Kato Round Three

Kato Round Three PhL 2.2.09 I intended this round to be my final meta notes, and started off in a very organized manner. But then I ended up getting way down into details, so this then became Round Three and meta notes were put off to the next and last document. Section 15 and Section 16 1 Definition of G(z) 1 Definition of ψi,n and Ei and Pi 2 The projector-pole expansion of G(z) 2 Projector Pi is the residue of the pole of G(z) at z-Ei 2 Definition of ψa,n and Ea,n ; of φa,i and Ea0 ; and of N and Ea and Ea0 and Γa 2 The Kato formulas for P and HP 3 Integral equation for G(z) and its series solution which is then installed into the Kato formulas 3 The projector-pole expansion of G0(z): definition of Pi0 and P0 ≡ Pa0 4 Examination of a sample Kato integral 5 Kato's Bag of Tricks 6 A Taylor Series Expansion 6 Use this expansion in our formula for G0(z) 7 Rewriting the G0(z) in terms of the operator Q0a 8 A mini-theorem about powers of G0(z) 9 Evaluation of a demonstration Kato integral 10 Evaluation of all the Kato integrals 11 Compute P(n) for n = 0,1,2,3 13 Compute B(n) ≡ (HP– Ea0P)(n) and (HP)(n) for n = 0,1,2,3 14 Compute energy corrections for the non-degenerate theory. 16 The Degenerate Case with Kato 20 The Level 1 Problem. 22 The Level 2 Problem. 23 The Level 3 Problem. 24 Section 15 and Section 16 Definition of G(z) 1. A bound state problem has some Hamiltonian H, and the resolvent is defined simply as G(z) = (z-H)-1 (1) an operator function of complex variable z. Definition of ψi,n and Ei and Pi 2. The problem has a state and energy spectrum which can be decomposed into a direct sum of subspace problems in the usual manner. For subspace Si we have eigenvectors {ψi,n} such that Hψi,n = Eiψi,n HPi = EiPi (2) where Pi is the usual projector for Si. The equation on the right may be verified by applying it to all states in the Hilbert Space (HS) of the entire problem. For states not in Si it says 0 = 0. Otherwise, it repeats the left equation. We know that ΣiPi = 1 if we add up over all subspaces of the problem. Any subspace Si could be non-degenerate (dimension = 1) or could be degenerate (dimension > 1). The projector-pole expansion of G(z) 3. Applying (2) to (1) we find that G(z) Pi = (z-H)-1Pi = (z-Ei)-1Pi => G(z) = Σi (z-Ei)-1Pi (3) where we just applied Σi over subspaces to both sides of the first equation to get (3). Projector Pi is the residue of the pole of G(z) at z-Ei 4. Now apply a contour integral to both sides of (3) and use the usual residue formula for a simple pole to get Pi = (2πi)-1 ∫Γi dz G(z) (4) where the contour Γi captures the pole at z = Ei. We could imagine another contour which captures some set of poles {Ei}, thus making a projector which projects onto the subspace spanned by the corresponding set of states {ψi}. We will be doing that next. Definition of ψa,n and Ea,n ; of φa,i and Ea0 ; and of N and Ea and Ea0 and Γa 5. Now assume H = H0 + λV and we are doing perturbation theory. When the perturbation is turned on, imagine that the equation Hψ = Eψ has a certain set of N eigenvectors ψa,n with eigenenergies Ea,n which, when the perturbation is gradually turned off (λ→0), become N degenerate eigenvectors φa,n of H0, all having the same eigenenergy Ea0. This is the key concept. We then want to define two distinct N dimensional spaces: space Ea is that spanned by the set of ψa,n just mentioned, while Ea0 is that spanned by the degenerate φa,n . As λ→0. we can certainly claim that Ea → Ea0. Clearly Ea0 is a subspace of the H0 problem. The space Ea is the union of at most N subspaces Si of the H problem: the number would be N only if it happened that all the ψa,n were non-degenerate. It does not matter to us whether any of the subspaces whose union is Ea is degenerate. Now imagine we create a contour Γa in the z plane which is carefully designed to circulate (in the usual sense) all the locations z = {Ea,n} as well as the location z = Ea0. This contour is also selected so as to not include in its circulation any z locations associated with the eigenenergies of other subspaces Ei0 or Ei where i ≠ a. In the following picture, I show a few eigenenergies of some Ei0 i ≠ a that lie outside the contour: So clearly we can think of the Γa contour as equivalent to a sum of N Γn contours around pole locations z = {Ea,n} of the operator function G(z). Each of these contours yields a Pn for that pole residue, and the sum of the Pn we shall call P. P then is the projection operator which projects out of any ket that portion lying in space Ea : P = Σn=1N Pn = Σn=1N { (2πi)-1 ∫Γn dz G(z) } = (2πi)-1 ∫Γa dz G(z) The Kato formulas for P and HP 6. Since G(z) ≡ (z-H)-1, we know that (z-H)G = 1 which says zG = 1+HG. Therefore (2πi)-1 ∫Γa dz zG(z) = (2πi)-1 ∫Γa dz [1+HG(z)] = (2πi)-1 H ∫Γa dz G(z) = HP. since the contour integral of 1 is 0, and H is not a function of z, and using (4) above for P. So let us restate this and the previous result, which are the main weapons in the Kato arsenal: P = (2πi)-1 ∫Γa dz G(z) (5a) HP = (2πi)-1 ∫Γa dz z G(z) (5b) The expressions are identical apart from an extra factor of z in the HP integrand. Integral equation for G(z) and its series solution which is then installed into the Kato formulas 7. We can define both G(z) ≡ (z-H)-1 and G0(z) ≡ (z-H0)-1 where H = H0 + λV. Very simple algebra then shows that G = Go(1 + λVG) (6) which is an operator equation for G. Our perturbation problem presumably concerns the motion of a single particle under Hamiltonian H. Were we to work in the coordinate space representation, for example, and were we to define G(r,r') ≡ <r|G|r'> and similarly for Go and V, we would write (6) as G(r,r') = G0(r,r') + λ ∫d3r" V(r,r")G(r",r') and we would refer to this as "an integral equation for G". The equation certainly resembles integral equations we commonly see in scattering theory, but here we are dealing with a bound state problem. 