degeneracy scraps

Scraps from Degen Pert theory Well, this first set of "scraps" is from my failed attempts to understand Schiff's 31.6 Plan A: fails. Here I was hypothesizing that the new adjusted ψ called ψ' (which is what must be appearing in 31.6) was the original ψ minus its projection onto the state ψ(m)0. Now I know this is the wrong relationship between ψ' and ψ. Schiff gave me no advice, so of course this was the first thing I tried. I have moved these notes to the scraps document. Here is a quick review of what happens if you go down this wrong path. Now let's define new ψ'(m)s states as follows (for s = 1,2,3...) ψ'(m)s ≡ ψ(m)s – < ψ(m)s | ψ(m)0> ψ(m)0 = ψ(m)s – a(m)s ψ(m)0 // define a Then we have < ψ'(m)s | ψ(m)0> = < ψ(m)s | ψ(m)0> – < ψ(m)s | ψ(m)0> <m|m> = 0 where we assume our uk states are orthonormal so that <m|m> = 1 . Now let's sub in for ψ(m)s the value ψ'(m)s + a(m)s ψ(m)0 into our all-s equation to get (H0 - E0) [ ψ'(m)s + a(m)s ψ(m)0 ] = ( W(m)1 - H' ) [ ψ'(m)s-1 + a(m)s-1 ψ(m)0 ] + Σj=2,3..s W(m)j [ψ'(m)s-j + a(m)s-j ψ(m)0 ] But we know that (H0 - E0) ψ(m)0 = 0 on the LHS, so we are left with (H0 - E0) ψ'(m)s = ( W(m)1 - H' ) [ ψ'(m)s-1 + a(m)s-1 ψ(m)0 ] + Σj=2,3..s W(m)j [ψ'(m)s-j + a(m)s-j ψ(m)0 ] We can now write the RHS as two separate pieces = ( W(m)1 - H' ) ψ'(m)s-1 + Σj=2,3..s W(m)jψ'(m)s-j ] + { ( W(m)1 - H' ) a(m)s-1 + Σj=2,3..s W(m)j a(m)s-j } ψ(m)0 Suppose we name this {...} object to be A(m,s). Then we arrive at this result (H0 - E0) ψ'(m)s = ( W(m)1 - H' ) ψ'(m)s-1 + Σj=2,3..s W(m)jψ'(m)s-j + A(m,s) ψ(m)0 where of course A depends on H' and lots of other stuff. I think Schiff has cheated us by pretending this last term does not exist. I am willing now to rename the primed functions to be unprimed, but this extra piece really is present. (H0 - E0) ψ(m)s = ( W(m)1 - H' ) ψ(m)s-1 + Σj=2,3..s W(m)jψ(m)s-j + A(m,s) ψ(m)0 < ψ(m)s | ψ(m)0> = 0 both for s = 1,2,3... We shall see if this extra piece affects any of his later calculations. What happens if we close our big equation above from the left with ψ(m)0 ? It seems to me that we get this result: 0 = – <ψ(m)0 | H' | ψ(m)s-1> + W(m)s + A(m,s) which then tells us that W(m)s = – A(m,s) + <ψ(m)0 | H' | ψ(m)s-1> so right off the bat we have a problem Houston. For each expansion level s, A(m,s) is different, so we cannot treat it as a constant energy offset. Plan C: fails. This plan made no sense at all, see scraps. Well, maybe there is a better way out. Suppose you don't impose this condition from the start, and suppose we do our perturbation theory and somehow come up with a convergent series for ψ(m) that solves our problem. Then we certainly can rescale that final result ψ(m) such that < ψ(m)0 | ψ(m) > =1. No question about it. In so doing, we are rescaling each term in the series. Suppose we came up with < ψ(m)0 | ψ(m) > = < ψ(m)0 | Σn=0 λn ψ(m)n > = 5 < ψ(m)0 | ψ(m)0 > + λ < ψ(m)0 | ψ(m)1 > + λ2 < ψ(m)0 | ψ(m)2 > + .... = 5 which says λ < ψ(m)0 | ψ(m)1 > + λ2 < ψ(m)0 | ψ(m)2 > + .... = 4 But this is not taking me anywhere useful. What do I do next here? I could scale λ to get the RHS = 4, but how do I know I can get each term to be 0, etc etc etc. ____________________________________________________________________ In fact, it is useful to express the above state ψmi in this manner: ψm,i= φi χi where φi H(m) and χi H φi = um,i + Σj Bij um,j χi = Σk≠m Bik uk Now suppose we form another basis in the m subspace by doing the following