (1a) s = 2 0 = P(1,1) (ε1 - V) |2> + ε2 P(1,1) |1> + ε3 |0> (1b) s = 3 We also have our s=1 "Qa0 projected equations" [ see ( ) above ] Q0 |1> = Q0aV |0> (2a) s = 1 Q0 |2> = Q0aV |1> – ε1 Q0a |1> (2b) s = 2 We are now faced with the problem of eliminating both |2> and |1> from (1b) above, and we have the same need to cause some Q0 factors to "appear" on the right. Let's blindly tread down the same path as before, making small changes. We can write our ε1 EV operator equation (still further restricted) as P(1,1)V P0 = ε1 P(1,1) (5) P(1,1)V (1 - Q0) = ε1 P(1,1) => P(1,1) (ε1-V ) = – P(1,1)V Q0 D' which is our new version of D'. This looks promising and we insert it at once into (1b) to get 0 = – P(1,1)V Q0 |2> + ε2 P(1,1) |1> + ε3 |0> and then we use (2b) to get 0 = – P(1,1)V Q0aV |1> – ε1 Q0a |1>) + ε2 P(1,1) |1> + ε3 |0> and we are happy to see the state |2> gone. We can further say, using (2a), Q0a |1> = Q0a Q0|1> = Q0a Q0aV |0> = Q0a2 V |0> and then we have 0 = – P(1,1)V (Q0aV |1> – ε1 Q0a2 V |0>) + ε2 P(1,1) |1> + ε3 |0> 0 = – P(1,1)V Q0aV |1> + ε2 P(1,1) |1> + ε3 |0> + ε1 Q0a2 V |0> 0 = P(1,1) (V Q0aV + ε2) |1> + ε3 |0> + ε1 Q0a2 V |0> Now we need to concoct some way to get Q0 on the right end for this first term. We do have one other weapon which is our ε2 EV operator equation, P(1)(V Q0aV) P(1) = ε1 P(1) which can restrict from just the left side as we did last time P(1,1)(V Q0aV) P(1) = ε1 P(1,1) Now blindly continuing, replace P(1) = 1 - Q0(1) where we define a new version of Q0 . Then P(1,1)(V Q0aV) (1 - Q0(1)) = ε1 P(1,1) And let's not forget our friendly s=2 equation from above which we alter slightly 0 = P(1,1)(ε1-V ) |1> + ε2 P(1,1) |0> Mix these things together a bit and we have P(1,1) (V Q0aV + ε2) = P(1,1)(V Q0aV) Q0(1) + ε1 P(1,1) Now, I think we can concoct a new version of (2a) above that says Q0(1) |1> = Q0a(1)V |0> (2a) s = 1 where Q0a(1) = Q0(1)(Ea0 - H0)-1 Q0(1). Let's just guess that this works and see what happens. We then have for our tentative EV equation precursor, 0 = P(1,1) (V Q0aV + ε2) |1> + ε3 |0> + ε1 Q0a2 V |0> 0 = { P(1,1)(V Q0aV) Q0(1) + ε1 P(1,1)} |1> + ε3 |0> + ε1 Q0a2 V |0> 0 = P(1,1)(V Q0aV) Q0(1)|1> + ε1 P(1,1) |1> + ε3 |0> + ε1 Q0a2 V |0> 0 = P(1,1)(V Q0aV) Q0a(1)V |0> + ε1 P(1,1) |1> + ε3 |0> + ε1 Q0a2 V |0> OK, I give up after a valiant effort. We still have some |1> left over in this equation, so it is not yet en EV equation. Finding the εs eigenvalue problem. It is abundantly clear that you cannot continue down the same path as the order increases, because the complexity exponentiates and you have to somehow delicately thread your way exactly right through each level to find the EV equation. This is why we need a more systematic and organized attack on the problem, and Kato did this in 1949 and Messiah reported out his results. P1. Finding the state corrections. As outlined above, Messiah gets us to the point of an eigenvalue problem at each ε level, which of course has eigenvectors at that level, but he does not talk at all about how you would compute the state corrections like |1>. Admittedly, it is usually the energies that are of most interest. Computation of |1; n,ε1> when N > 1 and |n,ε1> is non degenerate at the ε1 level Suppose a set of states were degenerate at the ε0 level, but at the ε1 level our "state n of interest" |0; n,ε1> is non-degenerate. This state is