Notes for Time-independent Perturbation Theory | PHY 4605, Study notes of Physics

Download Notes for Time-independent Perturbation Theory | PHY 4605 and more Study notes Physics in PDF only on Docsity! 3 Time-independent Perturbation Theory I 3.1 Small perturbations of a quantum system Consider Hamiltonian H0 + V̂ , (1) where H0 and V̂ both time-ind., and V̂ represents small perturbation to Hamiltonian H0, whose eigenfctns and eigenvals. |n〉 and En are known, H0|n〉 = En|n〉. (2) Nondegenerate perturbation theory ?? Suppose for now |n〉 nondegenerate. ?? Reasonable to assume if V̂ is small, perturbations to |n〉 and En should be small. So we have a small dimensionless param. in which to expand, replace V̂ by ²V̂ , take ² ¿ 1. Assume soln to H|ψ〉 = E|ψ〉 can be expanded E = En + ²δE(1) + ² 2δE(2) + . . . (3) |ψ〉 = |n〉 + ²|ψ(1)〉 + ²2|ψ(2)〉 . . . (4) Since |n〉 form complete set, the states |ψ(1)〉 and |ψ(2)〉 may be expanded |ψ〉 = |n〉 + ² ∑ m Cm|m〉 + ²2 ∑ m Dm|m〉 + . . . . (5) Now substitute (1-5) into Hψ = Eψ, (H0 + ²V̂ )(|n〉 + ² ∑ m Cm|m〉 + · · ·) = (En + ²δE(1) + · · ·)(|n〉 + ² ∑ m Cm|m〉 + · · ·), (6) or, using (2), ² ( V̂ |n〉 + ∑ m CmEm|m〉 ) + ²2 (∑ m CmV̂ |m〉 + ∑ m DmEm|m〉 ) + · · · 1 = ² ( δE(1)|n〉 + En ∑ m Cm|m〉 ) (7) +²2 ( δE(2)|n〉 + δE(1) ∑ m Cm|m〉 + En ∑ m Dm|m〉 ) . Now equate powers of ² to find (O(1)): V̂ |n〉 + ∑ m CmEm|m〉 = δE(1)|n〉 + En ∑ m Cm|m〉. (8) Now take inner product with 〈n|, use 〈n|m〉 = δnm: δE(1) = 〈n|V̂ |n〉 (9) so 1st-order shift in energy level for state |n〉 is just expectation value of V̂ in that state. Now take inner product with 〈m|, find Cm = 〈m|V̂ |n〉 En − Em , m 6= n (10) so energy eigenstate is, to 1st order, |ψ〉 = |n〉 + ²  Cn|n〉 + ∑ m 6=n |m〉〈m|V̂ |n〉 En − Em   +O(²2). (11) What is the coefficient Cn? Actually we can’t determine it! Note we could write instead of (9), |ψ〉 = (1 + ²Cn)  |n〉 + ² ∑ m 6=n |m〉〈m|V̂ |n〉 En − Em   +O(²2) (12) the same up to corrections ofO(²2). Here see that Cn just is part of overall normalization factor; can’t determine since S.-eqn. is linear. Same applies to Dn, etc. Go back and expand to 2nd order, using (9) and (10). Find: δE(2) = ∑ m 6=n 〈n|V̂ |m〉〈m|V̂ |n〉 En − Em , so (13) E = En + ²〈n|V̂ |n〉 + ²2∑m 6=n〈n|V̂ |m〉〈m|V̂ |n〉En − Em +O(² 3) (14) 2 H0|n, α〉 = En|n, α〉 (26) 〈n, α|m,β〉 = δmnδαβ . (27) Let’s again expand energy and wave fctn., E = En + δE(1) + δE(2) (28) |ψ〉 = |n, α〉 + ∑ mβ Cmβ|m,β〉 + · · · (29) where I supressed the ²’s for compactness. Completeness says V̂ |n, α〉 = ∑ m,β |m,β〉〈m,β|V̂ |n, α〉, (30) so generalization of (7) to O(²) is V̂ |n, α〉 + ∑ m,β CmβEm|m,β〉 = ∑ m,β |m,β〉 ( 〈m,β|V̂ |n, α〉 + CmβEm ) = δE(1)|n, α〉 + En ∑ mβ Cmβ|m,β〉. (31) Coefficicent of |m,β〉 on both sides of (31) must agree, so 〈m,β|V̂ |n, α〉 + CmβEm = EnCmβ + δE(1)δnmδαβ. (32) Two cases again: m 6= n Cmβ = 〈mβ|V̂ |nα〉 En − Em (33) m=n δE(1)δαβ = 〈nβ|V̂ |nα〉 (34) If 〈nβ|V̂ |nα〉 = 0 when α 6= β, the equation (34) makes sense, and we get δE(1) = 〈nα|V̂ |nα〉 (35) 5 Note this means possibly a different splitting for each of the