Part VI, Chapter 2 | Page 1 of 3

Degenerate Perturbation Theory

Special treatment when unperturbed energy levels share the same energy

When two or more unperturbed states share the same energy, the standard non-degenerate perturbation theory breaks down catastrophically. The denominators $E_n^{(0)} - E_m^{(0)}$ vanish, producing infinite corrections. Degenerate perturbation theory resolves this by finding the "good" basis within the degenerate subspace before applying perturbation theory.

Why Non-Degenerate Theory Fails

Recall the first-order state correction from non-degenerate perturbation theory:

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

If states $|n^{(0)}\rangle$ and $|m^{(0)}\rangle$ are degenerate, meaning $E_n^{(0)} = E_m^{(0)}$, then:

$$\frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}} = \frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{0} \to \infty$$

The denominator vanishes! The perturbation correction diverges. The same problem afflicts the second-order energy correction and all higher orders. We need a fundamentally different approach for the degenerate subspace.

The Degenerate Subspace

Suppose the energy level $E_n^{(0)}$ has $g$-fold degeneracy. There are $g$ linearly independent states all with the same energy:

$$\hat{H}_0|n^{(0)},\alpha\rangle = E_n^{(0)}|n^{(0)},\alpha\rangle, \quad \alpha = 1, 2, \ldots, g$$

These $g$ states span a $g$-dimensional degenerate subspace. The key observation is that any linear combination of degenerate states is also an eigenstate of $\hat{H}_0$ with the same energy:

$$|\phi\rangle = \sum_{\alpha=1}^{g} c_\alpha |n^{(0)},\alpha\rangle \quad \Rightarrow \quad \hat{H}_0|\phi\rangle = E_n^{(0)}|\phi\rangle$$

This freedom is both the source of the problem and the key to its solution. We must choose the "right" linear combinations that work well with the perturbation.

Finding the "Good" Basis

The resolution is to choose the basis within the degenerate subspace that diagonalizes the perturbation $\hat{H}'$. In this "good" basis, the off-diagonal matrix elements of $\hat{H}'$ within the degenerate subspace vanish, eliminating the problematic terms.

Define the $g \times g$ perturbation matrix (also called the $W$-matrix):

$$\boxed{W_{\alpha\beta} = \langle n^{(0)},\alpha|\hat{H}'|n^{(0)},\beta\rangle}$$

This is the matrix representation of $\hat{H}'$ restricted to the degenerate subspace. It is Hermitian ($W_{\alpha\beta} = W_{\beta\alpha}^*$) and therefore diagonalizable.

The Secular Equation

The first-order energy corrections are the eigenvalues of the $W$-matrix, found by solving the secular (characteristic) equation:

$$\boxed{\det\left(W - E^{(1)}\mathbf{I}\right) = 0}$$

Equivalently, we solve the eigenvalue problem:

$$\sum_{\beta=1}^{g} W_{\alpha\beta}\, c_\beta^{(i)} = E^{(1),i}\, c_\alpha^{(i)}, \quad i = 1, 2, \ldots, g$$

The $g$ eigenvalues $E^{(1),1}, E^{(1),2}, \ldots, E^{(1),g}$ are the first-order energy corrections for the $g$ originally degenerate states. The corresponding eigenvectors give the "good" zeroth-order states:

$$|n^{(0)},i\rangle_{\text{good}} = \sum_{\alpha=1}^{g} c_\alpha^{(i)}|n^{(0)},\alpha\rangle$$

Splitting of Degeneracy

The perturbation typically lifts the degeneracy, splitting the originally degenerate level into distinct energies:

$$E_n^{(i)} = E_n^{(0)} + E^{(1),i} + E^{(2),i} + \cdots, \quad i = 1, 2, \ldots, g$$

Three scenarios can occur:

Complete lifting: All $g$ eigenvalues are distinct. The degeneracy is fully resolved at first order.
Partial lifting: Some eigenvalues coincide. Residual degeneracy remains, which may be lifted at higher orders.
No lifting: All eigenvalues are equal ($W$ is proportional to identity). The perturbation does not split the degeneracy at first order.

Two-Fold Degeneracy in Detail

The most common case is $g = 2$. The $W$-matrix is $2 \times 2$:

$$W = \begin{pmatrix} W_{11} & W_{12} \\ W_{21} & W_{22} \end{pmatrix} = \begin{pmatrix} W_{11} & W_{12} \\ W_{12}^* & W_{22} \end{pmatrix}$$

The secular equation is:

$$(W_{11} - E^{(1)})(W_{22} - E^{(1)}) - |W_{12}|^2 = 0$$

Solving the quadratic:

$$\boxed{E^{(1)}_{\pm} = \frac{W_{11} + W_{22}}{2} \pm \sqrt{\left(\frac{W_{11} - W_{22}}{2}\right)^2 + |W_{12}|^2}}$$

The splitting between the two levels is:

$$\Delta E = E^{(1)}_+ - E^{(1)}_- = 2\sqrt{\left(\frac{W_{11} - W_{22}}{2}\right)^2 + |W_{12}|^2}$$

Note that if $W_{11} = W_{22}$ (diagonal elements equal), the splitting is simply $\Delta E = 2|W_{12}|$, determined entirely by the off-diagonal matrix element.

Symmetry and the Good Basis

Often, symmetry tells us the good basis without doing any calculation:

If $\hat{H}'$ commutes with an operator $\hat{A}$: Choose the degenerate basis to be simultaneous eigenstates of $\hat{H}_0$ and $\hat{A}$. Then $W$ is automatically diagonal in this basis.
Parity: If $\hat{H}'$ is odd under parity, the good basis states should have definite parity. Then $W_{\alpha\beta} = 0$ unless $|\alpha\rangle$ and $|\beta\rangle$ have opposite parity.
Angular momentum: If $[\hat{H}', \hat{L}_z] = 0$, use eigenstates of $\hat{L}_z$. States with different $m$ values don't mix.

Accidental vs Essential Degeneracy

Essential degeneracy is required by symmetry:

The $(2l+1)$-fold degeneracy in $m_l$ for a central potential (rotational symmetry)
Kramers degeneracy for half-integer spin systems (time-reversal symmetry)
Cannot be fully lifted without breaking the underlying symmetry

Accidental degeneracy is not required by obvious symmetry:

The $n^2$-fold degeneracy in hydrogen (hidden SO(4) symmetry of Coulomb problem)
Can be lifted by perturbations that respect the apparent symmetries
Often signals a hidden symmetry of the unperturbed problem

Looking Ahead

On the next page, we apply degenerate perturbation theory to the Stark effect in the $n=2$ level of hydrogen -- the classic example that reveals the linear Stark effect and demonstrates the full power of this formalism.

← Time-Independent PT Stark Effect in Hydrogen →

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