Kato Round One

Kato Round One Notes PhL 1.8.09 This is the first set of notes I took on the incredibly complicated Kato sections 15,16,17 in Messiah's Chapter 16. I now call this set of notes "round one" because I then did two more "rounds" and then finally some meta notes. There is so much sub-detail in this subject, you cannot see the forest for the very dense trees. It is hard to even remember what is generally going on here as you get buried in some piece of it. This round one is 30 pages long! By the way, I had just read Stakgold Chapter 2 and was interested in the Hilbert Space operator aspect of things when I started off, but that never was essential to Kato. Topic III. Explicit Forms for the Perturbation Expansion in all Orders 1 15. The Hamiltonian and its Resolvent G(z). (712). 1 16. Expansion of G(z), P and HP into power series in λV. (714). 3 Derivation of page 716 A. 4 Review of derivation of page 716 A. [ Again, skip this and see Kato's Bag in Round Three ] 8 Derive (67) and (68). 10 Derive (70) and (71). 12 Summary: 14 17. Calculation of Eigenvalues and Eigenvectors (717). 15 Case ga = 1. 15 Case ga > 1. 22 Comments on the Generalized Eigenvalue Problem. 26 Topic III. Explicit Forms for the Perturbation Expansion in all Orders 15. The Hamiltonian and its Resolvent G(z). (712). So here we go, starting 1.20.09 after having read Chap 2 Stakgold, and then having reviewed pretty completely all the perturbation notes I have made up to this point, including the notes above. Recall that you might want to solve the equation Lu - λu = f, you write it as (L-λ)u = f, and then u = (L-λ)-1f is your formal solution. In our case, I think we want to use L = H for our differential operator, the Hamiltonian of some problem. Then Hu - λu = f => u = (H-λ)-1f and we can call (H-λ)-1 by the name "resolvent" and this is what appears in (52) except λ = z. Footnote 1: The bound is ||A|| = max ||Ax||2 / ||x||2 as I now well know , so ||A|| = M. Messiah claims that the various notions of the space En pass to L2 only if we restrict to bounded = continuous linear operators. He does not say d/dx is a bounded operator (it is not), he says that the usual algebraic things we are used to doing, such as d/dx and limits and convergence can only be applied to a bounded operator. The claim is that (H-λ)-1 is bounded! Question: Is H a self-adjoint operator, or is it just symmetric in Stakgold lingo? Recall this theorem: Theorem 4: If A symmetric on all of H, then A is self-adjoint. (this is the case in En) I am betting that H acts on our entire L2 Hilbert Space, so it really is self-adjoint, and therefore also symmetric. Then here is another theorem we can have: Theorem 4: The entire spectrum of a self-adjoint operator lies on the real axis, and there is no residual spectrum, meaning that case 4 never occurs. So our H Hamiltonian will have only a point spectrum and a continuous spectrum, at most. I think the conditions of a bound state problem which we are studying right now make our eigenvalue range of interest rule out the continuous spectrum. [ but figure on page 713 shows some continuous spectrum.] Therefore, for our problem, we have only the point spectrum of H, and all other λ are in the resolvent set. In this set, B = H-λI is regular, and our definition of regular means that B-1 is bounded! If it were unbounded, we would be talking the continuous spectrum. But how would you prove directly that B-1 was bounded? max ||B-1x||2/||x||2 = some finite number. I have to admit, I am right now not sure how I would show this for our B = H-λ, even knowing that H is self-adjoint. Is H bounded in our problem? For only bound state solutions, I guess we could show that ||Hψ||/||ψ|| was less than the most negative eigenvalue which we know exists for a QM problem (the ground state). So even though d/dx is an unbounded operator in general, I suppose H is bounded for our problem. And then H* is also bounded. And then B is bounded. But I am still hard pressed to show that B-1 is bounded as Messiah claims. I will presume that B-1(λ) = his G(z) is analytic except at the eigenvalues where we know H is singular so we are in the nullspace of B. From my Chap 2 Stak notes: Note: Is Stakgold's continuous spectrum associated with the scattering type solutions in QM where the eigenvalue E is continuous? I assume this is true. Is the term eigenvalue then appropriate in this case? [ In any event, we avoid the continuous part of the spectrum when we draw our special contour Γa. ] Fact: If we write B(λ) = (A - λI), then we know that B(λ) has zeros at the eigenvalues of A. We know that for any eigenvector ψ, we have Bψ = 0 since ψ is in the nullspace of B. For the point spectra, these things really are zeros, they are isolated points. Perhaps if there are degenerate eigenvalues, we have multiple zeros. In any event, if we define the resolvent as G(λ) = B-1(λ), then the G(λ) has poles at the point spectra eigenvalues. So this Messiah set-up is the following: For the full H we have a set of energy poles E0 and corresponding eigenfunctions which we want to find. Probably as we turn off H' = V, we will see poles move together and merge in the E plane. [ yup ] Now his H is the full H, and I guess even this full H has some degeneracy left, so we can talk about Pi for the final problem to all orders, where Pi goes with the subspace of energy Ei. Page 714 A: First, recall our theorem from above: P0 f(H0) P0 = f(H0) P0 = P0 f(H0) = replace Ho in any of these by Ea0. Now just replace P0 by Pi and we can say Pi f(H) Pi = f(H) Pi = Pi f(H) = replace H in any of these by Ei [ See Round Two notes for more detail why the above is true. ] As an example, we say G(H) Pi = G(Ei) Pi where G is our resolvent, and this is page 714 A. Then just sum both sides on i and get 1 on the left, so interestingly we now have an expansion for the resolvent: G(z) = Σi (z-Ei)-1 Pi // this is page 714 (56) This says that the projector is the residue of the pole, fascinating, and we have (57). What about degeneracy and these poles? Suppose we have G(z) = P1'/(z-E1) + P1"/(z-E1) for degenerate eigenvalues where we give each one its own projector. We can write this as G(z) = (P1'+ P1")/(z-E1). And we would just combine the two projectors and write G(z) = P1/(z-E1), and this P1 would be the projector onto this little subspace of 2 eigenvalues. So I might write this: G(z) = Σs ( Σi Psi/(z-Es) ) = Σs (1/(z-Es)) Σi Psi = Σs (1/(z-Es)) Ps = Σs (Ps /(z-Es)) where Σs is a sum over multiple degenerate subspaces "s", and within each subspace we sum over the degenerate set of poles as shown, they all have energy Es so no i index needed there. We then process the result as shown above and define Ps as the entire subspace projector. So this then is how I think we should be interpreting result (56) (a subspace sum as shown above). So now all the derivation so far is justified for degenerate eigenvalues. He should have said something about that I think. The contour picking up a single eigenvalue pole is called Γi. He then allows as you could make a contour Γ which encircles some set of poles, and then the integral is the sum of the projectors for those poles, so we then have (58). I agree with B from the definition of G (since H commutes with H), and in pencil at the bottom of page 714 I derive (59) which is simple. 