---
title: Accidental Degeneracy and the Runge-Lenz Symmetry
module: The Quantum Hydrogen Atom
moduleNumber: 2
lessonNumber: 4
order: 204
summary: >
  Hydrogen energies depend only on n, so states of different ℓ at the same n are
  degenerate. This is not a coincidence but the mark of a hidden symmetry: the
  quantum Runge-Lenz vector is conserved for the 1/r potential alone, and
  together with angular momentum it generates the group SO(4). The Casimir
  invariant of that group reproduces E = −Z²Ry/n² and its representations count
  the n² states. Any departure from 1/r breaks the symmetry and lifts the
  ℓ-degeneracy.
topics: [The Quantum Hydrogen Atom]
sources:
  - book: Bransden & Joachain
    ref: "Ch. 3 — The Hydrogen Atom; the ℓ-degeneracy and its symmetry origin"
  - book: Griffiths & Schroeter
    ref: "Ch. 4 — Quantum Mechanics in Three Dimensions; §4.2 and problems on the Runge-Lenz vector"
  - book: Foot
    ref: "Ch. 2 — The Hydrogen Atom; degeneracy of the levels"
draft: false
---

The [radial solution](/atomic-physics/quantum-hydrogen-atom/radial-equation-in-full)
produced energies $E_n = -Z^2\mathrm{Ry}/n^2$ that ignore $\ell$. For a general
central potential the energy depends on both quantum numbers, $E_{n\ell}$,
because rotational symmetry alone guarantees only that levels of the same $\ell$
share their $2\ell+1$ values of $m$. Hydrogen collapses the $\ell$ label as well:
the $2s$ and $2p$ states, the $3s$, $3p$, and $3d$ states, are exactly
degenerate. A degeneracy beyond what the obvious symmetry requires is called
**accidental**, and in quantum mechanics an accidental degeneracy is almost
always the fingerprint of a symmetry that was not obvious. For hydrogen the
extra symmetry is generated by the Runge-Lenz vector, and the full symmetry
group is $SO(4)$ rather than the rotational $SO(3)$.

## The classical constant of motion

In the Kepler problem a particle of reduced mass $\mu$ in the potential
$V = -\gamma/r$, with $\gamma = kZe^2$, conserves energy and angular momentum
$\vec L = \vec r \times \vec p$. It conserves one more vector, the
**Laplace-Runge-Lenz vector**

$$
\vec A = \vec p \times \vec L - \mu\gamma\,\hat r.
$$

Direct differentiation using Newton's equation $\dot{\vec p} = -\gamma\hat r/r^2$
gives $\d\vec A/\d t = 0$ for the $1/r$ force and only for it. The vector lies in
the orbital plane (since $\vec A\cdot\vec L = 0$), points from the focus toward
perihelion along the major axis, and has magnitude

$$
|\vec A| = \mu\gamma\,e,
$$

with $e$ the orbital eccentricity. Its conservation is the statement that the
ellipse does not precess: the perihelion direction is fixed in space. Any
perturbation away from $1/r$ — an oblate central mass, a relativistic
correction, the screening of an alkali core — makes $\vec A$ rotate slowly, and
the orbit becomes a precessing rosette.

$$
% caption: For the 1/r force the orbit is a closed ellipse and the Runge-Lenz
% vector A stays fixed along the major axis; a non-Coulomb force precesses the
% ellipse and A rotates with it.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% left: closed ellipse
\begin{scope}
\fill[black!70] (0,0) circle (1.6pt);
\node[black, anchor=north east] at (0,-0.05) {focus};
\draw[black, very thick] (0,0) ++(0.6,0) ellipse (1.6 and 1.05);
\draw[acc, thick, ->] (0.6,0) -- (2.5,0);
\node[acc, anchor=south] at (2.0,0.05) {A};
\node[black!70, anchor=north] at (0.8,-1.35) {closed ellipse};
\end{scope}
% right: precessing rosette
\begin{scope}[xshift=6.2cm]
\fill[black!70] (0,0) circle (1.6pt);
\draw[black, very thick, rotate=0] (0.55,0) ellipse (1.5 and 0.95);
\draw[black, thick, rotate=40] (0.55,0) ellipse (1.5 and 0.95);
\draw[black, thick, rotate=80] (0.55,0) ellipse (1.5 and 0.95);
\draw[black, thick, ->] (0,0) -- (2.3,0);
\draw[black, thick, ->] (0,0) -- ({2.3*cos(40)},{2.3*sin(40)});
\node[black, anchor=south] at (2.0,0.05) {A};
\node[black!70, anchor=north] at (0.8,-1.35) {precessing rosette};
\end{scope}
\end{tikzpicture}
$$

## The quantum Runge-Lenz operator

Promoting $\vec A$ to an operator requires symmetrizing the $\vec p\times\vec L$
term, which is not Hermitian as written because $\vec p$ and $\vec L$ do not
commute. Pauli's Hermitian form is[^bj-3][^gs-rl]

$$
\hat{\vec A} = \tfrac{1}{2}\big(\hat{\vec p}\times\hat{\vec L} - \hat{\vec L}\times\hat{\vec p}\big) - \mu\gamma\,\hat r.
$$

Two properties carry the whole argument. First, $\hat{\vec A}$ commutes with the
Hamiltonian,

$$
[\hat H, \hat{\vec A}] = 0,
$$

so $\hat{\vec A}$ is a genuine constant of the motion and maps each energy
eigenspace into itself. Second, its components do not commute with each other or
with $\hat{\vec L}$; they close into an algebra. Working out the commutators
gives the three families

$$
[\hat L_i, \hat L_j] = i\hbar\,\epsilon_{ijk}\hat L_k,
\qquad
[\hat L_i, \hat A_j] = i\hbar\,\epsilon_{ijk}\hat A_k,
\qquad
[\hat A_i, \hat A_j] = i\hbar\,\epsilon_{ijk}\!\left(-2\mu \hat H\right)\hat L_k.
$$

The middle relation says $\hat{\vec A}$ transforms as a vector under rotations,
as it must. The last relation is the crucial one: two Runge-Lenz components
close back onto angular momentum, but with a coefficient $-2\mu\hat H$ that is
the energy operator. On a bound eigenspace, where $\hat H = E_n < 0$, that
coefficient is a positive constant, and rescaling absorbs it.

> **Definition (Scaled Runge-Lenz operator).** On the eigenspace of energy
> $E_n < 0$, define
> $$
> \hat{\vec D} = \frac{1}{\sqrt{-2\mu E_n}}\,\hat{\vec A},
> $$
> which has the dimensions of angular momentum. With this rescaling the three
> commutator families become $[\hat L_i,\hat L_j]=i\hbar\epsilon_{ijk}\hat L_k$,
> $[\hat L_i,\hat D_j]=i\hbar\epsilon_{ijk}\hat D_k$, and
> $[\hat D_i,\hat D_j]=i\hbar\epsilon_{ijk}\hat L_k$.

Those six operators $(\hat{\vec L}, \hat{\vec D})$ with those commutators are the
generators of the rotation group in four dimensions, $SO(4)$. The Coulomb
problem has a four-dimensional rotational symmetry that acts on no visible
fourth spatial axis; it lives in phase space, mixing position and momentum.

## Two decoupled angular momenta

The $SO(4)$ algebra untangles into two independent copies of the angular-momentum
algebra by the linear combinations

$$
\hat{\vec I} = \tfrac{1}{2}\big(\hat{\vec L} + \hat{\vec D}\big),
\qquad
\hat{\vec K} = \tfrac{1}{2}\big(\hat{\vec L} - \hat{\vec D}\big).
$$

A short computation from the commutators above shows that $\hat{\vec I}$ and
$\hat{\vec K}$ each satisfy the $su(2)$ algebra and commute with each other:

$$
[\hat I_i, \hat I_j] = i\hbar\,\epsilon_{ijk}\hat I_k,
\qquad
[\hat K_i, \hat K_j] = i\hbar\,\epsilon_{ijk}\hat K_k,
\qquad
[\hat I_i, \hat K_j] = 0.
$$

Each generates its own spin-like representation with eigenvalues
$I^2 = i(i+1)\hbar^2$ and $K^2 = k(k+1)\hbar^2$, where $i$ and $k$ run over
$0, \tfrac12, 1, \tfrac32, \ldots$. Two constraints tie them together. The
orthogonality $\hat{\vec A}\cdot\hat{\vec L} = \hat{\vec L}\cdot\hat{\vec A} = 0$
carries over to $\hat{\vec D}\cdot\hat{\vec L} = 0$, which forces

$$
\hat I^2 = \hat K^2
\quad\Longrightarrow\quad i = k.
$$

The two spins are equal. The physical Hilbert space of a bound level is the
product of two identical spin-$i$ multiplets.

