---
title: Time-Independent Perturbation Theory
module: Approximation Methods for Bound States
moduleNumber: 10
lessonNumber: 1
order: 1001
summary: >
  Almost no realistic Hamiltonian can be solved exactly. Perturbation theory
  treats a hard Hamiltonian as a solvable one plus a small correction and
  expands the eigenvalues and eigenstates in powers of that correction. We
  derive the first- and second-order energy shifts and the first-order state
  correction for a nondegenerate level, expose the small-denominator failure
  that degeneracy forces, and fix it by diagonalizing the perturbation inside
  the degenerate subspace to find the "good" zeroth-order states.
topics: [Approximation Methods for Bound States]
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 7; §7.1 Nondegenerate Perturbation Theory, §7.2 Degenerate Perturbation Theory"
  - book: Shankar
    ref: "Ch. 17 — The Variational and WKB Methods... §17.1 (perturbation series)"
  - book: Sakurai & Napolitano
    ref: "Ch. 5; §5.1 Time-Independent Perturbation Theory: Nondegenerate Case, §5.2 Degenerate Case"
draft: false
---

The catalogue of exactly solvable Hamiltonians is short: the free particle, the
[square well](/quantum-mechanics/wave-mechanics-1d/particle-in-infinite-and-finite-square-wells),
the [harmonic oscillator](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator),
the [hydrogen atom](/quantum-mechanics/central-potentials/the-hydrogen-atom),
and a handful of others. Every real system deviates from these idealizations: a
charged oscillator feels a stray field, hydrogen has relativistic and spin
corrections, an atom sits in a laboratory magnet. Perturbation theory computes
the response of a solved problem to a small addition without solving the new
problem from scratch. The strategy is to write the Hamiltonian as a solvable
piece plus a correction scaled by a bookkeeping parameter, expand the
eigenvalues and eigenstates as power series in that parameter, and match order
by order.

## The perturbation expansion

Split the Hamiltonian into an unperturbed part with known spectrum and a
perturbation,

$$
\hat H = \hat H^0 + \lambda \hat H', \qquad
\hat H^0\ket{n^0} = E_n^0\ket{n^0},
$$

where the $\ket{n^0}$ are orthonormal and complete, and $\lambda$ is a
dimensionless parameter carried through the algebra to track orders; it is set
to $1$ at the end, so $\hat H'$ is the physical perturbation. Assume the exact
eigenvalues and eigenstates deform smoothly from the unperturbed ones and admit
power series in $\lambda$,

$$
E_n = E_n^0 + \lambda E_n^1 + \lambda^2 E_n^2 + \cdots, \qquad
\ket{n} = \ket{n^0} + \lambda\ket{n^1} + \lambda^2\ket{n^2} + \cdots .
$$

The superscript counts the order in $\lambda$, not a power. Substituting both
series into $\hat H\ket{n} = E_n\ket{n}$ and collecting terms of equal order
produces one equation per power of $\lambda$:

$$
\begin{aligned}
\lambda^0:&\quad \hat H^0\ket{n^0} = E_n^0\ket{n^0}, \\
\lambda^1:&\quad \hat H^0\ket{n^1} + \hat H'\ket{n^0} = E_n^0\ket{n^1} + E_n^1\ket{n^0}, \\
\lambda^2:&\quad \hat H^0\ket{n^2} + \hat H'\ket{n^1} = E_n^0\ket{n^2} + E_n^1\ket{n^1} + E_n^2\ket{n^0}.
\end{aligned}
$$

The zeroth-order equation is the solved problem. The higher-order equations
each determine one more term in the two series. A choice of phase and
normalization removes the ambiguity in $\ket{n}$: fix
**intermediate normalization** $\braket{n^0 | n} = 1$, which forces every
correction to be orthogonal to the unperturbed state,
$\braket{n^0 | n^k} = 0$ for $k \ge 1$. The physical state is renormalized to
unit length only at the end.