16. Expansion of G(z), P and HP into power series in λV. (714). This sounds very heavy duty! We first define a separate Go that goes with Ho. Page 715 A is seen to be the identity 1 = 1 if you right-mult through by the inverse of the LHS. And then B follows at once, very simple. And sure enough, we get (61) which is an operator equation which has G on both sides. If we regard these operators as being in coordinate space, say, then the operator products are spatial integrals, and this is an integral equation! This looks just like scattering theory eg BD vol 1. Obviously if λ=0, we get G = Go. Here is the usual iterative solution: G = Go(1+λV[Go(1+λV{ Go(1+λVGo} } ) = Go + λ Go VGo + λ2 Go VGoVGo + λ3 Go VGoVGoVGo so yes G = Σn=0 λn Go(VGo)n At this point, he suddenly starts talking about an operator P without defining it. But we are referred to Section 8 which does have such a P. This is the section on the degenerate theory stuff. P is the projector onto the union of subspaces i=1,2....a which subspaces become Ea0 when λ=0. When λ=0, the projector P approaches P0, the projector onto Ea0 which has degeneracy g1 + g2 + .... + ga. We are supposed to imagine a contour that captures all these eigenvalues both when λ=0 and when λ = small where they have moved a little. This contour is called Γa. So page 715 C is our version of (58). OK, now take C and insert into it the (62) expansion for G. The integral over Go is of course the projector with λ=0 and that already has the name P0. Now if we restrict our interest to subspace Ea0, we can say that G0 = 1/(z-Ea0) so not an operator, and then we can say that Go(VGo)n = Vn G0n+1 and yes, this has a pole of order n+1 at z = Ea0. Suppose we wrote X = (1/2πi) dz f(z)/(z-zo)n+1 . The residue theorem then says X = 1/n! * ∂nf(zo) so I guess you could call X "the residue of the multiple pole". Now he is going to do a Laurent series, but I am not sure how he is going to apply this, not sure what f(z) is. [ But the idea of Laurent is that you include some negative powers. Really below we are just going to do a regular Taylor series! ] ___________________________________________________________________________________ Derivation of page 716 A. [ Skip the raw notes here, see summary about 5 below! This whole subject is summarized in the Round Three notes, section "Kato's Bag of Tricks". I recommend looking there and skipping here unless there is some particular detail of interest. ] This I think is the key equation of this whole section, but Messiah only gives a tiny hint as to where it comes from. I could not derive it quickly in a half hour of scratch paper, so now have to be more careful. He draws our attention to (56) which I can write like this: (the λ=0 limit of (56), that is to say) G0(z) = P0/(z-E0) + Σi≠0 Pi/(z-Ei) where these other "i" things are for the other (possibly degenerate) subspaces that make up the full spectrum of Ho when λ=0. The union of all these spaces is the perp space to Ea0. I refer to Ea0 as E0 (apologies). Notice that we have already produced the first isolated term in 716A by starting in this manner. So we want now to show this: Σi≠0 Pi/(z-Ei) = Σk=1∞ (-1)k-1 (z-E0)k-1 Q0ak First, notice that these sums are not the same. The sum on i goes over the other subspaces of Ho , while the k sum is some kind of Laurent series thing. So we now recall from earlier that Q0a ≡ Q0(E0 - H0)-1Q0 = Q0(E0 - H0)-1 = (E0 - H0)-1Q0 // subject to warning! [ The warning is that the above equalities are only true if you restrict to action on (Ea0) ] where of course Q0 = 1 - P0 = Σi≠0 Pi = sum of perp space projectors. The Laurent power series ingredient is something like this f(z) = Σk=0,∞ (z-E0)k f(k)(E0)/ k! but I don't yet know what function f(z) we are going to apply this to. Let's rewrite our desired result Σi≠0 Pi/(z-Ei) = Σk=1∞ (-1)k-1 (z-E0)k-1 Q0ak = - 1/(z-E0) [ Σk=1∞ (z-E0)k (-Q0a)k ] where now we have exposed something that at least "fits" the above power series in some sense. So, where do we begin? (a common question!) Note Added 1.31.09: skip over all this scratch paper like Plan A stuff and just read the Review which follows! Plan A. One approach would be to try to identify f(z). We might write the above sum as [ Σk=1∞ (z-E0)k (-Q0a)k ] = - 1 + Σk=0∞ (z-E0)k (-Q0a)k Then we might try to identify f(z) = Σk=0,∞ (z-E0)k f(k)(E0)/ k! = Σk=0∞ (z-E0)k (-Q0a)k which would imply that f(k)(E0)/ k! = (-Q0a)k = [Q0(H0 - E0)-1Q0]k Can we reverse engineer this somehow? Maybe start with k = 1, f(1)(E0)/ k! = (-Q0a) = [Q0(H0 - E0)-1Q0] = df/dz|z=E0 / k! Suppose f(1)(z) = [Q0(H0 - z)-1Q0] . That would agree with the above for k = 1. Then scratch shows we would have f(k)(z) = (k-1)! Qo (H0-z)-kQo Now what are our algebra rules for playing with these Q0 packed operators? For example, consider: [Q0(H0 - E0)-1Q0]2 = Q0(H0 - E0)-1Q0 Q0(H0 - E0)-1Q0 = Q0(H0 - E0)-1Q0(H0 - E0)-1Q0 Now we can use the fact that Q0 is all of unity except subspace 0, so with crude notation write Q0 = Σk≠0 |k><k| If we insert this three times on the right above, we find that (this is trivial to show, k are eigenstates of H0 so you just write it out). [Q0(H0 - E0)-1Q0]2 = Σk≠0 |k>(Ek- E0)2<k| Now we wonder if this might the same as [Q0(H0 - E0)-2Q0] . If we do the same thing here, we do in fact get the same result. You can just see this. What is the general theorem we are showing here? Well, we already know this theorem Q0 f(H0) Q0 = f(H0) Q0 = Q0 f(H0) = Σk≠0 |k> f(Ek) <k| because we get <k| f(Ho)|k'> = δk,k' f(Ek). So consider [Q0 f(H0) Q0]k = Q0 f(H0) Q0 Q0 f(H0) Q0 Q0 f(H0) Q0.... Q0 f(H0) Q0 First, we replace each Q0 Q0 by a single Q0 then when we expand each Q0, we get the <k| f(Ho)|k'> = δk,k' f(Ek). happening at each point and this we get = Σk,k',k"... |k><k| f(H0) |k'><k'| f(H0) |k"><k"| ...... = Σk,k',k"... |k><k| f(Ek') |k'><k'| f(Ek") |k"><k"| ...... = Σk,k',k"... |k> δk,k' f(Ek) δk',k" f(Ek")<k"| ...... = Σk |k> f(Ek)k <k| Therefore, we may conclude that [Q0 f(H0)k Q0] will be exactly this same sum. Thus we find that [Q0 f(H0) Q0]k = [Q0 f(H0)k Q0] and this interestingly says that we can do this: Q0 f(H0) Q0 Q0 f(H0) Q0 Q0 f(H0) Q0.... Q0 f(H0) Q0 = Q0 f(H0) Q0 f(H0) Q0 f(H0) Q0.... Q0 f(H0) Q0 = Q0 f(H0) 1 f(H0) 1 f(H0) 1.... 