messiah R1 SSPT Chap 16

Messiah on Perturbation Theory PhL 1.8.09 symbol grab: ΣkE H(m) H Ea0 (Ea0) (1.8.09) I have been confused and unhappy about degenerate perturbation theory (stationary state) since I started working on it on Jan 2, so my unhappiness has continued for about 6 days. I read Schiff and Saxon and could not find happiness, so now I will see if Messiah can bail me out. This will of course require going into a whole new notation with all the overhead that involves, but that is the price of consultation. [ It is now Jan 27 and I have just finished my Round Two notes, and things are still not crystal clear, but I was at least able to answer every question I asked. So as of today I have invested 25 days of my life on this topic (but also did Stakgold Chapter 2). My notes here and in Round Two don't deal with the Kato resolvent method, just the Messiah stuff in earlier sections, as carried on by me. At least I was able to produce expressions for those mysterious B'ij,1 coefficients from my personal exposition of SSPT. That happened for the first time today. ] [ It is now Feb 8 and I just finished up all Kato notes, 5 documents! ] Overview: All of Chapter 16 is on SSPT, pp 685-721. I read only the "theory sections" of this chapter and skipped over "applications" like Stark Effect, at least for now. The degenerate problem is fully solved (in a sense) in the tour de force Topic III Kato section, which I will need to review in some meta notes. This was about the hardest stuff to understand that I have ever read, and is based on a 1949 paper of Kato with reference on page 712. That work was updated in 1958 by Messiah's friend C. Bloch and must have come out just as this book was being wrapped up. There are of course newer treatments of the complicated degeneracy problem. I found a saved a 1981 paper by Silverstone and Moats that seems pretty good and references Kato and Bloch. Newer papers still exist, with an eye to practical calculation in higher orders. Messiah Volume II Chapter XVI (16): Stationary Perturbations. (p 685) 1 1. General introduction to Part Four (methods of approximation). 2 Topic I. The Non-Degenerate Case (p 686) 2 2. Expansion in Powers of the Perturbation (p 686) 2 Digression on the operators Qo and Q0a. 3 3. First-order perturbations (688). 5 6. Higher order corrections. (694). 6 Topic II. The Degenerate Case 7 8. Elementary Degenerate Theory (698). 7 Rederive facts about Q0 and Q0a . 10 Clarification of my Theorem B confusion. 17 Clarification of my B'ij confusion. 19 Messiah Volume II Chapter XVI (16): Stationary Perturbations. (p 685) 1. General introduction to Part Four (methods of approximation). You approximate in the world of bound states and in the world of collisions, that latter being messier. In the bound state world, Messiah has already done the WKB method in Chapter 6. In this chapter we focus on SSPT. Topic I. The Non-Degenerate Case (p 686) 2. Expansion in Powers of the Perturbation (p 686). He is going to refer to H' as V, but the smallness parameter is still called λ. The notation Ei0 are the H0 energy eigenvalues. States are | Ei0 α> where the α is needed only if you have degeneracy. So in this section we have just | Ei0 > and he favors using "a" for the index, so we have | Ea0 > . This "a" is like the "m" of Schiff, we are picking out some particular state to think about. As a shorthand notation, he will use |0> = | Ea0 > since we know we are talking all the time about "a" and we want to reduce notational clutter. The perturbed version of |0> will be called |ψ> which again agrees with Schiff's choice of notation for this state. Messiah will be "ket oriented" which is fine by me. Now we come to page 687 equation 16.4 (Roman numerals are horrible, Messiah should not have done that to my way of thinking). It says <0|ψ> = 1 which is Schiff's statement that < um | ψ ' (m)> = 1. Oddly, Messiah says nothing about this and I have a little pencil "WHY?" from some earlier time I looked at this chapter. This is explained as Plan D in my Schiff Chapter 8 notes, and also as Theorem B of my Degenerate Perturbation Theory doc. I am quite happy with it. Next we come to the power series shown in 16.5 and 16.6, so let's do more notational linkage: Schiff Messiah m a // selected state (or space of N degenerate states) H(m) Ea0 // space containing N unperturbed state(s) of interest Ea // space containing the N perturbed states H' V // perturbing Hamiltonian Wm = Σn=0 λn Wm,n E = Ea0 + Σn=1 λn εn // energy expansion Wm Em 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 |r> ( PhL) |Ea0α> // starting base states inside Ea0 We now come to Messiah's version of Schiff 31.4, but Messiah moves Schiff's RHS terms to the LHS. I notice Messiah's attempt to give a separate number for equations in a group using a little super. Next, Schiff's 31.6 which said <um| ψm,n> = 0 for n>0 appears as Messiah 16.8 page 688. Of course it is very nice in Messiah notation : <0|n> = 0 for n>0. Next, Schiff's 31.7 which said Wm,s = <um|H'|ψm,s-1> becomes εs = <0|V|s-1> but Messiah uses n instead of s, very reasonable, and thus we have Messiah 16.9. In Messiah, this result comes from "closing with <0| "onto his 31.4 version. Where is equation 688.A coming from? When he writes |E0α >, he just means some unperturbed eigenstate other than the |0> one which we are focusing on. If we close his 31.4 version with this, and then divide through by the energy difference that then appears, we get A page 688, fine. Messiah now defines Q0 to be the identity operator in our entire Hilbert