8. Just as we do in scattering theory, we can derive a series solution of (6) which has this form: G = Σn=0∞ λnG0(VG0)n = G0 + λ G0VG0 + λ2 G0VG0VG0 + ... (7) In any event, it is trivial to show that the series (7) solves our operator equation (6): RHS = Go(1 + λVG) = Go(1 + λV Σn=0∞ λnG0(VG0)n) = Go + λ Σn=0∞ λnGoVG0(VG0)n = Go + Σn=0∞ λn+1Go(VG0)n+1 = Go + Σm=1∞ λmGo(VG0)m = Σm=0∞ λmGo(VG0)m = LHS 9. We now use for G(z) in (4) and (5) our series solution (7) to get these new Kato formulas: P = (2πi)-1 ∫Γa dz G(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz G0(VG0)n (8) HP = (2πi)-1 ∫Γa dz zG(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz z G0(VG0)n (9) We are now going to interpret (8) and (9) as expansions in smallness parameter λ for our two operators P and HP. As noted earlier, P projects things onto space Ea , while H is the full Hamiltonian including the perturbing term. So the above pair of equations become P = Σn=0∞ λn P(n) P(n) ≡ (2πi)-1 ∫Γa dz G0(VG0)n (8') HP = Σn=0∞ λn (HP)(n) (HP)(n) ≡ (2πi)-1 ∫Γa dz z G0(VG0)n (9') We want to evaluate these contour integrals which means we need to learn about the z-plane structure of G0, which we now ponder. The projector-pole expansion of G0(z): definition of Pi0 and P0 ≡ Pa0 10. We are now going to repeat the work of sections 1,2,3 above, but for G0(z) instead of G(z). The first two sections yield these equations G0(z) ≡ (z-H0)-1 (1)0 H0φi,n = Ei0φi,n H0Pi0 = Ei0Pi0 (2)0 where the φi span the (possibly degenerate) subspace of eigenvalue Ei0, and where Pi0 projects out of any ket that portion which lies in the subspace Si0 spanned by the φi. We apologize for the fact that some of the 0's are subscripts while others are superscripts. For equation 3 then we have G0(z) Pi0 = (z-H0)-1Pi0 = (z-Ei0)-1Pi0 => G0(z) = Σi (z-Ei0)-1Pi0 (3)0 where Σi is over all the subspaces of operator H0. One of the subspaces in this sum is Ea0 above, having projector Pa0 ≡ P0 . We can break out this one subspace and say G0(z) = (z-Ea0)-1P0 + Σi≠a (z-Ei0)-1Pi0 (10) where the remaining sum is over other subspaces of H0. Examination of a sample Kato integral 11. Were we to simply install this expression for G0(z) into the Kato series expansions (8) and (9) above, we would have terms of this general form (we use the simple n=1 term of (8) as an example) ∫Γa dz G0VG0 = ∫Γa dz {(z-Ea0)-1P0 + Σi≠a (z-Ei0)-1 Pi0 } V{(z-Ea0)-1P0 + Σi'≠a (z-Ei'0)-1Pi'0} The contour Γa was selected to include the Ea,n eigenenergies as well as the eigenenergy Ea0. The eigenenergies Ea,n were some of the poles of G(z). Now that we have done a series expansion for G(z), we see that the expansion terms like the one shown above do not have poles at Ea,n. Instead, a particular term has a pole at Ea0 and also has poles at other Ei0 which lie outside our contour: Therefore we may deform our contour to become a simple circulation about the pole at z = Ea0. The integrand has four terms. The first has a double pole at z = Ea0 and integrates to 0. The fourth term has poles which all lie outside our contour and so also do not contribute. We are left then with the two cross terms, the first of which is this: (2πi)-1∫Γa dz {(z-Ea0)-1P0 }V{Σi≠a (z-Ei0)-1Pi0} = Σi≠a P0 V (Ea0- Ei0)-1Pi0 where we have written out the residue of the pole at z = Ea0. Below we will show this fact [ Σi≠a (Ea0- Ei0)-1 Pi0] = Q0a which allows this term to be written as simply P0VQ0a and the other term is Q0aV P0. This shows the general idea, but we want to follow Kato's systematic presentation of this idea. Kato's Bag of Tricks A Taylor Series Expansion 12. Suppose some function f(z) is analytic at z=a. We can expand f(z) in the following Taylor series around the point z = a as follows: f(z) = f(a) + Σn=1,∞ [ f(n)(a)/ n! ] (z-a)n Let's apply this to f(z) = (z-a)/(z-b). A quick calculation shows that f(0)(z) = (z-a)/(z-b) n=0 f(n)(z) = (a-b) (-1)n-1 n! (z-b)-(n+1) n>0 so that f(a) = (a-a)/(a-b) = 0 n=0 f(n)(a) = (-1)n-1 n! (a-b)-n n>0 So we then have derived this Taylor expansion (z-a)/(z-b) = Σn=1,∞ [(-1)n-1 (a-b)-n ] (z-a)n [ The LHS has a pole at z=b. The RHS equals –Σn=1,∞ 1 at this point and thus diverges. The LHS has a zero at z=a and the RHS is also zero at this point. This series probably converges in any disk around z=a with radius less than |b-a|. ] Now let's apply our Taylor expansion above setting a = Ea0 and b = Ei0: (z- Ea0)/(z- Ei0) = Σn=1,∞ [(-1)n-1 (Ea0- Ei0)-n ] (z- Ea0)n (11) Use this expansion in our formula for G0(z) 13. Recall now our expansion above for G0(z), equation (10) above G0(z) = (z-Ea0)-1P0 + Σi≠a (z-Ei0)-1Pi0 = (z-Ea0)-1P0 + X The second term which we shall call X has poles at all the locations z = Ei0 (which lie outside our contour). Now let's process this second term X, X = Σi≠a (z-Ei0)-1Pi0 = (z-Ea0)-1 Σi≠a {(z-Ea0)/(z-Ei0)} Pi0 Then with the Taylor expansion (11) above, X becomes X = (z-Ea0)-1 Σi≠a { Σn=1,∞ [(-1)n-1 (Ea0- Ei0)-n ] (z- Ea0)n }Pi0 = Σn=1,∞ (-1)n+1 (z- Ea0)n-1 [ Σi≠a (Ea0- Ei0)-n Pi0] where Σi≠a sums over all the subspaces of H0 other than Ea0. So G0(z) = P0/ (z-Ea0) + Σn=1,∞ (-1)n-1 (z- Ea0)n-1 [ Σi≠a (Ea0- Ei0)-n Pi0] (12) We are now going to come up with a compact way to express the last factor [...]. But first just a comment. Compare the above with our previous expression for G0(z), G0(z) = (z-Ea0)-1P0 + Σi≠a (z-Ei0)-1Pi0 The operator function G0(z) really does have poles at the points z = Ei0 as this last expression shows. The series expansion we used to get the previous expression for G0 is not valid near these poles, and that is why we don't see any poles at z = Ei0 in that expression! But we don't intend to use this second expression near those poles, so it is OK. We only use this second expression near the point z = Ea0 since that is where our contracted contour lies. Rewriting the G0(z) in terms of the operator Q0a 14. We know that 1 = Σi Pi0 = Pa0 + Σi≠a Pi0 = P0 + Σi≠a Pi0 where we are summing over subspaces of H0. Suppose the eigenstates of subspace Si0 are called φi,n. Then we can write, Pi0 = Σni | φi,ni>< φi,ni| (13) where ni enumerates however many eigenstates span the subspace Si0. Let's now define two new operators as follows Q0 ≡ 1 - P0 = Σi≠a Pi0 = Σi≠a Σni | φi,ni>< φi,ni| (14) Q0a ≡ Q0 (Ea0 - H0)-1Q0 (15) The state sum appearing in Q0 includes all eigenstates of H0 except those in Ea0. So one can say that the operator Q0 projects out of any ket the part that is in the perp space (Ea0) . Note that Q02 = Q0 which is true for any projection operator. We now claim this little theorem for the Q0a operator (Q0a)n = Q0 (Ea0 - H0)-n Q0 (16) and demonstrate it for the case n=2, where we use H0 |φi,ni> = Ei0 |φi,ni> in both directions, (Q0a)2 = Q0 (Ea0 - H0)-1 Q0 Q0 (Ea0 - H0)-1 Q0 = = Q0 (Ea0 - H0)-1 Q0 (Ea0 - H0)-1 Q0 = Q0 (Ea0 - H0)-1 Σi≠a Σni | φi,ni>< φi,ni| (Ea0 - H0)-1 Q0 = Q0 (Ea0 - Ei0)-1 Σi≠a Σni | φi,ni>< φi,ni| (Ea0 - Ei0)-1 Q0 = Q0 (Ea0 - Ei0)-2 Σi≠a Σni | φi,ni>< φi,ni| Q0 = Q0 (Ea0 - H0)-2 Σi≠a Σni | φi,ni>< φi,ni| Q0 = Q0 (Ea0 - H0)-2Q0 Q0 = Q0 (Ea0 - H0)-2Q0 QED We are now going to show that our factor [...] in (12) is given by [ Σi≠a (Ea0- Ei0)-n Pi0] = (Q0a)n (17) We start with the RHS and use our theorem above (Q0a)n = Q0 (Ea0 - H0)-n Q0 = Σi≠a Σni | φi,ni>< φi,ni| (Ea0 - H0)-n Q0 = Σi≠a Σni | φi,ni>< φi,ni| (Ea0 - Ei0)-n Q0 = Σi≠a Σni | φi,ni>< φi,ni| (Ea0 - Ei0)-n Σj≠a Σnj | φj,nj>< φj,nj| = Σi≠a Σni Σj≠a Σnj (Ea0 - Ei0)-n < φi,ni| φj,nj> | φi,ni> < φj,nj| = Σi≠a Σni Σj≠a Σnj (Ea0 - Ei0)-n δi,j δni,nj | φi,ni> < φj,nj| = Σi≠a Σni (Ea0 - Ei0)-n | φi,ni> < φi,ni| = Σi≠a (Ea0 - Ei0)-n Σni | φi,ni> < φi,ni| = Σi≠a (Ea0 - Ei0)-n Pi0 where we have assumed that eigenfunctions are orthonormalized in each subspace Si0. We have now shown this fairly impressive and compact operator expansion for G0(z) G0(z) = P0/ (z-Ea0) + Σn=1,∞ (-1)n-1 (z - Ea0)n-1 (Q0a)n (18) Notice that all the z-dependence is of the form (z-Ea0). The first term has a pole at z = Ea0. The series part is analytic at z = Ea0 and the nth term in the series has a zero of degree n-1 at z = Ea0. It is now helpful to make (18) even more compact as follows: G0(z) = Σn=0,∞ Rn yn-1 (19) where y ≡ (z-Ea0) R0 ≡ P0 and Rn ≡ (-1)n-1 (Q0a)n for n>0 so R1 = Q0a R2 = - Q0a2 R3 = Q0a3 etc Notice that, in each term in the sum for G0(z), the index of the coefficient operator Rn is one larger than the exponent of the matching power of y. Normally we are used to seeing these numbers be the same in a series. This fact leads to the following mini-theorem. A mini-theorem about powers of G0(z) 15. Consider this operator F ≡ [G0(z)]n+1 = [Σi=0,∞ Ri yi-1]n+1 = { Σ i=0,∞ R i yi-1} { Σ i=0,∞ R i y i-1} .. { Σ i=0,∞ R i y i-1} where we have n+1 summation indices i0 through in . Theorem: If the exponents of y of a particular term in this (n+1)-fold product add up to J, then the sum of the indices on the corresponding Ri for that same term add up to J + (n+1). The proof is that there are (n+1) factors in a particular term, and for each such factor the index of R is one more than the exponent of y, so the sum of the exponents must exceed the sum of the indices of R by n+1. Corollary: Suppose the exponents of y add up to -1. Then the indices of R add up to -1 + (n+1) = n. Generalization: Both the theorem and corollary would be valid for an operator of the more general form F' ≡ G0(z) A1 G0(z) A2 .......An G0(z) where there are n operators Ai sandwiched between pairs of G0 factors. If we set all the Ai = V, then this is exactly the situation that arises in the Kato integrals, as we shall see. Evaluation of a demonstration Kato integral 16. Now let's look what happens when we use this expansion (18) to evaluate a sample Kato integral. We consider the same case we considered back in section (11), ∫Γa dz G0VG0 = ∫Γa dz { P0/ (z-Ea0) + Σn=1,∞ (-1)n+1 (z - Ea0)n-1 (Q0a)n } V { P0/ (z-Ea0) + Σm=1,∞ (-1)m+1 (z - Ea0)m-1 (Q0a)m } = y=0 dy { Σn=0,∞ Rn yn-1}V { Σm=0,∞ Rm ym-1} // y ≡ (z-Ea0) where now the contour encircles y = 0 instead of z = Ea0. Let's write out a few terms: = y=0 dy { R0y-1 + R1 + R2y + R3y3 + ...