$$
% caption: Each order of the expansion feeds the next: the solved zeroth-order
% level and state seed the first-order corrections, which seed the second, so
% the series is built one power of the coupling at a time.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=17mm, minimum height=9mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (z) at (0,0) {order $0$};
\node[box] (o) at (3.0,0) {order $1$};
\node[box] (t) at (6.0,0) {order $2$};
\node[box] (m) at (9.0,0) {order $k$};
\draw[->, thick] (z) -- (o);
\draw[->, thick] (o) -- (t);
\draw[->, thick, dashed] (t) -- (m);
\node[anchor=north, black!70] at (3.0,-0.75) {$E_n^0$ known};
\node[anchor=north, black!70] at (6.0,-0.75) {$E_n^1$ found};
\end{tikzpicture}
$$

## First-order corrections

Project the first-order equation onto $\bra{n^0}$. The term
$\bra{n^0}\hat H^0\ket{n^1}$ equals $E_n^0\braket{n^0 | n^1}$ because $\hat H^0$
is Hermitian and acts to the left on its own eigenbra, and it cancels the
matching term on the right. What survives is the first-order energy shift.

> **Theorem (First-order energy shift).** The leading correction to a
> nondegenerate level is the expectation value of the perturbation in the
> unperturbed state,
> $$E_n^1 = \bra{n^0}\hat H'\ket{n^0}.$$

This is the single most-used result of the theory: to first order, the energy
moves by the average of the perturbation over the unperturbed probability
density. No new wavefunction is needed. For an oscillator nudged by a weak
quartic term $\hat H' = \gamma\,\hat x^4$, the ground-state shift is
$E_0^1 = \gamma\braket{0 | \hat x^4 | 0} = 3\gamma\,(\hbar/2m\omega)^2$, computed
entirely from the known ground state.

The first-order **state** correction requires the components of $\ket{n^1}$
along the other unperturbed states. Expand $\ket{n^1} = \sum_{m\neq n} c_m\ket{m^0}$
(the $n$ component vanishes by intermediate normalization) and project the
first-order equation onto $\bra{m^0}$ with $m\neq n$:

$$
E_m^0 c_m + \bra{m^0}\hat H'\ket{n^0} = E_n^0 c_m
\;\Longrightarrow\;
c_m = \frac{\bra{m^0}\hat H'\ket{n^0}}{E_n^0 - E_m^0}.
$$

> **Theorem (First-order state correction).** For a nondegenerate level,
> $$\ket{n^1} = \sum_{m\neq n} \frac{\bra{m^0}\hat H'\ket{n^0}}{E_n^0 - E_m^0}\,\ket{m^0}.$$

The perturbation mixes into $\ket{n}$ a little of every other unperturbed state,
weighted by the coupling matrix element $\hat H'_{mn} = \bra{m^0}\hat H'\ket{n^0}$
and divided by the energy gap. States close in energy mix strongly; distant
states barely contribute. The denominator $E_n^0 - E_m^0$ is the seed of both
the method's power and its failure: it is finite only when no other state shares
the energy $E_n^0$ — the nondegeneracy assumption, restated.

$$
% caption: The perturbation admixes other levels into the state with weight set
% by the coupling divided by the energy gap; the nearest levels dominate and the
% far ones are suppressed by the large denominator.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% central state n
\node[anchor=east, acc] at (-0.2,0) {level $n$};
\draw[acc, very thick] (0,0) -- (2.2,0);
% neighbours at various gaps
\draw[black, thick] (0,1.0) -- (2.2,1.0);
\draw[black, thick] (0,-0.9) -- (2.2,-0.9);
\draw[black, thick] (0,2.3) -- (2.2,2.3);
\draw[black, thick] (0,-2.1) -- (2.2,-2.1);
% mixing arrows, thicker = stronger
\draw[line width=2.2pt, <->] (2.4,0.05) -- (2.4,0.95);
\draw[line width=2.2pt, <->] (2.6,-0.05) -- (2.6,-0.85);
\draw[line width=0.7pt, <->] (2.9,0.1) -- (2.9,2.25);
\draw[line width=0.7pt, <->] (3.1,-0.1) -- (3.1,-2.05);
\node[anchor=west, black!70] at (3.3,1.2) {near: strong mixing};
\node[anchor=west, black!70] at (3.3,2.3) {far: weak mixing};
\end{tikzpicture}
$$

