degenerate perturbation theory

Perturbation Theory with Degeneracy PhL 1.3.09 symbol grab area: H(m) H H(m) H umi H(m) Degenerate Case (Schiff page 248) 1 1. An Example. 1 2. Notation and Hilbert Spaces. 2 3. Most General Expansion for ψmi. 2 4. Perturbation theory set up. 2 5. Theorem A: { umi must diagonalize H' in H(m) } 3 6. Theorem B: { show 31.6: <umi| ψmi,n> = 0 for n > 0) } ( "the normalization condition") 4 7. Theorem C: { Show 31.7: Wmi,s = < umi| H' | ψmi,s-1> } 6 8. Finding the coefficients B'ik,1 7 9. Finding the coefficients B'ij,1 8 10. Summary of this section on first-order degenerate perturbation theory: 11 Summary of the Summary: 12 11. Systematic application of equations 31.4 13 A. Close with <uk'|. 13 B. Close with <ums| and then look at s = i and s≠i as separate cases 15 C. Summary of what we just did above: A systematic iterative method good to all orders 17 12. Digression: Schiff page 249 lower half of the page: 18 13. Review of the development so far. 21 14. The special case where Wms,1 = Wmi,1 . 22 15. Comments on Saxon's Approach. 24 16. Comments on my Theorem A and a way to think of the Saxon approach. 24 (1) Possible Saxon method. 24 (2) The 2x2 diagonalization of H' in Schiff. 26 (3) Question: <umj| ψ'mi,rest> = 0 for i≠j ? No! 26 (4) Pause to consider a new (and bogus!) Theorem D. 27 17. Relation between Kato Method and Phil Method for Degeneracy [ 2.9.09 ] 31 18. Comparison to Messiah in General [2.9.09] 32 Degenerate Case (Schiff page 248) I really want to "start all over" here (in the case with degeneracy) and assume nothing can be reused until it is verified. And I might as well assume some arbitrary degeneracy count N and do this all at once. But first, consider this example. 1. An Example. Suppose we have a system of two electrons that form S=0 and S=1 states, and our perturbation is a mag field. We know that the states |1,±1> will be pulled away from the other two states which remain where they were since they have mS = 0. At least I think that is what will happen in first order. But this illustrates the idea that when the perturbation is turned on, the symmetry of the original Ho is broken to some extent, and certain linear combinations of states that were degenerate will prove to "hold together" and move apart from the original energy, and will be eigenstates of the perturbed Hamiltonian. So a new issue here is that you have to get the right linear combinations somehow so that things are continuous as λ→0, just as Schiff says. [ What Schiff does not seem to say but maybe I missed it is that these right linear combinations are those that diagonalize H' in the m subspace.] 2. Notation and Hilbert Spaces. This is the big problem here! If you have too much notation, you get bogged down with lots of extra symbols which conceal the simplicity of things. If you have too little, you get confused as well. So let's start with some unperturbed states uk which span a Hilbert Space H. We then want to consider a subset of these states which are degenerate with energy Em. These states span a subspace I will call H(m). So we can write H = H(m) H where H is just the rest of the Hilbert Space H. It is perpendicular to H(m) because if you pick any state in H(m) and any state in H, their inner product will be 0. I will denote states in the H(m)subspace by the notation umi where i = 1,2...N. In fact, these umi will be assumed to span the subspace which we assume is of dimension N. States in H will just be uk with a single subscript. There may be other subspaces within H but our effort here concerns doing perturbation theory on the states (and their energies) in a particular subspace we call H(m). Note added 1.19.09: This space H(m) is one of the eigenmanifolds of a finite-dimensional complex symmetric linear operator H, a subject treated in Stakgold. His spectral theorem on page 160 formalizes the notion that each eigenvalue really does inhabit a subspace which he calls a linear manifold, and he talks about a projection theorem which says the space is partitioned into these manifolds and you can use the projection operators Pi to do manipulations. This is what Messiah does in his resolvent section. 3. Most General Expansion for ψmi. Now, when we first "set out" to construct our perturbation theory, we write this most general possibility for how the degenerate state umi moves into state ψmi when the perturbation is turned on, where i is a degeneracy label, ψmi = umi + Σj=1,N Bij umj + Σk≠m Bik uk This says that when we turn on our perturbation (by gradually increasing λ from λ=0), various B coefficients will appear as shown above and our original eigenstate umi acquires corrections. The summation Σk≠m excludes all states in the subspace H(m). We assume that we acquire corrections from both H(m) states and from H states. 4. Perturbation theory set up. Our first need is to develop a "degeneracy version" of 31.4. So: Ho umi = Em umi umi H(m) Ho uk = Ek uk uk H (Ho + λH')ψmi = Wmi ψmi Wmi = Σn=0 λn Wmi,n Wmi,0 = Em ψmi = Σn=0 λn ψmi,n ψmi,0 = umi If we now insert these λ expansions into the SE and do what we did in the Schiff Chapter 8 notes, we get this result. Nothing really is different, we have the extra label "m" stuck to everything. (H0 - Em) ψmi,s = (Wmi,1 - H' ) ψmi,s-1 + Σj=2,3..s Wmi,j ψmi,s-j s = 1,2,3... and now I will write them out: // this is the new 31.4 (H0 - Em) ψmi,0 = 0 s = 0 (H0 - Em) ψmi,1 = (Wmi,1 - H' ) ψmi,0 s = 1 (H0 - Em) ψmi,2 = (Wmi,1 - H' ) ψmi,1 + Wmi,2 ψmi,0 s = 2 (H0 - Em) ψmi,3 = (Wmi,1 - H' ) ψmi,2 + Wmi,2 ψmi,1 + Wmi,3 ψmi,0 s = 3 Basically I have replaced the former superscript (m) with a subscript mi. These are "the perturbation equations" upon which everything is based. At this point in our development, we state and prove three "theorems": __________________________________________________________________________________ 5. Theorem A: { umi must diagonalize H' in H(m) } 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 = Emumi 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 (see above). In particular, the equation which equates λ1 on both sides says this: (H0 - Em) ψmi,1 = (Wmi,1 - H' ) umi You can see clearly how these terms arise from (H0+λH')ψmi = Wmiψmi. There are two terms from the LHS that are order λ, and there are two terms from the RHS that are order λ. 