Space minus the contribution of our unperturbed state of interest, just fine by me. [ It is the "perp space projector" to this one state. ] Digression on the operators Qo and Q0a. Equation 16.10 is "less obvious" to me. Here is my derivation. First start with Q0 ≡ 1 - |0><0| = Σk≠a |Ek0>< Ek0| This thing is an operator, and on the right above I write it as a sum over all the unperturbed eigenstates (which I label by Ek0) except the state a, which is the idea that we are subtracting out |0><0|. Notice that I cannot express these other states by the notation |k> because that notation is already used to mean | ψm,k> = | ψa,k>. This thing Q0 is a "projection operator". Notice this fact: Q02 = [1 - |0><0|] [1 - |0><0|] = 1 - |0><0 - |0><0| + |0><0|0><0| = 1 - |0><0| - |0><0| + |0><0| = 1 - |0><0| = Q0 as appropriate for any projection operator. Notice that this implies that Q0-1 = Q0 What exactly does this operator Q0 do? If you have some ket |φ> which is an arbitrary linear combination of all unperturbed kets in the full Hilbert Space, this projects out all pieces except the component along the unperturbed state Messiah calls |0> which is his state with Ea and would be Schiff's state um. It is sort of a deletion operator in this sense. Now obviously if you apply this to some "other unperturbed state" like uk somewhere, k ≠ a, then Q0| uk> = | uk>. But we are going to apply this more significantly to a perturbed state which we should regard as a rotated ket in the space of unperturbed basis kets. We have for example, |ψ> = Σn=0 λn |n> which I might enhance just a bit and write as |ψa> = Σn=0 λn |n; a> so Q0|ψa> = Σn=0 λn Q0 |n; a> = Σn=1 λn |n; a> = |ψa> - |0; a> so in this case Q0 removes the entire |0;a> part of |ψa>. [ It is a perp space projector. ] In particular, in Messiah notation we have Q0| n> = | n> n > 0 In my own notation where I had ψ'mi = { umi + Σj≠iB'ij umj } Σk≠m B'ik uk Q0ψ'mi = { Σj≠iB'ij umj } Σk≠m B'ik uk which is to say, it projects out ALL of the full perturbed state other than the unperturbed part. So this operator gives you access to the part of the perturbed state that is caused by the perturbation. It is an interesting operator. Now consider (where all three objects are operators since H0 is an operator). Q0 (Ea01 - H0)-1 Qo = Σk≠a |Ek0>< Ek0|(Ea01 - H0)-1 Σk'≠a |Ek'0>< Ek'0| = Σk≠a Σk'≠a |Ek0>< Ek0|(Ea01 - H0)-1|Ek'0>< Ek'0| = Σk≠a Σk'≠a |Ek0>< Ek0|(Ea01 - Ek01)-1|Ek'0>< Ek'0| // H0 acting to the left = Σk≠a Σk'≠a |Ek0>(Ea0 - Ek0)-1< Ek0|Ek'0>< Ek'0| // the 1's are implicit = Σk≠a Σk'≠a |Ek0>(Ea0 - Ek0)-1δk,k'< Ek'0| // all unperturbed states ortho = Σk≠a |Ek0>(Ea0 - Ek0)-1< Ek0| // which is 16.10 so this verifies the equation shown in 16.10. His summation notation is a little strange, I will stick with mine, they are both OK. He then makes this very strange notational definition: Q0/a ≡ Q0 (Ea0 - H0)-1 Qo = Σk≠a |Ek0>(Ea0 - Ek0)-1< Ek0| ≡ Q0a where I think Q0/a is all one symbol, a certain operator. I will call it Q0a . So why is this a useful thing to have? Here is a mini-theorem: Mini Theorem #1: Q0a(H0 - Ea0) = Q0 Proof: Where we use the fact that Q0-1 = Q0 = Qo2 : Q0a = Q0 (Ea0 - H0)-1 Qo Q0a-1 = Q0 (Ea0 - H0) Qo so Q0a (Ea0 - H0) = { Q0 (Ea0 - H0)-1 Qo} (Ea0 - H0) = { Q0 (Ea0 - H0)-1 Qo } { Qo (Ea0 - H0) Qo } Qo = Q0a Q0a-1 Qo = Qo QED. Now go back to 16.7n which is this (n > 0) where |0> is assumed "in the perp space" : ( a |k> ie) (H0 - Ea0) |n> + (V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1> – εn|0> = 0 (H0 - Ea0) |n> – εn|0> + [(V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1>] = 0 (Ea0 - H0) |n> + εn|0> = [(V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1>] If we apply Q0a to both sides of the last line above, the εn|0> is killed off by the right-side Q0 in Q0a, and then the remaining operator (Ea0 - H0) is sort of "inverted" by Q0a as shown above since Q0|n> = |n> and we get |n> = Q0a [(V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1>] // which is 16.11 which is 16.11. This fancy operator Q0a creates the next order of perturbed state when it acts on the square bracket ket which involves only lower order kets. So this is a pretty nice compact operator form for doing this operation. Comment: The operator Q0a "exposes" the perp-space state |n> on the LHS of 16.7 by inverting the operator (H0 - Ea0) normally sitting there. Your thus get a formal equation for |n> in terms of lower state corrections, which is the key idea to this iterative perturbation theory. 3. First-order perturbations (688). He writes the first order energy correction 16.12 as expected. Then 16.13 shows the energy through order λ, and notice he has H in there and not just V = H' , so <0|H|0> generates the first two terms Ea0 + λ <0|V|0>. Now we get to practice with his little Q operator stuff. Let's try it on the lowest order iteration right now: |1> = Q0a [(V-ε1)|0> = Q0a V |0> // which is 16.14 so <Eα0| 1> = <Eα0| Q0a V |0> = <Eα0| Σk≠a |Ek0>(Ea0 - Ek0)-1< Ek0| V |0> = (Ea0 - Eα0)-1 < Eα0| V |0> // which is 16.16 4. Ground state of Helium atom. skip for now 5. Coulomb energy of atomic nuclei. skip for now. 6. Higher order corrections. (694). So now we resume the theory flow. Equation 16.26 is the usual formula I think of as Schiff 31.7, but it is Messiah 16.9. This formula is always simple in any order. |n> = Q0a [(V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1>] |2> = Q0a (V-ε1)|1> so this one has just a single term as |1> had. Now we know that |1> = Q0aV |0> so we have |2> = Q0a (V-ε1) Q0aV |0> with ε1 = <0| V |0> = Q0a VQ0a V |0> – <0| V |0> Q0a2 V |0> = [ Q0a VQ0a – <0| V |0> Q0a2 ] V |0> For some reason I do not comprehend, he writes this second term differently, using this fact Q0a2 <0| V |0> V |0> = Q0a2 V |0><0| V |0> = Q0a2 V |0><0| V |0> to get |2> = [ Q0a VQ0a –Q0a2 V |0><0| ] V |0> // Messiah p 194 (27) and for the second time he uses the odd notation Q0/a2 to mean Q0a2. He suggests that you can keep going with this idea to get complex formulas. Let's just try one more: |n> = Q0a [(V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1>] |3> = Q0a [(V-ε1)|2> – ε2|1>] where ε2 = <0|V|1> = <0|V Q0aV |0> so get |3> = Q0a [(V- <0| V |0>){ [ Q0a VQ0a - Q0a2 V |0><0| ] V |0>} – <0|V Q0aV |0> Q0aV |0>] = Q0a V Q0a V Q0a V |0> + lots of other terms It looks a little like scattering theory and I think we are just picking up more messy energy denominators. Now start over and suppose ε1 was zero. Then we have |2> = Q0aVQ0aV |0> = (Q0aV)2 |0> ε2 = <0|V|1> = <0|V (Q0aV)1 |0> And now suppose ε2 is also 0. Then |3> = Q0a [(V-ε1)|2> – ε2|1>] = Q0a V|2> = Q0a V Q0a VQ0aV |0> = (Q0aV)3 |0> ε3 = <0|V|2> = <0|V(Q0aV)2 |0> So it seems pretty reasonable that this pattern can continue, and in general |n> = (Q0aV)n |0> which means of course |n> = (Q0aV) |n-1> // Messiah p 694 (30) εn = <0|V(Q0aV)n-1 |0> both provided ε1 = ε2 = .... εn-1 = 0 // Messiah p 694 (29) In some sense, (Q0aV) is a "raising operator" taking you from the n-1st perturbation contribution to the nth perturbation contribution. This really looks like some kind of iterative Foldy-Wouthuysen type deal going on here. The point is that you always want to carry the thing until you get some non-zero energy correction, and in this case that first correction has a somewhat simple form in that those "lots of other terms" won't be present. I am now on page 695 and I don't follow his convergence comment where I have just put a pencil ? in the margin. And I am skipping his bound comments which finish out this section. Come back some day if interested. I want to get on to the degenerate stuff which is why I am here in the first place. 7. Stark Effect for a Rigid Rotor (695). This is a rather long 2-page example of computing the second order energy correction for a certain problem. As 16.31 shows, this really requires an infinite summation so one wonders how that is done, or perhaps how approximated. I would guess you ignore distant weakly coupled states somehow. I am skipping this section for the reason just noted. Topic II. The Degenerate Case OK, I had to do a lot of work and notational switching just to get to this point, see all of the above notes in this doc. 8. Elementary Degenerate Theory (698). We have some pretty solid setting up going on here. Here are some preliminary comparisons of notation. First, here is the degenerate subspace where we have a bunch of states of H0 having the same eigenvalue H(m) of dimension N Ea0 of dimension ga m = a Happily he uses the word "subspace". P acting on a general vector in H = H(m) H projects out that portion in H(m) But now he is going to handle another complication. Suppose the perturbation H' = V does not fully split all the states in H(m). So as λ comes on, the degenerate states wonder off in groups. Here is a picture suggesting this idea, where we are plotting let us say the first order energy correction ε1 versus λ. and I am sure this situation is very common. When λ > 0, you can regard the group of 4 states as a subspace. So when λ >0 at some value like λ1 shown, we have H(m) → H(4) H(2) H(1) H(1) where I am using a sloppy notation just to make the point. Messiah is going to assume that there are n such groups that fan out as λ is turned on, and that they have dimensions: g1, g2 ..... gn = degeneracies of the groups g1 + g2 + g3 + ... + gn = ga Subspaces are called E1 , E2, .... En E1 E2 E3 En are the full energy eigenvalues for these smaller subspaces I like how we are going. In this context, he says to let ψ be one of those groups on the right with energy E (= some Ei). Equation 16.35: P0 keeps only states that are in H(m) = Ea0. Certainly an unperturbed state |0> of energy Ea0 survives this projector, QED. Equation 16.36. How does P0 act on this equation 17.71 ? Why does the first term vanish? (and then the other term gives 16.36 trivially) he seems to be claiming that: P0 (H0 – Ea0)|1> = 0 How can we write this projector using kets? P0 = Σk |Ek0>< Ek0| ? sum over all states in Ea0 I think this is right. It is just "this part of unity". Then we would have P0 (H0 – Ea0)|1> = Σk |Ek0>< Ek0|(H0 – Ea0)|1> = Σk |Ek0>< Ek0|( Ek0 – Ea0)|1> = 0 So we can add this Mini Theorem #2: P0 (H0 – Ea0) = 0 proven above How should I interpret this theorem? Consider: [ I am temporarily reverting to notation |n> meaning |un>, an unperturbed eigenstate, not the nth correction to some particular state. ] H0 = Σn |n><n|Ho // sum over the entire full Hilbert Space = Σn |n><n|En // since states |n> are eigenstates of H0 = Σn En |n><n| // sum still over all states. P0 H0 = Σj Ej |j><j| // sum is now only over subspace Ea0 Ej = Ea0 Meanwhile we know that 1(full) = Σn |n><n| // we used this above 1(Ea0) = Σj |j><j| = P0 // unity's portion in subspace Ea0 So then we know that P0 Ea0 = Ea0 Σj |j><j| Thus P0 H0 = P0 Ea0 . So here are some words: Question: Why is P0 (H0 – Ea0) = 0? Answer: When you act on the ANY operator Ω with the operator P0, whatever you end up with must be entirely within Ea0. Thus, if Ω is an operator whose eigenvalue is ωa for every basis state of Ea0, then it follows that P0 (Ω – ωa) = 0. Thus, if you restrict your interest