1 f(H0) Q0 How do I explain this? Well, we start at the right and say |φ> ≡ Qo |ψ> projects out only that portion of ψ that lies in the perp space to Ea0. Then we know that Ho |φ> is linear combination of perp space basis functions. That is |ψ> = Σi ai ui |φ> = Σi≠0 ai ui Ho |φ> = Σi≠0 ai Ho ui = Σi≠0 ai Ei ui so Ho |φ> stays in the perp space. Then of course f(H0) |φ> is also in the perp space, and that is why all those internal Q0 are unnecessary. Even the left one is unnecessary as we already know from long ago. So here is a little way to say this: Q0A(Ho) Q0 B(Ho) Q0C(Ho) Q0D(Ho) Q0 = A(Ho) B(Ho) C(Ho) D(Ho) Q0 So the main point so far is this: (-Q0a)k = [Q0(H0 - E0)-1Q0]k = Q0(H0 - E0)-kQ0 and so we at least now have some simplification of our "desired result" Σi≠0 Pi/(z-Ei) = Σk=1∞ (-1)k-1 (z-E0)k-1 Q0ak = – Σk=1∞ (z-E0)k-1 (-Q0a)k = – Σn=1∞ (z-E0)n-1 Q0(H0 - E0)-nQ0 Now suppose we just expand both sides. For LHS get LHS = Σk≠0 Pk/(z-Ek) = Σk≠0 | k> (z-Ek)-1<k| Meanwhile, we know that Q0(H0 - E0)-nQ0 = Σk≠0 | k> (Ek - E0)-n <k| Thus, we seem to have Σk≠0 | k> (z-Ek)-1<k| = Σk≠0 | k> { – Σn=1∞ (z-E0)n-1 (Ek - E0)-n } <k| And this certainly would be true if we could show that (z-Ek)-1 = – Σn=1∞ (z-E0)n-1 (Ek - E0)-n = – (z-E0)-1 Σn=1∞ (Ek - E0)-n (z-E0)n which can be rewritten: (z-E0)/ (z-Ek) = – Σn=1∞ (Ek - E0)-n (z-E0)n Is this a valid Laurent series? Let f(z) = (z-E0)/ (z-Ek) f(1)(z) = [(z-Ek) – (z-E0)] / (z-Ek)2 = (Eo- Ek) (z-Ek)-2 f(2)(z) = (Eo- Ek) (-2) (z-Ek)-3 f(3)(z) = (Eo- Ek) (-2)(-3) (z-Ek)-4 f(4)(z) = (Eo- Ek) (-2)(-3)(-4) (z-Ek)-5 f(n)(z) = (Eo- Ek) (-1)n+1 n! (z-Ek)-(n+1) f(n)(E0) = (Eo- Ek) (-1)n+1 n! (Eo-Ek)-(n+1) = (-1)n+1 n! (Eo-Ek)-n n>0 f(0)(E0) = f(E0) = (E0-E0)/ (Eo-Ek) = 0 n=0 Then we would expect our Laurent expansion of f(z) to be f(z) = Σn=0,∞ (z-E0)n f(n)(E0)/ n! (z-E0)/ (z-Ek) = Σn=1,∞ (z-E0)n * (-1)n+1 n! (Eo-Ek)-n /n! = Σn=1,∞ (-1)n+1 (z-E0)n (Eo-Ek)-n = – Σn=1,∞ (Ek-E0)-n (z-E0)n and lo and behold this is in fact our desired Laurent series. Review of derivation of page 716 A. [ Again, skip this and see Kato's Bag in Round Three ] (1) Our first step is to do a Taylor series expansion of this function f(z) = (z-E0)/(z-Ek) We know the Taylor series, expanding around the point z-Eo, is this: f(z) = Σn=0,∞ (z-E0)n f(n)(E0)/ n! By direct computation, we find that f(n)(E0) = (-1)n+1 n! (Eo-Ek)-n for n = 1,2,3.... = 0 for n=0 Thus, our Taylor series is this: (z-E0)/(z-Ek) = – Σn=1,∞ (Ek-E0)-n (z-E0)n which we then rewrite as (z-Ek)-1 = – (z-E0)-1 Σn=1∞ (Ek - E0)-n (z-E0)n (*) We are expanding around the point z = E0. Exactly at this point, the sum on the RHS = 0 and this somehow cancels the pole factor to give the finite factor on the left. More generally we have merely shown this fact (z-a)/(z-b) = – Σn=1,∞ (b-a)-n (z-a)n or 1/(z-b) = – Σn=1,∞ (b-a)-n (z-a)n-1 which is how you would expand the function on the left about the point z = a. (2) We now take our result (*) derived above (z-Ek)-1 = – (z-E0)-1 Σn=1∞ (Ek - E0)-n (z-E0)n (*) and we use this as a weight function in a projection sum where k≠0 means we are summing over all kets in our Hilbert Space except those in the Ea0eigenmanifold. Thus: Σk≠0 | k> (z-Ek)-1<k| = Σk≠0 | k> { – Σn=1∞ (z-E0)n-1 (Ek - E0)-n } <k| = – Σn=1∞ (z-E0)n-1 Σk≠0 | k> (Ek - E0)-n <k| (3) Next, we show these facts to be true (if this is unclear, see notes above in Plan A ) (-Q0a)n = [Q0(H0 - E0)-1Q0]n = Q0(H0 - E0)-nQ0 = Σk≠0 | k> (Ek - E0)-n <k| Thus, we arrive at this point: Σk≠0 | k> (z-Ek)-1<k| = – Σn=1∞ (z-E0)n-1(-Q0a)n = Σn=1∞ (-1)n+1(z-E0)n-1Q0an Meanwhile, for the LHS, we know that Σk≠0 | k> (z-Ek)-1<k| = Σk≠0 Pk / (z-Ek) but according to (56) page 714 we know that G0(z) = Σk Pk / (z-Ek) = P0 / (z-E0) + Σk≠0 Pk / (z-Ek) Therefore we have Σk≠0 | k> (z-Ek)-1<k| = G0(z) – P0 / (z-E0) and then we have finally G0(z) – P0 / (z-E0) = Σn=1∞ (-1)n+1(z-E0)n-1Q0an or G0(z) = P0 / (z-E0) + Σn=1∞ (-1)n+1(z-E0)n-1Q0an // which verifies page 716 A → This derivation is a pretty big jump for the reader I think, since it took me about 3 hours to do it. ___________________________________________________________________________________ OK, equations (65) and (66) are obvious given 716 A, so we are now at the pencil line on page 716. Derive (67) and (68). First we insert our new expansion (66) twice into (64): G0 = Σk=0 (-1)k+1(z-E0)k-1Sk // our new expansion Then we have from (68) A(n) = (1/2πi) dz { Σk=0 (-1)k+1(z-E0)k-1Sk }V {Σk'=0 (-1)k'+1(z-E0)k'-1Sk'} ..... V {Σk"=0 (-1)k"+1(z-E0)k"-1Sk"} where we have n+1 occurrences of { } and n occurrences of V. Now, how do we do this contour integral? Recall that it surrounds all the poles for the Ea0 subspace both before and after they move a little which some λ is applies. But of course when λ=0, the poles are all at the same place, namely, z = E0. This is the case we are now considering. Let's schematize the above equation like this: A(n) = (1/2πi) dy { a/y + b + cy + dy2 +...} V{ a/y + b + cy + dy2 +...} V .... V { a/y + b + cy + dy2 +...} Each combination of terms for all the sums gives a certain power of y times a constant. How do we list off the terms which have the form C/y? There are lots of them! Here are some of them a/y * b * b * b * b.....b a/y * cy * a/y * b * b.....b a/y * dy2 * a/y * a/y * b.....b So now that we see the point, let's reorder our integral: A(n) = Σk1=0 Σk2=0 .... Σkn+1=0 (1/2πi) dz (-1)k1+k2+..kn+1 + (n+1) (z-E0)k1+k2 +...