marked "n" in the following picture (ignore the ε2 column for now) As a possible starting point, we know from above that Q0 |1> = Q0aV |0> s = 1 Q0 |2> = Q0aV |1> – ε1 Q0a |1> s = 2 Q0 |3> = Q0aV |2> – ε1 Q0a |2> – ε2 Q0a |1> s = 3 etc. (a) We want to know the projections of |1; n, ε1> on all states in H . For states |k> (Ea0) we have no problem, we just use the first line above to get: <k| Q0 |1; n, ε1> = <k| Q0aV |0; n, ε1> <k| 1; n, ε1> = <k| Q0 (Ea0 - H0)-1Q0V |0; n, ε1> = <k| (Ea0 - H0)-1Q0V |0; n, ε1> = (Ea0 - Ek)-1<k|Q0V |0; n, ε1> = (Ea0 - Ek)-1<k|V |0; n, ε1> = (Ea0 - Ek)-1Vkn ≡ ck (to be used below) which we can compare to our "personal theory" result which was ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk B'ik,1 = <uk| H' |umi >/ (Em - Ek) (b) But we also need projections of |1; n, ε1> onto the states |0; n', ε1> for n' ≠ n. None of the above equations can help us because if we close with < 0; n', ε1|, each equation gives 0 = 0. We must therefore go back to the raw starting equations which were these: (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (H0 - Ea0) |3> = (ε1 - V) |2> + ε2 |1> + ε3 |0> s = 3 What happens if we close these equations with <0; n', ε1| ? The LHS is always 0. Let's write out the RHS's in more detail: 0 = < 0; n', ε1| (ε1,n - V ) |0; n, ε1> s = 1 0 = < 0; n', ε1| (ε1,n - V ) |1; n, ε1> + ε2,n < 0; n', ε1|0; n, ε1> s = 2 0 = < 0; n', ε1| (ε1,n - V) |2; n, ε1> + ε2,n< 0; n', ε1| 1; n, ε1> + ε3,n< 0; n', ε1| 0; n, ε1> s = 3 and we just pause to take note that if n' = n, the first says < 0; n', ε1| V |0; n, ε1> = ε1,n , a familiar result but we want to write it down. Now assume n' ≠ n so the δn,n' terms vanish, 0 = < 0; n', ε1| -V |0; n, ε1> s = 1 0 = < 0; n', ε1| (ε1,n - V ) |1; n, ε1> s = 2 0 = < 0; n', ε1| (ε1,n - V) |2; n, ε1> + ε2,n< 0; n', ε1| 1; n, ε1> s = 3 The first equation just reminds us that states |0; n, ε1> diagonalize V, so that equation will give us no further help beyond that fact. The third equation gets us tangled up with |2; n, ε1> which just makes things worse, so it is hard to imagine it will be helpful. So all we really have is the second equation: ε1,n < 0; n', ε1| 1; n, ε1> = < 0; n', ε1| V |1; n, ε1> Now consider this expansion of unity: 1 = Σn"=1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| where we list off the states in Ea0 using not the |r> states but the |0; n',ε1> states. Suppose we insert this unity in our equation above just to the right of the V operator. ε1,n < 0; n', ε1| 1; n, ε1> = < 0; n', ε1| V { Σn"=1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k|} |1; n, ε1> = Σn"=1,N < 0; n', ε1| V |0; n",ε1> <0; n",ε1 |1; n,ε1> + ΣkEoa < 0; n', ε1| V |k><k| 1; n,ε1> = < 0; n', ε1| V |0; n',ε1> <0; n',ε1 |1; n,ε1> + ΣkEoa < 0; n', ε1| V |k><k| 1; n,ε1> n'≠n Now we want to compact this stuff down to see what it is saying. We need to be very careful with notation here. Let's define: Vn'n' ≡ < 0; n', ε1| V |0; n',ε1> Vn'k ≡ < 0; n', ε1| V |k> Vkn ≡ <k|V |0; n, ε1> Cn' ≡ < 0; n', ε1| 1; n, ε1> // note that Cn = < 0; n, ε1| 1; n, ε1> = <0|1> = 0 ck ≡ <k| 1; n,ε1> = (Ea0 - Ek)-1<k|V |0; n, ε1> = (Ea0 - Ek)-1 Vkn We have to be careful because these V matrix elements are not the raw Vrr' type matrix elements! They are exactly what they say they are! Then we have ε1,n Cn' = Vn'n' Cn' + ΣkEoa Vn'k ck n'≠n (ε1,n – Vn'n') Cn' = ΣkEoa Vn'k ck n'≠n (Vnn – Vn'n') Cn' = ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn n'≠n We have said that state |0; n,ε1> is non-degenerate, so we then know that (Vnn – Vn'n') ≠ 0, and we have Cn' = (Vnn – Vn'n')-1 ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn = < 0; n', ε1| 1; n, ε1> So, finally we have our desired projections of | 1; n, ε1> onto the states | 0; n', ε1> which is what we set out to do above. So here then is our result for the state | 1; n, ε1> 1 = Σn"=1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| | 1; n, ε1> = Σn"=1,N |0; n",ε1> <0; n",ε1| 1; n, ε1> + ΣkEoa |k><k| 1; n, ε1> = Σn"=1,N Cn" |0; n",ε1> + ΣkEoa ck |k> so we get this result | 1; n, ε1> = Σn'=1,N Cn' |0; n',ε1> + ΣkEoa ck |k> where ck = (Ea0 - Ek)-1 Vkn Cn' = (Vnn – Vn'n')-1 ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn Cn = 0 If we compare this to the personal theory where ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk we do in fact find that B'ij,1 = Cj B'ik,1 = ck where we must recall that the states umj in the personal theory were ones which diagonalized V ! So finally after a very long time, I have found these coefficients B'ij,1. Now that we know the state | 1; n, ε1>, we can go back to our s=2 equation above 0 = < 0; n', ε1| (ε1,n - V ) |1; n, ε1> + ε2,n < 0; n', ε1|0; n, ε1> s = 2 and set n' = n to get 0 = < 0; n, ε1| (ε1,n - V ) |1; n, ε1> + ε2,n s = 2 0 = ε1,n – < 0; n, ε1| V |1; n, ε1> + ε2,n ε2,n = – ε1,n + < 0; n, ε1| V |1; n, ε1> where |1; n, ε1> = Σn"=1,N Cn" |0; n",ε1> + ΣkEoa ck |k> Notice that this is different from the result you get if your state of interest was non-degenerate in the left column of our picture Review of Computation of |1; n,ε1> when N > 1 1. We reminded ourselves that the states |0; n', ε1> for n' = 1,N are the eigenstates of the ε1 level EV equation and that we computed these states earlier in this document. These N states are those appearing in the center of our figure above. We then decided that we will just use these states as so computed, and won't ever be writing them out as | 0; n,ε1> = ΣrEa0 |r><r | 0; n,ε1> = ΣrEa0 U(1)r,n |r>. In other words, we assume that we have pre-computed these states | 0; n,ε1> and we will fly with them as a basis for the space Ea0. We then decided to focus on a non-degenerate state in this set : | 0; n,ε1>, shown as "n" in the above figure. Our goal was then to compute the state |1; n, ε1>. 2. Corresponding to (1) above, we decided to define the meaning of matrix elements like Vnn' and Vnk to be such that a state with an n type index refers to a state like |0; n', ε1> and not to one of our original Ea0 base states |r> . 3. We used the s=1 Q0-equation to find the projections of state |1; n, ε1> onto the |k> in (Ea0) . ck ≡ <k| 1; n,ε1> = (Ea0 - Ek)-1<k|V |0; n, ε1> = (Ea0 - Ek)-1 Vkn 4. We concluded that none of the Q0-equations was useful in finding the projections of |1; n, ε1> onto the states |0; n', ε1>. 5. We then wrote out the full-bore s=1equation, closed it with < 0; n', ε1| , and got 0 = < 0; n', ε1| (ε1,n - V ) |0; n, ε1> which told us that V is diagonal in the |0; n, ε1> basis. 