5.1.3 Degenerate Perturbation Theory#

Prompts

Why does non-degenerate perturbation theory break down when the unperturbed spectrum has degeneracies? What diverges, and why?
What is the effective Hamiltonian \(H^{\text{eff}} = PVP\), and how does diagonalizing it within the degenerate subspace resolve the divergence?
Apply degenerate perturbation theory to a spin-1 particle. How does the perturbation split the degenerate level, and what determines the splitting magnitude?
When does the perturbation lift a degeneracy completely versus partially? What role do symmetries and selection rules play?

Lecture Notes#

Overview#

When the unperturbed spectrum has degeneracies, the non-degenerate perturbation formulas diverge because energy denominators vanish. Degenerate perturbation theory resolves this by first diagonalizing the perturbation within the degenerate subspace, identifying the correct zeroth-order basis and the first-order energy splittings.

The Problem: Divergent Denominators#

Suppose \(H_0\) has a degenerate level with energy \(E_n^{(0)}\) and \(d\) linearly independent eigenstates \(|n,\alpha\rangle\) (\(\alpha = 1,\ldots,d\)). The non-degenerate first-order state correction from \S5.1.2,

\[|\psi_n^{(1)}\rangle = \sum_{m \neq n} \frac{\langle m^{(0)} | V | n^{(0)} \rangle}{E_n^{(0)} - E_m^{(0)}} |m^{(0)}\rangle\]

diverges when summing over states \(|m\rangle\) in the same degenerate subspace, since \(E_n^{(0)} - E_m^{(0)} = 0\).

Root Cause: Ambiguous Labeling

Non-degenerate perturbation theory assumes that each label \(n\) uniquely tracks an eigenstate as \(\lambda\) increases from zero. With degeneracies, the perturbation mixes degenerate states, and there is no unique correspondence between perturbed and unperturbed states. The fix: diagonalize \(V\) within the degenerate subspace to find the basis that does not mix at first order.

Generalized Hellmann-Feynman Approach#

Recall from \S5.1.2 the Hellmann-Feynman theorem: \(\mathrm{d}E_n/\mathrm{d}\lambda = \langle n(\lambda)|V|n(\lambda)\rangle\). For a non-degenerate level, applying this at \(\lambda=0\) directly gives \(E_n^{(1)} = \langle n^{(0)}|V|n^{(0)}\rangle\). For a degenerate level, \(|n(\lambda)\rangle\) at \(\lambda\to 0\) can be any linear combination within the subspace, so the Hellmann-Feynman condition becomes:

Within the degenerate subspace, \(V\) acts as a \(d\times d\) matrix. The correct zeroth-order states are its eigenvectors, and the first-order energies are its eigenvalues.

This is the essence of degenerate perturbation theory: diagonalize the perturbation restricted to the degenerate subspace.

Effective Hamiltonian#

Define the projection operator onto the degenerate subspace and its complement:

\[P_n = \sum_{\alpha=1}^d |n,\alpha\rangle\langle n,\alpha|, \qquad Q_n = \hat{I} - P_n\]

Effective Hamiltonian

The effective Hamiltonian acting within the degenerate subspace is:

(85)#\[H_n^{\text{eff}} = E_n^{(0)}\hat{I}_d + \lambda\, P_n V P_n + \lambda^2\, P_n V Q_n \frac{1}{E_n^{(0)} - Q_n H_0 Q_n} Q_n V P_n + O(\lambda^3)\]

First-order (\(P_n V P_n\)): direct matrix elements of \(V\) within the subspace.
Second-order: virtual transitions out of the subspace and back, mediated by the resolvent.

In the basis \(\{|n,1\rangle,\ldots,|n,d\rangle\}\), the first-order effective Hamiltonian is a \(d\times d\) matrix:

\[(P_n V P_n)_{\alpha\beta} = \langle n,\alpha|V|n,\beta\rangle\]

Diagonalizing this matrix yields the first-order energy shifts \(\Delta E_{n,\alpha}^{(1)}\) (eigenvalues) and the correct zeroth-order states \(|n,\alpha'\rangle\) (eigenvectors).

Derivation: Second-Order Virtual Transitions

The second-order term arises from expanding the eigenvalue equation to \(O(\lambda^2)\). The intermediate states live in the \(Q_n\) subspace (outside the degenerate manifold), so the energy denominators \(E_n^{(0)} - E_m^{(0)} \neq 0\) are safe. The result is the same structure as non-degenerate second-order corrections, but now acting as a matrix within the degenerate subspace rather than giving a scalar shift.

Application: Spin-1 Model#

Example: Spin-1 Particle

Setup. A spin-1 system with \(H_0 = (S^z)^2\) and perturbation \(V = \lambda(S^x + S^z)\).

The unperturbed spectrum:

\(E = 1\): doubly degenerate (\(|{+}1\rangle\), \(|{-}1\rangle\))
\(E = 0\): non-degenerate (\(|0\rangle\))

Effective Hamiltonian for the \(E=1\) subspace (\(d=2\)):

\[\begin{split}P_1 V P_1 = \begin{pmatrix} \langle{+}1|V|{+}1\rangle & \langle{+}1|V|{-}1\rangle \\ \langle{-}1|V|{+}1\rangle & \langle{-}1|V|{-}1\rangle \end{pmatrix} = \begin{pmatrix} 1 & 1/\sqrt{2} \\ 1/\sqrt{2} & -1 \end{pmatrix}\end{split}\]

(using \(\langle{\pm}1|S^z|{\pm}1\rangle = \pm 1\), \(\langle{+}1|S^x|{-}1\rangle = 1/\sqrt{2}\), and \(\langle{\pm}1|S^x|{\pm}1\rangle = 0\).)