+kn+1 - (n+1){Sk1} V {Sk2}V .... V Skn+1 Now, where is the simple pole? It requires that k1 + k2 + ..... + kn+1 - n -1 = - 1 or k1 + k2 + ..... + kn+1 = n // which is result (68) For any set of ki satisfying this rule, the (-1) power is just (-1), so we get A(n) = – Σk1=0 Σk2=0 .... Σkn+1=0 (rule) (1/2πi) dz (z-Eo)-1 {Sk1} V {Sk2}V .... V Skn+1 = – Σk1,k2...kn+1(rule) {Sk1} V {Sk2}V .... V{Skn+1} // which is result (67) and this is result (67). He writes Σk1,k2...kn+1(rule) = Σ(k,n) as a compact notation (I modify it) Now looking back at (63) we have finally achieved our goal of having "an expansion of P in powers of λ", where P is defined above in bold: it is the projector onto all the subspaces into which Ea0 breaks when we turn on λ. So we have P = P0 – Σn=1 λn Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1} Next, we want to look at various terms in this sum! The λ=0 term is just P0. We then have P,λ=0 = P0 P,λ=1 = – λ Σ(k,1) {Sk1} V {Sk2} = -λ [ S0VS1 + S1VS0] = + λ [ P0 V Q0a + Q0a V P0 ] P,λ=2 = -λ2 Σ(k,2) {Sk1} V {Sk2}V {Sk2} = -λ2 [ S0 V S0 V S2 +S0 V S2 V S0 + S2 V S0 V S0 + S0 V S1 V S1 + S1 V S0 V S1 + S1 V S1 V S0 ] = -λ2 [P0 V P0 V Q0a2 + P0 V Q0a2 V P0 + Q0a2 V P0 V P0 - P0 V Q0a V Q0a - Q0a V P0 V Q0a - Q0a V Q0a V P0 ] = +λ2 [ (P0 V Q0a V Q0a + Q0a V P0 V Q0a + Q0a V Q0a V P0) -( P0 V P0 V Q0a2 + P0 V Q0a2 V P0 + Q0a2 V P0 V P0) ] In this way, I see I could write the terms for each order in λ, very clear. Derive (70) and (71). We are told to "repeat the process we did for P for HP". We are supposed to start with (59), but not clear where to go next. This would give page 715 C with extra factor of z. If we replaced P by HP in (63) we could get (64) with an extra factor of z. Our equation for the A(n) object would then have this extra z in it: A'(n) = (1/2πi) dz z { Σk=0 (-1)k+1(z-E0)k-1Sk }V {Σk'=0 (-1)k'+1(z-E0)k'-1Sk'} ..... V {Σk"=0 (-1)k"+1(z-E0)k"-1Sk"} so now the coefficient of each power of (z-E0) is no longer a constant. I think we would then pick up both the single poles and the double poles! Here is the regrouping A'(n) = Σk1=0 Σk2=0 .... Σkn+1=0 (1/2πi) dz z (-1)k1+k2+..kn+1 + (n+1) (z-E0)k1+k2 +...+kn+1 - (n+1){Sk1} V {Sk2}V .... V Skn+1 The contribution from the single pole is the same as last time but we have an extra factor of E0 from the extra z (which matches change in dimensions going from P to HP). So this set of contributions would yield A'(n) = – E0 Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1} The other set of contributions will come from double poles due to the z in the numerator. This requires then that (z-E0)k1+k2 +...+kn+1 - (n+1) = (z-E0)-2 => k1 + k2 + ..... + kn+1 - n -1 = - 2 => k1 + k2 + ..... + kn+1 = n - 1 In this case the (-1) power is just +1. So then we have A'(n) = Σ(k,n-1) (1/2πi) dz z (z-E0)-2{Sk1} V {Sk2}V .... V {Skn+1} The residue rule creates no new factors, and we now get A'(n) = Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} So here is our complete result at this point: HP = H0P0 + Σn=1 λn A'(n) where [HP]0 is what you have when λ = 0, so that must be H0P0. And then A'(n) has these two terms: A'(n) = – E0 Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1} + Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} What about dimensions? We know that Q0a is 1/E but P0 and Q0 have no dimensions. Now consider Sk = -P0 k = 0 dim = 1 = Qoak k > 0 dim = E-k So we can say dim(Sk) = E-k for all k. Then our terms are like this: E1 * E-(sum of k's)En + E-(sum of k's)En = E1 E-nEn + E-(n-1)En // En from the n factors of V = E + E = E Looks good. Now we have for LHS of (70) (H-Eo)P = HP - E0P = { H0P0 + Σn=1 λn A'(n)} – Eo{ P0 + Σn=1 λn A(n)} = (H0 - Eo) P0 + Σn=1 λn A'(n) – E0 Σn=1 λn A(n) = Σn=1 λn [A'(n) - Eo A(n)] // since (H0 - Eo) P0 = 0 = Σn=1 λn [ [– E0 Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1} + Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1}] - [ – E0 Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1}] ] = Σn=1 λn Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} // first and third terms cancel = Σn=1 λn B(n) // verifies (71) and (70) I probably could have gotten there faster, but at least we have the results. Now as before, let's write out the first few terms, (H-Eo)P = Σn=1 λn Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} n = 1: = λ [ {S0} V {S0} ] = λ P0 V P0 n=2: = λ2 [ {S1} V {S0}V {S0} + permutations = λ2 [Q0a V P0 V P0 + P0 V Q0a V P0 + P0 V P0 V Q0a ] and this verifies (72). Summary: This was a bunbuster little section, what did we actually do here? We expanded the following four operators in powers of λ: G P HP (H-Eo)P G = Σn=0 Go(VGo)n P = P0 – Σn=1 λn Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1} HP = H0P0 + Σn=1 λn A'(n) where A'(n) = – E0 Σ(k,n) {Sk1} V {Sk2}V .... V{Skn+1} + Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} (H-Eo)P = Σn=1 λn Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} = Σn=1 λn B(n) where Sk = -P0 k = 0 = Qoak k > 0 where G(z) is the full resolvent = 1/(z-H). H is the full Hamiltonian H = H0 + λV P is the projector that grabs all states in the Ea0 spectrum when λ = 0 and also grabs them after they move when λ > 0, where you have to write some E1 + E2 + .. En as on page 698. To the lowest few orders we found that: P = P0 + λ [ P0 V Q0a + Q0a V P0 ] + λ2 [ (P0 V Q0a V Q0a + Q0a V P0 V Q0a + Q0a V Q0a V P0) –( P0 V P0 V Q0a2 + P0 V Q0a2 V P0 + Q0a2 V P0 V P0) ] (H-Eo)P = 0 + λ P0 V P0 + λ2 [Q0a V P0 V P0 + P0 V Q0a V P0 + P0 V P0 V Q0a ] I suspect that this last item somehow replaces that famous set of equations Schiff 31.4. And this is a pure operator expansion, there are no wavefunctions anywhere. 17. Calculation of Eigenvalues and Eigenvectors (717). Case ga = 1. Here our entire Ea0 space has only one state (because that is what ga= 1 says), and it will move somehow with λ. It ends up in space Ea which has dimension 1. [ Note: All of Messiah's mysterious statements in the opening paragraph of this section are verified and explained in my Kato Meta notes, section 2B and environs. ] Verify the following sentence: " In this case, the eigenvector of H is P|0> " Proof: We know that H|ψi> = Ei|ψi> for our entire problem in H. Consider the expansion |ψi> = |0; ψi> + λ |1; ψi> +λ2 |2; ψi> ... where we show the various state corrections at various λn levels. Since the |ψi> form a complete basis in H (not just in Ea or some other subset of H) , we can certainly expand |0; ψi> on this basis: |0; ψi> = Σj aj |ψj> Now suppose some specific |ψ0> is our non-degenerate eigenstate in question and it has a projector P which we might really call P(0) since it only passes state |ψ0> . Then we know that P|0; ψ0> = Σj aj P |ψj> = Σj aj δ0,j |ψj> = a0 |ψ0> Therefore, since |ψ0> is an eigenstate