linear combinations vm,s = Σi cs,i um,i um,s = Σi c-1s,i vm,i Now apply the operation Σi cs,i to the φi equation shown above to get Σi cs,i φi = Σi cs,i um,i + Σi cs,i Σj Bij um,j φs' = vm,s + Σi cs,i Σj Bij um,j where we have thus created now a new state φs' inside H(m). We would now like to select the coefficients c such that the following is true, where Bs vanishes when λ→0, just like all the other B things, φs' = vm,s + Bs vm,s = (1 + Bs) vm,s If we can find some states vm,s that make this be true, then the effect of the perturbation within H(m) is solely to add a multiple of vm,s, and none of the other vm,r are mixed in. Then we will have this situation for our perturbation theory "set up" : ψ 'm,i= φi' χi = (1 + Bs) vm,s Σk≠m Bik uk which greatly resembles the non-degenerate perturbation theory starting point. 4. Finding the c coefficients and Bs eigenvalues. Now how can we select the c coefficients to make this happen? We need this to be true: Σi cs,i Σj Bij um,j = Bs vm,s Now let's just start processing this thing: Σi cs,i Σj Bij um,j = Bs (Σj cs,j um,j) Σj { Σi cs,i Bij } um,j = Σj { Bs cs,j } um,j Since the um,j span the m subspace, the above can only be true if: Σi cs,i Bij = Bs cs,j Σi cs,i Bij = Σs' [δs,s'Bs'] cs',j = Σs'Λss' cs',j where Λ is the diagonal matrix shown. We then have the following matrix equation Σi cs,i Bij = Σs'Λss' cs',j c B = Λ c B c-1 = c-1 Λ c B c-1 = Λ Thus, our matrix of coefficients c is seen to be the unitary transformation that diagonalizes B. So our problem of finding the c coefficients boils down to finding this unitary transformation. The first step we know is to find the eigenvalues Bs which are the diagonal elements of Λ. This comes from the secular equation, and to find that equation, we have to write things out as an eigenvector problem: B c-1 = c-1 Λ Σj Bij c-1jk = Σj c-1ij Λjk = Σj c-1ij δjkBk = c-1ikBk = Bkc-1ik Σj Bij c-1jk = Bkc-1ik Σj Bij [c-1k]j = Bk [c-1k]i B [c-1k] = Bk [c-1k] Here our matrix is B, our N eigenvectors are [c-1k] with k as a label, and components [c-1k]i = c-1ik, and our eigenvalues are the Bk. So rewrite this as [ B - Bk 1 ] [c-1k] = 0 and the secular equation is (which allows non-trivial solution vectors [c-1k] ) det(B - Bk 1) = 0 and this determines our N eigenvalues Bk. The rows or columns of c or c-1 are then the coefficients we want. So let's assume that we can manually solve that whole problem. Before continuing, let's just review what we did here. We had within the m subspace: φs' = vm,s + Σi cs,i Σj Bij um,j φs' = vm,s + Σi cs,i Σj Bij Σk c-1j,k vm,k φs' = vm,s + Σk [ Σi Σj cs,i Bij c-1j,k ] vm,k We then sought c coefficients to have this be true [ Σi Σj cs,i Bij c-1j,k ] = Bs δs,k c B c-1 = Λ Then we ended up with φs' = vm,s + Bs vm,s = (1 + Bs) vm,s Now let's continue with our perturbation theory. ************************** We select our states vm,s carefully as discussed above and we then have the following starting point for doing perturbation theory, ψ ' mi= φi' χi = (1 + Bi) vm,i Σk≠m Bik uk where we now simplify by writing m,i = mi, ie, we remove this baggage comma. We now want to follow our Plan D over Schiff Chapter 8 notes. First, rescale the solution as follows ψ " mi = ψ' mi / (1 + Bs) = vmi Σk≠m aik uk where aik = Bik/ (1 + Bs) Let's now drop the double-prime on ψ since this is the thing we will be "working with" , ψmi = vmi Σk≠m aik uk where remember that these uk shown on the right are all in H . Consider then < vmi | ψmi > = 1 < vmj | ψmi > = 0 j ≠ i < uk | ψmi > = aik < ψ mi | ψmi > = some positive number that is most probably not unity! < vmi | Σk≠m aik uk > = 0 The trick now is to expand the correction coefficients in λ : aik = Σn=0 λn aik,n Since we know that ψ mi = vmi when λ = 0, we know that aik,0 = 0 so we could write aik = Σn=1 λn aik,n Now go way back to our original perturbation problem which is summarized in these statements: **************************** Derive a new (31.6). Above we had aik = Σn=1 λn aik,n ψmi = vmi Σk≠m aik uk Σn=0 λn ψmi,n = vmi Σk≠m (Σn=1 λn aik,n) uk Σn=1 λn ψmi,n = Σk≠m (Σn=1 λn aik,n) uk = Σn=1 λn { Σk≠m aik,n uk } => ψmi,n = Σk≠m aik,n uk n = 1,2,3.... Therefore we have < vmj| ψmi,n> = 0 n = 1,2,3... which is our new 31.6 This was the thing that got me so mystified in the Chap 8 notes but which now makes sense. All the corrections ψmi,n lie in H , so the corrections are orthogonal to anything in H(m). Notice that our new version of 31.6 now applies for any i,j in H(m), so the result is a little different that the old 31.6. Derive a new (31.7). Start with our "big equation", (H0 - Em) ψmi,s = (Wmi,1 - H' ) ψmi,s-1 + Σj=2,3..s Wmi,j ψmi,s-j s = 1,2,3... Let's rewrite this now like so (H0 - Em) ψmi,s = - H' ψmi,s-1 + Σj=1,2,3..s Wmi,j ψmi,s-j s = 1,2,3... Close from the left with < vmj| = < ψmj,0| . The LHS gives 0 due to the left-action of (H0 - Em). We pick up an H' term, and then we pick up a hit from the term on the right which has s-j=0 or j = s. Thus 0 = – < vmj| H' |ψmi,s-1> + Wmi,s < vmj | vmi> => (1) Wmi,s = < vmi| H' |ψmi,s-1> j = i 31.7 A s = 1,2,3... (2) 0 = < vmj| H' |ψmi,s-1> j ≠ i 31.7 B s = 1,2,3... Is this item (2) new, or is it already expected? It seems new to me. It implies several things: < vmj| H' | vmi > = 0 j ≠ i if s = 1 < vmj| H' |ψmi,s-1> = 0 j ≠ i if s > 1 If this is really true, it seems to me that this must be true: < vmj| H' |ψmi> = 0 j ≠ i because it is true in each λ order. Recall that ψmi = vmi Σk≠m aik uk . The implication then is that < vmj| H' | Σk≠m aik uk > = 0 j ≠ i Σk≠m aik < vmj| H' | uk > = 0 j ≠ i This really does seem new to me and I am not sure whether or not it is true. We expect that H' can cross between the two spaces H(m) and H . Let's let this ride for the time being. We have successfully arrived at a new 31.7 which is this Wmi,s = < vmi| H' |ψmi,s-1> s = 1,2,3... *************************************************** ψmi,1 = Σk≠m aik,1 uk so the above says (H0 - Em) Σk≠m aik,1 uk = (Wmi,1 - H' ) vmi Close this from the left with some uk' in H to get Σk≠m aik,1< uk'| (H0 - Em) |uk > = < uk'| (Wmi,1 - H' ) |vmi> = – < uk'| H' |vmi> The LHS allows Ho to be replaced by Ek' and then <k' |k> = δk',k which kills off the sum, so (Ek' - Em) aik',1 = – < uk'| H' |vmi> (Ek - Em) aik,1 = – < uk| H' |vmi> k ≠ m aik,1 = < uk| H' |vmi> /(Em - Ek) k ≠ m which is 31.10. So there we have it. We have computed the first order energy correction, and the first order wavefunction correction. Recall that ψmi,1 = Σk≠m aik,1 uk = Σk≠m { < uk| H' |vmi> /(Em - Ek) } uk ψmi ≈ vmi + λ ψmi,1 Wmi,1 = < vmi |H'| vmi > Happily, we know that Σk≠m is only over k in H so we don't get any denominator blow ups. Again, it depends on the nature of H' whether or not degeneracy is lifted for a state |mi> in first order. Note also that we have to know what the vmi are in order to compute this matrix element. We looked into that issue in a general way above, but now how would be actually do it in practice? 