to the space Ea0, you can always replace operator Ω by number ωa. H0 is just an example of such an operator, so P0 H0 = P0 Ea0. I can see this could be a powerful tool that I have not been using. Now, suppose we were to define a new Q0 operator: Q0 = 1 - P0 It seems pretty clear that if we think of H = H(m) H then it follows that Q0 = Σk |k><k| where k spans the perp space H This is similar to the Q0 operator we had on page 688, but there we subtracted out just one state's contribution to unity which we wrote as |0><0|, whereas here we are subtracting out all states in the subspace Ea0. Probably all the facts we showed about Q0 are still true but I want to rederive them here: [ P0 is the subspace projector, Q0 is the perp (to that subspace) projector ] Rederive facts about Q0 and Q0a . Define: Q0 = 1 – ΣjEo |j><j| = ΣkE |k><k| = "the perp space projector" Consider: F0a ≡ Q0 f(H0) Q0 Theorem: You can remove either of the Q0 operators and the result is still the same. Proof: Let's just evaluate all three objects and show they are the same: (1) Q0 f(H0) = ΣkE |k><k| f(H0) = ΣkE |k> f(Ek) <k| (2) f(H0) Q0 = f(H0) ΣkE |k><k| = ΣkE f(Ek) |k><k| = ΣkE |k> f(Ek) <k| (3) Q0 f(H0) Q0 = Q0 [f(H0) Q0 ] = Q0 ΣkE |k> f(Ek) <k| = ΣkE |k> f(Ek) <k| Application: Let f(H0) = (Ea0 - H0)-1 . Then we know that: Q0a ≡ Q0(Ea0 - H0)-1Q0 = Q0(Ea0 - H0)-1 = (Ea0 - H0)-1Q0 // see warning! WARNING: Two operators Ax=f and Bx=f are equal if they have the same domain and range, and if on that common domain they have the same values in the range. If we look at the operators above acting on the normal ket space, we find that Q0(Ea0 - H0)-1Q0 : (Ea0) => (Ea0) (1) using my special domain/range notation. Any point outside this domain is mapped to 0. So we can consider Q0(Ea0 - H0)-1Q0: H => (Ea0) (2) as a Stakgold "extension" of the first mentioned operator, where the mapped value is 0 for points outside (Ea0) . If we compare this last operator to Q0(Ea0 - H0)-1: H => (Ea0) (3) we see that operators (2) and (3) are not the same. In the extension space Ea0 , the operator (2) gives 0, but operator (3) diverges! So our proof of equality shown above is only valid for (Ea0) => (Ea0) . From the last two forms we then know that Q0a(Ea0 - H0) = (Ea0 - H0)Q0a = Q0 Now we can, as before, apply this to 16.7n which says (Ea0 - H0) |n> = [+ (V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1> – εn|0> ] where the state here being corrected could be either in the perp space or the subspace. We know that in either case we can apply Q0a to both sides and we get Q0 |n> = Q0a [ (V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1> – εn|0> ] (1) Suppose |0> lies in the perp space. Then Q0|0> = |0> and we just get Q0 |n> = Q0a [ (V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1> ] Now we don't know which space the various corrections lie in, so we cannot say Q0 |n> = |n> as we could before when we know that all corrections to |0> were in the "perp space due to the <n|0> = 0 normalization convention. So here we have to just let Q0 sit there. (2) Suppose |0> lies in the sub space. Then Q0|0> =0 and we just get Q0 |n> = Q0a [ (V-ε1)|n-1> – ε2|n-2> – .... – εn-1|1> – εn|0> ] so our only change is that we can throw out the last term, since |0> lies in the subspace for sure. We might as well then not bother with these two special cases and just include the εn|0> term. Suppose n = 1. We then get (think of this as a version of equation 16.71 ) Q0 |1> = Q0a(V-ε1)|0> = Q0a V|0> // agrees with 16.37 After much fiddling, we are now starting the second paragraph on page 699. Just as a reminder, from 16.36 we also know that P0(V-ε1)|0> = 0 Rewrite this as (we are now assuming that |0> is in the subspace somewhere) P0V|0> = ε1|0> P0VP0 |0> = ε1|0> He points out that this latter (either really) is an eigenvalue equation. If we close with some arbitrary state in the subspace, <Ea0α | , we get ( using Po = Σα' |Eaα'>< Eaα'| ) <Ea0α |P0VP0 |0> = ε1 <Ea0α |0> <Ea0α |V P0 |0> = ε1 <Ea0α |0> <Ea0α |V Σα' |Eaα'>< Eaα'| 0> = ε1 <Ea0α |0> = Σα' <Ea0α |V |Eaα'>< Eaα'| 0> = ε1 <Ea0α |0> // which is page 699 A which I might abbreviate as H'αα'ψ0α' = ε1 ψ0α' = Vαα'ψ0α'l // again, this is "a version of" 16.71 which is my famous subspace diagonalization problem! In Messiah, we have N = ga = the subspace dimension. So there will be ga eigenvalues for ε1. It is likely that a particular eigenvalue ε1 will have some multiplicity, but it could also be non-degenerate. Suppose a particular ε1 is non-degenerate. I think he is saying that this is a "singleton" in my picture above (copied below), and it will "come in for a landing" on some well defined state in Ea0. He claims that the state |0> is "fully determined to the zeroth order". He says this follows from 16.70,1. Why is this?? I would have thought that 16.70 would be enough, since it just says H0|0> = Ea0|0> . Let's put this question on hold for a moment. // Well, as I have commented above, the matrix equation itself comes from 16.71 and so that is where ε1 and its non-degeneracy (if true) comes from. Comment: This text is at the nub of my various problems and I have to study each phrase very carefully. Q0|1> = "the perp space component of the correction |1> " = determined by 16.37 // I agree But, we don't know what P0|1> is, the subspace projection. It is "undetermined" so far simply because we have not seen any expression for this. We do know that <1|0> = 0 from our norm convention. Now suppose instead that ε1 has multiplicity g1. In this case, we know that the landing state |0> is in the subspace E1 = Ea(1) ( as were defined on page 698), and in this case |0> is not known! He claims that if you want to know what |0> is, you have to "go to higher orders". I don't really know what he means by this. // But see below. At some order you hope to get an EV equation which you can solve for the actual states so you know what they are. Review: I took equation 71 and applied P0 to it and used my now-obvious fact that P0 (H0 – Ea0) = 0 to get (3) which said P0(V-ε1) |0> = 0. I first wrote this as an EV equation (P0V)|0> = ε1|0>. I then made the assumption that |0> was some state in Ea0 -- up to now that fact was not needed. This then allowed me to write the new EV equation P0VP0 |0> = ε1|0>. I then closed this with an arbitrary Ea0 state to get <Ea0α |V P0 |0> = ε1 <Ea0α |0>, where now on the left you see that the "right side P0" was necessary in the triplet P0VP0. I could have just "pulled Po out of the vacuum" so to speak, knowing that |0> Ea0. I then used the sum expansion for P0 which is Po = Σα' |Eaα'>< Eaα'| and I ended up with a matrix equation Vαα'ψ0α' = ε1 ψ0α and this then applied to any state |0> in Ea0. This thing is an NxN matrix EV problem involving all of the Ea0 space, dim = N = ga. ψ0α is the component of our |0> state onto the base state called α or <Ea0α | in longer notation. Our matrix equation is then V ψ0 = ε1 ψ0. In our notation, we don't have any label that says which of the Ea states that |0> is, we could say V ψ0i = ε1i ψ0i to indicate it is the ith base state (and maybe |0; i>). Then at least we have a way to enumerate the eigenvalues ε1i. In the simplest case to consider, all N of the ε1i are different, and then we could solve this EV equation for the unique eigenvectors ψ0i. This is thing I like to refer to as the "pre-diagonalization" process. In this case that all ε1i are different for i = 1..N, our picture would show that all states in Ea are "breaking out into singletons in the first order energy correction", and here is the picture: The solution of equation V ψ0i = ε1i ψ0i will determine the "correct" 0th order wavefunctions and the 1st order energy correction. If we solve this equation and find the four unique ψ0i eigenvectors, then those are the four states that are going to fan out as shown when λ > 0. This is an energy plot, ε1 on the vertical axis. So the point is that we know what the correct fanning out states are! So, if at the very start of your problem, you pre-diagonalize H' = V, then you will have the correct fanning out states if the eigenvalues are all different. I think all is well here. Now, suppose we instead find something like this: so that two of the ψ0i have the same value for ε1i. Then I suppose at this point, any linear combination of the two eigenstates ψ0i with this eigenvalue would be acceptable as your starting states. Now comes the big point: In the above picture, perhaps the 2 states stay together at the ε1 level. They may however split apart at the ε2 correction level! Or if not there, at some higher level. This then explains the comment I underlined in red (p 699) saying you then must "proceed to higher orders". In my example above, we have g1 = 2. We want to call this sub-sub space by the name Ea(1) with dimension g1. Clearly we have that Ea(1) Ea. So let's try to mess around at the ε2 level a bit. We obviously will deal with equation 72 which is the lowest on the ladder which shows ε2 somewhere. Messiah then suggests that we consider a projection operator P0(1) which projects out only components in Ea(1). I like it. Then let P0 = P(1) + P' where P' projects onto the states in the perp space of Ea(1) within Ea These projection operators are maybe better thought of as "filters" which block the wrong stuff and only pass the right stuff. So I now agree finally with equations B on page 699 and we can finally turn the page! Now we are going to do a little algebra top of page 700. Recall our little theorem above which said this: Consider: F0a ≡ Q0 f(H0) Q0 = ΣkE |k> f(Ek) <k| Q0 = ΣkE |k><k| Theorem: F0a ≡ Q0 f(H0) Q0 = f(H0) Q0 = Q0 f(H0) where Q0 was the perp space projector. New Theorem. I think we can make a similar theorem for P0 like this: G0a ≡ P0 f(H0) P0 = f(H0) P0 = P0 f(H0) // similar warning as noted above! As soon we you write P0 = Σkεa0 |k><k| , the H0 converts to Ea0 and then f(H0) becomes a number and you can put it anywhere you want. This theorem would of course apply in any subspace Ea(1) of Ea0 where you might use P(1)as the projector onto that subspace. So we can repeat the above claims with any superscript on P0 we like. In particular, we can say P(1) H0 = Ea0 P(1) B Another obvious fact is this one: P(1) Po = P(1) A since if you first project something into Ea0and thence into Ea(1), it is the same as if you go in one shot. So I have now justified all equations A,B,C on page 700, the last arising because (..) = 1 from 699 B. How then do we get D? We have three terms to look at. The third term passes through. The first two terms are this: P(1)V (P(1) + P') = P(1)VP0 He is claiming that P(1)VP0 = ε1 P(1) but I don't offhand see why that is true, another roadblock of many, which is why I move one sentence per hour in Messiah. We are not applying to a state here as we did in 16.36. We could apply this to an arbitrary state in our H to check its validity. We get 0=0 for any state not in Ea0, so we can restrict to that. We did know from earlier that P0VP0 |0> = ε1|0> for |0> being any state in Ea0. So to test the validity of the above claimed equation, we can start with this fact we know: P0VP0 |0> = ε1|0> |0> in