messiah R2 SSPT Chap 16 bad

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. Topic I. The Non-Degenerate Case (686) 1 2. Expansion in Powers of the Perturbation (686) 1 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. 11 Express the ε1 eigenvalue problem in matrix notation: 12 Express the ε1 eigenvalue problem as an operator equation. 13 Finding the ε2 eigenvalue problem. 13 Express the ε2 eigenvalue problem in matrix notation: 14 Express the ε2 eigenvalue problem as an operator equation. 16 Finding the ε3 eigenvalue problem. 16 Finding the εs eigenvalue problem. 18 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 = 1,2,3... 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: 0 = <0;n| ( ε1 - V ) |s-1;n> + εs s = 1,2,3... so 0 = <0;n| (ε1 - V ) |0;n > = <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). 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 1 |2> = Q0aV |1> – ε1 Q0a |1> 2 s = 2 ε2 = <0| V |1> 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 } |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 |1; n> 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 through second order. 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 , which means they are eigenstates of the full H up through second order. 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). 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> = ε2 |0> (3) 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| <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: 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. I am happier to have the subscripts on things like V and Q0a be those of the original |r> basis, so let's go through some extra steps to achieve this goal. So we need to replace the states |0; m,ε2> with sums of our original base states, and we quote from above making some symbol changes | 0; m",ε1> = ΣrEa0 |r><r | 0; m",ε1> = ΣrEa0 |r> U(1)r, m" <0; m',ε1| = Σr'Ea0 <0; m',ε1| r'><r' | = ΣrEa0 |r'> U(1)-1 m',r' Putting these in we get Σm"=1,N1 ΣrEa0 Σr'Ea0 U(1)-1 m',r'(V Q0aV)r'r U(1)r, m"<0;m",ε1 |0; m,ε2> = ε2,m <0; m',ε1|0; m,ε2> Now think of our solution eigenvectors in this way, ψ(m,ε2)m" = <0;m",ε1 |0; m,ε2> so we then have Σm"=1,N1 ΣrEa0 Σr'Ea0 U(1)-1 m',r'(V Q0aV)r'r U(1)r, m" ψ(m,ε2)m" = ε2,m ψ(m,ε2)m' (U(1)-1 V Q0aV U(1)) ψ(m,ε2) = ε2,m ψ(m,ε2) As you can see, all the information from the ε1 solution is needed [ in the form of U(1)] in order to solve the ε2 problem. Things are pretty complicated, but at least we can see how we would do this computationally if we had to. The eigenvalues ε2 come from det (U(1)-1 V Q0aV U(1) - ε2) = 0 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 the |1> level state corrections. 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 in the following picture (ignore the ε2 column for now) Let the projector for this one state be P(1) = 1 - Q(1) . Then define Q0a(1) ≡ Q0(1)(Ea0 - H0)-1Q0(1) Q0(1) = Σk≠1 |k><k| This sum over k includes all states in H0 except our one state |0; n,ε1> . The sum does not just include states that were in the original degenerate group of N states in Ea0 . So as a practical matter, the sum could to be computed as follows: Q0(1) = Σn'=1,N(n'≠n) |0; n',ε1> <0; n',ε1| + ΣkEoa |k><k| Here, the first sum includes all the other states in the ε1 column of the above picture, and then the second sum include all the rest of the states in (Ea0) . Notice that the different elements of the first sum in general have different values for ε1 (suggested by different vertical position in the picture). In theory, we could have carried out our perturbation theory analysis on all states in the left column, so that the