7. Comment on the invalidity of our first starting point. That starting point was this: ψmi= { umi + ΣjBij umj } Σk≠m Bik uk Our fundamental equations 31.4 come only from the SE and don't depend on "starting point" and the s=1 equation says this: CONTRADICTION BETWEEN ASTERISK BARS ****************************************************************** Schiff page 245 equation 31.4 second line says this: (H0 - Em) ψmi,1 = (Wmi,1 - H' ) ψmi,0 s = 1 Now given our umi starting point, this says that (H0 - Em) ψmi,1 = (Wmi,1 - H' ) umi since ψmi,0 = umi Now suppose we close this with umj where j ≠ i. The (H0 - Em) kills to the left giving 0. We get 0 = – < umj | H' | umi > But WHY should this be true?? It seems very likely it would NOT be true. ***************************** We can then evaluate (where Xxxx,n represents the λ order n of something) ψmi,1 = ΣjBij,1 umj Σk≠m Bik,1 uk If we insert this into our s=1 31.4 we get (H0 - Em) { ΣjBij,1 umj Σk≠m Bik,1 uk } = (Wmi,1 - H' ) umi 0 Σk≠m Bik,1 (Ek - Em) uk = (Wmi,1 - H' ) umi Close this equation with <mj| where j ≠ i. We get 0 0 = < umj |H' | umi> = 0 I am having a LOT of trouble understanding this result. Why should H' be diagonal inside the m subpace?? *********************************************88 Now close this first with some uk' in H Bik',1 (Ek' - Em) = – < k' | H'| mi> and this seems a reasonable thing and lets us compute Bik,1 = < k | H'| mi>/(Em - Ek) But now close the same equation with some umj in H(m) 0 0 = <mj| (Wmi,1 - H' ) mi> In the case that j = i, we get this reasonable result: Wmi,1 = <mi | H' |mi > In the case that j ≠ i, we get instead this result <mj | H' |mi > = 0 i ≠ j BUT: this is a contradiction because in general there is no reason for some H' not to link different states within the m subspace. Therefore, our "starting position" is inconsistent with doing perturbation theory except maybe in special cases where H'ji just happens to be 0 for all unequal state pairs in the subspace. Now for comparison, let's examine our second starting point which is this: ψmi= φi χi = vmi + Bi vmi Σk≠m Bik uk and let's again look at our s=1 version of 31.4 (H0 - Em) ψmi,1 = (Wmi,1 - H' ) ψmi,0 s = 1 where now we have ψmi,0 = vmi. Then we find: ψmi,1 = Bi,1 vmi Σk≠m Bik,1 uk So insert this into the previous equation to get 0 Σk≠m Bik,1 (Ek – Em) uk = (Wmi,1 - H' ) vmi Close with <k'| and get the same result as before. Close with vmj and also get same results as before. In particular we will still get 0 = <vmj | H' |vmi > when j≠i Now is THIS more reasonable than before for some reason? Is it possible to shuffle the umi states into vmi states such that there are no off diagonal elements of H' ? Aha!! So this transformation must be the one which diagonalizes H' ! Light bulb goes off. Only if you use states |vmi > which diagonalize H' is your perturbation theory self-consistent! Theorem 1: The states vmj = Σi cji umi must be chosen to diagonalize H' . Then perturbation theory can be done. Proof: if H' is not diagonal, we get the inconsistency noted above. Question: is the diagonality of H' in the v basis connected with our "second starting point" somehow? Go back again to the "first" starting point: ψmi= { umi + ΣjBij umj } Σk≠m Bik uk Does this form really cause any problems other than the inconsistency above if H' is non-diagonal? Maybe it is OK. What happens if we just try this thing in our Hψmi = Wmiψmi equation. Does anything bad happen? Let's