}V { R0y-1 + R1 + R2y + R3y3 + ...} If we think of this integral in terms of our mini-theorem above, there are n+1 = 2 factors of G0 involved, so n = 1. Our corollary tells us that any term which has a sum of y exponents adding up to -1 will have R indices adding up to -1 + (n+1) = +1. Of course it is only terms whose y exponents add up to -1 which contribute to the contour integral because only these terms are simple poles. All other terms in the integrand result in double poles, constants, or zeros, all of which yield 0. Using this rule (or just staring at the example above), one may evaluate the integral "by inspection": ∫Γa dz G0VG0 = (2πi) (R0VR1 + R1VR0 ) = (2πi) (P0VQ0a + Q0aVP0 ) Evaluation of all the Kato integrals 17. We now return to (8) and (9) above which we copy here P = (2πi)-1 ∫Γa dz G(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz G0(VG0)n (8) HP = (2πi)-1 ∫Γa dz zG(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz z G0(VG0)n (9) We are now going to interpret (8) and (9) as expansions in smallness parameter λ for our two operators P and HP. As noted earlier, P projects out of kets the portion lying in space Ea, while H is the full Hamiltonian H = H0 + λV. The reason we want to do these expansions will be revealed later. So the above pair of equations become P = Σn=0∞ λn P(n) P(n) ≡ (2πi)-1 ∫Γa dz G0(VG0)n HP = Σn=0∞ λn (HP)(n) (HP)(n) ≡ (2πi)-1 ∫Γa dz z G0(VG0)n We don't have any wonderful interpretation for operators like P(n)or (HP)(n) other than they are the λn component of P and HP expanded as shown. We start with the simpler P(n) integral. P(n) ≡ (2πi)-1 ∫Γa dz G0(VG0)n = (2πi)-1∫Γa dz G0V G0V.....VG0 There are n factors of V (which is why this term is order λn), and n+1 factors of G0(z). If we convert from dz to dy as in our sample integral, this becomes, using (19) G0(z) = Σi=1,∞ Ri yi-1 , = (2πi)-1∫Γa dy { Σ i=0,∞ Ri yi-1}V { Σ i=0,∞ Ri y i-1}V ...... V{ Σ i=0,∞ Ri y i-1} where summation indices are i0, i1.....in , a total of n+1 indices. This integrand exactly fits our generalized theorem and corollary above, so we may again evaluate the integral "by inspection". The contributing terms all have the same form ( n V's and n+1 R's) [ Ri V Ri V ...... V Ri] such that the sum of the R indices add up to n ! So we are done with this Kato integral: P(n) ≡ (2πi)-1 ∫Γa dz G0(VG0)n = Σ(n) Ri V Ri V ...... V Ri where R0 ≡ P0 and Rk ≡ (-1)k-1 (Q0a)k k>0 and where the notation Σ(n) means "sum over all n+1 indices ik such that the sum of these indices add up to n". 18. We now turn to the HP integrals. These have an extra factor of z in the integrand, and we would certainly like to turn this into a factor of (z-Ea0) = y so we have only powers of y to deal with. To this end, let's go back to P = (2πi)-1 ∫Γa dz G(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz G0(VG0)n (8) HP = (2πi)-1 ∫Γa dz zG(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz z G0(VG0)n (9) P = Σn=0∞ λn P(n) P(n) ≡ (2πi)-1 ∫Γa dz G0(VG0)n HP = Σn=0∞ λn (HP)(n) (HP)(n) ≡ (2πi)-1 ∫Γa dz z G0(VG0)n and let's subtract Ea0P from the HP equation (9) to get HP– Ea0P = (2πi)-1 ∫Γa dz (z- Ea0)G(z) = Σn=0∞ λn (2πi)-1 ∫Γa dz (z- Ea0) G0(VG0)n (10) = Σn=0∞ λn (HP– Ea0P)(n) where (HP– Ea0P)(n) ≡ (2πi)-1 ∫Γa dz (z- Ea0) G0(VG0)n = (HP)(n) – Ea0 P(n) We shall evaluate this last integral using our standard inspection method (modified slightly), and then we can say that (HP)(n) =(HP– Ea0P)(n) + Ea0 P(n) We now evaluate the integral (2πi)-1 ∫Γa dz (z- Ea0) G0(VG0)n by inspection as follows. It is exactly the same as our P(n) integral treated above, but there is one extra factor of y. Thus, the contributing terms are now those whose y exponents add up to one less than they did before, and that means that the sum of the Ri indices will add up to one less than they did before. Before they added up to n, so now they add up to n-1. So here is our conclusion: (HP– Ea0P)(n) ≡ (2πi)-1 ∫Γa dz (z- Ea0) G0(VG0)n = (HP)(n) – Ea0 P(n) = Σ(n-1) [ RiV RiV ...... V Ri] ≡ B(n) in Messiah p 717 (71) { we will use this short name later } using the same Σ notation as above. Therefore, we find that P(n) = Σ(n) [ RiV RiV ...... V Ri] (HP– Ea0P)(n) = Σ(n-1) [ RiV RiV ...... V Ri] (HP)(n) = Σ(n-1) [ RiV RiV ...... V Ri] – Ea0 P(n) = Σ(n-1) [ RiV RiV ...... V Ri] – Ea0 Σ(n) [ RiV RiV ...... V Ri] where R0 ≡ P0 and Rk ≡ (-1)k-1 (Q0a)k k>0 In every [..] factor, there are n factors of V, and n+1 factors or Ri . Let's look at dimensions for a moment. R0 = P0 is a dimensionless projection operator (as is any projection operator including Q0 = 1 - P0). The operator Q0a = Q0(Ea0- H0)-1Q0 has dimensions of 1/energy. Thus, dim(Rk) = (energy)-k. Each term in the sum for P(n)has