latter two columns of our picture would then be "fully populated". That requires of course a lot of work, because we have to solve the ε1 level EV problem for all groups of degenerate states in the left column ( in Ea0, Eb0, Ec0...) in order to fully populate the ε1 column. But if we did all this work, then we could say Q0(1) = Σn'=all (n'≠n) |0; n',ε1> <0; n',ε1| where now our sum would be over all "horizontal lines" in the ε1 column that we get by tracking over all the states from the left column. Of course not all these lines are shown in our picture above. In other words, if we do all this work, then we have a new basis |0; n,ε1> for the entire H, and then we are just knocking one state out of our sum to get Q0(1). Further developing Q0(1) using the first method above, we could of course use our expansions | 0; n',ε1> = Σr'Ea0 |r'><r' | 0; n',ε1> = Σr'Ea0 U(1)r',n'|r'> <0; n',ε1| = ΣrEa0 <0; n',ε1| r><r | = ΣrEa0 U(1)-1n',r <r| and then we would have Q0(1) = Σn'≠n { Σr'Ea0 U(1)r',n'|r'>}{ ΣrEa0 U(1)-1n',r <r| } + ΣkEoa |k><k| = Σn'≠n Σr'Ea0 ΣrEa0 U(1)r',n' U(1)-1n',r |r'><r| + ΣkEoa |k><k| which is an interesting formula. It says that if you insist on working in "the original basis" (whatever it was), then in that basis Q0(1) has lots of off diagonal elements in the first summation above. The form is a lot simpler (and diagonal) if you find out what the kets |0; n,ε1> are, and then work in that basis! This Q0a(1) above fits my general projector sandwich model so it should follow that Q0a(1) (Ea0 - H0) = Q0(1) Now consider our s=1 equation (H0 - Ea0) |1; n,ε1> = (ε1 - V ) |0; n,ε1> and apply Q0a(1) to both sides and get Q0(1) |1; n,ε1> = Q0a(1) (ε1 - V ) |0; n,ε1> Close with <0; n',ε1| where n' ≠ n, <0; n',ε1| Q0(1) |1; n,ε1> = <0; n',ε1| Q0a(1) (ε1 - V ) |0; n,ε1> But we know that <0; n',ε1| Q0(1) = <0; n',ε1| , so write this as <0; n',ε1|1; n,ε1> = <0; n',ε1| Q0a(1) (ε1 - V ) |0; n,ε1> n' ≠ n This tells us the projections of our state |1; n,ε1> onto all the N-1 states |0; n',ε1> for n' ≠ n, and we know that the projection onto the one state |0; n,ε1> is zero from our normalization condition. But we also have to worry about projections of our state |1; n,ε1> onto all the other states k in (Ea0) . We can find these exactly as above, Q0(1) of course "passes" these states through as well, so <k| Q0(1) |1; n,ε1> = <k| Q0a(1) (ε1 - V ) |0; n,ε1> <k |1; n,ε1> = <k| Q0a(1) (ε1 - V ) |0; n,ε1> Now we know the projection of |1; n,ε1> onto ALL states in our problem, we can now construct the first order corrected state as follows: |1; n,ε1> = Σn'≠n |0; n',ε1> <0; n',ε1|1; n,ε1> + ΣkEoa |k> <k |1; n,ε1> = Σn'≠n |0; n',ε1> <0; n',ε1| Q0a(1) (ε1-V) |0; n,ε1> + ΣkEoa |k> <k| Q0a(1) (ε1 - V) |0; n,ε1> where Q0(1) = Σn'=1,N, n'≠n |0; n',ε1> <0; n',ε1| + ΣkEoa |k><k| = 1 - P0(1) Q0a(1) = Q0(1)(Ea0 - H0)-1Q0(1) Within Ea0 we really need to work in the ε1 basis |0; n,ε1>, otherwise the above equations expand out into a huge mess. Computation of the |2> level state corrections. We have this EV equation for the bundle of N1 states degenerate in the ε1 column P(1)(V Q0aV) |0; m,ε2> = ε2 |0; m,ε2> m = 1,2...N1 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> Imagine that we solve this problem for the energies ε2 and the coefficients <0; m',ε1|0; m,ε2>. We then know all about the kets in the right column (at least those of interest to us and shown in the figure) |0; m,ε2> = Σm'=1,N1 |0; m',ε1> <0; m',ε1|0; m,ε2> They are just certain linear combinations of ε1 level states | 0;m,ε1> And suppose at this level, some state |0; m,ε2> becomes a singlet. This appears as the top state on the right, We can define a