Put a hydrogen atom in a uniform electric field. The ground state has definite parity, so the first-order shift vanishes, and the energy drops only as the square of the field: the quadratic Stark effect, measuring the atom’s polarisability. The $n=2$ level is different. 2s and 2p have the same energy and opposite parity, the field connects them, and degenerate perturbation theory splits the level linearly in the field — a signature of hydrogen’s accidental degeneracy.

A ball resting at the centre of a round bowl moves only slightly when the bowl is tilted a little, in proportion to the tilt squared. A ball on a flat tray rolls at once, in proportion to the tilt. The 1s level is the bowl; the degenerate $n=2$ level is the tray.

The $n=2$ level of hydrogen in a field $F$, in units of $eFa_{0}$ at the largest field drawn: the combinations of 2s and 2p$_{0}$ move up and down by , linearly in the field, while the two 2p states with stay put at first order.

A field $F$ along $z$ adds $\hat{V}=eF\hat{z}$ to the Hamiltonian of hydrogen’s electron. Parity and decide which matrix elements survive.

• $\hat{z}$ is odd, so $\bra{n\ell m}\hat{z}\ket{n\ell m}=0$: no first-order shift for any state of definite parity.

Build the $4\times4$ matrix of the perturbation in the degenerate level. Parity kills every diagonal element and every element between two $p$ states; conservation of $L_{z}$ kills every element between different $m$. One pair survives, between 2s and 2p, and its eigenvalues are the splitting.

$\hat{z}$ is odd, and states with equal $\ell$ have equal parity $(-1)^{\ell}$.

$[\hat{L}_{z},\hat{z}]=0$, so $$ cannot change ; 2s has .

With $z=r\cos\theta$ the element splits into a radial integral and an angular one, each done with the functions of the hydrogen chapter.

The level’s perturbation matrix.

The eigenvalues: two shifted states and two unshifted ones.

The perturbation $eFz$ is odd and the 1s state has definite parity, so its first-order Stark shift is zero. The ground state shifts only at second order, as the field squared.

In $n=2$ the degenerate states 2s and $2p_{0}$ are mixed by the field, and their combinations shift by $\pm3eFa_{0}$.

The perturbation $eFz$ commutes with $\hat{L}_{z}$, so it only connects states of equal $m$. The 2p states with $$ are not coupled to anything in the level.

Of the four states of $n=2$, two combinations shift up and down and two states stay put: three distinct levels at first order.

Great work! You now know how to:

• explain the quadratic Stark effect of the ground state
• derive the linear Stark splitting of $n=2$
• use parity and $m$ to decide which matrix elements vanish