## Second-order energy

The first-order shift can vanish by symmetry, and even when it does not, the
next term measures how the admixed states pull on the energy. Project the
second-order equation onto $\bra{n^0}$. The $\hat H^0$ and $E_n^1$ terms drop by
the same cancellation and orthogonality used before, leaving
$E_n^2 = \bra{n^0}\hat H'\ket{n^1}$. Insert the first-order state:

> **Theorem (Second-order energy shift).**
> $$E_n^2 = \sum_{m\neq n} \frac{\bigl\lvert \bra{m^0}\hat H'\ket{n^0}\bigr\rvert^2}{E_n^0 - E_m^0}.$$

Every term is a squared magnitude over a signed gap. Two structural facts follow
at once.

- **The ground state always moves down.** For $n$ the lowest level every
  denominator $E_0^0 - E_m^0$ is negative, so $E_0^2 \le 0$: second-order
  perturbation theory can only depress the ground-state energy, consistent with
  the [variational bound](/quantum-mechanics/approximation-methods/the-variational-method).
- **Levels repel.** A pair of coupled levels each push the other away. The lower
  of the two receives a negative shift (its partner sits above, negative
  denominator) and the upper a positive shift, so second order increases their
  separation. This **level repulsion** is why exact crossings of coupled levels
  are avoided.

$$
% caption: Second order pushes coupled levels apart: the lower member is
% depressed and the upper raised by equal and opposite amounts set by the
% coupling squared over the gap, so the pair repels rather than crosses.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={align=center}]
\definecolor{acc}{HTML}{4A6FA5}
% unperturbed pair (close together) on left
\draw[black, thick] (0,1.6) -- (1.6,1.6);
\draw[black, thick] (0,1.1) -- (1.6,1.1);
\node[anchor=east, black!70] at (0,1.35) {$\hat H^0$};
% perturbed pair (spread apart) on right
\draw[acc, very thick] (3.6,2.2) -- (5.2,2.2);
\draw[acc, very thick] (3.6,0.5) -- (5.2,0.5);
\node[anchor=west, acc] at (5.3,2.2) {raised};
\node[anchor=west, acc] at (5.3,0.5) {lowered};
% connecting dashed
\draw[black, dashed] (1.6,1.6) -- (3.6,2.2);
\draw[black, dashed] (1.6,1.1) -- (3.6,0.5);
\draw[acc, <->] (2.6,0.7) -- (2.6,2.0);
\node[anchor=west, black!70] at (2.65,1.35) {repulsion};
\end{tikzpicture}
$$

The full physical eigenvalue through second order, with $\lambda = 1$, reads

$$
E_n = E_n^0 + \hat H'_{nn} + \sum_{m\neq n}\frac{\lvert \hat H'_{mn}\rvert^2}{E_n^0 - E_m^0} + \cdots .
$$

> **Worked example.** An infinite square well of width $a$ carries a thin spike
> at its center, $\hat H' = \alpha\,\delta(x - a/2)$. The unperturbed states are
> $\psi_n^0(x) = \sqrt{2/a}\,\sin(n\pi x/a)$ with $E_n^0 = n^2\pi^2\hbar^2/2ma^2$.
> The first-order shift is the perturbation evaluated at the spike,
> $$
> E_n^1 = \alpha\,\lvert\psi_n^0(a/2)\rvert^2 = \frac{2\alpha}{a}\sin^2\!\frac{n\pi}{2}
> = \begin{cases} 2\alpha/a, & n \text{ odd},\\[2pt] 0, & n \text{ even}.\end{cases}
> $$
> Odd states have an antinode at the center and feel the spike; even states have
> a node there and are untouched to first order. A repulsive spike ($\alpha>0$)
> raises every odd level and leaves the even ladder in place, a splitting pattern
> that a symmetry argument (parity about the center) predicts before any integral
> is done.