P0(1,1) and corresponding Q0 for this one state Q0a(1,1) ≡ Q0(1,1) (Ea0 - H0)-1Q0(1,1) Q0(1,1) = Σm'=1,N1(m'≠m) |0; m',ε2><0;m',ε2| + Σn=N1+1,N |0; n,ε1> <0; n,ε1| + ΣkEoa |k><k| It is worth discussing the various sums here. Because P0(1,1) "passes" only the state |0; m,ε2> (top right in our picture), we know that Q0(1,1) must pass all other states. Now, this projector Q0(1,1) includes three groups. First, we need to include all the other states in the ε2 rightmost column shown above. There are N1-1 such states. Just to be able to include them means that we must know what they are! Luckily, we know what they are from the ε2 level EV problem solution. Second, we must include all the other states in the ε1 column below the N1 degenerate group. Only when we include these states will we have included a full basis of Ea0. There are N - N1 such states. Just to be able to include them means that we must know what they are! Luckily, we know what they are from the ε1 level EV problem solution. Third and last, we have to include the states in (Ea0) . Reminder: everything in the picture is at the "0 level" in terms of state representations. The picture is just showing the right linear combinations of our initial 0 level states. Now let's assume we have first solved the ε1 EV problem for the bundle of N degenerate states in the left column, and that we have then solved the ε2 EV problem for the bundle of N1 still degenerate states in the middle problem. We now recall our s=2 perturbation equation (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (*) and we use the fact that ( we have seen this many times now, same idea over and over again) Q0a(1,1) (Ea0 - H0) = Q0(1,1) Then we have Q0(1,1) |2; m,ε2> = Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 Q0a(1,1) |0; m,ε2> We are now going to close this with various groups of <0| level states, and for each group, the Q0(1,1) on the LHS will vanish when we apply it to the left, since all states will be "other" states relative to |0;m,ε2>: <0, m', ε2| Q0(1,1) |2; m,ε2> = <0, m', ε2 |2; m,ε2> = <0, m', ε2| Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 <0, m', ε2| Q0a(1,1) |0; m,ε2> m' ≠ m <0, n', ε1| Q0(1,1) |2; m,ε2> = <0, n', ε1 |2; m,ε2> = <0, n', ε1 | Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 <0, n', ε2 | Q0a(1,1) |0; m,ε2> n = N1+1,N <k| Q0(1,1) |2; m,ε2> = <k|2; m,ε2> = <k | Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 <k | Q0a(1,1) |0; m,ε2> k (Ea0) Now finally, using all these projection coefficients, we can construct our ε2 level state correction: |2,m,ε2 > = Σm'=1,N1(m'≠m) |0; m',ε2><0, m', ε2 |2; m,ε2> + Σn=N1+1,N |0; n,ε1> <0; n,ε1| 2; m,ε2> + ΣkEoa |k><k| 2; m,ε2> ********************* scraps ************************************************ everything here is junk, all of yesterday's work was wrong. Typical. Definition of P(1) and Q(1) . We want to define P(1) to project out only the state "n" in the above picture. The corresponding Q(1) is then given by Q0(1) = Σn'=1,N(n'≠n) |0; n',ε1> <0; n',ε1| + ΣkEoa |k><k| Here, the first sum includes all the other states in the ε1 column of the above picture (just those drawn), and then the second sum include all the rest of the states in (Ea0) . Notice that the different elements of the first sum in general have different values for ε1 (suggested by different vertical position of lines in the picture). Note 1: In theory, we could have carried out our perturbation theory analysis on all states in the left column, so that the latter two columns of our picture would then be "fully populated". That requires of course a lot of work, because we have to solve the ε1 level EV problem for all