initially de- generate states |nα〉, but is otherwise more or less same as nondegenerate case. But if 〈nβ|V̂ |nα〉 6= 0 when α 6= β, Eq. (34) doesn’t have a solution, we get some kind of contradiction. What’s going on? ?? Recall what we said about eigenvectors corresponding to distinct eigen- values being orthogonal. If eigenvalues are degenerate, various eigenvectors need not be orthogonal—however if they aren’t, we can construct a new set corresponding to same e’value which are! ??. Can pull similar trick here: construct N linearly independent combina- tions (N is the degeneracy of En, e’value in question) of form N eigenstates    |A〉 = ∑α Aα|nα〉 |B〉 = ∑α Bα|nα〉 |C〉 = ∑α Cα|nα〉 ... (36) Now choose {Aα, Bα, . . .} such that 〈A|V̂ |B〉 = 0, etc. (37) Redo previous calc. leading to (32), find now no inconsistency, rather δE(1) = 〈A|V̂ |A〉 if |A〉 is unperturbed state, = 〈B|V̂ |B〉 if |B〉 is unperturbed state, (38) ... If (37) holds, means V̂ is diagonal on subspace of degenerate eigenvectors, so can write V̂ |A〉 = VA|A〉, etc. In terms of the original basis this is ∑ α AαV̂ |n, α〉 = VA ∑ α Aα|n, α〉, (39) and inner product with 〈n, β| is ∑ α 〈n, β|V̂ |n, α〉Aα = VAAβ, (40) or, in suggestive matrix form, 6 V ~A = VA ~A, etc. (41) So we have reduced the degenerate state perturbation problem to the problem of diagonalizing V̂ and finding its eigenvalues and eigenvectors. Once found, the N eigenvalues VA, VB, etc. give the energy perturbations, E = En + VA, En + VB, . . .. Summary I know this is confusing. Here’s what you have to do in practice. If you have N degenerate states |α〉 corresponding to one eigenvalue E, construct the matrix 〈α|V |α′〉. If it happens to be diagonal, you’ve got it easy: for each α, read off the 1st order energy shifts, δEα = 〈α|V |α〉. The wave function will have no 1st order shift arising from the 〈α|V |α′〉, but beware it can still have corrections coming from the 〈α|V |m〉, where the |m〉 are the states outside the degenerate manifold you’re looking at. If 〈α|V |α′〉 isn’t diagonal, you need to find a new basis |α′〉 in which it is, i.e. diagonalize it! You get eigenvalues VA, VB, ... (the new energy shifts) and eigenvectors |A〉, |B〉..., which are linear combinations of the old |α〉. If you calculate the 1st order correction to the wavefunction with the new basis, the terms containing 〈A|V |B〉 will again vanish for A 6= B, but don’t forget you will still have terms like 〈m|V |A〉 which are nonzero. Linear Stark effect Note for nondegenerate case 1st order Stark shift vanished, i.e. δE ∼ E2. If two or more atomic states degenerate, need to go back & compute matrix elements 〈n, α|V̂ |n, β〉 (42) and diagonalize. While diagonal elements in original basis will still vanish due to parity (recall V̂ ∝ Er), off-diagonal ones needn’t. Thus when V̂ is nondiagonal there will in general be linear shift ∝ E. Linear dependence on applied field therefore =⇒ degenerate states. Equivalent in classical case to atom with permanent dipole moment, i.e. in a field dipole moment 7