The Stark Effect

An atom in an electric field

Place an atom in a uniform external electric field $\vec{\mathcal{E}} = \mathcal{E}\,\hat{z}$ . The electron of charge $-e$ acquires a potential energy that adds to the atomic Hamiltonian. The perturbation is

\hat{H}' = e\,\mathcal{E}\,\hat{z},

where $\hat{z}$ is the electron's position operator along the field. The shift of atomic energy levels in this field is the Stark effect, the electric analogue of the magnetic Zeeman effect.

First order usually vanishes

For a non-degenerate state $|n\rangle$ with definite parity, the first-order shift is $E^{(1)} = e\mathcal{E}\,\langle n|\hat{z}|n\rangle$ . But $\hat{z}$ is odd under parity while $|\langle n|^2$ is even, so the integrand is odd and the expectation value vanishes:

\langle n|\hat{z}|n\rangle = 0 \quad \text{for any parity eigenstate.}

The hydrogen ground state is non-degenerate and has definite parity, so it shows no linear Stark shift. Its leading effect is quadratic in $\mathcal{E}$ (the quadratic Stark effect), computed from the second-order formula and describing the induced electric dipole / polarizability.

The linear Stark effect needs degeneracy

The story changes when a level is degenerate and the degenerate states have opposite parity, so that $\hat{z}$ can connect them. The $n = 2$ level of hydrogen is the textbook case: it contains the $2s$ state (even parity) and three $2p$ states (odd parity). The operator $\hat{z}$ couples $2s$ to $2p_z$ only; matrix elements to $2p_x, 2p_y$ vanish by the magnetic quantum number selection rule.

Within the $\{|2s\rangle, |2p_z\rangle\}$ subspace the perturbation matrix is purely off-diagonal,

W = \begin{pmatrix} 0 & W_0 \\ W_0 & 0 \end{pmatrix}, \qquad W_0 = e\mathcal{E}\,\langle 2s|\hat{z}|2p_z\rangle = -3\,e a_0 \mathcal{E},

where $a_0$ is the Bohr radius and the value $-3ea_0\mathcal{E}$ comes from evaluating the hydrogen radial and angular integrals (a standard result). Diagonalizing $W$ gives eigenvalues

E^{(1)}_\pm = \pm 3\,e a_0 \mathcal{E}.

So the $n = 2$ level splits linearly with the field: one combination rises by $3ea_0\mathcal{E}$ , one falls by the same amount, and the remaining $2p_x, 2p_y$ states stay put. This linear splitting — absent for the ground state — is the signature of the degenerate, mixed-parity manifold.

Try it

For the hydrogen $n = 2$ linear Stark effect, the perturbation matrix in the $\{2s, 2p_z\}$ subspace is $\begin{pmatrix} 0 & -3 \\ -3 & 0 \end{pmatrix}$ in units of $e a_0 \mathcal{E}$ . Compute the magnitude of the first-order energy shift (the size of either eigenvalue) and return it as a number.

Run your code to see the quantum state.