Degenerate Perturbation Theory

Transcription

1 Physics G6037 Professor Christ 12/05/2014 Degenerate Perturbation Theory The treatment of degenerate perturbation theory presented in class is written out here in detail. 1 General framework and strategy We begin with a Hamitonian H which can be decomposed into an operator H 0 with known eigenvectors and eigenvalues and a second perturbing piece V : H = H 0 + λv. (1) The eigenvectors and eigen values of H and H 0 are defined as follows: H 0 φ n = E n (0) φ n (2) H ψ n = E n ψ n. (3) For simplicity we will assume that all of the eigenvalues E n (0) are distinct with the exception of a group of N deg vectors which share the same eigenvalue E n (0) = E deg. (0) We will label these N deg degenerate eigenvectors as { φ ni } 1 i Ndeg and define to the set of N deg indices of these degenerate states as S deg = {n i } 1 i Ndeg. We should recognize that there may be many possible choices for the basis of eigenvectors which lie in this N deg subspace of degenerate states. However, this subspace itself is well-defined and is denoted V deg as is the projection operator P deg onto this subspace: P deg = N deg i=1 φ ni φ ni. (4) Our objective is to express the exact eigenvalues and eigenstates of H in terms of those of H 0 as a power series in the small parameter λ. In order to easily express such a relationship we need to coordinate the labeling of the perturbed ( ψ n ) and unperturbed ( φ n ) eigenstates. For n S deg this is 1

2 easy. We simply use the label n for the perturbed ψ n whose eigenvalue E n obeys: lim E(λ) n = E n (0). (5) λ 0 Since each E n (0) corresponds to a unique index n there is no ambiguity. However, for the N deg perturbed eigenstates whose energies approach deg as λ 0 there is no guarantee that their λ 0 limit will match up with a particular unperturbed eigenstate φ n. In general we expect that if we were to label the N deg states whose energies approach deg as { ψ ni } 1 i Ndeg,we would find: N deg lim ψ n i = C ji φ nj. (6) λ 0 j=1 Later in our procedure we will go back and redefine the unperturbed states φ ni so that the C matrix above is the unit matrix: C ji = δ ij which implies a smooth limit as λ 0 relating our ψ n and φ n, consisent with the perturbative expansion that we were able to derive in the case of no degeneracy. However, at present we do not have enough information to make this redefinition of the φ ni states and must take a less convenient approach in order to make an expansion in λ possible. We introduce a new basis for the N deg -dimensional subspace spanned by the perturbed eigenvectors ψ ni, denoted ψ ni which are no longer eigenvectors of H but instead obey the conditions: lim n λ 0 i = φ nj (7) ψ ni φ ni is real and positive (8) These conditions are analogues our standard condition that φ n ψ n is real and positive, needed to define the phase of the perturbed states ψ n. For simplicity of notation we will also define: ψ n = ψ n (9) when n S deg. We will learn more about the relationship between the two sets of states ψ ni and φ ni for 1 i N deg as we develop the perturbation series which expresses the former in terms of the latter. Our new states ψ ni are now easily related to our unperturbed states and almost diagonalize H: H ψ n = H n ψ, (10) 2

3 where { δn n or n H n = S deg (11) H n otherwise Now we have a problem that can be solved in perturbation theory and which has reduced H to a matrix which is entirely diagonal except for a finite N deg N deg block which must be diagonalized by hand. We will now show how the states ψ n and the nearly diagonal matrix H n can be worked out to first and second order in λ. The final diagonalization of the N deg N deg matrix H ni must be carred out by some method other than perturbation j theory. Just as in the non-degenerate case we start with the eigenvalue equation, Eq. (10), take its matrix element on the left with φ n and insert a complete sum over a complete set of unperturbed states between (H + λv ) and ψ n : φ ( H 0 + λv )( ) φ φ ψn = φ ψ H n. (12) Next we introduce expansions in λ for both H n and φ ψ n = S n: H,n = δ ne n (0) + λe (1) n + λ2 n (13) φ ψ n = δ n + λs (1) n + λ2 λs (2) n (14) and substitute them into Eq. (12). Note that the matrices E (1) n and n will be diagonal except for the same N deg N deg non-diagonal block which appears in H n. We obtain the following equation: ( (0) E n δ + λv )( δ n + λs (1) n + λ2 S (2) ) n (15) = ( δ + λs(1) n + λ2 S (2) )( δ ne n (0) + λe (1) n + λ2 n )). 2 First-order degenereate perturbation theory We can next determine the matrices S (1) n and collecting all terms that are first order in λ: and E(1) nn by starting with Eq. (15) S(1) n + V n = E (1) n + S(1) n E(0) n. (16) 3