Ea0 We don't have a general operator equation P0VP0 = ε1 here because it would not be true for a ket in the perp space. But we could fix that by saying P0VP0 = ε1 P0 . So this IS a correct operator equation. So here we go: P(1)VP0 = P(1) P0VP0 // by our "obvious fact" quoted above = P(1) ε1 P0 // using our operator equation P0VP0 = ε1 P0 just quoted = ε1 P(1) // using the "obvious fact" again D' and thus we have shown p 700 D. We can rewrite this equation D as: P(1)V – ε1 P(1) = P(1)( V – ε1) = P(1)VQ0 D Now finally we are more prepared to mess with the 72 equation which starts out saying (H0 - Ea0) |2> + (V-ε1)|1> – ε2|0> = 0 Now apply P(1) from the left. Our usual state expansion idea for P(1) = Σkε1 |k><k| kills off the first term because it sets H0 = Ea0. We then get P(1) (V-ε1)|1> – ε2 P(1) |0> = 0 But we now recognize P(1) (V-ε1) = P(1)VQ0 from our alternate D form above, so we have P(1)VQ0 |1> – ε2 P(1) |0> = 0 E which is result E. BUT now look back at our earlier result 37 which said Q0 |1> = Q0a V|0> so we then have P(1)V Q0a V|0> – ε2 P(1) |0> = 0 P(1) [ V Q0a V – ε2 ] |0> = 0 // which is 16.39 version of 16.72 and this was certainly a long time coming. This equation resembles 16.36. So as we did before, we can write this as an EV equation where ε2 is the EV: P(1)V Q0a V |0> = ε2P(1) |0> Now let's assume that |0> is in our E(1) part of Ea0. Then we are free to insert P(1) on the left and remove it on the right to get P(1)(V Q0a V) P(1) |0> = ε2 |0> E1 which reminds us of our earlier equation P0 (V) P0 |0> = ε1|0> So here is what we are supposed now to do. We now have an eigenvalue problem within E(1) and we are going to get a matrix equation like this: (V Q0a V) ψ0i = ε2i ψ0i where i = 1,2.....gi within E(1) Suppose now that these ε2i are all different. Then we get unique ψ0i and these are then the correct states back in Ea0 for us to "start with". We could not find these states here until we went to second order in the energy correction! So I presume in his section to come, he will show that you just keep doing this process to as many orders as you need until you get a splitting of the energies. Now we backtrack to say this: suppose our original ε1 was non-degenerate. Then we were able to identify |0> cleanly. Then consider: (V Q0a V)|0> = ε2 |0> ε2 = <0| V Q0a V|0> // which is result F on page 700 so in that case ( ε1 non-degenerate), we have a formula for the ε2 correction. And if ε1 was degenerate with multiplicity g1, then we solve the above matrix equation (V Q0a V) ψ0i = ε2i ψ0i to get the ε2 energies, and if they are all different, we are done and we know the states ψ0i. Otherwise, we have to continue on to third order! He comments that you may retain degeneracy through all orders of perturbation theory. Note added 1.28.09. When I first was reading this, I did not correctly interpret the meaning of a state |0>. One must understand that this is not "just any state" in Ea0 or E(1) or what have you. The state |0> is the "zeroth order component" of a state that solves the full SE to all orders in λ. So only certain linear combinations of the "starting base states" (which I like to call |r> and he calls |E0aα>) are "legal" for |0>. To find the legal linear combinations, you have to process things out to a point where degeneracy is broken! So when you solve the EV problem at some level, such as (V Q0a V)|0> = ε2 |0 above, you not only discover the eigenvalues ε2 , but you also discover the correct states |0> that are "legal". You might say that these are the shadows or projections into the ε0 level of full SE solutions. Comment: In my personal work on this problem, I just assumed that, when I did my pre-diagonalization, all the eigenvalues were different in the H(m)space, and thus I was "done" with the problem of identifying the correct states to start with. Here Messiah has made the point that if this is not the case, you try to get the right states by going to higher and higher order just as he described for second order. I think he is going to present a fancy formalism showing exactly how you do this. More Comments: Messiah always writes an unperturbed state as |0>. Although you might then have formulas involving this state, he points out that you cannot use those formulas unless you actually know what the state |0> is. When you have a degeneracy situation, you may have to go to something like fourth order ε4 before you can know what your states are. Otherwise all you know is that the state |0> is some as-yet-undetermined vector in a degenerate subspace, and then you cannot do a calculation with a formula involving |0>, such as ε1 = <0|V|0>. // good, 1.28.09 Clarification of my Theorem B confusion. One of my main motivations for even looking at Messiah was to understand my Theorem B which was this: You cannot "do" a perturbation analysis on the states umi unless they diagonalize H' within H(m). In the proof of that theorem I start with the ladder equations (H0 - Em) ψmi,1 = (Wmi,1 - H' ) umi i = 1,2....N " level 1" (Ea – H0) |1,i> = (V-ε1i)|0,i> // Messiah notation for same equation where I just assume that the m subspace is spanned by some orthonormal basis umi i = 1,2..N. I assume this without regard to whether or not that basis diagonalizes H'. I know from Hilbert Space theory that in any finite subspace I can always construct such an orthonormal basis by doing GSO. Now, just staring at either equation above, you don't "see" a matrix eigenvalue equation. In Messiah, we make this matrix equation appear by applying P0 to both sides. This kills the LHS and gives P0 (V-ε1)|0> = 0 which we convert to our desired EV equation. In Schiff, we instead "close" from the left to get this <umj| (Wmi,1 - H' ) |umi> = 0 H'ij = Wmi,1 (1)ij and this also "kills" the LHS. What we now "see" is a diagonal EV equation! How come I don't ever "see" the non-diagonalized equation first? That is my complaint. [ a good complaint I think ] OK, go back to the starting point again, (H0 - Em) ψmi,1 = (Wmi,1 - H' ) umi Suppose we "restrict our interest" to the m subspace. Then can I set the LHS = 0? Well, if by "restrict our interest" we mean that "we promise only to close from the left with subspace states", then I guess you can set LHS = 0. We then have within H(m) this set of equations, for i = 1,2,3.... 