Time-independent degenerate perturbation theory

Suppose a level $E^{(0)}$ is $g$-fold degenerate, its subspace spanned by the orthonormal states $\ket{\phi_1},\dots,\ket{\phi_g}$. A “good” zeroth-order state is some combination of them:

Step 1: write the first-order equation. Just as in the last section, the $\lambda^1$ order gives

Step 2: project into the degenerate subspace. Hit both sides with $\bra{\phi_b}$. Acting to the left, the Hermitian $\hat H_0$ gives $E^{(0)}\braket{\phi_b}{\psi^{(1)}}$, which cancels the first term on the right — exactly the same cancellation as last section, except this time what survives is a matrix equation:

The left side is the $g\times g$ matrix $W_{ba}\equiv\bra{\phi_b}\hat H'\ket{\phi_a}$ acting on the column vector of coefficients. This is a standard eigenvalue problem:

Step 3: solve the eigenvalue problem. A non-trivial solution exists only if

This determinant equation is called the secular equation (the name comes from celestial mechanics: the same mathematics was once used to compute the “secular”, long-term perturbations of planetary orbits). Its $g$ roots are the $g$ first-order energy shifts — the degeneracy splits into at most $g$ levels — and the eigenvector $\vec c$ belonging to each root gives one “good” zeroth-order state.

Boiled down to an operating manual:

1. Identify the degenerate subspace and pick any basis for it;
2. Compute the $g\times g$ perturbation matrix $W$ in that basis;
3. Diagonalise $W$: eigenvalues = first-order shifts, eigenvectors = good zeroth-order states.

Note that this step is not an approximation: diagonalising within the subspace is exact. The “perturbative” approximation lies in neglecting the coupling between the subspace and the levels outside it (that is second-order business).

Integrate directly with the Chapter 6 hydrogen wavefunctions (radial part × angular part):

(No need to memorise the number — just remember the scale is the Bohr radius $a_0$: that is simply how big the $n=2$ orbitals are.) So within the $\ket{2s},\ket{2p_0}$ block, the $2\times2$ matrix $W$ is

The secular equation $\det(W-E^{(1)}I)=(E^{(1)})^2-(3e\mathcal E a_0)^2=0$ solves to

$n=2$ splits into three lines: $-3e\mathcal E a_0$, $0$ (doubly: $2p_{\pm1}$), $+3e\mathcal E a_0$ — matching the symmetric splitting pattern Stark saw, and proportional to $\mathcal E$.

Plug in numbers: at $\mathcal E=10^5\ \text{V/cm}=10^7\ \text{V/m}$ (a strong laboratory field),

corresponding to about 13 cm⁻¹ in wavenumbers — easy work for a spectrometer.

Why is the response linear here? The “good” states $(\ket{2s}\mp\ket{2p_0})/\sqrt2$ are superpositions of opposite-parity states — they carry a built-in electric dipole moment ($\langle z\rangle=\pm3a_0\neq0$), with no need to wait for the field to polarise them. With a ready-made dipole, the energy is naturally proportional to $\mathcal E$; the ground state has no degenerate partner to combine with, so it can only be polarised slowly by the field, hence $\mathcal E^2$. The linear splitting of the Stark lines is direct evidence that “a ready-made dipole is hiding in the degenerate subspace”.