Degenerate Perturbation Theory

Degenerate perturbation theory is the correct first step when a perturbation acts inside a subspace of states with the same unperturbed energy. Ordinary nondegenerate formulas fail because they divide by zero energy differences and because the “right” zeroth-order states are not determined until the perturbation is considered.

The core rule is:

$$\text{Diagonalize }V\text{ inside the degenerate subspace.}$$

Required background. Use matrix diagonalization to find the eigenvectors of the projected perturbation and nondegenerate perturbation theory for mixing with remote levels.

The perturbation within a degenerate eigenspace

Let $H_0$ have a degenerate eigenspace $\mathcal D$ with energy $E^{(0)}$:

$$H_0\lvert a\rangle = E^{(0)}\lvert a\rangle, \qquad a=1,\ldots,d.$$

Let $P$ be the projector onto this subspace:

$$P=\sum_{a=1}^d\lvert a\rangle\langle a\rvert.$$

The Hamiltonian is

$$H(\lambda)=H_0+\lambda V.$$

Take $\lambda$ real and dimensionless, $H_0$ and $V$ self-adjoint, and the displayed basis orthonormal. Assume that $\mathcal D$ is finite dimensional and isolated by a positive gap from the complementary spectrum. For unbounded operators, the operator or quadratic-form family must also permit the local expansion; spectral isolation alone is insufficient.

Inside $\mathcal D$, define the perturbation matrix

$$W_{ab} = \langle a|V|b\rangle.$$

Equivalently,

$$W=PVP$$

as an operator on the degenerate subspace.

Why the projected perturbation selects the states

Seek one branch in the form

$$\begin{aligned} |\psi_\alpha(\lambda)\rangle &=|\alpha\rangle+\lambda|\alpha^{(1)}\rangle+O(\lambda^2),\\ E_\alpha(\lambda) &=E^{(0)}+\lambda w_\alpha+\lambda^2 E_\alpha^{(2)}+O(\lambda^3), \qquad P|\alpha\rangle=|\alpha\rangle. \end{aligned}$$

The coefficient of $\lambda$ in the eigenvalue equation is

$$(H_0-E^{(0)})|\alpha^{(1)}\rangle =(w_\alpha-V)|\alpha\rangle.$$

Since $P(H_0-E^{(0)})=0$, its $P$ projection is a solvability condition:

$$W\lvert \alpha\rangle = w_\alpha\lvert \alpha\rangle, \qquad \lvert\alpha\rangle\in\mathcal D.$$

Then the first-order energies are

$$E_\alpha = E^{(0)}+\lambda w_\alpha+O(\lambda^2).$$

Thus diagonalizing $W$ is required for a continuous eigenstate branch to start from $|\alpha\rangle$; it is not an extra approximation. The perturbation splits the degeneracy at first order when the eigenvalues $w_\alpha$ differ. If some remain equal, $W$ has not yet selected a basis within that residual subspace: higher orders or additional symmetry information must do so.

Why Nondegenerate Formulas Fail

The nondegenerate first-order state correction contains terms

$$\frac{V_{mn}} {E_n^{(0)}-E_m^{(0)}}.$$

For states $m$ inside the degenerate subspace, the denominator is zero. This is not a removable technical problem. It says the perturbation can produce order-$1$ rotations inside the degenerate subspace even when $\lambda$ is small.

The right basis must be chosen before expanding outside the subspace.

Two-State Example

Suppose two states $\lvert1\rangle,\lvert2\rangle$ share energy $E^{(0)}$ and the perturbation matrix is

$$W = \begin{pmatrix} a&b\\ b^*&c \end{pmatrix}.$$

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

$$w_\pm = \frac{a+c}{2} \pm \sqrt{ \left(\frac{a-c}{2}\right)^2 + \lvert b\rvert^2 }.$$

If $a=c$ and $b\ne0$, the good zeroth-order states are symmetric and antisymmetric combinations up to the phase of $b$, and the physical splitting is $2|\lambda|\lvert b\rvert$ to first order. The two coefficient eigenvalues differ by $2|b|$.

