130_notes.dvi

22.3.2 Hydrogen Atom Ground State in a E-field, the Stark Effect.

We have solved the Hydrogen problem with the following Hamiltonian.

H 0 =

p^2

2 μ

−

Ze^2

r

Now we want to find the correction to that solution if anElectric field is applied to the atom.
We choose the axes so that the Electric field is in the z direction. Theperturbtionis then.

H 1 =eEz

It is typically a small perturbation.For non-degenerate states, the first order correction to the
energy is zero because the expectation value of z is an odd function.

E(1)nlm=eE〈φnlm|z|φnlm〉= 0

We therefore need to calculate thesecond order correction. This involves a sum over all the
other states.

E

(2)

100 =e

(^2) E 2 ∑
nlm 6 =100
|〈φnlm|z|φ 100 〉|^2
E 1 (0)−E(0)n
We need to compute all the matrix elements of z between the groundstate and the other Hydrogen
states.
〈φnlm|z|φ 100 〉=

∫

d^3 rR∗nl(rcosθ)R 10 Ylm∗Y 00

We can do the angular integral by converting the cosθterm into a spherical harmonic.

Y 00 cosθ=

1

√

4 π

√

4 π

3

Y 10

The we can just use the orthonormality of the spherical harmonicsto do the angular integral, leaving
us with a radial integral to do.

〈φnlm|z|φ 100 〉 =

1

√

3

∫

r^3 drRnl∗R 10

∫

dΩYlm∗Y 10

=

δℓ 1 δm 0

√

3

∫

r^3 R∗nlR 10 dr

The radial part of the integral can be done with some work, yielding.

|〈φnlm|z|φ 100 〉|^2 =

1

3

28 n^7 (n−1)^2 n−^5

(n+ 1)^2 n+5

a^20 δℓ 0 δm 0 ≡f(n)a^20 δℓ 0 δm 0

We put this back into the sum. The Kronecker deltas eliminate the sums overℓandm. We write
the energy denominators in terms of the Bohr radius.

E 100 (2) = e^2 E^2

∑∞

n=2

f(n)a^20

−e^2

2 a 0 +

e^2

2 a 0 n^2