write it out: (H0 + H') ψmi = Wmiψmi for ψmi which "starts out" in H(m) (H0 + H') ψk = Wkψk for ψk which "starts out" in H Now consider only the first equation above and insert our expansions (H0 + H') [ { umi + ΣjBij umj } Σk≠m Bik uk ] = Wmi [{ umi + ΣjBij umj } Σk≠m Bik uk] There are two terms to consider on the LHS, the H0 term and the H' term: LHS Ho term = Em { umi + ΣjBij umj } Σk≠m Bik Ekuk LHS H' term = H' [ { umi + ΣjBij umj } Σk≠m Bik uk ] RHS = {Wmi umi + ΣjBij Wmi umj } Σk≠m Bik Wmi uk] Notice that the " LHS H' term" is going to "break" the a b structure. That is to say, probably H' applied to some umj is going to result in components in both H(m) and H . Thus it would be wrong to attempt a naive separate subspace equality. Let's write our equation this way: LHS H' term = RHS – LHS Ho term Then we have RHS – LHS Ho term = {[Wmi-Em] umi + ΣjBij [Wmi-Em] umj } Σk≠m Bik [Wmi-Ek] uk where we do have subspace decomposition here. But then LHS H' term = { H'umi + ΣjBij H' umj } + Σk≠m Bik H'uk and here there is no longer a separation, so we se use a regular + to indicate this fact. So we end up with this huge equation that I don't know if we can even make use of (I am looking for inconsistency right now) { H'umi + ΣjBij H' umj } + Σk≠m Bik H'uk = {[Wmi-Em] umi + ΣjBij [Wmi-Em] umj } Σk≠m Bik [Wmi-Ek] uk All I know how to do is close this with states from the left. (1) So let's first close with uk' in H to get { <k'|H'|mi> + ΣjBij <k'|H' |mj> } + Σk≠m Bik <k'|H'|k> = Bik' [Wmi-Ek'] This messy result links together lots of H' matrix elements and Wmi and Ek' and all the Bij and Bik. You might ask: why not just diagonalize H' in the entire Hilbert Space H(m) H. But if we knew how to do that, we would not be doing perturbation theory. So I guess all I can say about the above messy equation is this: I see no glaring inconsistencies. (2) Next, let's close with uml in H(m) and see what happens in that case: ml { <ml |H'|mi> + ΣjBij <ml |H' |mj> } + Σk≠m Bik <ml |H'|k> = [Wmi-Em] { δli + Bil } I see nothing glaringly inconsistent at this point. But now let's assume we have chosen a basis that diagonalizes H' within H(m). Then this says δli <mi |H'|mi> + Bil <ml |H'|ml> + Σk≠m Bik <ml |H'|k> = [Wmi-Em] { δli + Bil } which we write out in the two cases. (2: l=i) First, here is the l = i case: <mi |H'|mi> + Bii <mi |H'|mi> + Σk≠m Bik <mi|H'|k> = [Wmi-Em] { 1 + Bii } (1 + Bii) <mi |H'|mi> + Σk≠m Bik <mi|H'|k> = [Wmi-Em] (1 + Bii) [Wmi – Em – <mi |H'|mi>] (1 + Bii) = Σk≠m Bik <mi|H'|k> This looks OK. The bracket on the LHS is probably the corrected energy 2nd order and beyond. Notice that the H(m) numbers Bij for i≠j don't even enter here, so would not matter whether you used the u or the v function basis, so either starting point seems OK in the case that H' is diagonal in H(m). We could rewrite that above in this way (1 + Bii) = Σk≠m Bik <mi|H'|k> / [Wmi – Em – <mi |H'|mi>] (2: l≠ i) Now the l ≠ i case says this Bil <ml |H'|ml> + Σk≠m Bik <ml |H'|k> = [Wmi-Em] Bil [Wmi–Em – <ml |H'|ml>] Bil = Σk≠m Bik <ml |H'|k> This seems to give an expression for the Bil as follows: Bil = Σk≠m Bik <ml |H'|k> [Wmi–Em – <ml |H'|ml>] i ≠ l Theorem 2: Diagonalizing H' does not in general require that Bil be diagonal within H(m). Proof: above we have a formula for Bil and it certainly does not seem to vanish. Conclusion: My "second starting point" with the v basis is no better than my "first starting point" with the u basis, so all that effort of getting to this second starting point was a complete waste. So here I sit, having gotten nowhere at all in my private effort to do degenerate perturbation theory. Too bad, nice try. I did seem to obtain two theorems however. The first says you have to diagonalize H' in the m subspace if you expect to do perturbation theory. The second says the first starting point is OK provided you diagonalize H' with your basis in the m subspace. 