6. We then wrote out the full-bore s=2 equation, closed it with < 0; n', ε1|, and got for n' ≠ n ε1,n < 0; n', ε1| 1; n, ε1> = < 0; n', ε1| V |1; n, ε1> n' ≠ n We then inserted (this I think was the crucial step) 1 = Σn"=1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| to the right of V, and then used the diagonality of V to get our final result for | 1; n, ε1>: | 1; n, ε1> = Σn'=1,N Cn' |0; n',ε1> + ΣkEoa ck |k> where ck = (Ea0 - Ek)-1 Vkn Cn' = (Vnn – Vn'n')-1 ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn Cn = 0 // from the normalization condition 7. We then compared this to our personal theory and found that ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk B'ij,1 = Cj B'ik,1 = ck and this gave us the long-sought solution of the problem of finding the B'ij,1. 8. We then rewrote the full-bore s=2 equation, closed it with < 0; n, ε1| and got 0 = < 0; n, ε1| (ε1,n - V ) |1; n, ε1> + ε2,n < 0; n, ε1|0; n, ε1> s = 2 so that ε2,n = ε1,n + < 0; n, ε1| V |1; n, ε1> with state |1; n, ε1> as given in (6) above. Computation of |1; n,ε1> when N = 1 ( |n,ε1> is in this case non-degenerate at ε1 level) We want to do the same calculation as above, but now with N = 1 and this means we have state |0; n,ε1> and states |k> and there are no other states. So we can just refer to |0; n,ε1> as |n> because we know that the linear combination here is just |0; n,ε1> = |n> where |n> is one of the base states |r>. As before, start with Q0 |1> = Q0aV |0> s = 1 We want to know the projections of |1; n, ε1> on all states in H . For states |k> (Ea0) we have no problem, we just use the line above and close with <k| to get: <k| Q0 |1; n, ε1> = <k| Q0aV |0; n, ε1> <k| 1; n, ε1> = <k| Q0 (Ea0 - H0)-1Q0V |0; n, ε1> = <k| (Ea0 - H0)-1Q0V |0; n, ε1> = (Ea0 - Ek)-1<k|Q0V |0; n, ε1> = (Ea0 - Ek)-1<k|V |0; n, ε1> = (Ea0 - Ek)-1Vkn ≡ ck But this is all the states there are! And the projection on itself is 0, so we are done. Thus, | 1; n, ε1> = Σk (Ea0 - Ek)-1<k|V |0; n, ε1> |k> = Σk (Ea0 - Ek)-1Vkn |k> which we can compare to our "personal theory" result which was ψmi,1 = { 0 } Σk≠m B'ik,1 uk B'ik,1 = <uk| H' |umi >/ (Em - Ek) We can now go back to the raw starting equations which were these: (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 What happens if we close these equations with <0; n, ε1| ? The LHS is always 0. Let's write out the RHS's in more detail: 0 = < 0; n, ε1| (ε1,n - V ) |0; n, ε1> s = 1 0 = < 0; n, ε1| (ε1,n - V ) |1; n, ε1> + ε2,n < 0; n, ε1|0; n, ε1> s = 2 The first equation tells us that ε1,n = Vnn as expected. So all we really have is the second equation: ε1,n < 0; n, ε1| 1; n, ε1> = < 0; n, ε1| V |1; n, ε1> – ε2,n But < 0; n, ε1| 1; n, ε1> = 0 so this says ε2,n = < 0; n, ε1| V |1; n, ε1> = <0; n, ε1| V { Σk (Ea0 - Ek)-1Vkn |k> } = Σk (Ea0 - Ek)-1 Vkn <0; n, ε1| V |k> = Σk (Ea0 - Ek)-1 Vkn Vnk = Σk (Ea0 - Ek)-1 |Vkn|2 // Messiah p 695 (31) So here is a summary of all our results here: ε1,n = Vnn | 1; n, ε1> = Σk (Ea0 - Ek)-1Vkn |k> ε2,n = < 0; n, ε1| V |1; n, ε1> = Σk (Ea0 - Ek)-1 |Vkn|2 Comments: in the case N = 1, there are no states < 0; n' , ε1| which are in Ea with n' ≠ n, so all our stuff involving Cn' goes completely away. It has no meaning at all. You cannot take a limit as N→1 of those results. For example, when N > 1 we get this result: (Vnn – Vn'n') Cn' = ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn n'≠n You cannot just set n' = n here and conclude that 0 = ΣkEoa Vnk (Ea0 - Ek)-1 Vkn . That result is in fact wrong, because we know this sum is in fact ε2,n. The above equation for n' = n does not even exist! Computation of |2; n,ε1> in the case N > 