Franz Schwabl Quantum Mechanics 4th Edit · Problem 11.9

Relativistic Corrections

Physics — Quantum Mechanics · degenerate-perturbation-theory-second-order

Problem

The energy levels following from the secular equation (11.14c) are non degenerate in most cases. Consider a Hamiltonian

H_0

with two degenerate eigenstates

|a\rangle

and

|b\rangle

. (a) Show that these remain degenerate iff

\langle a|H_1|a\rangle = \langle b|H_1|b\rangle

and

\langle a|H_1|b\rangle = 0

. (b) To lift the degeneracy then one has to turn to second order. Derive from (11.5c) a secular equation of the type (11.14b) with the replacement

H_{1ij} \to H_{1ij} + \sum_{n \neq a,b} \frac{\langle i|H_1|n\rangle\langle n|H_1|j\rangle}{E_i^0 - E_n^0} \quad ,

where

E_i^0 \equiv E_a^0 = E_b^0