> **Worked example.** A charged oscillator in a uniform field tests the second-order
> formula against an exact answer. Take $\hat H^0 = \hat p^2/2m + \tfrac12 m\omega^2\hat x^2$
> and $\hat H' = -qE\hat x$. The first-order shift vanishes for every level, since
> $\hat x$ is odd and $\braket{n^0 | \hat x | n^0} = 0$. The second-order shift needs
> the [ladder matrix elements](/quantum-mechanics/oscillator-and-symmetry/ladder-operators-and-the-number-states)
> of $\hat x$, nonzero only for $m = n\pm 1$, so with $E_n^0 - E_{n\pm1}^0 = \mp\hbar\omega$,
> $$
> E_n^2 = q^2E^2\left[\frac{\lvert\braket{n{+}1 | \hat x | n}\rvert^2}{-\hbar\omega}
> + \frac{\lvert\braket{n{-}1 | \hat x | n}\rvert^2}{+\hbar\omega}\right]
> = q^2E^2\,\frac{\hbar}{2m\omega}\left[\frac{n+1}{-\hbar\omega} + \frac{n}{\hbar\omega}\right]
> = -\frac{q^2E^2}{2m\omega^2}.
> $$
> Every higher order vanishes. Completing the square,
> $\hat H = \hat p^2/2m + \tfrac12 m\omega^2(\hat x - qE/m\omega^2)^2 - q^2E^2/2m\omega^2$,
> confirms the exact spectrum is the unperturbed ladder shifted rigidly by
> $-q^2E^2/2m\omega^2$. Perturbation theory is exact to second order here, a rare
> case where the series terminates.

## The validity condition

The expansion is an approximation, useful when successive terms shrink. The
ratio of the first-order state correction's typical coefficient to unity, or
equivalently of the second-order energy to the first-order gap, gives the
criterion.

> **Definition (Perturbative validity).** Nondegenerate perturbation theory
> converges rapidly when every coupling matrix element is small against the
> corresponding level spacing,
> $$
> \bigl\lvert \bra{m^0}\hat H'\ket{n^0}\bigr\rvert \ll \bigl\lvert E_n^0 - E_m^0\bigr\rvert
> \quad (m\neq n).
> $$

The perturbation need not be small in absolute terms; it must be small compared
to the gaps it bridges. A weak coupling between two nearly degenerate levels can
still be "large" in this sense, and the series then fails. The remedy for the
extreme case of exact degeneracy is the subject of the rest of the lesson.

## Degeneracy breaks the formula

Suppose $E_n^0$ is $g$-fold degenerate, with an orthonormal basis
$\ket{n^0_1}, \dots, \ket{n^0_g}$ of the degenerate subspace. The nondegenerate
formulas contain denominators $E_n^0 - E_m^0$ that vanish whenever $\ket{m^0}$ is
another state of the same energy. Unless the offending numerator
$\bra{m^0}\hat H'\ket{n^0}$ also vanishes, the first-order state correction is
infinite and the derivation collapses.

The failure is not physical but a symptom of a bad starting point. When a level
is degenerate, the unperturbed problem does not single out a preferred basis
inside the degenerate subspace: any orthonormal combination diagonalizes
$\hat H^0$ equally well. The perturbation does select a preferred basis, and
starting the expansion in the wrong one produces the singular denominators.
The task is to identify the **good states**, the particular zeroth-order
combinations toward which the exact eigenstates collapse as $\lambda \to 0$.

$$
% caption: A degenerate level admits any basis for its subspace under the
% unperturbed Hamiltonian; the perturbation lifts the degeneracy and picks out
% the unique good combinations that the exact states approach as the coupling
% is switched off.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% degenerate level
\draw[black, line width=2.4pt] (0,1.4) -- (2.4,1.4);
\node[anchor=east, black!70] at (0,1.4) {$g$-fold};
% arrow
\draw[->, acc, thick] (2.9,1.4) -- (4.3,1.4);
\node[anchor=south, black!70] at (3.6,1.5) {perturb.};
% split levels
\draw[acc, very thick] (4.8,2.2) -- (6.6,2.2);
\draw[acc, very thick] (4.8,1.4) -- (6.6,1.4);
\draw[acc, very thick] (4.8,0.6) -- (6.6,0.6);
\node[anchor=west, black!70] at (6.7,2.2) {good state $1$};
\node[anchor=west, black!70] at (6.7,1.4) {good state $2$};
\node[anchor=west, black!70] at (6.7,0.6) {good state $3$};
\end{tikzpicture}
$$