1 where |n,ε1> is non-degenerate at ε1 level Ignore the rightmost ε2 part of this picture. Our state of interest is "n" in the middle column. We know that Q0 |2; n, ε1> = Q0aV |1; n, ε1> – ε1 Q0a |1; n, ε1> s = 2 (a) We could get the projections of |2; n, ε1> onto our |k> states by closing with <k| <k|Q0 |2; n, ε1> = <k|Q0aV |1; n, ε1> – ε1 <k|Q0a |1; n, ε1> = <k|Q0a (V-ε1) |1; n, ε1> <k|2; n, ε1> = <k| Q0 (Ea0 - H0)-1Q0 (V-ε1) |1; n, ε1> = <k| (Ea0 - H0)-1Q0 (V-ε1) |1; n, ε1> = (Ea0 - Ek)-1<k| Q0 (V-ε1) |1; n, ε1> = (Ea0 - Ek)-1<k| (V-ε1) |1; n, ε1> ≡ dk and to finish this off, we would have to insert our solved |1; n, ε1> expansion. Note the dk definition. We know that no Q0-equation will help us to the other required projections, so we state the full-bore equations in hopes of finding a path to follow: (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (H0 - Ea0) |3> = (ε1 - V) |2> + ε2 |1> + ε3 |0> s = 3 The last equation looks promising, so write it out in more detail (H0 - Ea0) |3; n, ε1> = (ε1,n - V) |2; n, ε1> + ε2,n |1; n, ε1> + ε3,n |0; n, ε1> s = 3 (b) Let's try closing with <0; n',ε1| . Then of course LHS = 0 and we get 0 = <0; n',ε1| (ε1,n - V) |2; n, ε1> + ε2,n <0; n',ε1| 1; n, ε1> + ε3,n <0; n',ε1| 0; n, ε1> 0 = <0; n',ε1| (ε1,n - V) |2; n, ε1> + ε2,n Cn' + ε3,n δn,n' where we recognize our coefficient Cn' from previous work above, and we also know ε2,n. Now let's consider only n' ≠ n so the last term vanishes, leaving us with 0 = <0; n',ε1| (ε1,n - V) |2; n, ε1> + ε2,n Cn' n' ≠ n Now seems the right time to insert to the right of (ε1,m - V) a unity written as follows: 1 = Σn"=1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| We then have 0 = Σn"=1,N <0; n',ε1| (ε1,n - V) |0; n",ε1> <0; n",ε1 |2; n, ε1> + ΣkEoa <0; n',ε1| (ε1,n - V) |k><k |2; n, ε1> + ε2,n Cn' n' ≠ n = Σ n"=1,N (ε1,n - V)n'n" <0; n",ε1 |2; n, ε1> + ΣkEoa (ε1,n - V)n'k <k |2; n, ε1> + ε2,n Cn' = Σ n"=1,N (ε1,n - V)n'n" <0; n",ε1 |2; n, ε1> + ΣkEoa Vn'k dk + ε2,n Cn' = (ε1,n - V)n'n' <0; n',ε1 |2; n, ε1> + ΣkEoa Vn'k dk + ε2,n Cn' n' ≠ n Let's then define Dn' ≡ <0; n',ε1 |2; n, ε1> and let's take note that (ε1,n - V)n'n' = ε1,n δn'n' – Vn'n' = ε1,n - Vn'n' = ( Vnn – Vn'n') so the above can be written 0 = ( Vnn – Vn'n') Dn' + ΣkEoa Vn'k dk + ε2,n Cn' n' ≠ n Now since we said our state "n" is non-degenerate, we know that all the other Vn'n' are different from Vnn so the factor ( Vnn – Vn'n') is not 0 for any n' ≠ n, and we then find that Dn' = – ( Vnn – Vn'n')-1 ΣkEoa Vn'k dk + ε2,n Cn' n' ≠ n (c) We have now computed all the required projections, and we find then that |2; n, ε1> = { Σn"=1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| } |2; n, ε1> = ( Σn"=1,N |0; n",ε1> <0; n",ε1|2; n, ε1> + ΣkEoa |k><k|2; n, ε1> = ( Σn'=1,N Dn' |0; n',ε1> + ΣkEoa dk |k> where Dn' = <0; n',ε1 |2; n, ε1> = – ( Vnn – Vn'n')-1 ΣkEoa Vn'k dk + ε2,n Cn' n' ≠ n Dn = 0 dk = <k|2; n, ε1> = (Ea0 - Ek)-1 <k| (V-ε1) |1; n, ε1> |1; n, ε1> = Σn'=1,N Cn' |0; n',ε1> + Σk'Eoa ck' |k'> ck = (Ea0 - Ek)-1 Vkn Cn' = (Vnn – Vn'n')-1 ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn Cn = 0 ε2,n = ε1,n + < 0; n, ε1| V |1; n, ε1> (d) Just for fun we can insert the | 1; n, ε1> expansion into the dk expression: dk =( Ea0 - Ek)-1 <k| (V-ε1){ Σn'=1,N Cn' |0; n',ε1> + Σk'Eoa ck' |k'>} = (Ea0 - Ek)-1 [ Σn'=1,N Cn'<k| (V-ε1) |0; n',ε1> + Σk'Eoa (Ea0 - Ek')-1 Vk'n <k| (V-ε1) |k'> ] But now <k |0; n',ε1> = 0 and <k| (V-ε1) |k'> = Vkk' - ε1δkk' so get dk = (Ea0 - Ek)-1 [ Σn'=1,N Cn'Vkn' + Σ k'Eoa (Ea0 - Ek')-1 Vkk'Vk'n – (Ea0 - Ek)-1 Vnn Vkn ] which maybe one could interpret in a Feynman graph manner, but not today please. Computation of |2; n, ε1> in the case N = 1 ( |n,ε1> is non-degenerate at the ε1 level) (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (a) We could get the projections of |2; n, ε1> onto our |k> states by closing with <k| <k|Q0 |2; n, ε1> = <k|Q0aV |1; n, ε1> – ε1 <k|Q0a |1; n, ε1> = <k|Q0a (V-ε1) |1; n, ε1> <k|2; n, ε1> = <k| Q0 (Ea0 - H0)-1Q0 (V-ε1) |1; n, ε1> = <k| (Ea0 - H0)-1Q0 (V-ε1) |1; n, ε1> = (Ea0 - Ek)-1<k| Q0 (V-ε1) |1; n, ε1> = (Ea0 - Ek)-1<k| (V-ε1) |1; n, ε1> ≡ dk which is the same dk appearing in the N > 1 case, and to finish this off, we would have to insert our solved |1; n, ε1> expansion. But we have that expansion from above in the N = 1 case, and it is this: |1; n, ε1> = Σk' (Ea0 - Ek')-1Vk'n |k'> so insert this to get dk = <k|2; n, ε1> = (Ea0 - Ek)-1<k| (V-ε1) |1; n, ε1> = (Ea0 - Ek)-1 Σk' (Ea0 - Ek')-1Vk'n <k| (V-ε1) |k'> = (Ea0 - Ek)-1 Σk' (Ea0 - Ek')-1 Vkk'Vk'n + (Ea0 - Ek)-1 (Ea0 - Ek)-1<k| (V-ε1) |k> = (Ea0 - Ek)-1 { Σk' (Ea0 - Ek')-1 Vkk'Vk'n + (Ea0 - Ek)-1<k| (V-ε1) |k> } = (Ea0 - Ek)-1 Σk' (Ea0 - Ek')-1 Vkk'Vk'n + (Ea0 - Ek)-2(Vkk - ε1) And then our result is just |2; n, ε1> = Σk dk |k> = Σk { (Ea0 - Ek)-1 Σk' (Ea0 - Ek')-1 Vkk'Vk'n + (Ea0 - Ek)-2(Vkk - ε1) } |k> (a') The Q0a method. Going back to our start here we could also write our Q0-equation Q0 |2; n, ε1> = Q0aV |1; n, ε1> – ε1 Q0a |1; n, ε1> = Q0a(V-ε1) |1; n, ε1> But since Q0 = 1 - | 0; n, ε1><0; n, ε1| and since <0; n, ε1| 2; n, ε1>, we can remove the Q0 : |2; n, ε1> = Q0a(V-ε1) |1; n, ε1> // Messiah p 694 (27) which is a formal solution to our question. We can then put a unity to the left of (V-ε1) 1 = |n><n| + Σk |k><k| // 1 = P0 + Q0 |2; n, ε1> = Q0a {|n><n| + Σk |k><k| } (V-ε1) |1; n, ε1> But the right end of Q0a is Q0 which kills off |n> and passes |k> so we get |2; n, ε1> = Σk Q0 (Ea0 - H0)-1 { |k><k| } (V-ε1) |1; n, ε1> = Σk Q0(Ea0 - Ek)-1 { |k><k| } (V-ε1) |1; n, ε1> = Σk (Ea0 - Ek)-1 { |k><k| } (V) |1; n, ε1> = Σk (Ea0 - Ek)-1 <k|V |1; n, ε1> |k> But from above we had (Ea0 - Ek)-1<k| (V-ε1) |1; n, ε1> ≡ dk so our result is then |2; n, ε1> = Σk dk |k> which replicates our previous result. Computation of |1; m,ε2> in the case N > 1 and N1> 0 where |m,ε2> is non-degenerate at ε2 level This is the case that our picture was really meant to describe. We certainly want to at least try to use the above "review" as a guide here. 1. We remind ourselves that the states |0; m', ε2> for m' = 1, N1 are the eigenstates of the ε2 level EV equation and that we (in theory) computed these states earlier in this document. These N1 states are those appearing at the right of our figure above. We decide that we will just use these states as so computed. We then decide to focus on a non-degenerate state in this set : | 0; m, ε2>, shown as "m" in the above figure. Our goal is then to compute the states |1; m, ε2> and |2; m, ε2>. 2. Corresponding to (1) above, we decide to define the meaning of matrix elements like Vnn' and Vnk to be such that a state with an n type index refers to a state like |0; n', ε1> and not to one of our original Ea0 base states |r> . Furthermore, if a V index is m', we will be referring to a state |0; m', ε2>. 