groups of degenerate states in the left column ( in Ea0, Eb0, Ec0...) in order to fully populate the ε1 column. But if we did all this work, then we could say Q0(1) = Σn'=all (n'≠n) |0; n',ε1> <0; n',ε1| where now our sum would be over all "horizontal lines" in the ε1 column that we get by tracking over all the states from the left column. Of course not all these lines are shown in our picture above. In other words, if we do all this work, then we have a new basis |0; n,ε1> for the entire H, and then we are just knocking one state out of our sum to get Q0(1). Note 2: Further developing Q0(1) using the first method above, we could of course use our expansions | 0; n',ε1> = Σr'Ea0 |r'><r' | 0; n',ε1> = Σr'Ea0 U(1)r',n'|r'> <0; n',ε1| = ΣrEa0 <0; n',ε1| r><r | = ΣrEa0 U(1)-1n',r <r| and then we would have Q0(1) = Σn'≠n { Σr'Ea0 U(1)r',n'|r'>}{ ΣrEa0 U(1)-1n',r <r| } + ΣkEoa |k><k| = Σn'≠n Σr'Ea0 ΣrEa0 U(1)r',n' U(1)-1n',r |r'><r| + ΣkEoa |k><k| which is an interesting formula. It says that if you insist on working in "the original basis" (whatever it was), then in that basis Q0(1) has lots of off diagonal elements in the first summation above. The form is a lot simpler (and diagonal) if you find out what the kets |0; n,ε1> are, and then work in that basis! Definition of Q0a(1) . In my first go at this, I tried using Q0a(1) ≡ Q0(1)(Ea0 - H0)-1Q0(1), but then I realized that this does not work and is singular because we don't have full Q0 "protection". For example, if we apply the above to some state |0; n',ε1> with n' ≠ n, we get (Ea0 - H0)-1Q0(1) |0; n',ε1> = (Ea0 - H0)-1|0; n',ε1> = ∞ since |0; n',ε1> is a linear combination of states which span Ea0. A better candidate is this: Q0a(1) ≡ Q0(1)Q0 (Ea0 - H0)-1 Q0Q0(1) = Q0(1)Q0a Q0(1) This Q0a(1) above fits my general projector sandwich model so it should follow that Q0a(1) (Ea0 - H0) = Q0(1) Now consider our s=1 equation (H0 - Ea0) |1; n,ε1> = (ε1 - V ) |0; n,ε1> and apply Q0a(1) to both sides and get Q0(1) |1; n,ε1> = Q0a(1) (ε1 - V ) |0; n,ε1> The ε1 term vanishes because Q0(1) |0; n,ε1> = 0, so things simplify to Q0(1) |1; n,ε1> = – Q0a(1)V |0; n,ε1> Close with <0; n',ε1| where n' ≠ n, <0; n',ε1| Q0(1) |1; n,ε1> = – <0; n',ε1| Q0a(1)V |0; n,ε1> = But we know that <0; n',ε1| Q0(1) = <0; n',ε1| , so write this as <0; n',ε1|1; n,ε1> = <0; n',ε1| Q0a(1)V ) |0; n,ε1> n' ≠ n = <0; n',ε1| Q0(1)(Ea0 - H0)-1Q0(1)V ) |0; n,ε1> = <0; n',ε1| (Ea0 - H0)-1Q0(1)V ) |0; n,ε1> = ΣrEa0 U(1)-1n',r <r| (Ea0 - H0)-1Q0(1)V ) |0; n,ε1> = diverges !!!! This tells us the projections of our state |1; n,ε1> onto all the N-1 states |0; n',ε1> for n' ≠ n, and we know that the projection onto the one state |0; n,ε1> is zero from our normalization condition. But we also have to worry about projections of our state |1; n,ε1> onto all the other states k in (Ea0) . We can find these exactly as above, Q0(1) of course "passes" these states through as well, so <k| Q0(1) |1; n,ε1> = <k| Q0a(1)V |0; n,ε1> <k |1; n,ε1> = <k| Q0a(1)V |0; n,ε1> = <k| Q0(1)(Ea0 - H0)-1Q0(1)V |0; n,ε1> = <k| (Ea0 - H0)-1Q0(1)V |0; n,ε1> = (Ea0 - Ek)-1 <k| Q0(1)V |0; n,ε1> = (Ea0 - Ek)-1 <k| V |0; n,ε1> Now we know the projection of |1; n,ε1> onto ALL states in our problem, we can now construct the first order corrected state as follows: |1; n,ε1> = Σn'≠n |0; n',ε1> <0; n',ε1|1; n,ε1> + ΣkEoa |k> <k |1; n,ε1> = Σn'≠n |0; n',ε1> <0; n',ε1| Q0a(1)V |0; n,ε1> + ΣkEoa |k> <k| Q0a(1)V |0; n,ε1> where Q0(1) = Σn'=1,N, n'≠n |0; n',ε1> <0; n',ε1| + ΣkEoa |k><k| = 1 - P0(1) Q0a(1) = Q0(1)(Ea0 - H0)-1Q0(1) Within Ea0 we really need to work in the ε1 basis |0; n,ε1>, otherwise the above equations expand out into a huge