Past exam of the advanced quantum physics course of the University of Cambridge / 2013 / Paper 1 / 2 / c

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- Solution c

Solution

The Stark effect changes atomic energy levels in a static electric field. For

E = E \hat{z}

, the electric-dipole interaction is

H^{'} = - d_{z} E = e E z

for an electron, with

e > 0

. A nondegenerate state of definite parity has

⟨ z ⟩ = 0

, so its leading quadratic Stark effect is

Δ E_{n}^{(2)} = e^{2} E^{2} \sum_{r \neq = n} \frac{∣ ⟨ r ∣ z ∣ n ⟩ ∣ ^{2}}{E _{n} - E _{r}} .

The ground state shift is negative, conventionally

- α E^{2} /2

with positive electric polarizability.^[ref]

Degenerate opposite-parity states require degenerate perturbation theory. In ideal Coulomb hydrogen, the

2 s, 2 p_{0}

subspace has an off-diagonal matrix element of magnitude

3 e a_{B} E

, where

a_{B}

is the Bohr radius. Its mixed states shift by

Δ E = \pm 3 e a_{B} E, Δ E_{2 p_{\pm 1}} = 0 to first order .

This Linear Stark effect in the first excited hydrogen eigenspace assumes fine-structure and Lamb splittings are negligible compared with the mixing. In an isolated parity eigenstate with such degeneracies resolved, sufficiently weak fields restore quadratic behaviour. Strong fields require further mixing and can ionize the atom.^[ref]

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