3. The calculation: (a) We use the s=1 Q0-equation to find the projections of state |1; m, ε2> onto the |k> in (Ea0) . Q0 |1> = Q0aV |0> s = 1 <k| Q0 |1; m, ε2> = <k| Q0aV |0; m, ε2> <k| 1; m, ε2> = <k| Q0 (Ea0 - H0)-1Q0V |0; m, ε2> = <k| (Ea0 - H0)-1Q0V |0; m, ε2> = (Ea0 - Ek)-1<k|Q0V |0; m, ε2> = (Ea0 - Ek)-1<k|V |0; m, ε2> = (Ea0 - Ek)-1Vkm ≡ ek (to be used below) (b) Next, let's go for the projections of |1; m, ε2> onto the states |0; m',ε2> for m' ≠ m . Start with (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 and close with <0; m',ε2| which are of course in Ea0 so as usual have LHS's = 0 0 = < 0; m', ε2| (ε1,m - V ) |0; m, ε2> s = 1 0 = < 0; m', ε2| (ε1,m - V ) |1; m, ε2> + ε2,m < 0; m', ε2|0; m, ε2> s = 2 If we set m' = m in the first equation we find ε1,m = Vmm which seems at least reasonable. Now assume that m' ≠ m so the δm,m' terms vanish, 0 = < 0; m', ε2| (- V ) |0; m, ε2> m ≠ m' s = 1 0 = < 0; m', ε2| (ε1,m - V ) |1; m, ε2> s = 2 The first equation just tells us that states |0; m, ε2> diagonalize V, so that equation will give us no further help beyond that fact. But let's write the fact down! Vm'm= δm'm Vmm = δm'm ε1,m The second equation above then says: ε1,m < 0; m', ε2| 1; m, ε2> = < 0; m', ε2| V |1; m, ε2> Now consider this expansion of unity: 1 = Σm"=1,N1 |0; m",ε2> <0; m",ε2| + Σn"=N1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| where now we have three different pieces of unity. I will try it this way and see where we get. Insert on the right side of V as we did before ε1,m < 0; m', ε2| 1; m, ε2> = < 0; m', ε2| V |1; m, ε2> = Σm"=1,N1< 0; m', ε2| V |0; m",ε2> <0; m",ε2| 1; m, ε2> + Σn"=N1,N < 0; m', ε2| V |0; n",ε1> <0; n",ε1| 1; m, ε2> + ΣkEoa< 0; m', ε2| V|k><k| 1; m, ε2> = Σm"=1,N1Vm'm" <0; m",ε2| 1; m, ε2> + Σn"=N1,NVm'n" <0; n",ε1| 1; m, ε2> + ΣkEoaVm'kck This equation involves two kinds of | 1; m, ε2> projections so let's give them names: Em" ≡ <0; m",ε2| 1; m, ε2> Fn" ≡ <0; n",ε1| 1; m, ε2> so the above line becomes ( note that Em = 0) = Σm"=1,N1Vm'm" Em" + Σn"=N1,NVm'n" Fn" + ΣkEoaVm'kek m' ≠ m Now the LHS of this equality is ε1,m < 0; m', ε2| 1; m, ε2> = ε1,m Em' so we arrive at: ε1,m Em' = Σm"=1,N1Vm'm" Em" + Σn"=N1,NVm'n" Fn" + ΣkEoaVm'kek m' ≠ m But we know from above that Vm'm" = δm'm"ε1,m' so this becomes ε1,m Em' = ε1,m' Em' + Σn"=N1,NVm'n" Fn" + ΣkEoaVm'kek m' ≠ m Em'(ε1,m – ε1,m') = Σn"=N1,NVm'n" Fn" + ΣkEoaVm'kek m' ≠ m But I would argue that the states m and m' are degenerate at the ε1 level so LHS = 0 ! We are then left with 0 = Σn"=N1,NVm'n" Fn" + ΣkEoaVm'kek m' ≠ m N1-1 equations Fn" ≡ <0; n",ε1| 1; m, ε2> N-N1 unknowns (c) I think we will now obtain something perhaps similar by looking for the projections of |1; m, ε2> onto the states |0; n',ε1> for n' = N1...N. Start with (H0 - Ea0) |1> = (ε1 - V ) |0> s = 1 (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 and close with <0; n',ε1| which are of course in Ea0 so as usual have LHS's = 0 0 = <0; n',ε1| (ε1,m - V ) |0; m, ε2> s = 1 0 = <0; n',ε1| (ε1,m - V ) |1; m, ε2> + ε2,m <0; n',ε1|0; m, ε2> s = 2 Looking at our picture, it seems clear that |0; m, ε2> is orthogonal to any <0; n',ε1| in the group of interest here. This kills part of the first equation and part of the second, so we get 0 = <0; n',ε1| (- V ) |0; m, ε2> s = 1 n' = N1...N 0 = <0; n',ε1| (ε1,m - V ) |1; m, ε2> s = 2 This first equation tells us something that seems pretty new to me: Vn'm = 0. Why would this be the case I wonder? It seems wrong. Well go back to (H0 - Ea0) |1,x> = (ε1 - V ) |0,x>. If you close this with ANY state in Ea0 you are going to get 0 = <0,y | (ε1 - V ) |0,x>, Of course both states have to be "legal" in that they have to be the 0 components of states which solve the SE, and in our case all these states are very special states which solve certain EV problems at ε1 and ε2 levels. So I guess it could be true. So as we always do, write out the second equation like so: ε1,m<0; n',ε1 |1; m, ε2> = <0; n',ε1 | V | 1; m, ε2> = ε1,m Fn' n' = N1....N where we recognize one of our defined factors above. So now insert our same unity as above, 1 = Σm"=1,N1 |0; m",ε2> <0; m",ε2| + Σn"=N1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| and as before write out all the terms. The difference here is the left state <0; n',ε1| was previously the state < 0; m', ε2|, so let's copy down the above result and just edit it ε1,m<0; n',ε1 |1; m, ε2> = <0; n',ε1 | V | 1; m, ε2> ε1,m <0; n',ε1| 1; m, ε2> = <0; n',ε1| V |1; m, ε2> = Σm"=1,N1<0; n',ε1| V |0; m",ε2> <0; m",ε2| 1; m, ε2> + Σn"=N1,N <0; n',ε1| V |0; n",ε1> <0; n",ε1| 1; m, ε2> + ΣkEoa<0; n',ε1| V|k><k| |1; m, ε2> = Σm"=1,N1Vn'm" <0; m",ε2| 1; m, ε2> + Σn"=N1,NVn'n" <0; n",ε1| 1; m, ε2> + ΣkEoaVn'kek ε1,m Fn' = Σm"=1,N1Vn'm" Em" + Σn"=N1,NVn'n" Fn" + ΣkEoaVn'kek n' = N1...N (d) Now let's gather both sets of equations right here: 0 = Σn"=N1,NVm'n" Fn" + ΣkEoaVm'kek m' ≠ m N1-1 equations ε1,m Fn' = Σm"=1,N1Vn'm" Em" + Σn"=N1,NVn'n" Fn" + ΣkEoaVn'kek n' = N1...N = N-N1 eq Em" ≡ <0; m",ε2| 1; m, ε2> Em = 0 N1 - 1 unknowns Fn" ≡ <0; n",ε1| 1; m, ε2> N - N1 unknowns. Overall, then, we seem to have N-1 equations in N-1 unknowns which are the Em and the Fn. So in theory we can solve for these coefficients. Let's imagine we have done that. We then have 1 = Σm"=1,N1 |0; m",ε2> <0; m",ε2| + Σn"=N1,N |0; n",ε1> <0; n",ε1| + ΣkEoa |k><k| | 1; m, ε2> = Σm"=1,N1 |0; m",ε2> <0; m",ε2| 1; m, ε2> + Σn"=N1,N |0; n",ε1> <0; n",ε1| 1; m, ε2> + ΣkEoa |k><k| 1; m, ε2> = Σm"=1,N1 |0; m",ε2> Em" + Σn"=N1,N |0; n",ε1> Fn" + ΣkEoa |k>ek and this is our desired result. Computation of |2; m,ε2> in the case N > 1 and N1> 0 where |m,ε2> is non-degenerate at ε2 level I could go on and try to find the solution for | 2; m, ε2> . Things would be similar to the above, but the starting point is more complicated, being (H0 - Ea0) | 2; m, ε2> = (ε1 - V ) | 1; m, ε2> + ε2 | 0; m, ε2> s = 2 As before, we would close with states in the three different classes and end up with a set of equations we could hopefully solve. I see how this is probably doable, but I will decline the fun, we have done enough I think. Comments Having done all this now, we certainly see the need for a more systematic methodology! But I suspect that even with that methodology, any actual calculation you try to do with have all the messiness you see above! Notice that everything we have done here has been based on states like | 0; m, ε2> which are fancy linear combinations of our base states in Ea0. We never dealt with other possible Hilbert spaces. We played here with the "shadows" of the various states that are cast into Ea0. The trick was to find the right base-state linear combinations of the shadows that are correct.