of H, P|0; ψ0> must also be an eigenstate of H. The meaning of the notation |0> is this |0; ψ0> for our state of interest. It is just a confusing coincidence here that we have used the numeral 0 to label our state ψ0 and then we talk about the level 0 component of this state. Now if we ignore normalization, we have a very detailed solution to our problem: |ψ0> = P |0> = P0 |0> + λ [ P0 V Q0a + Q0a V P0 ] |0> + λ2 [ (P0 V Q0a V Q0a + Q0a V P0 V Q0a + Q0a V Q0a V P0) –( P0 V P0 V Q0a2 + P0 V Q0a2 V P0 + Q0a2 V P0 V P0) ] |0> We can simplify slightly using these two rules: P0 |0> = |0> Q0 |0> = 0 = Q0a |0> the perp space operator |ψo> = P |0> = |0> + λ [Q0a V ] |0> + λ2 [ (Q0a V Q0a V) –(P0 V Q0a2 V + Q0a2 V P0 V) ] |0> = |0> + λ Q0a V |0> + λ2 [(Q0a V Q0a V) –(P0 V Q0a2 V + Q0a2 V P0 V) ] |0> | ... This implies that, in Messiah notation, |1> = Q0a V |0> |2> = [(Q0a V Q0a V) –(P0 V Q0a2 V + Q0a2 V P0 V) ] |0> [ But this is the "Kato series" which has the extra renormalization terms like the middle one seen here. See Kato Meta for an explanation of all this stuff. ] _____________________________________________________________________________ Explain the extra middle term. The first equation appears as (14) on page 680. The second appears as (27) on page 694, but I don't see that we agree. What appears in (27) are the first and third terms, if we say P0 = |0><0| . Is there some reason that the middle term is 0? P0 V Q0a2 V |0> =?= 0 Let's try to compute this thing: Use 1 = |0><0| + Σk |k><k| where second sum is over the perp space. P0 V Q0a2 V|0> = P0 V Q0a2 { |0><0| + Σk |k><k| }V |0> = P0 V Q0a2 { Σk |k><k| }V |0> // because Q0 kills all but the perp space = P0 V Q0 (Ea0- H0)-2 { Σk |k><k| }V |0> = Σk P0 V Q0 (Ea0- Ek)-2 { |k><k| }V |0> = Σk (Ea0- Ek)-2 P0 V { |k><k| }V |0> = Σk (Ea0- Ek)-2 P0 { |0><0| + Σk' |k'><k'| }V { |k><k| }V |0> = Σk (Ea0- Ek)-2 P0 { |0><0| }V { |k><k| }V |0> // because P0 kills the perp space = Σk (Ea0- Ek)-2 { |0><0| }V { |k><k| }V |0> = Σk (Ea0- Ek)-2 { |0>VokVk0 = Σk (Ea0- Ek)-2 |Vk0|2 |0> = (a number) |0> In passing, we note that ε2,n = < 0; n, ε1| V |1; n, ε1> = Σk (Ea0 - Ek)-1 |Vkn|2, so our "number" is not equal to this energy correction. So OK, I think I see the light here. This extra term that does not appear in (27) just gives a multiple of |0> and we would interpret this term as <0|2> using the above definition of |2>. So this is a question of normalization! If you normalize states as in (27), where you have <0|2> = 0, you don't get this kind of term. So I conclude that we are NOT now working with the normalization scheme <0|n> = 0! How might we identify terms that just give renormalization like this? Why doesn't the last term do this same thing? Let's compute it: Q0a2 V P0 V|0> = Q0a2 V P0 { |0><0| + Σk |k><k| }V |0> = Q0a2 V P0 { |0><0| }V |0> // because P0 kills off the perp space = Q0a2 V { |0><0| }V |0> = Q0a2 { |0><0| + Σk |k><k| }V { |0><0| }V |0> = Q0a2 { Σk |k><k| }V { |0><0| }V |0> = Q0(Ea0- H0)-2 { Σk |k><k| }V { |0><0| }V |0> = Σk Q0(Ea0- Ek)-2 { |k><k| }V { |0><0| }V |0> = Σk (Ea0- Ek)-2 { |k>Vk0V00 = Σk [ (Ea0- Ek)-2 Vk0V00 ] |k> ≠ (a number) * |0> !!! It seems that if you end up with a perp space sum close to the left end of things, you won't be getting a normalization situation. Let's go back to the previous case again P0 V Q0a2 V|0> No matter what we get from V Q0a2 V|0>, the left end P0 will kill everything except |0> ! Theorem: So, any term with a P0 on the left end is going to involve normalization only. So, let's go back to our expansion: |2> = [(Q0a V Q0a V) –(P0 V Q0a2 V + Q0a2 V P0 V) ] |0> I know by inspection that P0 V Q0a2 V|0> = (number) |0>, whereas I know that the other two terms can only give perp space results and cannot therefore contain any |0> component. Thus, if I want to impose the condition <0|2> = 0, I just drop this middle term. |2'> = |2> + c |0> is what our result above says, |2'> be our resolvent section result OK, let this ride for now. [ this issue is reviewed and resolved in Kato Meta Notes section 2A ] _____________________________________________________________________________ The point is that we have in principle an exact expression for the state |ψo> to any order we want, and this comes from being able to do the operator expansion for P. |ψ0> = k'|0> + λ Q0a V |0> + λ2 [Q0a V Q0a V – Q0a2 V P0 V ] |0> | + λ3 P3 |0> + ... _____________________________________________________________________________ Next mystery: " eigenenergy comes from equation HP = EaP " This is a pure operator equation. I am not use to a "pure operator eigenvector problem" since there are no vectors here. If we apply this to state |0> we get H |ψo> = Ea |ψo> so it is at least consistent. So I will allow this one tentatively. Note added 1.31.09: Well, having reread earlier parts, I think this idea is common. If you apply an operator equation to the state |0> and that produces an EV problem, as in the last paragraph, then you can make the claim quoted above. Next mystery: Tr(P) = Tr(P0) = 1 Why on earth would this be true, and what size matrices are these? Well, let's say our problem has some dimension N. If we go into the perturbed basis |ψi>, call it, then for |ψ0> the matrix P is NxN with a single 1 on the diagonal at the 0,0 entry, and all other elements are 0 (in this basis). So in this basis, Tr(P) = 1. But that is then true in any basis and is independent of basis. A similar argument for Po in the unperturbed basis says Tr(P0) = 1, so that is true as well in any basis. Fine. This would work fine even for N=∞ because there is only one non-zero diagonal element. Now apply this to our equation above which says HP = EaP, then tr(HP) = Ea ! Next, apply this to (7) to get tr(HP) - Ea0 = Σn=1λn tr(B(n)) = Ea - Ea0 Wow, we suddenly have a full expansion for the energy correction, and we learn that εn = the correction at the λn level = tr(B(n)) Next mystery: Tr(P0A) = <0|A|0> Write out the trace like so, in the unperturbed basis, Tr(P0A) = (P0A)ii = (P0)ijAji But P0 = |0><0| so (P0)ij = <i |0><0| j> Then (P0)ijAji = <i |0><0| j>Aji = <0| j>Aji<i |0> = <0| j><j|A|i> <i |0> = <0|A|0> QED Maybe this is faster Tr(P0A) = Tr(|0><0|A) = <i |0><0| A | i > = <0| A | i ><i |0> = <0|A|0> Now lets look at Ea - Ea0 = Σn=1λn tr(B(n)) = Σn=1λn tr [ (Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} ] 0 + λ tr(P0 V P0) + λ2 tr [Q0a V P0 V P0 + P0 V Q0a V P0 + P0 V P0 V Q0a ] + λ3 ... The lowest