130_notes.dvi

= a^30 E^2

∑∞

n=2

2 f(n)

−1 +n^12

= − 2 a^30 E^2

∑∞

n=2

n^2 f(n)

n^2 − 1

This is all a little dissatisfying because we had to insert the general formula for the radial integral
and it just goes into a nasty sum. In fact, we could just start with the first few states to get a good
idea of the size of the effect. The result comes out to be.

E

(2)

100 =−^2 a

3

0 E

(^2) (0.74 + 0.10 +...) =− 2. 25 a 3
0 E
2
The first two terms of the sum get us pretty close to the right answer. We could have just done
those radial integrals.
Now we computed, theelectric dipole momentof the atom which is induced by the electric field.
d=−

∂∆E

∂E

= 4(1.125)a^30 E

The dipole moment is proportional to the Electric field, indicating thatit is induced. The E field
induces the dipole moment, then the dipole moment interacts with theE field causing a energy shift.
This indicates why the energy shift is second order.

22.3.3 The Stark Effect for n=2 Hydrogen.

The Stark effect for the n=2 states of hydrogen requires the useof degenerate state perturbation
theory since there are four states with (nearly) the same energies. For our first calculation, we will
ignore the hydrogen fine structure and assume that the four states are exactly degenerate, each with
unperturbed energy ofE 0. That isH 0 φ 2 ℓm=E 0 φ 2 ℓm. The degenerate statesφ 200 ,φ 211 ,φ 210 , and
φ21(−1).

The perturbation due to an electric field in the z direction isH 1 = +eEz. So our first order
degenerate state perturbation theory equation is

∑

i

αi

〈

φ(j)|H 0 +eEz|φ(i)

〉

= (E 0 +E(1))αj.

This is esentially a 4X4 matrix eigenvalue equation. There are 4 eigenvalues (E 0 +E(1)), distinguished
by the indexn.

Because of the exact degeneracy (H 0 φ(j)=E 0 φ(j)),H 0 andE 0 can be eliminated from the equation.

∑

i

αi(E 0 δij+

〈

φ(j)|eEz|φ(i)

〉

) = (E 0 +E(1))αj

E 0 αj+

∑

i

αi

〈

φ(j)|eEz|φ(i)

〉

= E 0 αj+E(1)αj

∑

i

αi

〈

φ(j)|eEz|φ(i)

〉

= E(1)αj