Role of Symmetry

Symmetry often explains degeneracy and constrains $W$. If $V$ respects the symmetry responsible for the degeneracy, $W$ may be proportional to the identity inside an irreducible multiplet, and the degeneracy may remain at first order.

If $V$ breaks that symmetry, $W$ usually splits the multiplet according to the remaining symmetry.

This is why perturbation theory and symmetry should be used together. Diagonalizing a large perturbation matrix without recognizing block structure is both inefficient and less informative.

Mixing outside and inside the original eigenspace

Set $Q=I-P$. The $Q$ projection of the first-order equation gives

$$(E^{(0)}-QH_0Q)Q|\alpha^{(1)}\rangle=QV|\alpha\rangle.$$

The inverse acts only on $Q\mathcal H$. In a discrete complementary spectrum this determines

$$\lvert\alpha^{(1)}_\perp\rangle = \sum_{r\notin\mathcal D} \frac{\langle r|V|\alpha\rangle} {E^{(0)}-E_r^{(0)}} \lvert r\rangle.$$

Continuum states, if present, contribute the corresponding spectral integral. This equation fixes only the component outside $\mathcal D$; it cannot determine $P|\alpha^{(1)}\rangle$ because $H_0-E^{(0)}$ vanishes there.

To find that remaining component, project the second-order eigenvalue equation onto $P$. Substituting the complementary component just found yields

$$(W-w_\alpha I_P)P|\alpha^{(1)}\rangle =-K|\alpha\rangle+E_\alpha^{(2)}|\alpha\rangle,$$

where

$$K=PVQ(E^{(0)}-QH_0Q)^{-1}QVP.$$

For distinct $w_\alpha$, projection onto $\langle\alpha|$ and $\langle\beta|$ gives, respectively,

$$E_\alpha^{(2)}=K_{\alpha\alpha}, \qquad \langle\beta|\alpha^{(1)}\rangle =\frac{K_{\beta\alpha}}{w_\alpha-w_\beta} \quad(\beta\ne\alpha).$$

Choose intermediate normalization $\langle\alpha|\psi_\alpha(\lambda)\rangle=1$ to set $\langle\alpha|\alpha^{(1)}\rangle=0$. The full first-order correction then contains both the external sum and the internal sum over $\beta$. The internal rotation is still first order in the state: a second-order induced coupling $\lambda^2K_{\beta\alpha}$ is divided by a first-order splitting $\lambda(w_\alpha-w_\beta)$. Zwiebach’s notes, §1.2.2, derive this distinction explicitly.

If slopes coincide, the projected equation instead requires diagonalizing $K$ within their common eigenspace. If they are distinct but $|\lambda K_{\beta\alpha}|$ is comparable with $|w_\alpha-w_\beta|$ at the desired coupling, diagonalize the retained matrix $E^{(0)}I_P+\lambda W+\lambda^2K$ rather than trusting a small internal rotation. Its truncation still needs control of the omitted orders and the gap to $Q$.

A three-level model checked exactly

Let two states of energy zero couple to one remote level $|r\rangle$ of energy $\Delta>0$, with real $g_1,g_2$. Specify the entire model by

$$H(\lambda)= \begin{pmatrix} 0&0&\lambda g_1\\ 0&0&\lambda g_2\\ \lambda g_1&\lambda g_2&\Delta \end{pmatrix}.$$

Here $PVP=0$, but

$$K=-\frac1\Delta \begin{pmatrix}g_1^2&g_1g_2\\g_1g_2&g_2^2\end{pmatrix}.$$

For $g=\sqrt{g_1^2+g_2^2}>0$, its normalized eigenvectors are

$$|B\rangle=\frac{g_1|1\rangle+g_2|2\rangle}{g}, \qquad |D\rangle=\frac{-g_2|1\rangle+g_1|2\rangle}{g}.$$

