The Time-Independent

Field Note: corrections to energy levels when the Hamiltonian is static and the perturbation does not switch on and off.

— Trolla

Time-independent perturbation theory is the workhorse. It answers the question every experimentalist asks: my spectrum has lines, and now I've turned on a field—where do they move?

Start with the non-degenerate case. The unperturbed Hamiltonian $\hat{H}_0$ has eigenstates $|\psi_n^{(0)}\rangle$ and eigenvalues $E_n^{(0)}$. The perturbation $\hat{H}'$ is small and time-independent. The goal: find the corrected energies and wavefunctions.

First-order energy

$$E_n^{(1)} = \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle$$

This is the expectation value of the perturbation in the unperturbed state. Simple. Physical. Intuitive. The perturbation adds, on average, this much to the energy. If $\hat{H}' = -e\mathcal{E}z$ (an electric field), then $E_n^{(1)} = -e\mathcal{E}\langle z \rangle_n$. For states with definite parity, $\langle z \rangle = 0$, so the first-order Stark shift vanishes. The hydrogen $n=1$ state has no linear Stark effect. That's not a bug—it's symmetry protecting the spectrum.

First-order wavefunction

$$|\psi_n^{(1)}\rangle = \sum_{m \neq n} \frac{\langle \psi_m^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle}{E_n^{(0)} - E_m^{(0)}} |\psi_m^{(0)}\rangle$$

The perturbed wavefunction is a mixture of the other unperturbed states. The coefficients depend on the matrix elements (how strongly the perturbation couples) and the energy differences (how hard it is to mix). The state deforms—it's no longer a pure eigenstate of $\hat{H}_0$. It picks up "admixture" from the nearby states.

Second-order energy

$$E_n^{(2)} = \sum_{m \neq n} \frac{|\langle \psi_m^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle|^2}{E_n^{(0)} - E_m^{(0)}}$$

This is the one that matters most. Second-order energy corrections capture the indirect influence of every other state. State $n$ can't mix with state $m$ directly (at first order, the wavefunction changes), but at second order, the energy shifts because the perturbation virtually visits state $m$ and comes back.

For the ground state, all denominators are negative, so $E_0^{(2)} < 0$ always. The perturbation always lowers the ground-state energy. This is the quantum-mechanical version of "the system relaxes when disturbed." The polarizability of any bound state is positive because of this.

Degenerate perturbation theory

Here's where it gets interesting. Suppose states $|a\rangle$ and $|b\rangle$ share the same energy $E^{(0)}$. The denominator vanishes. The formula breaks.

What actually happens? The perturbation splits the degeneracy. In the subspace of degenerate states, you diagonalize $\hat{H}'$—the perturbation itself. The eigenvalues of that small matrix are your first-order energy shifts. The eigenstates are the "good" linear combinations that the perturbation selects.

Example: the linear Stark effect in hydrogen $n=2$. Fourfold degeneracy. The electric field breaks it. You build the $4 \times 4$ matrix of $\hat{H}' = -e\mathcal{E}z$ in the $n=2$ subspace. Three zero eigenvalues (unaffected states) and one pair $\pm 3e\mathcal{E}a_0$ (split states). The degeneracy is partially lifted. You get a line splitting into three—central line unchanged, two shifted symmetrically. That's the linear Stark effect, and it only exists because of the degeneracy.

For $n=3$, you get nine states, five distinct first-order shifts. More splitting. The pattern encodes the symmetry of the perturbation relative to the unperturbed state.

Degeneracy and symmetry

The whole degenerate perturbation procedure is really just symmetry in disguise. When a perturbation commutes with some symmetry operator of $\hat{H}_0$, the good quantum numbers survive. When it breaks a symmetry, the degeneracy lifts. The matrix you diagonalize lives in the degenerate subspace and transforms according to the irreducible representations of the remaining symmetry group.

This is why group theory is inseparable from perturbation theory. The selection rules for which matrix elements vanish, which states mix, and which splittings occur—all of it is governed by the transformation properties of $\hat{H}'$ under the symmetry group of $\hat{H}_0$.

When to stop

In practice, you compute as many orders as the physics demands. First order captures the dominant shift when it's nonzero (degenerate states, or when the perturbation has a large diagonal matrix element). Second order captures the dominant shift when first order vanishes by symmetry. Third order and beyond are rarely needed unless you're hunting for tiny effects—the Lamb shift, for instance, which is actually computed with quantum electrodynamics, not simple perturbation theory.

The convergence is the real question. Is $\hat{H}'$ truly "small"? How small is small enough? In hydrogen, the fine-structure perturbation is small because $\alpha^2 \ll 1$ (the fine-structure constant squared). In molecules, the Born-Oppenheimer approximation treats the ratio of electron to nuclear mass as a small parameter. The perturbation parameter is always physical—mass ratios, coupling strengths, field strengths.

That's the power of time-independent perturbation theory: it turns the intractable into a systematic expansion around the known, with each order encoding more physics, and each physical parameter of the problem appearing explicitly in the perturbation.