Diagonalization: \(\det(P_1VP_1 - \delta E\,\hat{I}) = 0\) gives \((\delta E)^2 = 3/2\), so:

\[\Delta E_{1,\pm}^{(1)} = \pm\sqrt{3/2}\]

The doubly degenerate level is completely lifted into two distinct levels separated by \(2\lambda\sqrt{3/2}\).

Example: Linear Stark Effect in Hydrogen

Problem. Apply a uniform electric field \(\boldsymbol{E} = \mathcal{E}_0\hat{z}\) to the hydrogen atom \(n=2\) level (fourfold degenerate, ignoring spin): \(|2,0,0\rangle\), \(|2,1,0\rangle\), \(|2,1,{\pm}1\rangle\).

Solution. The perturbation \(V = e\mathcal{E}_0 z = e\mathcal{E}_0 r\cos\theta\) has selection rules \(\Delta\ell = \pm 1\), \(\Delta m_\ell = 0\). Only the pair \(|2,0,0\rangle \leftrightarrow |2,1,0\rangle\) couples; the \(m_\ell = \pm 1\) states decouple completely. The \(4\times 4\) effective Hamiltonian is block-diagonal: a \(2\times 2\) block that splits linearly in \(\mathcal{E}_0\) (the linear Stark effect), plus two zero eigenvalues. The linear splitting occurs because the \(n=2\) subspace contains states of opposite parity (\(s\) and \(p\)), allowing first-order mixing.

Discussion

In degenerate perturbation theory, you diagonalize \(V\) within the degenerate subspace to find the “good basis.” But what if \(V\) itself has degenerate eigenvalues within that subspace? Can you determine the energy levels from first-order alone, or must you go to second order? Give an example where residual degeneracy is lifted only at second order.

Comparison: Degenerate vs Non-Degenerate#

Feature	Non-Degenerate	Degenerate
Unperturbed spectrum	All levels distinct	Contains degenerate subspaces
First-order energy	\(\langle n \vert V \vert n \rangle\) (scalar)	Eigenvalues of \(P_n V P_n\) (\(d\times d\) matrix)
Zeroth-order basis	Given uniquely	Must diagonalize \(V\) within subspace
Divergence	None (denominators finite)	Would diverge; prevented by diagonalization
Physical effect	Level shifts	Degeneracy lifting (full or partial)

Summary#

Non-degenerate perturbation theory diverges when the unperturbed spectrum has degeneracies (zero energy denominators).
The fix: diagonalize \(V\) within the degenerate subspace via the effective Hamiltonian \(H_n^{\text{eff}} = P_n V P_n\) to find the correct zeroth-order states and first-order energy splittings.
Virtual transitions to states outside the subspace enter at second order through the resolvent \(Q_n(E_n^{(0)} - H_0)^{-1}Q_n\).
The degeneracy may be fully or partially lifted depending on the symmetry of the perturbation.

Homework#

1. Show that the projection operator \(P = \sum_\alpha |n,\alpha\rangle\langle n,\alpha|\) onto a degenerate subspace satisfies \(P^2 = P\), \(Q^2 = Q\), and \(PQ = 0\), where \(Q = \hat{I} - P\).

2. A spin-1 particle has \(H_0 = (S^z)^2\) and perturbation \(V = \lambda(S^x + S^z)\). (a) Identify the degenerate subspace. (b) Construct the \(2\times 2\) effective Hamiltonian \(P_1VP_1\) in the \(\{|{+}1\rangle, |{-}1\rangle\}\) basis. (c) Diagonalize it to find the first-order energy shifts and the corrected eigenstates.

3. In the hydrogen atom \(n=2\) manifold, the perturbation \(V = e\mathcal{E}_0 z\) has selection rules \(\Delta\ell = \pm 1\), \(\Delta m_\ell = 0\). (a) Write the \(4\times 4\) matrix \(\langle 2,\ell',m'|V|2,\ell,m\rangle\) using \(\langle 2,0,0|z|2,1,0\rangle = -3a_0\). (b) Diagonalize to find the first-order energy shifts. (c) Which states remain degenerate, and why?

4. When does a perturbation partially lift a degeneracy (some eigenvalues of \(P_nVP_n\) remain equal)? Explain using symmetries and selection rules, and give a concrete example.

5. For the corrected zeroth-order states \(|n,\alpha'\rangle\) obtained by diagonalizing \(P_nVP_n\), show that the first-order state correction from non-degenerate perturbation theory can now be applied without divergences. What role does the diagonalization play in ensuring finite denominators?

6. A two-fold degenerate level has effective Hamiltonian \(H^{\text{eff}} = \begin{pmatrix} a & b \\ b^* & c \end{pmatrix}\). (a) Find the eigenvalues in terms of \(a\), \(b\), \(c\). (b) Under what condition does the perturbation fail to lift the degeneracy at first order? (c) Interpret geometrically on the Bloch sphere.

7. The second-order effective Hamiltonian includes the term \(P_n V Q_n (E_n^{(0)} - H_0)^{-1} Q_n V P_n\). (a) Explain why this represents “virtual transitions” out of and back into the degenerate subspace. (b) Show that this term is Hermitian. (c) For the spin-1 model of Problem 2, identify which outside state (\(|0\rangle\)) contributes and estimate the magnitude of the second-order correction relative to first order.