7. Review the question of how the | vmi > are actually computed. It took me a whole day to just write down all of the above, after getting totally confused in the morning. The Hilbert Space idea helped a lot. But now I don't really see how you would actually solve a problem. I need to somehow relate things like Bik to H' which I have not done at all. This will be where I continue next time. Recall from above that aik = Bik/ (1 + Bs) k in H where Bs are the eigenvalues of Bij within the m subspace. Theorem A: Consider some eigenvalue problem H0ur = Erur where the ur span some Hilbert Space H. Suppose H contains a degenerate subspace H(m)of dimension N>1, such that H = H(m) H, where H(m) is spanned by an orthonormal subset of the ur, which we call umi, such that H0umi = Emum for i = 1,2....N (the m label just tells us the umi are uk within H(m)). Suppose we want to do a perturbation theory analysis for the Hamiltonian H = Ho + λH' where λ is a smallness parameter. Theorem A says: You cannot "do" a perturbation analysis on the states umi unless they diagonalize H' within H(m). We might be able to do perturbation analysis for states uk outside H(m), but not for states within H(m). By "do a perturbation analysis on a state" we mean find how a state and its energy smoothly change from their λ=0 values of umi and Em to some altered values which form a power series in λ. Proof: This is a proof by contradiction. Assume that we can form a power series like so ψmi = umi + λ ψmi,1 + λ2 ψmi,2 + .... ψmi,0 = umi Wmi = Em + λ Wmi,1 + λ2 Wmi,2 + .... Wmi,0 = Em and assume that H' is not diagonal within H(m). When we insert the above expansions into the equation (H0+λH')ψmi = Wmiψmi and equate like powers of λ on both sides, we get a series of equations that Schiff calls 31.4 on page 245. In particular, the equation which equates λ1 on both sides says this: (H0 - Em) ψmi,1 = (Wmi,1 - H' ) umi Suppose we close from the left with umj for j ≠ i. We get < umj| (H0 - Em) = 0 acting to the left. Since the basis functions are orthogonal and Wmi,1 is just a number, we get only the H' term on the right. We thus find that: 0 = <umj | H' | umi> for any j ≠ i But this says that H' must be diagonal, which contradicts our assumption. Thus, if the umi are selected in such a way that H' has off-diagonal elements, perturbation theory on umi is inconsistent. We must first find new basis functions vmi which diagonalize H' within umi, and then we can do perturbation theory on these vmi. On page 249 Schiff says that if you pick the wrong linear combinations in H(m), the perturbation theory "breaks down". This theorem shows exactly how it breaks down. ___________________________________________________________________________________ Theorem B: It is possible to normalize ψmi such that <umi| ψmi,n> = 0 for n > 0. ( This is our degeneracy version of Schiff 31.6 on page 246. ) Proof. Consider the most general change that can happen to ψmi as it leaves umi, ψmi = { umi + ΣjBij umj } Σk≠m Bik uk ψmi H(m) H where the Brs