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. [ 2.9.09. This bothered me a lot for a while, but resolved when I realized H' must be diagonal in the sense above if umi are to satisfy the perturbation equations (the EV problems derived from these equations). It is those equations that force it. If you pick some other vmi, they don't satisfy the pert equations so you have to throw them out. You can start with some vmi, but right away you arrive at the level 1 diagonalization problem H'|ψ> = Wmi,1|ψ> and you have to find eigenfunctions and eigenvalues of this problem, which then leads to the diagonalizing |ψ>= umi. ] ___________________________________________________________________________________ Notational detail: since H = H(m) H, we can decompose x in H as x = x1+ x2 etc. Below I keep writing x = x1 x2 as a reminder of this fact. I only use this when the thing on the left really is in H(m) and the thing on the right is in H . This notation is not official of course. ___________________________________________________________________________________ 6. Theorem B: { show 31.6: <umi| ψmi,n> = 0 for n > 0) } ( "the normalization condition") That is to say, 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, noted above: 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 likely also depend on H0 when H'≠0, I am not sure because I have never tried to make the connection. ] [ Note added 1.29.09. Later in Messiah and below we get results like this Cn' = (Vnn – Vn'n')-1 ΣkEoa Vn'k (Ea0 - Ek)-1 Vkn for the Bij type coefficients at level 1. They "depend on H0" because all states here are eigenstates of H0 and of course the energies are all eigenenergies of H0 ] 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 first 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 n > 0 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) [ umi is presumed normalized] the normalization of ψ'mi is fixed by the above choice, and we cannot then also 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 adjust the normalization of ψ'mi to get unity. So the expansion we shall be using is this: ψmi = { umi + Σj≠iB'ij umj } Σk≠m B'ik uk // all orders of λ at once ψmi,n = { Σj≠iB'ij,n umj } Σk≠m B'ik,n uk // n > 0 Notice that there are two distinct portions of the matrix B'rs . One portion applies within the H(m) subspace. Here, B'ij is an NxN matrix. The diagonal elements of this matrix never enter our theory due to the Σj≠i and the way we normalized ψmi, so if we want, we can just set B'ii = 0. So, this inner B matrix is completely off-diagonal inside H(m) . The other portion appearing as B'ik has the first index in H(m) and the second in H. Here are some pictures showing various parts of our B matrix. If we are only INTERESTED in doing perturbation theory on states within H(m), then the part of the matrix marked "don't care" is never used. But on the right we show the larger picture where there might be multiple degenerate blocks. We have shown a few singleton blocks here where "non degenerate" perturbation theory could be used. In general, you think one strip of the B matrix at a time. ________________________________________________________________________________ 7. Theorem C: { Show 31.7: Wmi,s = < umi| H' | ψmi,s-1> } ( Ie, the 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 higher 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. Thus, 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. ___________________________________________________________________________________ 8. Finding the coefficients B'ik,1 Apply new 31.7 with s = 1 to get Wmi,1 = < umi | H' | umi > = <mi |H'|mi > which is now true for each state "i" in the subspace. We might have vanishing matrix elements for some i and not for others, as in our example at the start of this document. Meanwhile, our 31.4 for s=1 says: (H0 - Em) ψmi,1 = (Wmi,1 - H' ) ψmi,0 s = 1 but from above we know that for order n = 1 we can write ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk so the above says (H0 - Em) [{ Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk ] = (Wmi,1 - H' )umi (*) Close (*) from the left with some <uk'| in H . We know that H0 acting to the left produces Ek' uk'. So we then have <uk'| (Ek' - Em) [{ Σj≠iB'ij,1 |umj> } Σk≠m B'ik,1 |uk> ] = <uk'| (Wmi,1 - H' )|umi > <uk'| (Ek' - Em) [Σk≠m B'ik,1 |uk> ] = <uk'| (Wmi,1 - H' )|umi > (Ek' - Em) B'ik',1 = <uk'| (Wmi,1 - H' )|umi > (Ek' - Em) B'ik',1 = – <uk'| H' |umi > (Ek - Em) B'ik,1 = – <uk| H' |umi > B'ik,1 = <uk| H' |umi >/ (Em - Ek) Notice that if we close (*) from the left with some <umj| in H(m) the LHS = 0 and we then learn is this: 0 = <umj| (Wmi,1 - H' )|umi > If i = j, this duplicates our first order energy correction result. If i ≠ j, it duplicates our result that H' must be diagonal within H(m). 9. Finding the coefficients B'ij,1 What we are missing now is an expression for B'ij,1 which are the correction coefficients inside H(m). It seems to me that we need to learn what these are if we want to fully learn what ψmi,1 looks like. Schiff somehow dodges this question completely in his page 249-251 discussion. How might we get "access" to the B'ij coefficients? In the s=1 equation 31.4 access is blocked by the protective operator (H0 - Em) on the left hand side, so closing with a state in H(m) always gives 0. What if we look at the next equation in the ladder of equations, (H0 - Em) ψmi,2 = (Wmi,1 - H' ) ψmi,1 + Wmi,2 ψmi,0 (**) s = 2 ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk ψmi,2 = { Σj≠iB'ij,2 umj } Σk≠m B'ik,2 uk Insert these things into (**) then close with <ums| in H(m). As usual, LHS = 0 due to the projection, but this just blocks our access to the B'ij,2 coefficients, not the B'ij,1 coefficients! We then get 0 = <ums| (Wmi,1 - H' ) [{ Σj≠iB'ij,1 |umj> } Σk≠m B'ik,1 |uk>] + <ums| Wmi,2 |umi> On the RHS, let's first evaluate the Wmi,1 terms: Wmi,1 <ums| [{ Σj≠iB'ij,1 |umj> } Σk≠m B'ik,1 |uk>] = Wmi,1 B'is,1 ( 1 - δs,i) because the second term gives nothing and, if s = i, the first term also gives nothing because the sum does not include the state |umi>. Here we have finally "accessed" a B'is,1 coefficient. Next, let's look at the H' terms: – Σj≠iB'ij,1 <ums| H' |umj> – Σk≠m B'ik,1<ums| H' |uk> now we use the fact that <ums| H' |umj> = δs,j <ums| H' |ums> so the first term above becomes – Σj≠iB'ij,1 δs,j <ums| H' |ums> = – B'is,1<ums| H' |ums> ( 1 - δs,i) Again, if s = i, then the j sum does not include s and we get no hit. So the total H' terms are – B'is,1 Wms,1 ( 1 - δs,i) – Σk≠m B'ik,1<ums| H' |uk> Now we can put back our pieces to get 0 = Wmi,1 B'is,1( 1 - δs,i) – B'is,1 Wms,1 ( 1 - δs,i) – Σk≠m B'ik,1<ums| H' |uk> + Wmi,2 δs,i where the last term comes from <ums| Wmi,2 |umi> . We can rewrite this as: B'is,1(Wms,1 – Wmi,1 ) ( 1 - δs,i) = – Σk≠m B'ik,1<ums| H' |uk> + Wmi,2 δs,i (***) Small digression: But from our theorem above we know that Wmi,s = < umi| H' | ψmi,s-1> => Wmi,2 = < umi| H' | ψmi,1> So this tells us that Wmi,2 = < umi| H' | [{ Σj≠iB'ij,1 |umj> } Σk≠m B'ik,1 |uk > ] = Σj≠iB'ij,1 < umi| H' | umj> + Σk≠m B'ik,1 < umi| H' |uk > = Σj≠iB'ij,1 δi,j < umj| H' | umj> + Σk≠m B'ik,1 < umi| H' |uk > = Σk≠m B'ik,1 < umi| H' |uk > (****) The first term gives nada because the sum does not include j = i. Noted added 1.29.09: The gray box above shows very nicely that the Wmi,2 energy corrections have no dependence whatsoever on the B'ij,1 coefficients. That is why it is easy to compute Wmi,2. Above we already found that B'ik,1 = <uk| H' |umi >/ (Em - Ek) so we can stuff this in to get a simple result for Wmi,2, see below The lack of dependence arises from Σj≠i H'ij = Σj≠i δi,j H'ii = 0, ie, from the fact that we have pre-diagonalized (if you will) the states umi, and the fact that we have normalized things as in Theorem B above. Now go back to our (***) equation above. First, consider it in the case the s ≠ i. We then get B'is,1(Wms,1 – Wmi,1 ) = – Σk≠m B'ik,1<ums| H' |uk> Second, consider it in the case the s = i, which gives 0 = – Σk≠m B'ik,1<umi| H' |uk> + Wmi,2 This seems to indicate that Wmi,2 = Σk≠m B'ik,1<umi| H' |uk> Thus, we have replicated our digression calculation above for Wmi,2, which is a confidence booster. We can use our earlier result above for the B'ik,1 to write this as Wmi,2 = Σk≠m <umi| H' |uk> <uk| H' |umi >/ (Em - Ek) which says that Wmi,2 ≠ 0 only if |umi > "communicates" to itself through the amplitude product shown. Now the s ≠ i equation above was this: B'is,1(Wms,1 – Wmi,1 ) = – Σk≠m B'ik,1<ums| H' |uk> So IF two H(m)states pull apart in energy at first order (which is to say, H' when pre-diagonalized has different diagonal elements for s and i) , then we know that B'is,1 = [Σk≠m B'ik,1<ums| H' |uk>] /(Wmi,1 – Wms,1 ) i ≠ s Wmi,1 ≠ Wms,1 So we are making some progress here! This says that if H'mi,mi ≠ H'ms,ms, then we have a formula for our desired coefficient B'is,1 ! Recall, by the way, that earlier we found that B'ik,1 = <uk| H' |umi >/ (Em - Ek) for k in H so we can rewrite the above as B'is,1 = Σk≠m <ums| H' |uk><uk| H' |umi >/ [(Em - Ek)(Wmi,1 – Wms,1 )] = (Wmi,1 – Wms,1 )-1 (H'Q0mH')si for i ≠ s and Wmi,1 ≠ Wms,1 What does this tell us? It says to me this: if |umi> "communicates" with |ums> via the second order amplitude product shown (involving only intermediate states in H), THEN both the states |umi> and |ums> have "corrections" within H(m), and these are the first order correction coefficients. Note added 1.30.09. Remember what we are doing here: ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk We found the B'ik,1 easily, and we then found the B'ij,1 with more work when i ≠ j . We can think of the state "i" as being singles state "n" in our later Messiah picture below, and the various j as listing the other states in the center column. So in the topology below, at least, we have our full answer for the level 1 coefficients. 10. Summary of this section on first-order degenerate perturbation theory: The uk states must be selected such that H' is diagonal in H(m) { "prediagonalization"} [ You can choose not to do this, but then your first order of business is to solve the level 1 diagonalization problem to find the eigenenergies and states, and then "you have done it". ] The first order energy corrections are simple: Wmi,1 = < ui |H'| ui > = H'ii They are just the diagonal elements of H' in H(m). [ the index "i" is always assumed to be in H(m)] Our general first-order expansion for the wave function is this, where <umi| ψmi,1 > = 0: ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk We showed that the H coefficients are given by B'ik,1 = <uk| H' |umi >/ (Em - Ek) = H'ki / (Em - Ek) k in H [ 2.9.09: B'ik,1 = <k| Qa0V |0 > = <k|1> in Messiah p 689 notation. ] and we got this result from the s=1 equation closed on the left with a uk' state. In passing, we noted the second order energy correction for our states to be Wmi,2 = Σk≠m B'ik,1 < umi| H' |uk > = Σk≠m <umi| H' |uk> <uk| H' |umi >/ (Em - Ek) = Σk≠m B'ik,1 H'ik = Σk≠m H'ik H'ki/ (Em - Ek) = Σk≠m |H'ik|2 /(Em - Ek) [ 2.9.09: ε2 = <0|V|1> = <0|VQa0V|0> in Messiah p 700 equation F notation. ] and this comes from our "general formula" Wmi,2 = < umi| H' | ψmi,1>. This is exactly the same result as obtained in the non-degenerate theory, except you don't sum over the H(m) partners of umi. We saw how the diagonality of H' in the umi basis, and our normalization convention, causes this Wmi,2 to not depend on the B'ij,1 coefficients. So Wmi,2 depends on the coupling from the one state mi to the perp k states. We considered two states i and j within H(m). If their first order energies are different, meaning that the two diagonal elements of H' are different, then we found that B'ij,1 = [Σk≠m B'ik,1<umj| H' |uk>] /(Wmi,1 – Wmj,1 ) i ≠ j Wmi,1 ≠ Wmj,1 = Σk≠m <umj| H' |uk><uk| H' |umi >/ [(Em - Ek)(Wmi,1 – Wmj,1 )] [ 2.9.09: Not sure how this relates to Messiah. ] However, if the two states i and j have the same first order energies, then we learn nothing about the B'ij,1 at this level of computation. The above came from the s=2 equation. So, this "first order" perturbation theory gives us (1) the first order energy corrections Wmi,1 (2) the wave function corrections B'ik,1 with k in H (3) the wave function corrections B'ij,1 with j in H(m) only if Wmi,1 ≠ Wmj,1 Summary of the Summary: The name of the game is to squeeze out all you can from each equation in the equation ladder 31.4. In the above the lowest equation s=1 was this (H0 - Em) ψmi,1 = (Wmi,1 - H' ) ψmi,0 s = 1 and from this equation we obtained two facts: Wmi,1 = < umi | H' | umi > = H'ii // already known from "the general rule" B'ik,1 = <uk| H' |umi >/ (Em - Ek) = H'ki/ (Em - Ek) // for k in H We next examined the s=2 equation which was this: (H0 - Em) ψmi,2 = (Wmi,1 - H' ) ψmi,1 + Wmi,2 ψmi,0 s = 2 and from this equation we also obtained two facts: Wmi,2 = Σk≠m B'ik,1 < umi| H' |uk > // already known from the general rule B'is,1(Wms,1 – Wmi,1 ) = – Σk≠m B'ik,1<ums| H' |uk> and for Wmi,1 ≠ Wmj,1 this last result gave us B'ij,1 = [Σk≠m B'ik,1<umj| H' |uk>] /(Wmi,1 – Wmj,1 ) OK, let's now try to be a little more systematic. Note added 1.30.09. Consider from above that Wmi,2 = Σk≠m B'ik,1 < umi| H' |uk > = Σk≠m B'ik,1H'ik = Σk≠m H'ki/ (Em - Ek)H'ik = Σk≠m H'ik (Em - Ek)-1H'ki = (H'Q0mH')ii This says that the second order energy corrections are the diagonal elements of operator H'Q0mH'. I don't think this tells us that H'Q0mH' is a diagonal matrix in our "i" basis. 11. Systematic application of equations 31.4 (see summary a few pages below) Here I want to try to compute the "general-n" equation 31.4 so I don't have to keep redoing things over and over again. This equation reads as follows: (H0 - Em) ψmi,n = - H' ψmi,n-1 + Σt=1,n Wmi,t ψmi,n-t n = 1,2,3... Let's break off the very last sum term which has t = n: (H0 - Em) ψmi,n = - H' ψmi,n-1 + Σt=1,n-1 Wmi,t ψmi,n-t + Wmi,n umi If we restrict to n > 1, all wavefunctions shown will be at least first order, so we can use our general expansion which is this ψmi,s = Σj≠iB'ij,s umj + Σk≠m B'ik,s uk So let's now insert this in all three places: (H0 - Em) [ Σj≠iB'ij,n |umj> + Σk≠m B'ik,n |uk>] = – H' [ Σj≠iB'ij,n-1 |umj> + Σk≠m B'ik,n-1 |uk>] + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t |umj> + Σk≠m B'ik,n-t |uk>] + Wmi,n |umi> A. Close with <uk'|. A. First, let's close this with <uk'| in the perp space. We get (Ek' - Em) [ Σj≠iB'ij,n <uk'|umj> + Σk≠m B'ik,n <uk'|uk>] = – <uk'| H' [ Σj≠iB'ij,n-1 |umj> + Σk≠m B'ik,n-1 |uk>] + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t <uk'|umj> + Σk≠m B'ik,n-t <uk'|uk>] + Wmi,n <uk'|umi> Our first rewrite is as follows (Ek' - Em) [ 0 + Σk≠m B'ik,n δk,k'] = – <uk'| H' [ Σj≠iB'ij,n-1 |umj> + Σk≠m B'ik,n-1 |uk>] + Σt=1,n-1 Wmi,t [ 0 + Σk≠m B'ik,n-t δk,k'] Now use δk,k' to remove the Σk≠m sum in two places, and at the same time expand the middle line: (Ek' - Em) [ B'ik',n ] = – Σj≠iB'ij,n-1 <uk'| H' |umj> – Σk≠m B'ik,n-1<uk'| H' |uk> + Σt=1,n-1 Wmi,t [ B'ik',n-t] – Σj≠iB'ij,n-1 H'k'j – Σk≠m B'ik,n-1H'k'k + Σt=1,n-1 Wmi,t [ B'ik',n-t] This tells us the "out of strip" values of the B'ik',n in terms of the "strip" B' of lower orders! We are missing the n = 1 equation, so let's develop it as a special case (H0 - Em) ψmi,1 = - H' umi + Wmi,1 umi n = 1 Insert on the left ψmi,1 = Σj≠iB'ij,1 umj + Σk≠m B'ik,1 uk and we get (H0 - Em) [Σj≠iB'ij,1 |umj> + Σk≠m B'ik,1 |uk>] = - H' |umi> + Wmi,1 |umi> Now close with <uk'| to get (Ek' - Em) [Σj≠iB'ij,1 <uk'|umj> + Σk≠m B'ik,1<uk'|uk>] = - <uk'|H' |umi> + Wmi,1 <uk'|umi> (Ek' - Em) [0 + Σk≠m B'ik,1δk,k'] = - <uk'|H' |umi> + 0 (Ek' - Em) [B'ik',1] = - H'k'i B'ik',1 = H'k'i / (Em - Ek') So let's now summarize fully our result with closure with <uk'|: (Ek' - Em) B'ik',1 = - H'k'i n = 1 (Ek' - Em) B'ik',n = – Σj≠iB'ij,n-1 H'k'j n > 1 – Σk≠m B'ik,n-1H'k'k + Σt=1,n-1 Wmi,t B'ik',n-t So at least for the coefficients of the form B'ik',n we have a starting point with B'ik',1 given by the first equation. We do not yet, however, have any starting point for the coefficients of the type B'ij,n-1 . Notice how I am relying on k and j to keep track of which group of coefficients we are dealing with. B. Close with <ums| and then look at s = i and s≠i as separate cases B. Second, let's close this with <ums| in the m subspace. We start with (H0 - Em) [ Σj≠iB'ij,n |umj> + Σk≠m B'ik,n |uk>] = – H' [ Σj≠iB'ij,n-1 |umj> + Σk≠m B'ik,n-1 |uk>] + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t |umj> + Σk≠m B'ik,n-t |uk>] + Wmi,n |umi> n>1 Closure at once causes LHS = 0, so we are left with 0 = – <ums| H' [ Σj≠iB'ij,n-1 |umj> + Σk≠m B'ik,n-1 |uk>] + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t <ums|umj> + Σk≠m B'ik,n-t <ums|uk>] + Wmi,n <ums||umi> Our first round of simplification gives 0 = – Σj≠iB'ij,n-1 <ums| H' |umj> – Σk≠m B'ik,n-1<ums| H' |uk> + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t δs,j + 0] + Wmi,n δs,i Now use the fact that <ums| H' |umj> = δs,j <ums| H' |ums> = δs,j Wms,1. We then have 0 = – Σj≠iB'ij,n-1 δs,j Wms,1 – Σk≠m B'ik,n-1<ums| H' |uk> + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t δs,j + 0] + Wmi,n δs,i Now we want to claim this general thing Σj≠i Fij δs,j = Fis (1 - δs,i) and then use that idea twice to get, and also use 0 = – B'is,n-1 Wms,1 (1 - δs,i) – Σk≠m B'ik,n-1<ums| H' |uk> + Σt=1,n-1 Wmi,t B'is,n-t(1 - δs,i) + Wmi,n δs,i Next, break out the t=1 term from the sum to get 0 = – B'is,n-1 Wms,1 (1 - δs,i) – Σk≠m B'ik,n-1<ums| H' |uk> n>1 + Wmi,1 B'is,n-1(1 - δs,i) + Σt=2,n-1 Wmi,t B'is,n-t(1 - δs,i) + Wmi,n δs,i This allows us to combine the first and third terms together, and put them on the LHS: B'is,n-1(1 - δs,i) [Wmi,1 – Wms,1] n>1 = + Σk≠m B'ik,n-1<ums| H' |uk> – Σt=2,n-1 Wmi,t B'is,n-t(1 - δs,i) – Wmi,n δs,i This thing is valid for all i,s in the m subspace. As before, let's develop the n = 1 equation separately, (H0 - Em) ψmi,1 = - H' umi + Wmi,1 umi n = 1 Closing with <ums| makes 0 on the left and we get 0 = – <ums| H' | umi> + Wmi,1 δs,i This is of course where we first learn that H' must be diagonal in the umi and then this just gives us the first order energy correction, so the n = 1 equation is not so interesting here, we already "know" everything it has to offer us. Now go back to the previous n> 1 result B'is,n-1(1 - δs,i) [Wmi,1 – Wms,1] n>1 = + Σk≠m B'ik,n-1H'sk – Σt=2,n-1 Wmi,t B'is,n-t(1 - δs,i) – Wmi,n δs,i For i = s, it tells us this: 0 = Σk≠m B'ik,n-1 H'ik – Wmi,n Wmi,n = Σk≠m B'ik,n-1 H'ik = (H'Q0mH')ii so this is a result we already knew, and here we have rederived it. For i ≠ s it tells us this: n>1 B'is,n-1 [Wmi,1 – Wms,1] = Σk≠m B'ik,n-1H'sk – Σt=2,n-1 Wmi,t B'is,n-t And this equation in general tells us the H(m) B'is,n-1 coefficients in terms of the B'ik,n-1 of the same lower, and B'is,n-t of lower order. Question: What do we do when Wms,1 = Wmi,1 for a pair of states? It seems that this denies us a starting point for the B'is,1 coefficients! For n=2 our equation above is B'is,1 [Wmi,1 – Wms,1] = Σk≠m B'ik,1<ums| H' |uk> C. Summary of what we just did above: A systematic iterative method good to all orders We took the general ladder equation and closed it first with <k| and second with <ums| and got lots of interesting results -- here they are: ψmi,n = { Σj≠iB'ij,n umj } Σk≠m B'ik,n uk n = 1,2,3... Wmi,1 = H'ii (0) Closure with <uk'|: (Ek' - Em) B'ik',1 = - H'k'i n = 1 (1) (Ek' - Em) B'ik',n = – Σj≠iB'ij,n-1 H'k'j n > 1 – Σk≠m B'ik,n-1H'k'k + Σt=1,n-1 Wmi,t B'ik',n-t (2) Closure with <ums|: (H0 - Em) ψmi,1 = - H' umi + Wmi,1 umi n = 1 (3) B'is,n-1(1 - δs,i) [Wmi,1 – Wms,1] n>1 = + Σk≠m B'ik,n-1H'sk – Σt=2,n-1 Wmi,t B'is,n-t(1 - δs,i) – Wmi,n δs,i The last equation (not numbered) can be further broken down as to whether s ≠ i or s = i: B'is,n-1 [Wmi,1 – Wms,1] = Σk≠m B'ik,n-1H'sk – Σt=2,n-1 Wmi,t B'is,n-t s ≠ i (4) Wmi,n = Σk≠m B'ik,n-1 H'ik s = i (5) The very last equation states the famous result Wmi,n = <umi | H' | ψmi,n-1>. Now what kind of iterative scheme do these equations allow? I am now happy to consider a state "i" which is like the state "n" in our Messiah picture, one that has become a singlet after the first level EV equation. What then can we learn about this singlet state i's coefficients? First, we know B'ik',1 from equation (1), and we know the Wmi,1 from (0) We set n = 2 in equation (4) to find B'is,1 which has only the leading term on the RHS. This equation then says B'is,1 [Wmo,1 – Wms,1] = Σk≠m B'ik,1<ums| H' |uk> = (H'Q0mH')si We set n = 2 in equation (2) to get B'ik',2 in terms of the B'ik',1 We set n=3 in equation (4) to get B'is,2 in terms of B'ik,2 and B'is,1 and Wmi,2 And so on. So, for a singlet state "n", this set of equations allows us to compute the corrected state ψmi,n to arbitrarily high order. The iterative process seems totally well defined. There might be some "operator method" to put things into a more compact notation, but eventually we have to do all the calculations at each stage that this iterative method implies. ___________________________________________________________________________ 12. Digression: Schiff page 249 lower half of the page: In my current understanding of things, if we have an N = 2 degeneracy, we have to find uk that will diagonalize H' in order to get started. My usual approach is this S-1H'S = Λ H' S = SΛ H'ijSjk = SijΛjk = Sij δjkΛkk = ΛkkSik So we have an eigenvalue equation of this form: H'ijSjk = ΛkkSik where notice that the index k denotes the column of Sik. We can denote a column by [Sk]i where i reads off the components of the column. Then we have H'ij [Sk]i = Λkk [Sk]i or H' Sk = Λkk Sk and we see as usual that the columns of Sik are the eigenvectors of H'. Now let's start with basis column vectors (1 0) and (0 1) and write a general one as (a,b). Once we diagonalize H' in this process, the two diagonal elements which I call Λkk are really <umk| H'|umk> for the two values of k we can call k = 1,2. In my notation I would call these Wmk,1 for k = 1,2. Schiff just calls them "the two values of W1". I expect to find two solutions (a,b) and (a', b'), one for each Wmk,1 value. So here is one of our eigenvalue equations: H Sk = W1 Sk H' = H for short here = W1 H11 a + H12 b = W1 a H21 a + H22 b = W1 b ( H11 - W1) a + H12 b = 0 m =1 l = 2 a = am b = al H21a + (H22-W1)b = 0 And this is Schiff 31.15 on page 249 (finally). He gets it from the 31.4 second line, but I get it differently. I regard 31.4 second line as saying H' must be diagonal, and then above I find out how to make it be diagonal. Fine. The two W1 come from the secular equation which is this det(H - W11) = 0 det (H11-W1) (H22-W1) - H12H21 = 0 W12 - (H11 + H22)W1 + (H11H22 - H12H21) = 0 W1 = (H11 + H22)/2 ± [(H11 + H22)2 - 4 (H11H22 - H12H21)]1/2/2 = (H11 + H22)/2 ± [ (H11 – H22)2 +4 |H12|2 )]1/2/2 And this is Schiff 31.16 in his notation. The two W1 solutions will only be equal if both square root terms are 0, which means if H11 = H22 and H12 = 0, which is to say, if H' was a multiple of the identity! In this case of course it is already diagonal and we just get W1 = the equal diagonal elements. Now recall our result (Em - Ek)B'ik,1 = <uk| H' |umi > k in H I suppose we can write |umi > = a |1> + b|2> in ket notation and then I get (Em - Ek)B'ik,1 = <uk| H' |1 >a + <uk| H' |2 >b and this is what he means by equation 31.18. It is just pulling teeth, his notation just stinks to high heaven. Just as he has no distinguishing index on W1, he also has none on his a(1)k which is really my object B'ik,1. How can I interpret Schiff's obscure notation? me Schiff me Wmi,1 W1 B'ik,1 a k(1) B'ij,1 a l(1) , am(1) B'12,1 and B'21,1 ?? Schiff: ψ1 = Σn an(1)un Me: ψmi,1 = { Σj≠iB'ij,1 umj } Σk≠m B'ik,1 uk If I were to go with an "implied" i = 1,2 index, this last would become ψm,1 = { Σj≠iB'j,1 umj } Σk≠m B'k,1 uk = { Σj≠i aj(1) umj } Σk≠m ak(1) uk which at least looks like his ψ1 = Σn an(1)un . I just don't think a symbol a l(1) is well defined in his notation. I am now reminded of this Amazon reviewer's comments" "Legend tells that Leonard Schiff, while learning quantum mechanics from J. R. Oppenheimer, produced a careful set of notes which eventually became the first edition of this text. Soon followed a second, which was, for a long time, the unanimous choice of QM text-book in English. The third introduced things like group theory, a little of what was then called formal scattering theory (Lippman-Schwinger equation), and other new things. Schiff's text was a minimalist one, in the sense that you found everything, but in a very compact, no frills style. A teacher was required. The ratio of words to formulas was quite small. It entirely dominated the panorama. I, studying by myself, tried very hard, but could not understand, for instance, his treatment of perturbation of degenerate levels. Neither could I understand the infamous "box of very large dimensions" where the systems with a continuous spectrum were trapped so that their spectra became discrete. Even less could I acquire the much desired "global view" of quantum mechanics. I was very unhappy with Schiff. Someone suggested Bohm's textbook for me, but, at that moment, it seemed too leisurely paced for my impatience. Then I hit upon the newly arrived English translation of Landau, Lifshitz, "and all was Light". Nowadays I think there is no more place for "ole Schiff", but I must confess some of my colleagues still swear by their second edition, with the famous greenish McGraw-Hill cover. Well, I must recognize that the Dirac equation treatment was quite good, and that you could trust the equations as they appeared in the text: almost no typos." I really do think Schiff has badly dropped the ball here! His notation just cannot handle the problem he is looking at! ___________________________________________________________________________ 13. Review of the development so far. So now let's review what I have done on my own with what I regard as adequate notation: 1. My general expansion for ψ is as follows: ψmi,n = Σj≠iB'ij,n umj + Σk≠m B'ik,n uk n > 0 ψmi,0 = umi n = 0 and I assume that H' has been pre-diagonalized within H(m) so we have "the right" umi . 2. The general form of equation 31.4 may be written as follows: (H0 - Em) ψmi,n = - H' ψmi,n-1 + Σt=1,n-1 Wmi,t ψmi,n-t + Wmi,n umi n>1 (H0 - Em) ψmi,1 = - H' umi + Wmi,0 umi n=1 And if we insert the above expansions into our n>1 formula we get this for our 31.4 (H0 - Em) [ Σj≠iB'ij,n |umj> + Σk≠m B'ik,n |uk>] = – H' [ Σj≠iB'ij,n-1 |umj> + Σk≠m B'ik,n-1 |uk>] + Σt=1,n-1 Wmi,t [ Σj≠iB'ij,n-t |umj> + Σk≠m B'ik,n-t |uk>] + Wmi,n |umi> 3. If we close item 2 with <uk'| in H , we get these results: (Ek' - Em) B'ik',1 = - <uk'|H' |umi> n = 1 (Ek' - Em) B'ik',n = – Σj≠iB'ij,n-1 <uk'| H' |umj> n > 1 – Σk≠m B'ik,n-1<uk'| H' |uk> + Σt=1,n-1 Wmi,t B'ik',n-t and this does seem to provide a systematic way to build up B' coefficients from those of lower order. B. If we close item 2 with <ums| in H(m), we get these results: 0 = – <ums| H' | umi> + Wmi,1 δs,i n = 1 B'is,n-1(1 - δs,i) [Wmi,1 – Wms,1] n>1 = + Σk≠m B'ik,n-1H'sk – Σt=2,n-1 Wmi,t B'is,n-t(1 - δs,i) – Wmi,n δs,i We can then consider the two cases: s = i and s ≠ i. For s=i: Wmi,1 = <umi| H' | umi> n=1 Wmi,n = Σk≠m B'ik,n-1<umi| H' |uk> n > 1 For s≠i: <ums| H' | umi> = 0 n=1 B'is,n-1 [Wmi,1 – Wms,1] = Σk≠m B'ik,n-1<ums| H' |uk> – Σt=2,n-1 Wmi,t B'is,n-t n>1 The s=i results tell us how to construct the nth order energy correction if we are given the (n-1)th order coefficients of the perp space only, which I call B'ik,n-1 . The s≠i results tell us the values of the m-subspace B'is,n-1 in terms of the perp space B'ik,n-1 of the same order, and the m-space B'is,n-t of lower orders. 14. The special case where Wms,1 = Wmi,1 . However, if it happens that our initially diagonalized H' matrix in the m-subspace has identical diagonal elements for i and s, then we have Wms,1 = Wmi,1 and we have to deal with this as a special case. This is how far I have gotten today, and I don't know how to proceed in this case. Should I regard B'is,n-1 as blowing up, or should I regard B'is,n-1 as being undetermined? [ I think undetermined is correct.] If we assume this is the case, and if we don't imagine B'is,n-1 blowing up, then we have 0 = – Σk≠m B'ik,n-1<ums| H' |uk> + Σt=2,n-1 Wmi,t B'is,n-t n>1 Let's write this out for n = 2 and n=3 and n=4 ( all only valid for i≠s) 0 = – Σk≠m B'ik,1<ums| H' |uk> n=2 0 = – Σk≠m B'ik,2<ums| H' |uk> + Wmi,2 B'is,1 n=3 0 = – Σk≠m B'ik,3<ums| H' |uk> + Wmi,2 B'is,2 + Σt=2,3 Wmi,3 B'is,1 n=4 I think we may be getting somewhere. The n=2 equation is a peculiar fact I guess, only true when i≠s. What happens if I install into it the known coefficients B'ik,1 = <uk| H' |umi >/ (Em - Ek) = H'ki/ (Em - Ek) Then the n=2 equation seems to say 0 = – Σk≠m <ums| H' |uk><uk| H' |umi >/ (Em - Ek) s ≠ i 0 = – Σk≠m H'skH'ki/ (Em - Ek) = – (H'QomH')si s ≠ i (H'QomH')si = 0 when s ≠ i // 2.9.09 Ie, we have to diagonalize VQ0aV !! This is a fascinating fact now that I have done some Messiah work. At first I thought it was some useless sum rule, but now I see it is the solution to our whole problem!! We have now quietly exposed new conditions (from the perturbation equations) on our states "i" and "s" that they must be certain linear combinations of the basic states which already diagonalized H' and which now must diagonalize the operator H'QomH' within the N1 subspace! These new linear combinations continue to diagonalize H' because Wmi,1 is the same for all the eigenstates of our N state level 1 EV problem. These states i and s are the ones I called |0; m,ε2> in my Messiah notes. So from now on, we have to think of the states i and s as being these special states. Also, we already showed earlier in these notes that the diagonal elements are (H'QomH')ii = Wmi,2 . So right at this point we have found that: " The states in the N1 subspace which are degenerate and which I label with letters like i and s, are in fact eigenvectors of the EV problem H'QomH' |i,1> = Wmi,2|i,1> . " So we have stumbled upon the level 2 EV problem by doing our "general treatment", very good. Of course this is confirmed in Messiah. Now consider that |i,1> = Σj≠iB'ij,1 |umj> = the portion of our first order state correction that lies in the space Ea0. Our "solution for the B'ij,1" is the same as our solution for the eigenstates to the above EV problem, so we are suddenly done! We start with some starting states and diagonalize H'QomH' and then we know the B'ij,1 . There is no need for more elaborate equations here! (old comment retained) Wow! I wonder if it really is this complicated? This is far beyond Schiff, but perhaps one of my other books has some comments. This is just too messy for me to pursue without having a source to check things against. I think in practice one "ignores far away levels" and maybe things simplify. 15. Comments on Saxon's Approach. I have now read Saxon's "take" on this subject, and it is different from that of Schiff. I quote my own notes in my Saxon doc on some of the key differences: There are no explicit power series expansions, but we are given a successive approximations rule. He does not normalize ψ in the special way that makes <umi| ψmi- umi> = 0. He does not "require" that H' must start off being diagonal in the umi, though he does end up with a result that gives the required linear combinations. He tends to work with objects "to all orders" rather than "in particular orders" He uses what I call Dij coefficients in place of Schiff's B'ik coefficients. My other sources on this subject are Messiah and, I see, Pauling & Wilson. 16. Comments on my Theorem A and a way to think of the Saxon approach. (1) Possible Saxon method. This is the theorem that says H' must be diagonal within H(m) . It results from balancing the λ1 terms on both sides of the power-series-expanded full SE. Now consider this f(λ) = f0 + f1λ + f2λ2 + f3λ3 + g(λ) = g0 + g1λ + g2λ2 + g3λ3 + Such a balance would say that fi = gi . But suppose you were to write these in a different way: f(λ) = f0 + f1λ + F2λ2 F2 = f2 + f3λ + f4λ2 + ... g(λ) = g0 + g1λ + G2λ2 G2 = g2 + g3λ + g4λ2 + ... where now we include in F2 and G2 "the λ2 term and all higher orders". Now if someone says f(λ) = g(λ), you then have F2 = G2 . But we would only say F2 = f2 to zeroth order. Also, the equation F2 = G2 would not be the same as the equation f2 = g2, though both equations are true. I think maybe Saxon works with things like F2 because he subtracts out first order results. Look for example at Saxon page 191 equations 28 and 29. These both have the form I show above : f(λ) = f0 + f1λ + F2λ2 In the case of the coefficients, it happens that f0 = 0. Let's see what happens if we take the Saxon approach relative to our Theorem A. Consider this: ψmi = umi + λ ψmi,rest ψmi,0 = umi Wmi = Em + λ Wmi,1 + λ2 Wmi,rest Wmi,0 = Em // I could have used λ Wmi,rest where "rest" means "all the rest of the terms packed into one term". What happens if we plug this into our SE and just separate out the λ0 terms from the rest: (there are 4+6 = 10 terms) (H0 + λH') { umi + λ ψmi,rest} = (Em + λ Wmi,1 + λ2 Wmi,rest) { umi + λ ψmi,rest} H0 umi = Em umi λ0 λH' umi + λH0 ψmi,rest + λ2 H' ψmi,rest = λWmi,1 umi + λEm ψmi,rest + λ2 Wmi,rest umi + + λ2 Wmi,1 ψmi,rest + λ3 Wmi,rest ψmi,rest λ1 and higher Now cancel one λ to get ( note there really are 10 terms above and below) H' umi + H0 ψmi,rest + λ H' ψmi,rest = Wmi,1 umi + Em ψmi,rest + λ Wmi,rest umi + + λ Wmi,1 ψmi,rest + λ2 Wmi,rest ψmi,rest λ1 and higher or (H0 - Em) ψmi,rest = (Wmi,1 - H') umi + λ Wmi,rest umi + λ (Wmi,1 + λ Wmi,rest ) ψmi,rest Now close with mj ≠ mi and get 0 = - H'mj,mi + λ (Wmi,1 + λ Wmi,rest) < umj | ψmi,rest> NOW we would claim this: H'mj,mi = λ ( Wmi,1 + λWmi,rest) < umj | ψmi,rest> and then we would say "Oh, H'mj,mi = 0 only in zeroth order" but when higher orders are included, we do not get H'mj,mi = 0." Everything done above is EXACT! No approximations were made. H'mj,mi is not zero! ψmi = umi + λ ψmi,rest In general we don't know much about the component of ψmi,rest on umj for j≠i, but it is likely non-zero for a general H'. So now I have two conflicting facts: [1] my Schiff argument says H'mj,mi = 0 [2] my Saxon argument says H'mj,mi ≠ 0 Which argument is correct??? I could try a sample problem I suppose. Does this contradiction have anything to do with degeneracy? YES! In both cases, I have to "close" with some mj ≠mi, so I am assuming there are at least two states with energy Em. So this contradiction is specific to perturbation theory of degenerate states. Here is one possible resolution: "If the power series expansions exist, meaning they converge, then H'mj,mi = 0 in H(m) and it will then magically turn out that < umj | ψmi,rest> = 0, and there will be no contradiction". (2) The 2x2 diagonalization of H' in Schiff. Now, any text author who does the Schiff "series thing" is going to run into this H'mj,mi = 0 requirement, I think. I certainly did right away. I am amazed that Schiff does not seem to mention it. On the other hand, on page 249 when Schiff is faced with actually doing a degeneracy analysis with N=2, the first thing he does is he finds basis functions that diagonalize H' !! Recall from notes above that he does this: = W1 m =1 l = 2 a = am b = al H' Sk = Λkk Sk // my notation. (3) Question: <umj| ψ'mi,rest> = 0 for i≠j ? No! In Theorem B above I show this fact <umi| ψ'mi,n> = 0 n>0 which is 31.6 Now compare: ψmi = umi + λ ψmi,rest ψmi = umi + λ ψmi,1 + λ2 ψmi,2 + ... I have clearly proven that <umi| ψ'mi,rest> = 0 Here is the question: is there some equivalent theorem that says <umj| ψ'mi,rest> = 0 for j ≠ i ? If so, this would quickly resolve our contradiction above. Let's copy the previous proof and edit as needed: 6. Theorem B': { show 31.6: <umj| ψmi,n> = 0 for n > 0 and j ≠ i } Proof. Renormalize as before 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 n > 0 where we imagine expanding the B' just as we expand ψ' or W. Then we have for j' ≠ i <umj'| ψ'mi,n> = <umj'| Σj≠iB'ij,n umj> = Σj≠iB'ij,n δj'j = B'ij',n Therefore <umj| ψ'mi,n> = B'ij,n B'ij,n We know that B'ij,0 = 0 because there is it starts at the λ = 1 level. Thus we have shown that <umj| ψ'mi,rest> = B'ij and this is not necessarily 0 ! Therefore, here is what the Saxon argument is saying: H'mj,mi = λ ( Wmi,1 + λWmi,rest) B'ij Well, what about the first factor ( Wmi,1 + λWmi,rest). Could it be zero somehow? If H' were diagonal in H(m), then you might argue that there will be no energy corrections ever and so this factor should vanish. (4) Pause to consider a new (and bogus!) Theorem D. Here is the full Hψm1=Wm1ψm1 equation written in the unperturbed uk basis: First of all, recall that in this basis, the matrix H0 is completely diagonal. Now here I am assuming that we "prediagonalize" H'. I have positioned H' in the upper left of the H matrix for convenience. In that NxN square, both H0 and H' are diagonal in the uk basis. Outside of this NxN square, we have H elements everywhere. The diagonal elements along the dotted line are order 1 = λ0 but the off-diagonal elements all arise from λH' and are therefore O(λ). We are NOT in block diagonal form, which means we still have mixing between H(m) and H : we cannot pretend we can isolate the H(m) problem. For our eigenstate, we have chosen ψm1 and that is why there is a 1 at the top of the state vector. This column vector is a representation of this fact used often above: ψm1 = { um1 + Σj≠iB'1j umj } Σk≠m B'1k uk so the "1" here really stands for the leading um1 state, and the first few B' stand for B'1jum j = 2..N and then all the later B' below the line stand for B'1kuk for k > N. The diagonal elements marked D are really these Dii = (H0)ii + λ (H')ii = Em1 + λH'mi,mi i = 1...N Dij = 0 for i ≠ j where we put Em1 for all of H(m) since these states were degenerate before H' was applied Now let's write out the first N lines of the above matrix equation: (Em1 + λH'm1,m1) um1 + Σk≠m H1,k B'1k uk = Wm1 um1 (Em1 + λH'm2,m2) B'12um2 + Σk≠m H2,k B'2k uk = Wm1 B'12um2 etc My entire motivation here was to try to isolate coefficients like B'12 to learn something about them , so let's see if that has worked. Close the second equation above with um2 and we get (Em1 + λH'm2,m2) B'12 = Wm1 B'12 // since um2.uk = 0 (Wm1 – Em1 – λH'm2,m2) B'12 = 0 This is interesting. First of all, if it just happens that H'm2,m2 = H'm1,m1 ( ie, after we prediagonalize H', we find that the 11 and 22 elements are degenerate, sort of a second level of degeneracy in our problem), then the paren above is really Wm1,2 and higher which is order λ2. But this should not in general be 0. And if 11 ≠ 22, then again, no reason for it to be 0. So this does seem to imply that B'12 = 0 to all orders!!! This is a fact I have been long looking for, but I cannot see a simple reason why it should be true apart from the above derivation. For example, suppose paren = 5λ2 , then 5λ2B'12 = 0 and this says B'12 = 0 to all orders. Notice that in this entire Theorem D development, we have made NO approximations. Every single equation we have written above is exact to all orders of λ. This seems to tell us that if we prediagonalize H', then we shall get no mixing within H(m) to all orders, and we can then claim that really ψmi = { umi + Σj≠iB'ij umj } Σk≠m B'ik uk = um1 Σk≠m B'1k uk Is this really true? I have this sinking feeling that it is "too good to be true". Contradiction noted: Restate this again: Our Theorem D purports to show that B'ij = 0 to all orders for i ≠ j. Is this really true? If we look back earlier in these notes, I obtained using Schiff's power