dimensions as follows: (energy)-n (energy)+n = (energy)0 where the second factor comes from the n factors of V, while the first from the R's. Each term in the sum (HP– Ea0P)(n) has these dimensions: (energy)-(n-1) (energy)+n = (energy)1 as we would expect. Finally, looking at (HP)(n), we see that both terms are (energy)1, again as we would expect. Compute P(n) for n = 0,1,2,3 19. Remember that n is the number of V factors P(n) = Σ(n) [ RiV RiV ...... V Ri] n=0: R0 n = 1: R1VR0 + R0 VR1 n=2: R1VR1VR0 + R1VR0VR1 + R0VR1VR1 + R2VR0VR0 + R0VR2VR0 + R0VR0VR2 Since the permutations are pretty obvious, let's start over and just indicate the number of permutations to add to each first term. Remember that n is the number of V's: (see comments below) n=0: R0 n=1: R1VR0 + 1 perm n=2: (R0VR1VR1 + 2 perms) + (R0VR0VR2 + 2 perms) n=3: (R3VR0VR0VR0 + 3 perms) + (R2VR1VR0VR0 + 11 perms) + (R1VR1VR1VR0 + 3 perms) n=4 : (R4VR0VR0VR0VR0 + 4 perms) + (R3VR1VR0VR0VR0 + 19 perms) + (R2VR2VR0VR0VR0 + 9 perms) + (R2VR1VR1VR0VR0 + 29 perms) + (R1VR1VR1VR1VR0 + 4 perms) There are 19 perms of the second n=4 term as follows: (5,3) = 10 ways to place the three 0's, and for each of these, there are 2 ways to place the 1 and 3. There are 29 permutations of the 4th term as follows: There are (5,2) = 10 ways to place the two 0's, and for each of these ways, there are (3,2) = 3 ways to place the 1's. Let's now write things out through n=3 in our full notation, where recall R0 ≡ P0 and R1 = Q0a R2 = - Q0a2 R3 = Q0a3 etc Here then are our lowest terms for P (all terms in a permutation group have the same sign) P(0) = P0 P(1) = ( Q0aV P0 + 1 p) P(2) = – (Q0a2VP0VP0 + 2 p) + (Q0aVQ0aVP0 + 2 p) P(3) = (Q0a3VP0VP0VP0 + 3 p) – (Q0a2V Q0aVP0 VP0 + 11 p) + (Q0aVQ0aVQ0aVP0 + 3 p) Compute B(n) ≡ (HP– Ea0P)(n) and (HP)(n) for n = 0,1,2,3 [ It turns out that we don't really need the (HP)(n), just the B(n). ] 20. It is easiest to compute the terms for (HP– Ea0P)(n) then use (HP)(n) = (HP– Ea0P)(n) + Ea0 P(n), so using the same shorthand notation as above we find: [Note that Messiah on page 717 refers to (HP– Ea0P)(n) as simply B(n). (HP– Ea0P)(n) = Σ(n-1) [ RiV RiV ...... V Ri] // n = number of V's n=0: no terms because Ri indices would have to add up to -1 n=1: R0VR0 n=2: (R1VR0VR0 + 2 perms) n=3: (R2VR0VR0VR0 + 3 perms) + (R1VR1VR0VR0 + 5 perms) We can translate this according to our same rule R0 ≡ P0 R1 = Q0a R2 = - Q0a2 R3 = Q0a3 etc (HP– Ea0P)(0): no terms (HP– Ea0P)(1): P0VP0 (HP– Ea0P)(2): (Q0aVP0VP0 + 2 perms) (HP– Ea0P)(3): – (Q0a2VP0VP0VP0 + 3 perms) + (Q0aVQ0aV P0VP0 +5 perms) So now apply our rule to get the (HP)(n): (HP)(n) = (HP– Ea0P)(n) + Ea0 P(n) (HP)(0) = Ea0P0 (HP)(1) = P0VP0 + Ea0( Q0aV P0 + 1 p) (HP)(2) = (Q0aVP0VP0 + 2 p) + Ea0 { – (Q0a2V P0VP0 + 2 p) + (Q0aV Q0aVP0 + 2 p)} (HP)(3) = – (Q0a2VP0VP0VP0 + 3 p) + (Q0aVQ0aV P0VP0 + 5 p) + Ea0 { (Q0a3VP0VP0VP0 + 3 p) – (Q0a2V Q0aVP0 VP0 + 11 p) + (Q0aVQ0aVQ0aVP0 + 3 p) } Note: In Messiah page 717, Compute energy corrections for the non-degenerate theory. 21. We begin this section with a few simple mini-theorems: Theorem 1: tr(P) = 1 = tr(P0) for the non-degenerate case Proof: In the simplest basis, P is a diagonal matrix with everything 0 except for a single 1 on the diagonal at the location of the non-degenerate state |ψ> of interest where H|ψ> = Ea|ψ>. Thus, sum of diagonal elements = 1. Result is then basis independent since a trace. The exact same proof applies for P0 using state |φ> where H0|φ> = E0a|φ>. Note that |ψ>→ |φ> as λ → 0. Theorem 2: HP = EaP for the non-degenerate case Proof: Apply to state |ψ> and get H|ψ> = Ea|ψ>. Apply to all other states and get 0=0. Thus, claim is true as an operator equation. Theorem 3: tr(P0AB) = tr(ABP0) = tr(BP0A)= <φ|AB|φ> // traces in cyclic order where H0|φ> = E0a|φ> and |φ> is our non-degenerate state. Proof: tr(P0AB) = tr(BP0A) = tr(B|φ><φ|A) = Σi <i| B|φ><φ|A|i> = Σi <φ|A|i><i| B|φ> = <φ| AB |φ> where we use the fact that P0 = |φ><φ| and 1 = Σi|i><i| where all states of H0 are included in the sum. [ Note that we assume all these states are orthonormal.] From the first two theorems, we may conclude that tr(HP– Ea0P) = tr(HP) – tr(Ea0P) = tr(EaP) – tr(Ea0P) = Ea - Ea0 which is the total energy correction due to all powers of λ. Now recall that (HP– Ea0P) = Σn=0∞ λn (HP– Ea0P)(n) where (HP– Ea0P)(n) = Σ(n-1) [ RiV RiV ...... V Ri] ≡ B(n) Thus we can say Ea - Ea0 = Σn=1∞ λn εn = tr(HP– Ea0P) = tr(Σn=0∞ λn (HP– Ea0P)(n)) = Σn=0∞ λn tr(HP– Ea0P)(n) and thus our nth order energy correction εn is simply given by εn = tr[ (HP– Ea0P)(n)] = <φ| (HP– Ea0P)'(n) |φ> or shall we say εn = tr[B(n)] = <φ|B'(n)|φ> // |φ> must be unit normalized The left equalities follow from the previous equation. The right equalities follows from Theorem 3 where we mean by the notation B'(n) that we have "knocked out" one factor of P0 in each term in B(n) [which then let's us use Theorem 3] . For example, we have (HP– Ea0P)(1) = B(1) = P0VP0 Then to make B'(1), we knock out one of the P0 factors, so (HP– Ea0P)'(1) = B'(1) = VP0 Then we compute: ε1 = <φ| B'(1) |φ> = <φ| VP0 |φ> = <φ|V|φ> which is a pretty famous result, where we used P0|φ> = |φ>. The next