depend in some complicated way on H'. If we set H' = 0, all Brs = 0. [ the Brs may also depend on H0 when H'≠0, I am not sure because I have never tried to make the connection. ] We are showing things to all orders of the power series λ, and we are allowing that ψmi might leave umi in every direction at the same time as λ leaves the value 0. You cannot get more general that that because the ur span the Hilbert Space H. Now break out the umi term in the second sum so we have ψmi = { umi + Biiumi + Σj≠iBij umj } Σk≠m Bik uk = { (1 + Bii) umi + Σj≠iBij umj } Σk≠m Bik uk Then renormalize ψ into ψ'mi = ψmi/(1 + Bii) and define B'rs = Brs/(1 + Bii) to get ψ'mi = { umi + Σj≠iB'ij umj } Σk≠m B'ik uk We know that ψ'mi,0 = umi. For any higher order component we have ψ'mi,n = { Σj≠iB'ij,n umj } Σk≠m B'ik,n uk where we imagine expanding the B' just as we expand ψ' or W. Then we have <umi| ψ'mi,n> = 0 n>0 which is 31.6 because <umi| is orthogonal to all the states on the RHS. This object ψ'mi,n is the function we use in our perturbation theory and we drop the prime. Since Bii is fully determined by H' (and H0), since umi is presumed normalized, the normalization of ψ'mi is fixed by the above choice, and we cannot then further require that < ψ'mi | ψ'mi> = 1. This will in general not be true for the way we have defined ψ'mi. When we are all finished with our problem, we can then change the normalization of ψ'mi to get unity. ___________________________________________________________________________________ Theorem C: Wmi,s = < umi| H' | ψmi,s-1> ( degeneracy version of 31.7 Schiff p 246). This says that if you know the wave function to some order, you can compute the energy correction of the next order. Proof: When we expand the full SE in ψ and match powers, we get this result: (H0 - Em) ψmi,s = (Wmi,1 - H' ) ψmi,s-1 + Σj=2,3..s Wmi,j ψmi,s-j s = 1,2,3... which we can rewrite like so (H0 - Em) ψmi,s = - H' ψmi,s-1 + Σj=1,2,3..s Wmi,j ψmi,s-j s = 1,2,3... If we close from the left with umi, the LHS = 0 due to the left action of (H0 - Em) . On the right we pick up a term from the H' term. What about all those other terms? The terms have this form (last factor) ψmi,s-1 ψmi,s-2 ψmi,s-3 .... ψmi,1 ψmi,0 = umi From Theorem B we claim that <umi| ψmi,n> = 0 n>0 All the factors shown have n>0 except that very last one which has n = 0. This, we pick up only this last term of all the terms in the second sum. That is the term with j = s, so we then have 0 = – <umi| H' | ψmi,s-1> + Wmi,s QED Comment: What happens if we close from the left with some uml with l ≠ i ? We would get 0 = – <uml| H' | ψmi,s-1> + Wmi,s <uml | umi> + Σj=1,2,3..s-1 Wmi,j <uml |ψmi,s-j> => <uml| H' | ψmi,s-1> = Σj=1,2,3..s-1 Wmi,j <uml |ψmi,s-j> l ≠ i But now we know nothing about <uml |ψmi,s-j> for l ≠ i and in general all these factors will be non-zero and all we get is some kind of sum rule that is not very interesting. ___________________________________________________________________________________ ******************** (H0 + λH') { umi + λ ψmi,rest} = (Em + λ Wmi,1 + λ2 Wmi,rest) { umi + λ ψmi,rest} H0 umi = Em umi λ0 H' umi + H0 ψmi,rest = Wmi,1 umi + Em ψmi,rest λ1 This last says (H0 - Em) ψmi,rest = (Wmi,1 - H') umi which is like the corresponding Schiff equation but we have "rest" in here now. Now close with mj ≠ mi: 0 = - <umj| H' |umi> We STILL seem to have this diagonal H' appearing, I cannot make it go away! But suppose I go back and try it this way: