Degenerate Perturbation Theory

Why the simple formulas break

The first- and second-order formulas both contain energy denominators $E_n^{(0)} - E_m^{(0)}$ . If two unperturbed states share the same energy — a degeneracy — that denominator is zero and the expressions diverge. The failure is not physical; it is a sign that we picked the wrong starting basis.

When a level is $g$ -fold degenerate, any linear combination of its eigenstates is also an eigenstate of $\hat{H}^{(0)}$ with the same energy. The perturbation selects a particular set of combinations as the correct zeroth-order states — the ones that diagonalize $\hat{H}'$ inside the degenerate subspace. Ordinary perturbation theory assumed we already knew them; degenerate theory finds them.

The recipe

Let $\{|a\rangle, |b\rangle, \dots\}$ span the degenerate subspace with common energy $E^{(0)}$ . Build the matrix of the perturbation restricted to that subspace:

W_{ij} = \langle i | \hat{H}' | j \rangle, \qquad i, j \in \text{degenerate subspace}.

Then:

Diagonalize $W$ . Its eigenvalues are the first-order energy corrections $E^{(1)}_k$ ; the degeneracy is (generically) lifted.
The eigenvectors of $W$ are the good zeroth-order states — the correct combinations to use if you continue to higher order.

For a two-dimensional subspace with

W = \begin{pmatrix} W_{aa} & W_{ab} \\ W_{ab}^* & W_{bb} \end{pmatrix},

the eigenvalues are

E^{(1)}_\pm = \frac{W_{aa} + W_{bb}}{2} \pm \sqrt{\left(\frac{W_{aa} - W_{bb}}{2}\right)^2 + |W_{ab}|^2}.

The two levels, originally coincident, split by $E^{(1)}_+ - E^{(1)}_- = 2\sqrt{(\Delta/2)^2 + |W_{ab}|^2}$ where $\Delta = W_{aa} - W_{bb}$ .

A worked split

Suppose a level at $E^{(0)} = 5$ is doubly degenerate and the perturbation couples the two states symmetrically:

W = \begin{pmatrix} 0 & 0.2 \\ 0.2 & 0 \end{pmatrix}.

Here $W_{aa} = W_{bb} = 0$ and $W_{ab} = 0.2$ , so $E^{(1)}_\pm = \pm 0.2$ . The level splits into two: $5 + 0.2 = 5.2$ and $5 - 0.2 = 4.8$ . The good states are the symmetric and antisymmetric combinations $\tfrac{1}{\sqrt 2}(|a\rangle \pm |b\rangle)$ , the eigenvectors of $W$ .

Try it

For the degenerate level above with $W = \begin{pmatrix} 0 & 0.2 \\ 0.2 & 0 \end{pmatrix}$ , compute the larger first-order eigenvalue $E^{(1)}_+$ and return it as a number.

Run your code to see the quantum state.