case is a little more interesting: (HP– Ea0P)(2) = B(2) = (Q0aVP0VP0 + 2 perms) = (Q0aVP0VP0 + P0VQ0aVP0 + P0VP0VQ0a) Then (HP– Ea0P)'(2) = B'(2) = (Q0aVP0V + P0VQ0aV + VP0VQ0a) where we knocked one P0 off an end of each term. Then we have ε2 = <φ| B'(2) |φ> = <φ| Q0aVP0V + P0VQ0aV + VP0VQ0a |φ> = <φ| P0VQ0aV|φ> = <φ|VQ0aV|φ> which is another famous result. Again we used the fact that P0|φ> = |φ>, and we know that therefore Q0|φ> = (1-P0) |φ> = 0 and thus Q0a|φ> = 0 since Q0a ≡ Q0(Ea0– H0)-1Q0. So in this situation, any factor of Q0a which "touches" either <φ| or |φ> kills off that term. So only one of three terms survives. Just for fun, we can write ε2 out in more detail to show the meaning of Q0a. In these equations we use Q0 = Σk |k><k| where this sum is over all states of H0 except state |φ> ( sum over the perp space states) ε2 = <φ|VQ0aV|φ> = <φ|V Q0(Ea0– H0)-1Q0V|φ> = <φ|V Σk |k><k|(Ea0– H0)-1Q0V|φ> = Σk<φ|V |k><k|(Ea0– Ek)-1Q0V|φ> = Σk(Ea0– Ek)-1<φ|V |k><k| Q0V|φ> = Σk(Ea0– Ek)-1<φ|V |k><k| V|φ> = Σk(Ea0– Ek)-1VφkVkφ = Σk(Ea0– Ek)-1|Vφk|2 which is a more traditional form for ε2. In complicated formulas, we insert this expansion for Q0 in opportune places within each occurrence of Q0ak and then (Ea0– H0)-k becomes the number (Ea0– Ek)-k and we end up with a formula with one perp space sum for each isolated Q0ak factor. Before trying the ε3 term, let's look once more at this knock-out business. A "term" always begins and ends with a P0 or with a Q0a. So there are four types of terms: P0V....VP0 Q0a V....VP0 P0V....V Q0a Q0a V....V Q0a We know, by the way, that terms of the last type will have a P0 internal somewhere, because that is the nature of the terms in B(n). For the moment, let's ignore the fourth term type and look at the first three. In all three cases, we see the P0 we are going to knock out on one end or the other. In terms of the matrix element <φ| B'(n)|φ> [ where prime means we have done the knock out ] , knocking out an ending P0 has no effect on the matrix element, since P0|φ> = |φ> anyway. So for the first three term types, when we take the knocked-out matrix element, it is the same as the non-knocked-out matrix element, and thus we see at once that only the first of the three term types can survive because of the ending Q0a factors. So let's start over and dismiss the two central types of terms shown above, so we only need to worry about these two types of terms: P0V....VP0 Q0a V....V Q0a For the first term type we will say this tr(P0V....VP0) = <φ| V....VP0|φ> // knocking out the left P0 = <φ| V....V|φ> What happens with the other term type? Imagine it has this form Q0a VAP0BV Q0a where we know that somewhere inside is at least one factor of P0 and we have here exposed one such factor. For this term we would have tr(Q0aVAP0BVQ0a) = <φ| (BVQ0a)(Q0aVA)|φ> since tr(bP0a)= <φ|ab|φ> where the key fact here is that we have to reverse order of b and a when P0 appears out in the middle somewhere. Thus, we find that tr(Q0aVAP0BVQ0a) = <φ| BVQ0a2VA|φ> This term will be non-zero if the leftmost end of B and the rightmost end of A is a P0! Notice that it would have been a mistake to take the initial term type Q0a V....V Q0a and throw it out before doing the knockout, based on the fact that each end has a Q0a. Now let's apply this logic to our n=3 terms: B(3) = – (Q0a2VP0VP0VP0 + 3 perms) + (Q0aVQ0aV P0VP0 + 5 perms) Let's abbreviate Q0a = 1 and P0 = 0 and write out these terms symbolically as B(3) = – (12000 +01200 +00120 +00012) + (1100 + 1010 + 1001 + 0110 + 0101 + 0011) We now throw out all terms which have a 1 just one of the two ends B(3) ≈ – (01200 +00120) + (1001 + 0110) = – (P0VQ0a2VP0V + P0VP0VQ0a2V) + (P0VQ0aVQ0aVP0) + (Q0aVP0V P0VQ0a) Ignoring the last term which is of class 4, the other terms yield tr(first terms) = <φ| – (P0VQ0a2VP0V + P0VP0VQ0a2V) + (P0VQ0aVQ0aVP0)|φ> = <φ| – VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV|φ> For the last term 1001, we select one of the internal P0 to knock out, and we get tr(Q0aVP0V P0VQ0a) = <φ| VP0VQ0aQ0aV |φ> = <φ| VP0VQ0a2V |φ> Thus our total result from all four terms is this = <φ| – VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV + VP0VQ0a2V |φ> and we see that two terms cancel, giving us this final result tr(B'(3)) = <φ| – VQ0a2VP0V + VQ0aVQ0aV |φ> = <φ| VQ0aVQ0aV |φ> – <φ| VQ0a2VP0V and this (finally!) agrees with Messiah page 717 (74). We have dutifully followed the prescription of "knocking out" one end factor of P0. In retrospect, this only makes a difference with terms of the fourth type considered above. The conclusion here is that the Kato expansion method allows for a systematic evaluation of the energy correction of a degenerate state to arbitrary order. Based on our previous paragraph, the operative formula is this: εn = tr[ B(n)] = <φ|B'(n)|φ> where B(n) = Σ(n-1) [ RiV RiV ...... V Ri] and where the prime means we have knocked out a P0 in each subterm of B(n). It is a completely mechanical process to do the combinatorics to compute B(n), we then throw out non-contributing terms, and get our result as a sum of terms of the form <φ| ....