0 = (Wmi,1 - H' ) umi => H' umi = Wmi,1 umi H' |umi> = Wmi,1 |umi> This says that the |umi> are solutions to the EV equation shown, in addition to being an arbitrary set of orthonormal basis functions. So, when we start, we just say the |umi> span H(m) and are arbitrary but orthonormal. But then when we examine our order-λ1 ladder equations restricting to H(m), we learn that in fact these ladder equations say that the kets |umi> are eigenkets of H' so they cannot be completely "arbitrary" after all. You can certainly construct an orthonormal basis |vmi> which spans H(m) and which does NOT diagonalize H'. But these states then do not satisfy our ladder equations. We will have (Wmi,1 - H' ) vmi ≠ 0 at least for one or more i in 1..N Let's try saying it again and again until it sinks in: Fact: If you are looking for an orthonormal basis of kets in H(m) which solve the level-1 ladder equations in H(m), then those kets must be eigenkets of H' and so H' will be diagonal in that basis. In my notes, once again, I said this for Theorem B: You cannot "do" a perturbation analysis on the states umi unless they diagonalize H' within H(m). and I used the term "inconsistent". I would now state this differently. I would say that if you want a set of H(m) basis states that solve the ladder equations through level 1, then those basis states must diagonalize H'. So the first order of business is going to be to get H' diagonalized within H(m), and of course in so doing we learn the spectrum for ε1 = Wmi,1. And we can go further. Suppose we find that there is still some degenerate subspace after we diagonalize H' such that some ε1 has multiplicity g1. Then our claim will be this: If you want a set of basis states within H(1) H(m) that solve the ladder equations through level 2, then those basis states must diagonalize the operator H'QmoH' = VQa0V as Messiah shows page 700 (39). Here, the operator Q0m is given by Q0m ≡ Q0(Em - H0)-1Q0 where Q0 is the perp space projector relative to H(m) and of course Em is the eigenenergy of H(m). And presumably if we still have some degeneracy after this ε2 analysis, we use the level three equations to resolve things. Clarification of my B'ij confusion. In my own "personal" development of perturbation theory, I would write things like so ψmi = { umi + Σj≠iB'ij umj } Σk≠m B'ik uk ψmi,n = { Σj≠iB'ij,n umj } Σk≠m B'ik,n uk n > 0 and I would ask "how do I compute the coefficients B'ij " ? These tell you how the perturbed state picks up components of basis states within H(m). I kept finding that these coefficients were sort of undeterminable. In Messiah, we see a similar thing happening. When we consider the level 1 ladder equations, we find page 699 16.37 that Q0 |1> = Q0a H |0> but this is telling us information only about the part of the first order state correction B'ik,1 that is in the perp space to H(m)! We learn nothing about B'ij,n . So at least both Messiah and I have "the same problem" at this ladder equations level. Flailing Away with no real success: Now you would think that the level 2 ladder equations would tell you something about the B'ij,n. Suppose that we diagonalize H' and this removes all degeneracy, the simplest case. Then "if we restrict our interest" to H(m), our level 2 ladder equation set reads: (V-ε1)|1> - ε2|0> = 0 // "level 2" which in my notation is more like this: (H' - Wim,1)ψmi,1 - Wim,2 umi = 0 which becomes (H' - Wim,1)[ { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk ] = Wim,2 umi At this point, I "closed" with some ums inside H(m) ums (H' - Wim,1)[ { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk ] = Wim,2 δs,i Σj≠iB'ij,1 ums (H' - Wim,1) umj + Σk≠m B'ik,1 ums (H' - Wim,1) uk = Wim,2 δs,i But (H' - Wim,1) is a diagonal matrix within H(m) (only!) so this all becomes Σj≠iB'ij,1 (H'ss - Wim,1) δs,j + Σk≠m B'ik,1 (H'sk - Wim,1 δs,k) = Wim,2 δs,i Σj≠iB'ij,1 (Wsm,1 - Wim,1) δs,j + Σk≠m B'ik,1 H'sk = Wim,2 δs,i We then looked first at s = i and got then no hit from the first term and Σk≠m B'sk,1 H'sk = Wsm,2 and then we looked with s ≠ i and got B'is,1 (Wsm,1 - Wim,1) + Σk≠m B'ik,1 H'sk = 0 and then since we assumed that all the H' diagonal values are different, we get our answer: B'is,1 = [Σk≠m B'ik,1 H'sk] /(Wmi,1 – Wms,1 ) i ≠ s Wmi,1 ≠ Wms,1 But we learned earlier in my notes that B'ik,1 = H'ki/ (Em - Ek) so we end up then with B'is,1 = 1 /(Wmi,1 – Wms,1 ) * [Σk≠m H'sk H'ki/ (Em - Ek)] i ≠ s Wmi,1 ≠ Wms,1 and of course we know that B'ss,1 = 0 from the way we normalized. So this result says that you get some non-zero B'is,1 only if H'sk H'ki ≠ 0 for at least some k in the perp space (and even in this case it could in theory still be zero). Let's write this out a little: B'is,1 = 1 /(Wmi,1 – Wms,1 ) * [Σk≠m <s H'|k> (Em - Ek)-1 <k|H'|i> ] Now recall from Messiah above that Q0m = Σk≠a |Ek0>(Em0 - Ek0)-1< Ek0| Thus, we