## Degenerate perturbation theory

Work inside the degenerate subspace. A good state is a combination
$\ket{\phi} = \sum_{j=1}^{g} c_j \ket{n^0_j}$ whose first-order energy shift is
well defined. Repeat the first-order derivation, but now project the
first-order equation onto each degenerate basis bra $\bra{n^0_i}$. The
$\hat H^0$ terms still cancel, and because the exact state approaches $\ket{\phi}$
the equation becomes an eigenvalue problem for the coefficients:

$$
\sum_{j=1}^{g} W_{ij}\, c_j = E^1 c_i, \qquad
W_{ij} \equiv \bra{n^0_i}\hat H'\ket{n^0_j}.
$$

> **Theorem (Degenerate perturbation theory, first order).** The first-order
> energy shifts of a $g$-fold degenerate level are the eigenvalues of the
> $g\times g$ Hermitian matrix $W_{ij} = \bra{n^0_i}\hat H'\ket{n^0_j}$ of the
> perturbation restricted to the degenerate subspace. The corresponding
> eigenvectors give the good zeroth-order states.

Diagonalizing $W$ replaces the singular sum with a finite linear-algebra
problem. The perturbation is projected onto the subspace, the resulting matrix
is diagonalized, and its eigenvalues are the split levels. Degeneracy is lifted
completely when the eigenvalues are distinct; a repeated eigenvalue signals a
residual degeneracy that a higher order, or a second perturbation, may still
break.

$$
% caption: Degenerate perturbation theory restricts the perturbation to the
% degenerate subspace, forming the matrix of its matrix elements; the eigenvalues
% of that block are the first-order splittings.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% draw a 2x2 matrix W
\draw[thick] (0,0) rectangle (3.2,2.0);
\draw[black] (1.6,0) -- (1.6,2.0);
\draw[black] (0,1.0) -- (3.2,1.0);
\node at (0.8,1.5) {$W_{aa}$};
\node at (2.4,1.5) {$W_{ab}$};
\node at (0.8,0.5) {$W_{ba}$};
\node at (2.4,0.5) {$W_{bb}$};
\node[anchor=east] at (-0.2,1.0) {$W =$};
% arrow to eigenvalues
\draw[->, acc, thick] (3.7,1.0) -- (5.1,1.0);
\node[anchor=south, black!70] at (4.4,1.05) {diagonalize};
\draw[acc, very thick] (5.6,1.6) -- (7.0,1.6);
\draw[acc, very thick] (5.6,0.4) -- (7.0,0.4);
\node[anchor=west, black!70] at (7.1,1.6) {$E_{+}^1$};
\node[anchor=west, black!70] at (7.1,0.4) {lower shift};
\end{tikzpicture}
$$

For the two-fold case the eigenvalues have a closed form. With diagonal
elements $W_{aa}, W_{bb}$ and off-diagonal coupling $W_{ba} = W_{ab}^\ast$,

$$
E_\pm^1 = \frac12\left(W_{aa} + W_{bb}\right)
\pm \frac12\sqrt{\left(W_{aa} - W_{bb}\right)^2 + 4\lvert W_{ab}\rvert^2}.
$$

The splitting $E_+^1 - E_-^1 = \sqrt{(W_{aa}-W_{bb})^2 + 4\lvert W_{ab}\rvert^2}$ never
closes when the off-diagonal coupling $W_{ab}$ is nonzero: even states with equal
diagonal elements split by $2\lvert W_{ab}\rvert$. This is the two-level avoided crossing in
its simplest algebraic form, the same structure that governs the
[Zeeman and Stark effects](/quantum-mechanics/approximation-methods/the-zeeman-and-stark-effects)
and molecular bonding.

## Finding good states from symmetry

Diagonalizing $W$ is unnecessary when a symmetry already labels the degenerate
states. Suppose an observable $\hat A$ commutes with both $\hat H^0$ and
$\hat H'$, and the degenerate basis states are chosen as eigenstates of $\hat A$
with distinct eigenvalues $a_i \neq a_j$. Then

$$
(a_i - a_j)\,W_{ij} = \bra{n^0_i}[\hat A, \hat H']\ket{n^0_j} = 0,
$$

so $W_{ij} = 0$ for $i \neq j$: the matrix is already diagonal in the eigenbasis
of $\hat A$, and those states are the good ones.

> **Theorem (Good states from a shared symmetry).** If a Hermitian operator
> $\hat A$ commutes with $\hat H^0$ and with $\hat H'$, the simultaneous
> eigenstates of $\hat H^0$ and $\hat A$ are good zeroth-order states, and their
> first-order shifts are $E_i^1 = W_{ii} = \bra{n^0_i}\hat H'\ket{n^0_i}$.

This converts a diagonalization into an inspection. In hydrogen the perturbing
interactions of the [next lesson](/quantum-mechanics/approximation-methods/fine-structure-and-the-real-hydrogen-atom)
commute with the total angular momentum, so the coupled
$\ket{n,\ell,j,m_j}$ basis is good and the fine-structure shifts are read off as
diagonal expectation values, no $g\times g$ matrix required. Choosing the right
symmetry-adapted basis before perturbing is the practical heart of the degenerate
theory.

$$
% caption: When a symmetry operator commutes with both the unperturbed
% Hamiltonian and the perturbation, its distinct eigenvalues label the degenerate
% states and force the perturbation matrix diagonal, so those eigenstates are
% already the good ones.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=15mm, minimum height=9mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (a) at (0,0) {$\hat A$ commutes\\with $\hat H^0$};
\node[box] (b) at (3.4,0) {and with the\\perturbation};
\node[box, draw=acc, text=acc] (c) at (7.2,0) {$W$ diagonal};
\draw[->, acc, thick] (a) -- (b);
\draw[->, acc, thick] (b) -- (c);
\node[anchor=north, black!70] at (3.6,-0.8) {distinct eigenvalues of $\hat A$};
\end{tikzpicture}
$$

## Second order and the structure of the series

Beyond first order the degenerate and nondegenerate theories rejoin. Once the
good states diagonalize $W$ and split the level, the remaining states are
nondegenerate (assuming distinct eigenvalues) and the second-order formula
applies with the good states as the unperturbed basis, the sum running over all
states outside the original subspace. When a degeneracy survives to first order,
second-order degenerate perturbation theory diagonalizes an effective operator
built from couplings to the external states, but the same principle governs:
find the basis the perturbation prefers, then expand.

The perturbation series is asymptotic, not convergent, for most physical
Hamiltonians: the terms shrink at first and eventually grow, so truncating at
low order is accurate while summing to all orders diverges. For the systems of
this module the first two orders already capture fine structure, the Zeeman and
Stark shifts, and molecular binding to spectroscopic accuracy. The
[variational method](/quantum-mechanics/approximation-methods/the-variational-method)
supplies a complementary tool bounding the ground state from above, and the
[WKB approximation](/quantum-mechanics/approximation-methods/the-wkb-approximation)
handles the semiclassical regime where no small parameter multiplies a clean
perturbation.

[^griffiths-pt]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_ 3rd ed., §7.1 — the nondegenerate expansion, first- and second-order corrections, intermediate normalization; §7.2 — degenerate perturbation theory, the $W$ matrix, and good states from a commuting observable. Cambridge, 2018. [Publisher listing](https://www.cambridge.org/highereducation/books/introduction-to-quantum-mechanics/990799CA07A83FC5312402AF089777B7).
[^shankar-pt]: **Shankar**, _Principles of Quantum Mechanics_ 2nd ed., Ch. 17 — the perturbation series derived by projection, the validity criterion, and the degenerate case as a subspace diagonalization. Springer, 1994. [SpringerLink](https://link.springer.com/book/10.1007/978-1-4757-0576-8).
[^sakurai-pt]: **Sakurai & Napolitano**, _Modern Quantum Mechanics_ 3rd ed., §5.1–§5.2 — Rayleigh–Schrödinger perturbation theory, the second-order shift, level repulsion, and the degenerate case handled by diagonalizing the perturbation in the degenerate subspace. Cambridge, 2021.