order has tr(P0 V P0) = tr(P02 V ) = tr(P0 V ) = <0 | V | 0> = ε1 // our very familiar first order result For second order we can write: ε2 = tr [Q0a V P0 V P0 + P0 V Q0a V P0 + P0 V P0 V Q0a ] tr [P0 Q0a V P0 V + P0 V Q0a V P0 + P0 V P0 V Q0a ] = tr [ Po { Q0a V P0 V + V Q0a V P0 + V P0 V Q0a } ] = <0| Q0a V P0 V + V Q0a V P0 + V P0 V Q0a| 0> = <0| V Q0a V| 0> = Σk≠0 <0| V | k> (Ek - E0)-1 <k| V| 0> // since Q0 kills off |0> Maybe I can write down the general result! We always need this tr(B(n)) = tr [ (Σ(k,n-1) {Sk1} V {Sk2}V .... V {Skn+1} ] What terms are going to survive our process above? He claims that each term in the above contains P0 somewhere. We have n+1 exponents and they have to add up to n-1. If they were all ≥ 1, the sum would end up being ≥ n+1 . So at least some of them have to have the value ki = P0 , I agree. So P0 appears somewhere in every contributing term. We then think of this term as tr(P0AB) = tr(ABP0) = tr(BP0A)= <φ|AB|φ> So imagine processing each term in this way. I call this step above "knocking out a P0". Then if we have a Q0a power on either end of a term after doing the knockout, the term is 0 ! So OK, let's now try n = 3. The exponents have to add up to n-1 = 2. tr(B(3)), 0002 = tr [ {S0}V {S0}V{S0} V {S2} + permutations] Notice that this form does not have cyclic symmetry! We cannot say for example 0002 = 2000. The reason is that 0002 means 0v0v0v2 and if you rotate this, you get 20v0v0v ≠ 2v0v0v0. On the other hand, we do have reverse-ordering symmetry, so that 0002 = 2000 since 0v0v0v2 = 2v0v0v0. Which terms contribute? 0002 no S2 = (Q0a)2 and kills against |0> 0020 yes 0200 yes // and same as 0020 by reversal 2000 no S2 = (Q0a)2 and kills against <0| So only terms with S2 on the interior can contribute! So we get tr(B(3)), 0002 class = tr [ {S0}V {S2}V{S0} V {S0} + {S0}V {S0}V{S2} V {S0} ] = 2 tr [ {S0}V {S2}V{S0} V {S0} // reverse order rule = -2 tr [ {P0}V { Q0a2}V{ P0} V { P0} = - 2 <0 | V { Q0a2}V{ P0} V |0> = - 2 <0 | V Q0a2V P0V |0> = - 2 <0 | V P0VQ0a2V |0> // if we reverse the other term We can replace any residual interior P0 with |0><0| to simplify the matrix element into a product of two simpler ones, so the above becomes = -2 <0 | V Q0a2V |0><0| V |0> This agrees with the second term in (74) except for the factor of 2. But see below! Next, consider tr(B(3)), 0011 = tr [ {S0}V {S0}V{S1} V {S1} + permutations ] 0011 no 0101 no 0110 yes 1001 yes, see below 1010 no 1100 no First, here is the 0110 term: tr [ {S0}V {S1}V{S1} V {S0} ] = tr(P0VQoaVQoaV P0) = <0| VQoaVQoaV P0 |0> = <0| VQoaVQoaV |0> and this is the first term in (74). Now consider the 1001 term tr [ {S1}V {S0}V{S0} V {S1} ] = tr(QoaVP0VP0V Qoa) = <0| VP0V Qoa QoaV|0> = <0| VP0VQoa2V |0> = <0| V Qoa2V P0V |0> where again the two forms shown are equivalent, which is obvious if you insert P0 = |φ><φ| . If we now add up all terms, we get tr(B(3)), 2000 class = - 2 <0 | V { Q0a2}V{ P0} V |0> = - 2 <0 | V Q0a2V P0V |0> tr(B(3)), 1100 class = <0| VQoaVQoaV |0> + <0| V Qoa2V P0V |0> We now see that one of the 1100 class terms cancels one of the 2000 terms and our result is then tr(B(3)) = <0| VQoaVQoaV |0> – <0| V Qoa2V P0V |0> and this agrees with Messiah page 717 (74). Now you might ask: how do you compute something like <0 | V Q0a2V |0> ? I suppose you just do it this way: <0 | V Q0a2V |0> = Σk',k"<0|V|k'><k'| Q0a2|k"><k"|V |0> But then Q0a2 = Q0(H0 - E0)-2Q0 = Σk≠0 | k> (Ek - E0)-2 <k| so then have <0 | V Q0a2V |0> = Σk',k" <0 | V|k'> <k'| [Σk≠0 | k> (Ek - E0)-2 <k|] |k"><k"|V |0> = Σk',k" Σk≠0 <0 | V|k'> δk',k (Ek - E0)-2 δk,k" <k"|V |0> = Σk≠0 <0 |V|k> (Ek - E0)-2 <k|V |0> Easier to start over: <0 | V Q0a2V |0> = Σk≠0 <0 | V | k> (Ek - E0)-2 <k| V |0> So here would be the full 3rd order energy correction: ε3 = <0| VQoaVQoaV |0> – 2 <0 | V Q0a2V |0><0| V |0> = Σk≠0 Σk'≠0 <0| V | k> (Ek - E0)-1 <k|V | k'> (Ek' - E0)-1 <k'|V |0> - 2 <0| V |0> Σk≠0 <0 | V | k> (Ek - E0)-2 <k| V |0> where the first term involves a sort of triple communication channel from our state back to itself. So OK, I think everything is well-defined here, albeit not easy to compute in practice. Now we are ready for the Real McCoy: Case ga > 1. Preliminaries. Question: What does Messiah mean by the notations Ea and Ea0 ? He never defines Ea as far as I can see, anywhere in this chapter. Here is my interpretation, where I show a "perturbation situation" I know from my general theory that ψmi = umi + Σj=1,N Bij umj + Σk≠m Bik uk that a perturbed state picks up pieces from all eigenstates, not just from eigenstates within its initial degenerate group. But for very small λ, we know that the contributions of the local degenerate group are going to be larger, and I show these contributions by solid lines. The dashed lines represent small contributions. The above picture shows my interpretations of Ea and Ea0 . On page 718 Messiah says that Ea and Ea0 "have the same number of dimensions". In my picture above that number would be 5. But we cannot claim that one of the 5 states in Ea is a linear combination only of states from Ea0 due to the dotted line contributions. Regardless, I think we can regard the 5 states of Ea as being a Hilbert Space spanned by those 5 vectors (heavy dash = state = vector). The states on the right side are eigenstates of the full H, while those on the left are eigenstates of H0. Question: What is the meaning of the various "projection operators" like P = Pa and P0 = Pa0 with respect to our picture? Here are some examples: Pa | any of the states in Eb> = 0 Pa | any of the states in Ea> = | exactly that state in Ea> Pa0 | any of the states in Eb0> = 0 Pa0 | any of the states in Ea0> = | exactly that state in Ea0> Now he makes a statement which I have a lot of trouble with, so I will quote it: "Let us suppose that any vector in Ea may be considered as the projection in Ea of a certain well defined vector of Ea0." Well, I need maybe to build up to this with some claims of my own. In the above picture, if I pick one of the Ea states and apply P0a to it, I know I will get some linear combination of the 5 states in Ea0. I could also apply P0b and get the smaller contribution from states in Eb0 . Think ψmi = umi + Σj=1,N Bij umj + Σk≠m Bik uk So, certainly if you apply P0a to a specific state in Ea , you will get a unique and specific state in Ea0. Question: what are the characteristics of an operator like P0a? We know that P0a2 = 1. It's domain is