can write our result more compactly as B'is,1 = 1 /(ε1i – ε1s ) * <s | H' Q0m H' |i> Also, from above we just had Σk≠m B'sk,1 H'sk = Wsm,2 ε2i = Σk≠m B'sk,1 H'sk = Σk≠m H'sk B'sk,1 H'ki/ (Em - Ek) = <s| H' Q0m H'|s> // see M 700F because that same structure appears. So we can summarize all these results compactly as: ε1i = <i | H' |i> B'ik,1 = H'ki/ (Em - Ek) // level 1 ladder equations ε2i = <i| H' Q0m H'|i> B'is,1 = 1 /(ε1i – ε1s ) * <s | H' Q0m H' |i> // level 2 ladder equations which only applies to the case that ε1i ≠ ε1s for all i,s of interest. So I have replicated my work in my on deg doc, and have put it into more compact messiah notation, and I have a complete answer to the knowing all the B'ab,1 ! I then realized that we had a problem Houston if ε1i = ε1s and of course this is the very problem that Messiah attacks! Let's see now if I have really learned anything. Assume we have degeneracy g1 at the ε1 level, so we are dealing with a subspace H(1)say of dimension g1. Let's just try to run through the exact sequence above and see what happens. Cut, paste and edit: ______________________________ Then "if we restrict our interest" to H(m), our level 2 ladder equation set reads: (V-ε1)|1> - ε2|0> = 0 // "level 2" which in my notation is more like this: [ at this point, we know ε1 ] (H' - Wim,1)ψmi,1 - Wim,2 umi = 0 which becomes [ but we don't yet know exactly what ψmi,1 is, it is somewhere in H(1) ] (H' - Wim,1)[ { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk ] = Wim,2 umi [ Thus, we are saying we don't know what any of these B' coefficients are yet ] At this point, I "closed" with some ums inside H(1) ums (H' - Wim,1)[ { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk ] = Wim,2 δs,i Σj≠iB'ij,1 ums (H' - Wim,1) umj + Σk≠m B'ik,1 ums (H' - Wim,1) uk = Wim,2 δs,i But (H' - Wim,1) is a diagonal matrix within H(1) (only!) so this all becomes Σj≠iB'ij,1 (H'ss - Wim,1) δs,j + Σk≠m B'ik,1 (H'sk - Wim,1 δs,k) = Wim,2 δs,i Σj≠iB'ij,1 (Wsm,1 - Wim,1) δs,j + Σk≠m B'ik,1 H'sk = Wim,2 δs,i We then looked first at s = i and got then no hit from the first term and Σk≠m B'sk,1 H'sk = Wsm,2 = ε2s = <s | H' Q0m H' |s> [ is likely different for different s ] and then we looked with s ≠ i and get B'is,1 (Wsm,1 - Wim,1) + Σk≠m B'ik,1 H'sk = 0 => Σk≠m B'ik,1 H'sk = 0 => [Σk≠m <s H'|k> (Em - Ek)-1 <k|H'|i> ] = <s | H' Q0m H' |i> = 0 So what we learn at this level is that the states in H(1) which I am calling |i> [ a completely different notation than the |n> of Messiah!!! ] must diagonalize the operator H' Q0m H' if they are to solve the level 2 ladder equations (and all lower level ones). Once we actually carry out this diagonalization, we then know the various ε2s = <s | H' Q0m H' |s> which are just the diagonal elements of our diagonalized matrix. Since we now know exactly what these states are, we can actually compute something like this. Of course since Q0m involves the perp space H which is infinite dimensional, we have a rough calculation to do, since there are an infinite number of terms to add up, as Σk≠m above shows. We might eliminate certain terms in the sum by using symmetry to show that certain <k|H'|i> = 0. Or we might approximate by only including "nearby" states |k> which have therefore small energy denominators which dominate the sum. Now, never letting go like a good bulldog, I want to compute B'is,1 for these states I just supposedly found by diagonalizing H' Q0m H'. Obviously we are not going to get an answer from the above level 2 equations, so we go to the level 3 equations, and now I am a bit on my own. Let's try this in Messiah notation first. From 16.73 we have (H0 – Ea0) |3> + (V-ε1)|2> – ε2|1> – ε3|0> = 0 where |0> is a state in H(1). Now I can either "go formal" or "go informal" at this point. Doing the second, I just say "if we restrict our interest closing from the left only with states in H(1)", this equation becomes (V-ε1)|2> – ε2|1> – ε3|0> = 0 or I could be formal: P(1)(V-ε1)|2> – ε2 P(1)|1> – ε3|0> = 0 What do I do next??? Well, this seems to say that ε2 P(1)|1> = P(1)(V-ε1)|2> – ε3|0> and P(1)|1> is basically what I am looking for. So I guess I need to learn both |2> and ε3 before I have an answer to my question. But if I close the above with <0|, the ε2 term vanishes and I will get <0| P(1)(V-ε1)|2> = ε3 = <0| P(1)V|2> // because <0|2> = 0 so if I can just find |2>, then I can compute ε3, so we really have one unknown |2> that we need to find. Now go back to the level 2 equations which say (H0 – Ea0) |2> + (V-ε1)|1> – ε2|0> = 0 If I apply Q0a to both sides this becomes Q0 |2> = Q0a [ – (V-ε1)|1> + ε2|0> ] = – Q0a (V-ε1)|1> so this tells me for sure the perp space part of |2> Well, now I am just back to my big confusion as I was in degen doc working on this same problem. I need to thread a path somehow to get to the solution, but I don't know how to do it, and of course now I am caught halfway between two totally different notations. So now I think is the perfect time to read the systematic solution starting on Messiah page 712, because the author he quotes there has done this threading, so I should not spend 6 weeks trying to reinvent this wheel. 9. Atomic Levels in the Absence of Spin-Orbit Forces (700) skip for now. 10. Spin-Orbit Forces. LS and jj coupling (703) skip for now. 11. The Atom in LS coupling. Splitting due to spin-orbit coupling (705). skip for now. 12. The Zeeman and Paschen-Back Effects (706). skip for now. 13. Symmetry of H and Removal of Degeneracy (709). skip for now. 14. Quasi-Degeneracy (711). skip for now. review pause here