The Stark Effect in Hydrogen

At the $n=2$ level, hydrogen has a 4-fold degeneracy. The four states are:

All four states have the same unperturbed energy:

This degeneracy is accidental (specific to the Coulomb potential) and arises from the hidden SO(4) symmetry of hydrogen. The $2s$ state has $l=0$ (even parity), while the $2p$ states have $l=1$ (odd parity).

An external uniform electric field $\vec{\mathcal{E}} = \mathcal{E}_0\hat{z}$ produces the perturbation:

where we use $z = r\cos\theta$ in spherical coordinates. This perturbation is odd under parity ($z \to -z$) and preserves the $z$-component of angular momentum ($[\hat{H}', \hat{L}_z] = 0$).

Before computing the $4 \times 4$ $W$-matrix, symmetry eliminates most matrix elements:

Rule 1: $\Delta m = 0$

Since $[\hat{H}', \hat{L}_z] = 0$, the perturbation cannot change $m$. Therefore $\langle n,l',m'|r\cos\theta|n,l,m\rangle = 0$ unless $m' = m$.

Rule 2: $\Delta l = \pm 1$

Since $\cos\theta \propto Y_1^0$, the angular integral vanishes unless $l' = l \pm 1$ (from the properties of spherical harmonics and the Wigner-Eckart theorem).

Consequence:

The states $|2,1,1\rangle$ and $|2,1,-1\rangle$ have $m = \pm 1$ and cannot couple to any other $n=2$ state (since $|2,0,0\rangle$ has $m=0$ and $|2,1,0\rangle$ has $m=0$). Only $|2,0,0\rangle$ and $|2,1,0\rangle$ can mix.

In the basis $\{|2,0,0\rangle, |2,1,0\rangle, |2,1,1\rangle, |2,1,-1\rangle\}$, the $W$-matrix is:

The diagonal elements vanish because $\hat{H}'$ is odd under parity and each basis state has definite parity. The only non-zero element is:

Using the hydrogen wave functions:

The matrix element separates into radial and angular parts:

The angular integral gives $1/\sqrt{3}$ (using standard spherical harmonic identities). The radial integral evaluates to:

Combining everything:

This is real, so $W_{12} = W_{21}$.

The $4 \times 4$ matrix block-diagonalizes. The $|2,1,\pm 1\rangle$ states are unaffected ($E^{(1)} = 0$). The $2 \times 2$ block for $|2,0,0\rangle$ and $|2,1,0\rangle$ is:

The eigenvalues are:

The four originally degenerate $n=2$ states split into three levels:

The total splitting is:

This is the linear Stark effect: the energy splitting is proportional to the first power of the electric field. This is in sharp contrast to the quadratic Stark effect for non-degenerate levels.

The distinction between linear and quadratic Stark effects is fundamental:

The linear effect is much larger for typical field strengths. Physically, the "good" states $|\pm\rangle$ are superpositions of $s$ and $p$ states, which have permanent electric dipole moments:

These mixed-parity states possess permanent electric dipole moments that interact linearly with the applied field.