|φ> wherein we shall find exactly n factors of V (corresponding to λn) , and n-1 factors of either Q0a or P0 , where these will only occupy interior positions shielded from the bra and ket by V operators. Any interior P0 gets replaced by |φ><φ| , and for calculation any interior Q0a gets replaced by itself times the perp space sum Q0 = Σk |k><k| which then allows evaluation of each term as one or more sums over k. There is no need to compute the state corrections in this method, as there is in the traditional iterative method which is presented earlier in Messiah's Chapter 16. The main power of the Kato method, however, comes in solving the degenerate problem! This method is reviewed in detail in our Round Two Resolvent notes, but here we give a summary. The Degenerate Case with Kato In the case that Ea0 is a degenerate eigenvalue of H0 things are more complicated. We end up with a sequence of "generalized" eigenvalue equations as we go to each higher order in λ. To all orders, the eigenvalue equation is this: Ha |Ea0α> = Eα Ka |Ea0α> where Ha and Ka are defined as Ha ≡ P0HPP0 Ka ≡ P0PP0 where we see the presence of our HP and P operators, for which we now have λ expansions due to all our work done above. For example, we have P = Σn=0∞ λn P(n) P(n) ≡ (2πi)-1 ∫Γa dz G0(VG0)n HP = Σn=0∞ λn (HP)(n) (HP)(n) ≡ (2πi)-1 ∫Γa dz z G0(VG0)n and we computed the integrals above to be P(n) = Σ(n) [ RiV RiV ...... V Ri] (HP– Ea0P)(n) = Σ(n-1) [ RiV RiV ...... V Ri] (HP)(n) = (HP– Ea0P)(n) + Ea0 P(n) Thus we can write Ha ≡ P0HPP0 = Σn=0∞ λn P0 (HP)(n)P0 Ka ≡ P0PP0 = Σn=0∞ λn P0 P(n)P0 Eα = Ea0 + Σn=1∞ λnεn According to my Pause for Comments in the Round Two resolvent doc, you have to think of the above equation in the sense of "through some order", not "in" a particular order. Here is possible way to do things: Ha ≡ P0HPP0 Ka ≡ P0PP0 La ≡ P0(HP– Ea0P)P0 = P0BP0 B = (HP– Ea0P) Ha = La + Ea0 P0PP0 = La + Ea0 Ka Our EV equation then becomes Ha |Ea0α> = Eα Ka |Ea0α> = (La + Ea0 Ka) |Ea0α> or La|Ea0α> = (Eα – Ea0) Ka |Ea0α> Now maybe here are better names for our operators Ka = P' = P0PP0 // P sandwiched by P0's La = B' = P0BP0 // B sandwiched by P0's Then our EV equation becomes B'|Ea0α> = (Eα – Ea0) P' |Ea0α> Now let's look at this thing through various orders. But first, let's collect our data right here: P(0) = P0 P(1) = ( Q0aV P0 + 1 p) P(2) = – (Q0a2VP0VP0 + 2 p) + (Q0aVQ0aVP0 + 2 p) P(3) = (Q0a3VP0VP0VP0 + 3 p) – (Q0a2V Q0aVP0 VP0 + 11 p) + (Q0aVQ0aVQ0aVP0 + 3 p) B(0): 0 B(1): P0VP0 B(2): (Q0aVP0VP0 + 2 perms) B(3): – (Q0a2VP0VP0VP0 + 3 perms) + (Q0aVQ0aV P0VP0 +2 perms) We can do some precomputation here with our usual rules for throwing things out: P'(0) = P0 P'(1) = 0 P'(2) = – (P0V Q0a2VP0 ) P'(3) = (P0VQ0a3VP0VP0 + P0VP0VQ0a3VP0) – (P0VQ0a2V Q0aV P0 + P0VQ0aV Q0a2V P0) In the above, we deleted all terms with a Q0 on either end. Similarly we get B'(0): 0 B'(1): P0VP0 B'(2): (P0VQ0aVP0) B'(3): – (P0VQ0a2VP0VP0 + P0VP0VQ0a2VP0) + (P0VQ0aVQ0aV P0) Then we can just write : P' = P0 - λ2 (P0V Q0a2VP0 ) + λ3 { (P0VQ0a3VP0VP0 + P0VP0VQ0a3VP0) – (P0VQ0a2V Q0aV P0 + P0VQ0aV Q0a2V P0)} B' = λ P0VP0 + λ2P0VQ0aVP0 + λ3 { – (P0VQ0a2VP0VP0 + P0VP0VQ0a2VP0) + (P0VQ0aVQ0aV P0) } B'|Ea0α> = (Eα – Ea0) P' |Ea0α> The Level 1 Problem. I will here install B' through order λ1 on the LHS. Since (Eα – Ea0) is itself order λ, I will only install P' through order λ0. Thus: B'|Ea0α> = (Eα – Ea0) P' |Ea0α> λ P0VP0'|Ea0α> = (Eα – Ea0) P0 |Ea0α> Identify the energy difference with λε1 and do the usual so this becomes V|Ea0α>= ε1|Ea0α> which is the traditional result. The Level 2 Problem. I will here install B' through order λ2 on the LHS. Since (Eα – Ea0) is itself order λ, I will only install P' through order λ1. Thus: B'|Ea0α> = (Eα – Ea0) P' |Ea0α> { λ P0VP0 + λ2P0VQ0aVP0} |Ea0α> = (Eα – Ea0) { P0 }|Ea0α> { λ V + λ2VQ0aV} |Ea0α> = (Eα – Ea0)|Ea0α> { λ V + λ2VQ0aV} |Ea0α> = (λε1 + λ2ε2 + λ3ε3 + ....) |Ea0α> This is the equation we would solve to find exact eigenvalues (Eα – Ea0) as f(λ), which we could Taylor expand f(λ) as shown. We would also find the exact eigenkets |Ea0α> for the above problem. Now suppose we try to write the above single equation as two equations by "matching powers of λ". I think the right thing to say is this: λV |Ea0α> = λε1|Ea0α> + order(λ2) and higher λ2VQ0aV |Ea0α> = λ2ε2|Ea0α> + order(λ3) and higher So neither of these equations is an exact EV equation. Only the full equation written above is the real EV equation. BUT, our second equation here is an approximate EV equation if we throw out order λ3 and above. Similarly, the first equation is an approximate EV equation of we through out order λ2 and above. These two approximate EV equations of course have different solutions |Ea0α> ! If we are interested in computing the eigenenergies only, we can deal with these approximate equations, but if we want to know that true states |Ea0α>, we have to do the full equation, which with λ = 1 we write as { V + VQ0aV} |Ea0α> =ΔEα |Ea0α> ΔEα = (Eα – Ea0) I would call this a well-defined eigenvalue problem that one would actually solve for eigenvalues and eigenkets. In particular: det(V + VQ0aV - ΔEα1) = 0 gives the eigenenergies. It is not clear to me how the