all of the Hilbert Space H0 or H. But its range is the space Ea0. I am claiming that the spaces H and H0 are the same space in effect, and we just have different base states comparing λ = 0 to λ = λ1. So we can say this: Pa0 : H → Ea0 or Pa0 : H0 → Ea0 You can see that in general Pa0 has a huge nullspace which agrees with the partial range. So Pa0 is certainly not 1-to-1 on H0, and thus Pa0-1 does not exist. But then we could consider the following restriction of Pa0: Pa0 : Ea → Ea0 Pa0-1 : Ea0 → Ea If we define this operator to be so restricted, it might be 1-to-1 . I know that going in the direction shown, for each state in Ea you get exactly one unique state in Ea0, so it is at least a function. It has no nullspace as so defined, so I think the inverse mapping exists. I might then write that as on the right above. I now conjecture that Pa0-1 = Pa . So let's first talk about Pa a bit. I copy from above and edit. I think there is no question but that any state in Ea0 is a mixture of ALL states on the right side, not just those in Ea. So, certainly if you apply projector Pa to a specific state in Ea0 , you will get a unique and specific state in Ea. Question: what are the characteristics of an operator like Pa? We know that Pa2 = 1. It's domain is all of the Hilbert Space H0 or H. But its range is the space Ea. So we can say this: Pa : H → Ea or Pa : H0 → Ea You can see that in general Pa has a huge nullspace which agrees with the partial range. So Pa is certainly not 1-to-1 on H. We could consider the following restriction of Pa: Pa : Ea0 → Ea Pa-1 : Ea → Ea0 Now I will let my conjecture that Pa0-1 = Pa for the restricted operators ride, I think it is exactly correct. Now let's go back to Messiah's sentence above which we still need to get happy with: "Let us suppose that any vector in Ea may be considered as the projection in Ea of a certain well defined vector of Ea0." This then would go with my restricted Pa : Ea0 → Ea. OK, and we could also make the reverse statement, and it would involve Pa0 : Ea → Ea0 . I am now happy with the entire large paragraph at the start of page 718. Next Messiah sentence is this: " Thus any eigenvector of H in Ea can be put into the form Pa | Ea0α> " Well, in my picture all the states on the right are meant to be eigenstates of H. And I just said that any state on the right in Ea is the image via Pa of some unique state in Ea0. To find this unique state, we could just write out this eigenstate of H in the Ho basis and look at just the contributions from Ea0 states, the first terms in my usual ψ expansion, and that combination of states would be that unique state which Messiah is going to call | Ea0α>. I guess we just regard α as the label on this unique state. Note that this state | Ea0α> is not itself an eigenstate of H, though it is certainly an eigenstate of H0 since it is a mixture of degenerate states all with the same energy Ea0. So I agree that, for any eigenstate |φ> in Ea , we could find the unique state | Ea0α> in Ea0 such that |φ> = Pa | Ea0α> . Happy again! Comment: I just wonder what the typical reader of this section made of it! Then he says H { P|Ea0α> } = Eα { P|Ea0α> } where H is the full H and Eα is the fully corrected energy, etc. This is just fine, since we just said |φ> = P|Ea0α> is an eigenstate of H. Both sides of this equation are the same ket in Ea, easier perhaps to access this on the RHS. So here goes the processing: H P|Ea0α> = Eα P|Ea0α> // |Ea0α> is the back image of an Ea eigenstate P0H P|Ea0α> = Eα P0P|Ea0α> In this last step, we take our H eigenstate state |φ> = Pa | Ea0α> and apply P0 to it. I would have thought this would give you |Ea0α> with my idea above that Pa0-1 = Pa for the restricted operators. In any event, applying it to both sides of an equality certainly gives you another equality. Now one more process P0H P P0|Ea0α> = Eα P0P P0|Ea0α> Ha |Ea0α> = Eα Ka |Ea0α> Ha = P0H P P0 Ka = P0PP0 where we extracted P0 from the Ea0 state on both sides. No problem, and the new operators are as defined by comparison. Messiah now wants to talk about these operators acting on Ea0 . We can see that if you act on a ket in H0 outside this subspace, you will get 0 due to the right-side P0 , and the same is true if you act on a bra outside Ea0. So truly these operators really only act in Ea0. Theorem 1: Any projection operator is a "part of unity" and you can think of it as having a partial diagonal of 1's in some region. It is therefore symmetric and Hermitian in that basis. The Hermiticity will survive any unitary transformation of basis. So projection operators are Hermitian. Theorem 2: Projection operators have positive matrix elements. Proof: Let ψ = Σiaiui < ψ | Pj| ψ> = Σiai*ui† ajuj = |aj|2 This is probability of being in state j. So a projector is positive definite. This would be true also if we had a projector which projected out a set of base states. I assume orthonormal basis. Theorem 3: The operator Ka = P0PP0 is positive definite. Lemma 3: If you can write A = BB†, then A is positive definite. Proof of lemma: Aψ = λψ then λ ||ψ||2 = <ψ|Aψ> = <ψ| BB†|ψ> = <B†ψ|B†ψ> = ||B†ψ||2, so λ = ||B†ψ||2/ ||ψ||2 > 0. Could we have B†ψ = 0 ? Yes, I suppose that is possible, but somehow we I think that does not happen in our application below. This is a loose bolt. Proof of theorem: Ka = P0PP0 = P0PPP0 = (P0P) (P0P)† = BB†, QED. This would be true for any K = CPC where C was Hermitian and P a projection operator Now back to Ha |Ea0α> = Eα Ka |Ea0α> Ha = P0 HP P0 Ka = P0 P P0 Write this as (Ha - Eα Ka) |Ea0α> = 0 We expect this might have non-zero solutions provided that det(Ha - Eα Ka) = 0 Now our amazing claim: THIS secular equation determines the eigenvalues Eα for our states in Ea, to all orders! If you can solve this "generalized eigenvalue problem" also for the eigenvectors that go with these eigenvalues, then you have found eigenvectors |Ea0α> of this equation, and THEN you know that P |Ea0α> are the eigenvectors of H that go with these same Eα eigenvalues. This seems amazingly convoluted. ________________________________________________________________________ Comments on the Generalized Eigenvalue Problem. (1) Here is a Wolfram claim: "The determinant of a positive definite matrix is always positive, so a positive definite matrix is always nonsingular." Well in particular, if a matrix is Hermitian we know we can diagonalize it with a unitary similarity, and the diagonal