mess. In my "personal" degenerate perturbation theory notes I had this equation ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk // n > 0, but have set n=1 in 3 placs and I kept having trouble finding the values for B'ij,1 . I was able to get the second coefficient set as B'ik,1 = <uk| V |umi >/ (Em - Ek) Computation of the |2> level state corrections. We have this EV equation for the bundle of N1 states degenerate in the ε1 column P(1)(V Q0aV) |0; m,ε2> = ε2 |0; m,ε2> m = 1,2...N1 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> Imagine that we solve this problem for the energies ε2 and the coefficients <0; m',ε1|0; m,ε2>. We then know all about the kets in the right column (at least those of interest to us and shown in the figure) |0; m,ε2> = Σm'=1,N1 |0; m',ε1> <0; m',ε1|0; m,ε2> They are just certain linear combinations of ε1 level states | 0;m,ε1> And suppose at this level, some state |0; m,ε2> becomes a singlet. This appears as the top state on the right, We can define a P0(1,1) and corresponding Q0 for this one state Q0a(1,1) ≡ Q0(1,1) (Ea0 - H0)-1Q0(1,1) Q0(1,1) = Σm'=1,N1(m'≠m) |0; m',ε2><0;m',ε2| + Σn=N1+1,N |0; n,ε1> <0; n,ε1| + ΣkEoa |k><k| It is worth discussing the various sums here. Because P0(1,1) "passes" only the state |0; m,ε2> (top right in our picture), we know that Q0(1,1) must pass all other states. Now, this projector Q0(1,1) includes three groups. First, we need to include all the other states in the ε2 rightmost column shown above. There are N1-1 such states. Just to be able to include them means that we must know what they are! Luckily, we know what they are from the ε2 level EV problem solution. Second, we must include all the other states in the ε1 column below the N1 degenerate group. Only when we include these states will we have included a full basis of Ea0. There are N - N1 such states. Just to be able to include them means that we must know what they are! Luckily, we know what they are from the ε1 level EV problem solution. Third and last, we have to include the states in (Ea0) . Reminder: everything in the picture is at the "0 level" in terms of state representations. The picture is just showing the right linear combinations of our initial 0 level states. Now let's assume we have first solved the ε1 EV problem for the bundle of N degenerate states in the left column, and that we have then solved the ε2 EV problem for the bundle of N1 still degenerate states in the middle problem. We now recall our s=2 perturbation equation (H0 - Ea0) |2> = (ε1 - V ) |1> + ε2 |0> s = 2 (*) and we use the fact that ( we have seen this many times now, same idea over and over again) Q0a(1,1) (Ea0 - H0) = Q0(1,1) Then we have Q0(1,1) |2; m,ε2> = Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 Q0a(1,1) |0; m,ε2> We are now going to close this with various groups of <0| level states, and for each group, the Q0(1,1) on the LHS will vanish when we apply it to the left, since all states will be "other" states relative to |0;m,ε2>: <0, m', ε2| Q0(1,1) |2; m,ε2> = <0, m', ε2 |2; m,ε2> = <0, m', ε2| Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 <0, m', ε2| Q0a(1,1) |0; m,ε2> m' ≠ m <0, n', ε1| Q0(1,1) |2; m,ε2> = <0, n', ε1 |2; m,ε2> = <0, n', ε1 | Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 <0, n', ε2 | Q0a(1,1) |0; m,ε2> n = N1+1,N <k| Q0(1,1) |2; m,ε2> = <k|2; m,ε2> = <k | Q0a(1,1) (ε1 - V ) |1; m,ε2>+ ε2 <k | Q0a(1,1) |0; m,ε2> k (Ea0) Now finally, using all these projection coefficients, we can construct our ε2 level state correction: |2,m,ε2 > = Σm'=1,N1(m'≠m) |0; m',ε2><0, m', ε2 |2; m,ε2> + Σn=N1+1,N |0; n,ε1> <0; n,ε1| 2; m,ε2> + ΣkEoa |k><k| 2; m,ε2>