4 If either n or are not in the set S deg of degenerate states, then the consequeces are exactly as we found in non-degenerate perturbation theory. For n this equation can be solved for S (1) n without any need for a nonzero off-diagonal element E (1) n n. When n = E (1) nn is determined and S nn (1) is left unconstrained. However, as in the non-degenerate case our condition im(s nn (1) ) and the unit norm of the state ψ n combine to determine S nn (1) =0. Thus, to first order in λ and for either n S deg or / S deg we have: E (1) n = δ nv nn (17) S (1) V n n = n. (18) E n (0) For both n = n i and = n j in S deg, then S n (1) cancels between the left andn right sides of Eq. (16) and we are left with a formula for which there is a non-diagonal matrix E n (1) connecting degenerate states: E (1) n = V n (19) and through first order the matrix H n = ψ H ψ n is given by: H n = δ n n + λe (1) n, (20) where E (1) n is a diagonal matrix with diagonal element V nn when either n or is lies outside of S deg while if n, S deg then E (1) n is the N deg N deg matrix V n. The last step is to perform a unitary transformation on the N deg vectors ψ ni 1 i Ndeg which span this N deg -dimensional space to diagonalize this N deg N deg matrix. In the discussion of second order degenerate perturbation theory below we will assume that this diagonalization has been performed so that in our transformed basis: E n (1) = V n = V ni n i δ n. (21) for 1 i, j N deg. Because the matrix S n (1) drops out of Eq. (16), it has not been determined. For the first-order, non-degenerate case only S nn (1) was left undetermined at this step. However, our phase conventions for the perturbed states ψ n and the orthogonality of those states required S nn (1) = 0. The analogous choice is available here. As in the non-degenerate case we must ensure that 4

5 our new states ψ n are orthonormal which will be achieved if the transformation matrix S is unitary: S (S ) n = δ n. (22) To first order this equation becomes a condition on S (1) n: S (1) +(S ) n =0. (23) One can see that this equation is automatically obeyed if n or are not elements of S deg but must be imposed if both n and belong to S deg : S (1) n +(S ) n =0. (24) Thus, we are permited to use an arbitrary anti-hermitian matirx S (1) n,if we wish. This amounts to permitting a first-order unitary rotation among our N deg degnerate states φ ni. Since such a first order shift would serve no purpose, we adopt this simplest convention S (1) n =0 1 i, j N deg. (25) 3 Second-order degenereate perturbation theory Extending this proceedure to second order is a simple repetition of the steps just taken. However, the off-diagonal elements of H n which appear will now be second order terms term which must be computed and are less obvious than the simple off-diagonal elements of V n. Just as before we start with our general equation, Eq (15) and pick out the terms which are of order λ 2 : S(2) n + V S(1) n n = n + S (1) E(1) n n + S (2) n ne n (0). (26) Just as in the first-order case, this is easily solved for S (2) n and the diagonal, second order energy matrix n determined if either n or n is not an element of S deg : n = δ n V n S (1) n = δ n 5 n V n V n n (27)

6 S (2) n = = 1 n 1 n ( V S(1) n n ) S(1) E(1) n n n V n V n n V n E(0) n V nn. (28) For the case when both = n i and n = n j lie in the degenerate set S deg, the second order matrices S n (2) in Eq. (26) will cancel between the left- and right-hand sides and cannot be used to remove the off-diagonal terms. In this case Eq. (26) can be written as an equation for the N deg N deg, second-order matrix E n (2) : n = V ni S(1) n j S (1) n i E(1) n j (29) = V ni S(1) n j S (1) n V nj n j (30) using = n i and n = n j with 1 i, j N deg and assuming that the first order matrix E (1) n j has already been diagonalized as part of the first-order solution. We should consider three possibilities: 1. There are no second-order terms connecting the degenerate states. In this case the full matrix n will have only diagonal terms and we have succeeded in diagonalizing H through second order in λ. 2. The degeneracy has been lifted at first order. In this case all of the elements of the now diagonal matrix V n are distinct so V ni n i = V nj n j implies that i = j. In this case the fundamental problem posed by degenerate perturbation theory, how to determine which combination of the unperturbated states φ ni correspond to the perturbed states has been resolved and we should expect no further difficulties. In fact, this is the case because although we cannot use the matrix S n (2) to remove the second-order off-diagonal terms which have appeared, we can make use of the first order matrix S n (1) (which so far has not been used and simply set to zero for convenience) to remove the off-diagonal elements of E n (2). That this is possible can be seen by rewriting Eq. (30) to show explicitly these so far unconstrained elements of S n (1) : E n (2) = n j + V ni n i S n (1) S n (1) V nj n j. (31) S deg V ni S(1) 6

7 Since the matrix element S n (1) has not previously been contrained and was set to zero only for simplicity, it can now be chosen to make this off-diagonal term E n (2) =0: S (1) n = 1 V nj n j V ni n i S deg V ni S(1) n j (32) V n nj 1 = V ni n V nj n j V. (33) ni n i S deg E n (0) j This is indeed first-order in V as required but an interesting ratio of a second-order divided by a first-order piece. Thus, if unique, nondegenerate states are resolved at first order in λ then no further difficulties arise at order λ 2. The equations which must be obeyed to make diagonal can be solved and end up determining the previous un- n specified coefficients S (1) n. 3. The N deg degenerate states φ ni remain degenerate at first order. Under these circumstances both the matrix elements S n (1) and S n (2) cancel from the expression giving E n (2) which implies that no choices remain to remove the off-diagonal terms from E n (2). As a result to second order in perturbation theory the matrix ( ) H n = δ n E (0) (2) n + λv nn + λe n, (34) where n is diagonal except for the N deg N deg block: n = V ni S(1) n j = V ni S deg S deg V n nj n j (35) which must be diagonalized explicilty, without the further use of perturbation theory. Finally we should observe that the second order energy shift E n (2) given in Eq. (27) for states φ n where n is not in the degenerate set S deg are actually independent of the basis used for the degenerate set of states { φ ni } ni S deg so these energies are unaffected by any final diagonalization steps taken in this subspace to deal with the second order degeneracy: n = n V n V n n (36) 7

8 = = n S deg n S deg V n V n n + n S deg V n V n V n V n + φ n V P degv φ n E n (0) n n deg n (37). (38) 8