The coupled combination $|B\rangle$ has second-order coefficient $-g^2/\Delta$. The orthogonal combination $|D\rangle$ is dark: its two amplitudes to reach $|r\rangle$ cancel. In the ordered basis $(|D\rangle,|B\rangle,|r\rangle)$, the full Hamiltonian is

$$H=0\oplus \begin{pmatrix}0&\lambda g\\\lambda g&\Delta\end{pmatrix}.$$

Consequently $|D\rangle$ is an exact zero-energy eigenstate, while the branch from $|B\rangle$ has energy

$$E_B(\lambda) =\frac{\Delta-\sqrt{\Delta^2+4\lambda^2g^2}}2 =-\frac{\lambda^2g^2}{\Delta} +\frac{\lambda^4g^4}{\Delta^3}+O(\lambda^6).$$

This independently checks the negative sign, the collective coupling $g$, and the order of the neglected energy term. The full bright-branch state also contains a remote amplitude $-\lambda g/\Delta+O(\lambda^3)$ after unit normalization. Its remote probability is

$$p_r=\frac12\left(1-\frac{\Delta}{\sqrt{\Delta^2+4\lambda^2g^2}}\right) =\frac{\lambda^2g^2}{\Delta^2}+O(\lambda^4).$$

A vanishing first-order matrix therefore neither preserves the degeneracy nor prevents first-order leakage in the ket. The dark state stays exact because of the specified coupling structure; generic extra matrix elements need not preserve it.

Validity

Degenerate perturbation theory assumes that the degenerate subspace is well separated from states outside it:

$$\left| \frac{\lambda\langle r|V|\alpha\rangle} {E^{(0)}-E_r^{(0)}} \right| \ll1$$

for relevant outside states $r$, with a small total squared admixture as well. The cluster must be isolated from the full complementary spectrum, and the operator family must permit the expansion. If nearby outside states also mix strongly, enlarge the subspace as in quasi-degenerate perturbation theory.

Exercises

1. Diagonalize

$$W = \begin{pmatrix} 0&v\\ v&0 \end{pmatrix}$$

for real $v$ and find the first-order energy shifts.

Solution

The eigenvectors are

$$\frac{1}{\sqrt2} (\lvert1\rangle+\lvert2\rangle)$$

with eigenvalue $v$, and

$$\frac{1}{\sqrt2} (\lvert1\rangle-\lvert2\rangle)$$

with eigenvalue $-v$. Thus the first-order energies are

$$E_\pm = E^{(0)}\pm\lambda v+O(\lambda^2).$$

2. In the basis $(|1\rangle,|2\rangle,|r\rangle)$, take

$$H_0=\operatorname{diag}(0,0,\Delta), \qquad V=\begin{pmatrix}a&0&g_1\\0&c&g_2\\g_1&g_2&0\end{pmatrix},$$

where all entries are real, $\Delta>0$, and $a\ne c$. Find the energy through second order and the state through first order for the branch starting at $|1\rangle$. Explain why keeping only the remote-state admixture misses part of the answer.

Solution

The first-order slopes are $a,c$, and $K_{ij}=-g_i g_j/\Delta$. Thus

$$E_1(\lambda)=\lambda a-\lambda^2\frac{g_1^2}{\Delta}+O(\lambda^3),$$

and, in intermediate normalization,

$$|\psi_1(\lambda)\rangle =|1\rangle-\lambda\frac{g_1g_2}{\Delta(a-c)}|2\rangle -\lambda\frac{g_1}{\Delta}|r\rangle+O(\lambda^2).$$

The two visits to the remote state induce a coupling between $|1\rangle$ and $|2\rangle$, even though $V_{12}=0$. The resulting rotation is order $\lambda$ because their energy splitting is also order $\lambda$. At $a=c$, this expression is invalid: diagonalize $K$ inside the remaining degenerate pair. For distinct slopes, both $|\lambda g_i|/\Delta$ and $|\lambda g_1g_2/[\Delta(a-c)]|$ must be small for this state expansion to be useful.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• B. Zwiebach, “Non-degenerate and Degenerate Perturbation Theory”, Quantum Physics III, MIT OpenCourseWare 8.06, 2018, §§1.2.2–1.2.3 — state corrections when degeneracy is lifted at first or second order.