elements (eigenvalues) will all be positive, so the determinant will be positive. And the determinant is invariant under similarity, so certainly a positive definite Hermitian matrix is non-singular because the determinant cannot be 0. It can therefore be inverted. (2) We know from above that K is positive definite, so we may then conclude that K-1 exists. At first this seems odd since K = P0PP0 which is a product of three non-invertible projection operators, but the triplet shown must be invertible. (3) Then our generalized eigenvalue problem can be reduced to a regular eigenvalue problem Hψ = λKψ => (K-1H) ψ = λψ (4) The generalized eigenvalue problem is in itself a whole field of study. From http://www.cs.umu.se/~dacke/foweb00/new/NGEP/index.html we have this little introduction, including the new term "matrix pencil". we get this little introduction. The page goes on to show how you would solve the problem numerically. _____________________________________________________________________ The idea at this point is then to insert our perturbation expansions for HP and P to get Ha and Ka to some definite order, then you can find Eα to that order. Recall from earlier that we found P = P0 + λ [ P0 V Q0a + Q0a V P0 ] + λ2 [ (P0 V Q0a V Q0a + Q0a V P0 V Q0a + Q0a V Q0a V P0) –( P0 V P0 V Q0a2 + P0 V Q0a2 V P0 + Q0a2 V P0 V P0) ] (H-Eo)P = 0 + λ P0 V P0 + λ2 [Q0a V P0 V P0 + P0 V Q0a V P0 + P0 V P0 V Q0a ] Now when we compute Ha = P0 HP P0 or Ka = P0 P P0 , we can throw out terms where we will get QoaPo adjacent anywhere, so that really means we only keep terms that begin and end with P0 and then we don't have to replicate these leading and trailing values. So here we go: Ka = P0 P P0 = P0 - λ2 P0 V Q0a2 V P0 Ha = P0 HP P0 = P0 { (H-Eo)P } P0 + Eo P0 P P0 = P0 { (H-Eo)P } P0 + Eo Ka = Eo Ka + λ P0 V P0 + λ2 P0 V Q0a V P0 and so we have verified (77) and (78). So let's write our secular equation through order λ2: det(Ha - Eα Ka) = 0 det[ Eo Ka + λ P0 V P0 + λ2 P0 V Q0a V P0 – Eα Ka ] = 0 det [λ P0 V P0 + λ2 P0 V Q0a V P0 – (Eα– Eo)Ka ] = 0 det [λ P0 V P0 + λ2 P0 V Q0a V P0 – (Eα– Eo){ P0 - λ2 P0 V Q0a2 V P0} ] = 0 I would guess that since we are in our Ea0 space, we can now just throw out all the P0 things: det [λ V + λ2 V Q0a V – (Eα– Eo){ 1 - λ2 V Q0a2 V} ] = 0 det [λ V + λ2 V Q0a V – (Eα– Eo){ 1 - λ2 V Q0a2 V} ] = 0 If we keep only λ0 terms, we get (Eα– Eo)det(1) = 0 so that Eα = Eo . If we keep only λ0 and λ1 terms, this says det [λ V– (Eα – Eo) 1] = 0 Of course the size of this determinant is the size of the space Ea0. I could write this out as det = εijk .... [λ V1i– (Eα – Eo) δ1i] [λ V2j– (Eα – Eo) δ2j] [λ V3k– (Eα – Eo) δ3k] ... Let's do the 2x2 case please: det = [λ V11– (Eα – Eo) ] [λ V22– (Eα – Eo) ] – λ V12 λ V21 = 0 Absorb λ into V and set εα = (Eα – Eo) to get det = [V11– εα ] [V22– εα ] –V12 V21 = 0 det = [εα –V11] [εα –V22] –V12 V21 = 0 This is a quadratic equation for εα and εα2 – (V11+ V22)εα + (V11 V22 – V12 V21) = 0 which looks pretty much like Saxon page 205. Notice that B2 - 4AC = (V11+ V22)2 - 4 (V11 V22 – V12 V21) = (V11– V22)2 + 4 V12 V21 So the answer is this εα = (V11+ V22)/2 ± [(V11– V22)2/4 + V12 V21] 1/2 If you happen to have a problem where V11 = V22 = 0, then the result is this εα = ± λ [V12 V21] 1/2 and this is the famous result that the two states are pushed apart, equally up and down. They "repel" each other. Now it could happen that all four elements Vij in our 2x2 subspace are 0. In this case and this case only we get B2 - 4AC = 0 so there is no splitting at all, and no shift either. This is where we would have to go to higher order and that will involve the effect of distant matrix elements. So go back to det [V + V Q0a V – εα { 1 - V Q0a2 V} ] = 0 where again I have embedded the λ into the B . Recall now that Q0an = Q0(H0 - E0)-nQ0 = Σk≠0 | k> (Ek - E0)-n <k| so to solve for the energies at this level, you have to go off and first compute Aij = Σk≠0 <i| V | k> (Ek - E0)-1 <k|V|j> Bij = Σk≠0 <i| V | k> (Ek - E0)-2 <k|V|j> Then we have det [V + A – εα { 1 - B} ] = 0 and this is another messy quadratic in εα . In our λ1 situation, we have A = B = 0 and we recover our previous equation which was det [V – εα 1] = 0 Conclusion: Suppose we have a degenerate subspace Ea0 of dimension N. When we turn on the perturbation, we can compute the corrected energies through second order as follows: det [V + V Q0a V – εα { 1 - V Q0a2 V} ] = 0 εα = (Eα – Eo) [ I think we can discard the V Q0a2 V factor on the grounds that it is down one power of λ. ] and again the λ are absorbed into the V. This is an NxN determinant. What is involved in the computation is the matrix elements Vij within Ea0 and also Vik where k is outside the subspace, and we also need to know the Ek outside the subspace. This solves the problem completely for any N !!! Only when you actually compute things will you know if there is still some degeneracy left when λ is turned on. I should have said: The above det = 0 equation is a polynomial of degree N in εα and so there will be N roots. Degeneracy will still exist if there is some algebraic multiplicity > 1 for a root! Very good! Now back for a moment to this EV problem (Ha - Eα Ka) |Ea0α> = 0 Suppose we have found the eigenvalues Eα to some order, how do we then solve this thing? (H - λαK)vα = 0 within Ea0 Normally if there were an inverse for K, you would change this to (K-1H - λα)vα = 0 within Ea0 and then it is a standard problem. For a given λα, there are N unknowns to find. But we have N equations: I guess this is the same with the original equation: (H - λαK)vα = 0 within Ea0 Σj(H - λαK)ij(vα)j = 0 i = 1,2....N so N equations each in N unknowns. How do we know there are solutions? Well, this problem really imposes that fact somehow. But more generally, if both are Hermitian, you can probably show it. All of the above is the Kato method of 1949, reference is given p 712. Messiah concludes this section with a slight variation due to Bloch 1958 which brings in yet another operator U and the claim is that the expansions are a bit simpler. But not enough detail is given here to follow the work. For example, it is not obvious why you can "define" an operator U as shown in (79). Bloch must talk about this. Here is some information on the reference, etats lies means bound states. [ I see that Messiah was a friend of this Bloch at Saclay and write Bloch's obit in 1972 ] Notice his mention of "diagrams".