series approach this equation: B'is,1 = [Σk≠m B'ik,1<ums| H' |uk>] /(Wmi,1 – Wms,1 ) i ≠ s Wmi,1 ≠ Wms,1 This certainly does NOT seem to be zero. So I just said that B'ij = 0 to all orders, but Schiff's approach tells us right here that B'ij,1 ≠ 0. How can something be 0 to all orders, but be non-zero right in the very first order? Another contradiction in our 6 day saga! I seem to be stuck in a Dr. Who time loop, around and around we go with endless contradictions. Perhaps the time has come to look at Messiah. Maybe he will point out some little detail (or some large fact) that I am blinded to which is causing these contradictions. The biggest contradiction really is that my theory says you cannot do perturbation theory without prediagonalizing your states, but Saxon and Schiff happily do their theories without prediagonalizing. I can also look at Pauling and Wilson. // I carefully reread all of the above on 1.19.09 after doing a long tour of Stakgold Chap 2. Everything still makes sense, and the confusions still exist. I have gained the suspicion that projection operators can be useful in place of constantly "closing" things from the left all the time. _______________________________________________________________ Resolution of the above Theorem D mystery: H|ψm1> = Wm1|ψm1> |ψm1> = { |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk> H [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = Wm1[{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk> ] Now insert 1 = {Σs |ums><ums| + Σk|k><k|} for subspace plus perp space: H {Σs |ums><ums| + Σk|k><k|} [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = Wm1[{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk> ] now close with some <ur| which could be a ums or could be a k state. <ur| H {Σs |ums><ums| + Σk|k><k|} [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = <ur| Wm1[{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk> ] Suppose I just combine all into a uniform "unity" <ur| H {Σr' |ur'><ur'| } [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = <ur| Wm1[{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk> ] Σr'<ur| H |ur'> <ur'| } [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = <ur| Wm1[{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk> ] Σr'Hrr' <ur'| ψ1m> = Wm1 <ur| ψ1m> Σr'Hrr'( ψ1m)r' = Wm1 ( ψ1m)r So here finally is our "matrix equation". I can write the matrix as I wrote it above. But what do the column vectors really look like? ( ψ1m)1 = <um1| } [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = 1 ( ψ1m)2 = <um2| } [{ |um1> + Σj≠1B'1j |umj> } Σk≠m B'1k |uk>] = B'12 and so on. So our matrix equation drawn above is accurate exactly as it is written! There are no extra factors umj that you "add to it". So our "first few equations" are really these: (Em1 + λH'm1,m1) + Σk≠m H1,k B'1k = Wm1 (Em1 + λH'm2,m2) B'12 + Σk≠m H2,k B'2k = Wm1 B'12 etc So now this second equation says: (Wm1 – Em1 – λH'm2,m2) B'12 = – Σk≠m H2,k B'2k (Wm1 – Em1 – Wm2,1) B'12 = – Σk≠m H2,k B'2k and now there is no longer an "argument" that says we have proven that B'12 = 0. On the contrary, this seems to give us an expression for our B'12 to all orders: B'12 = – (Wm1 – Em1 – λH'm2,m2)-1 Σk≠m H2,k B'2k Of course we don't know what Wm1 is nor do we know what B'2k is (both needed to all orders), but this is nonetheless a relationship between the various coefficients to all orders that is just a result of our matrix equation shown above graphically. We could expand both sides in orders of λ, but of course this will just give us our usual "perturbation equation". 17. Relation between Kato Method and Phil Method for Degeneracy [ 2.9.09 ] In my "personal" approach above, I think of the "renormalized" expansion for ψ in Ea, ψmi = { umi + Σj≠iB'ij umj } Σk≠m B'ik uk // all orders <umj| ψmi> = B'ij <k| ψmi> = B'ik ψmi,n = { Σj≠iB'ij,n umj } Σk≠m B'ik,n uk // some order n > 0 <umj| ψmi,n> = B'ij,n <k| ψmi,n> = B'ik,n and then I write down the "perturbation equations" and try to solve them for the B'. I end up with a viable iterative scheme which handles degeneracy. In the Kato approach, we end up with an EV problem at some level, for example, V |φα : 1> = ε1 |φα : 1> // The 1 means eigenket computed through level 1 Once we solve this EV problem for the kets |φi : 1> which lie in Ea0, in order to get the kets in Ea we have to do this |ψi : 1> = P |φi : 1> Then at that point, we could find our coefficients: <φj|ψi : 1> = <φj|P|φi : 1> = B'ij,1 i,j in 1..N range and similarly for <k|. At level 2 we would say the following, for any degenerate subspace of N VQa0V |φi : 2> = ε2 |φi : 2> i = 1..M // OK M=1 too <φj|ψi : 2> = <φj|P|φi : 2> = (B'ij,1 + B'ij,2) i,j in 1..M range So that is how we can relate the Kato method to our coefficients. So even with Kato, we don't have any fancy formula for the B'ij,n in the degeneracy case. You cannot avoid doing the work in either notation. In the non-degenerate case, we do have a formula of sorts. |ψ ;n> = coefficient of λn in { <φ| Σm=1 λmP(m)|φ>-1 [ Σk=1 λk P'(k)|φ> ] } B'1k,n = <k|ψ ;n> = coefficient of λn in { <φ| Σm=1 λmP(m)|φ>-1 <k| [ Σk=1 λk P'(k)|φ> ] } 18. Comparison to Messiah in General [2.9.09] Messiah's great contribution is his introduction of the projection operators. In his pre-Kato notes, he deals with P0 and Q0 = 1 - P0 in both non-degenerate and then degenerate cases. Then we get the Q0a operator which causes everything to get very compact, no energy denominator sums needed. [ Messiah does not credit anyone for this Q0a thing, maybe Kato invented it or some earlier person. ] Of course H' = V in Messiah, so in the end, everything we do can be written in terms of P0 and Q0a and V. When we get into the Kato approach, we end up with a perturbation theory that looks a lot like scattering theory although Messiah never draws Feynman graphs (Bloch does, he says). In such graphs, each vertex would get a factor of λV, and each line would be a propagator (z-H0)-1. These things are not spacetime propagators, but are time-independent energy-space propagators. (z-H)-1= (z-H0)-1 + (z-H0)-1λV(z-H0)-1 + (z-H0)-1λV(z-H0)-1λV(z-H0)-1 When this expansion is put into the formula for projector objects P and B, we get similar expansions for these objects, but now the propagators" are replaced either by P0 or Q0a. Using these expansions for P and B, we are able to discover the "generalized EV equations" P0BP0 |φα> = (Eα – Ea0) P0PP0|φα> and these lead us to a solution of any degeneracy problem, albeit lots of work is still needed once the EV problem has been identified.