Degenerate perturbation theory

Intuition

When several states share an energy, the ordinary formulas divide by zero. The way out is to choose the unperturbed states wisely. Within the degenerate level any combination is an eigenstate of $\hat{H}_{0}$ , and the perturbation picks out particular combinations — those that diagonalise it within the level. On them the first-order shifts are the eigenvalues of the small matrix of the perturbation in the level, and the level splits.

Several identical balls resting at the bottom of a flat dish can sit anywhere. Tilt the dish slightly and they roll to definite places; the tilt chooses where they settle. A perturbation chooses the states of a degenerate level the same way.

A level holding three states at $E^{(0)}$ , split by a perturbation of strength $\lambda$ from 0 to 1. To first order each new level moves in a straight line, with a slope equal to an eigenvalue of the perturbation’s matrix within the level.

For a level of $g$ degenerate states $\ket{i}$ , form the $g\times g$ matrix of the perturbation within it. Its eigenvalues are the first-order shifts; its eigenvectors are the right unperturbed states.

W_{ij}=\bra{i}\hat{V}\ket{j},\qquad \sum_{j}W_{ij}c_{j}=E^{(1)}c_{i}

Properties

The eigenvectors of $W$ , the good states, are the limits of the perturbed states as $\lambda\to0$ .
If a symmetry of the perturbation labels the degenerate states differently, $W$ is already diagonal in them and the good states are those labels.
Eigenvalues of $W$ that coincide leave part of the degeneracy; the next order may split it.
The trace of $W$ does not depend on the basis, so the shifts add up to whichever states are used: the centre of gravity of the level moves by .

Diagonalising within the level

\begin{aligned}&\ket{n^{(0)}}=\sum_{i}c_{i}\ket{i},\quad \hat{H}_{0}\ket{i}=E^{(0)}\ket{i}\\ &\Longrightarrow\ Wc=E^{(1)}c\end{aligned}

Write the zeroth-order state as an unknown combination of the degenerate states and project the first-order equation on each of them. The left side vanishes, since each $\ket{i}$ has the same unperturbed energy, and what remains is the eigenvalue equation of the matrix $W$ .

Proof steps

\ket{n^{(0)}}=\sum_{j}c_{j}\ket{j}

Any combination of the degenerate states is an eigenstate of $\hat{H}_{0}$ .

(\hat{H}_{0}-E^{(0)})\ket{n^{(1)}}=(E^{(1)}-\hat{V})\ket{n^{(0)}}

The first-order equation.

\bra{i}(\hat{H}_{0}-E^{(0)})\ket{n^{(1)}}=0

$\bra{i}$ is an eigenbra with the same energy $E^{(0)}$ .

0=E^{(1)}c_{i}-\sum_{j}\bra{i}\hat{V}\ket{j}c_{j}

Project the right side on $\bra{i}$ .

\sum_{j}W_{ij}c_{j}=E^{(1)}c_{i}

The first-order shifts are the eigenvalues of $W$ .

Applications

Practice

A Small Matrix

Within a degenerate level, write the perturbation as a matrix. Its eigenvalues are the first-order shifts of the split level.

W_{ij}=\bra{i}\hat{V}\ket{j}

Try it

A doubly degenerate level has $W=\begin{pmatrix}1&2\\2&1\end{pmatrix}$ in units of $\lambda$ . What is the larger first-order shift?