eigenvalues so obtained would compare to the sum of the eigenvalues of the two approximate EV equations shown above. Would we find that ΔEα ≈ λε1 + λ2ε2 ? Another view of the Level 2 problem. Suppose we solve the Level 1 problem V|Ea0α>= ε1|Ea0α>, and suppose we find that all N eigenvalues are the same, so that we still have N degenerate kets. This would be the case for example if V were a multiple of the identity matrix. In this case, any set of kets |Ea0α> in the space Ea0 would be eigenkets of V. And in this radical case, VQ0aV = k Q0a and this would vanish against |Ea0α> and then our whole pyramid collapses into nothing. Now suppose instead we solve the Level 1 problem and find that at least one subset of the kets are still degenerate, so we still have some M degenerate kets (perhaps several sets, but focus on one set). Of course M < N now. How would this situation affect our Level 2 problem as stated above? { V + VQ0aV} |Ea0α> =ΔEα |Ea0α> ΔEα = (Eα – Ea0) If we now restrict |Ea0α> to this M dimensional subspace, I think we can say that V|Ea0α>= ε1|Ea0α> and in fact V = ε1 as an operator on this subspace Then our full Level 2 problem says: { ε1 + VQ0aV} |Ea0α> =ΔEα |Ea0α> (VQ0aV) |Ea0α> = (Eα – Ea0 - ε1) |Ea0α> = Ki |Ea0α> // See Messiah p 700 (39) This is now an exact MxM eigenvalue problem which we could then solve for M eigenvalues Ki. In the last equation above, ε1 is the common eigenvalue of all M states in the Level 1 problem, it is a single well-defined real number. Our secular equation is then det[VQ0aV – (Eα – Ea0 - ε1) 1] = 0 which gives us a polynomial of degree M in the quantity (Eα – Ea0 - ε1). We might then find that even after Level 2, we still have a degenerate subspace of some dimension J < M < N. Then I think we would look at the Level 3 problem. See below. The Level 3 Problem. I will here install B' through order λ3 on the LHS. Since (Eα – Ea0) is itself order λ, I will only install P' through order λ2. Thus: P' = P0 - λ2 (P0V Q0a2VP0 ) B' = λ P0VP0 + λ2P0VQ0aVP0 + λ3 { – (P0VQ0a2VP0VP0 + P0VP0VQ0a2VP0) + (P0VQ0aVQ0aV P0) } B'|Ea0α> = (Eα – Ea0) P' |Ea0α> Let's pre-remove the P0 on each end of everything P' = 1- λ2 (V Q0a2V) B' = λ V + λ2VQ0aV + λ3 { – (VQ0a2VP0V + VP0VQ0a2V) + (VQ0aVQ0aV ) } And next, set λ = 1 and do out the signs: (but remember each V is order λ) P' = 1- V Q0a2V B' = V + VQ0aV –VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV } B'|Ea0α> = (Eα – Ea0) P' |Ea0α> { V + VQ0aV –VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV }|Ea0α> = (Eα – Ea0){ 1- VQ0a2V}|Ea0α> Now this equation maintains the "generalized eigenvalue problem form", unlike our Level 1 problem. I don't know how to solve this, or whether { 1- VQ0a2V} has an inverse (probably not). But in principle this kind of problem can be solved and we would end up with eigenvalues for (Eα – Ea0) and eigenkets of the form |Ea0α>. Imagine the λ's still in place. We then have an equation of this form: { Aλ + Bλ2 + Cλ3 + ....} |Ea0α> = (Eα – Ea0){ 1+λ2D .....}|Ea0α> Since (Eα – Ea0) ~ λ, we see that "the next terms" on each side would be order λ4. Suppose we expanded the eigenvalue in a power series in the usual way, so then we have { Aλ + Bλ2 + Cλ3 + ....} |Ea0α> = [ λε1 + λ2ε2 + λ3ε3 + ....] { 1+λ2D .....}|Ea0α> Notice that the term λ3ε3 { 1} |Ea0α> makes an order λ3 contribution on the RHS, so it's presence is felt through this order. So I think I want to claim that if you could solve this equation, you would have a good estimate for the energy correction through λ3 order. It is not totally obvious how you make this claim. Now let's try our little test question again. Suppose we have a subspace of dimension J < M < N which is still degenerate after our level 2 processing. Then in the subspace of the J's, we have { V + VQ0aV} |Ea0α> =ΔEα,2 |Ea0α> where ΔEα,2 is the same number for all J eigenvectors in this space. Thus, if we restrict to the J space, we should be able to replace V + VQ0aV with the number ΔEα,2 that we found by solving the level 2 problem. This is a unique number for this subspace. Then our Level 3 equation becomes: { V + VQ0aV –VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV }|Ea0α> = (Eα – Ea0){ 1- VQ0a2V}|Ea0α> { ΔEα,2 –VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV }|Ea0α> = (Eα – Ea0){ 1- VQ0a2V}|Ea0α> Perhaps this makes the equation slightly simpler. Could also write as ΔEα,2|Ea0α> + V{ –Q0a2VP0 – P0VQ0a2 + Q0aVQ0a}V |Ea0α> = (Eα – Ea0){ 1- VQ0a2V}|Ea0α> I guess I really need to look at Kato someday to see how this is handled. Even without solving it, we could at least find (Eα – Ea0) through third order using our secular equation: det [ ΔEα,2 –VQ0a2VP0V – VP0VQ0a2V + VQ0aVQ0aV – (Eα – Ea0){ 1- VQ0a2V} ] = 0 This would still be a polynomial of degree J in the quantity (Eα – Ea0) . In theory, we can write out our EV to any order we want and try to solve it. We can always write the secular equation in any order, although the number of terms is going to increase fast. Kato has at least given us a framework in which we can attempt to solve nasty degeneracy problems, and in theory we could do it through perhaps 4th or 5th order without breaking the bank.