QuantumPhysics.dvi

10.5 Excited states and degenerate perturbation theory

The energy of a state|n,ℓ,m〉is

E^0 n=−

1

2 n^2

mec^2 α^2 (10.34)

and is independent of the angular momentum quantum numbersℓ,m. To investigate the

Stark effect on excited states, we begin by evaluating the matrix elements ofH 1 in the basis

|n,ℓ,m〉. For simplicity, we shall concentrate on the first excited states withn= 2. There

are 4 states, and we have the following matrix elements,

H 1 =eE







0 M 0 0

M∗ 0 0 0

0 0 0 0

0 0 0 0







| 2 , 0 , 0 〉

| 2 , 1 , 0 〉

| 2 , 1 ,+1〉

| 2 , 1 ,− 1 〉

(10.35)

Here, all diagonal matrix elements vanish becausezhas odd parity while the states| 2 ,ℓ,m〉

all have definite parity. The following 5 matrix elements vanish because of the Wigner-Eckard

theorem,

〈 2 , 0 , 0 |z| 2 , 1 ,± 1 〉=〈 2 , 1 , 0 |z| 2 , 1 ,± 1 〉=〈 2 , 1 , 1 |z| 2 , 1 ,− 1 〉= 0 (10.36)

The only remaining matrix elements are non-vanishing, and we set

〈 2 , 0 , 0 |z| 2 , 1 , 0 〉 = M

〈 2 , 1 , 0 |z| 2 , 0 , 0 〉 = M∗ (10.37)

This completes the calculation of the matrix elements ofH 1. The full Hamiltonian in this

basis is now given by putting together the contribution from the unperturbed Hamiltonian,

which givesE 20 for all 4 states, withH 1 , and we get

H=







E 20 eEM 0 0

eEM∗ E 20 0 0

0 0 E 20 0

0 0 0 E^02







| 2 , 0 , 0 〉

| 2 , 1 , 0 〉

| 2 , 1 ,+1〉

| 2 , 1 ,− 1 〉

(10.38)

Diagonalizing this matrix, we obtain the 4 energy eigenvalues,

E 2 , 1 = E 20 +eE|M|

E 2 , 2 = E 20 −eE|M|

E 2 , 3 =E 2 , 4 = E 20 (10.39)

Note that a two-fold degeneracy remains on the last two states.