$$
% caption: The n = 3 level as the product of two spin-1 multiplets: a 3×3 grid of
% states indexed by the two projections. Anti-diagonals collect states of common
% total ℓ, giving 1 + 3 + 5 = 9.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.4,0) -- (3.2,0) node[right, black!70] {$m_I$};
\draw[->, black] (0,-0.4) -- (0,3.2) node[above, black!70] {$m_K$};
\foreach \x in {0.7,1.5,2.3}{
  \foreach \y in {0.7,1.5,2.3}{
    \fill[black] (\x,\y) circle (2.4pt);
  }
}
% highlight anti-diagonals (constant x+y) with faint guides
\draw[black!70, thick] (0.5,0.9) -- (0.9,0.5);
\draw[black, thick, dashed] (0.5,1.7) -- (1.7,0.5);
\draw[black, thick, densely dotted] (0.5,2.5) -- (2.5,0.5);
\node[black!70, anchor=west] at (2.7,2.5) {s: 1};
\node[black, anchor=west] at (2.7,2.0) {p: 3};
\node[black, anchor=west] at (2.7,1.5) {d: 5};
\end{tikzpicture}
$$

## Energy and degeneracy from the algebra

The operator identity behind the Kepler relation $A^2 = \mu^2\gamma^2 + 2\mu E L^2$
acquires a quantum correction $\hbar^2$:[^bj-3]

$$
\hat{\vec A}\cdot\hat{\vec A} = \mu^2\gamma^2 + 2\mu\hat H\big(\hat L^2 + \hbar^2\big).
$$

Dividing by $-2\mu E_n$ and using $\hat D^2 = \hat A^2/(-2\mu E_n)$ gives, on the
level of energy $E_n$,

$$
\hat L^2 + \hat D^2 = -\frac{\mu\gamma^2}{2E_n} - \hbar^2.
$$

The left side is fixed by the representation: since
$\hat I^2 = \tfrac14(\hat L^2 + \hat D^2)$ when $\hat{\vec L}\cdot\hat{\vec D}=0$,

$$
\hat L^2 + \hat D^2 = 4\hat I^2 = 4\,i(i+1)\hbar^2.
$$

Equating the two expressions,

$$
4\,i(i+1)\hbar^2 + \hbar^2 = -\frac{\mu\gamma^2}{2E_n}
\quad\Longrightarrow\quad
(2i+1)^2\hbar^2 = -\frac{\mu\gamma^2}{2E_n}.
$$

Naming the positive integer $n \equiv 2i+1$ (integer because $2i$ is a
non-negative integer) solves for the energy without ever touching a differential
equation:

> **Theorem (Coulomb spectrum from $SO(4)$).** The bound-state energies of the
> hydrogenic Hamiltonian are
> $$
> E_n = -\frac{\mu\gamma^2}{2\hbar^2 n^2} = -\frac{\mu k^2 Z^2 e^4}{2\hbar^2 n^2}
> = -\frac{Z^2\,\mathrm{Ry}}{n^2},
> \qquad n = 2i+1 = 1,2,3,\ldots
> $$
> The level $n$ carries the representation $(i,i)$ with $i=(n-1)/2$, of dimension
> $$
> (2i+1)\times(2i+1) = n^2,
> $$
> reproducing the $n^2$ spatial degeneracy as the product of two spin-$i$
> multiplets.

The angular-momentum content of that $n^2$-dimensional space, decomposed back
into $SO(3)$ representations, is the direct sum
$\ell = 0, 1, \ldots, n-1$, exactly the allowed orbital values. The
Runge-Lenz vector is the operator that rotates one $\ell$ into another at fixed
energy — it is the ladder that connects, for example, $2s$ to $2p$.

$$
% caption: Each hydrogen level n is the (i,i) representation of SO(4) with
% i = (n−1)/2; the product of two spin-i multiplets has dimension n², which
% decomposes into the orbital values ℓ = 0 … n−1.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth,
  box/.style={draw, minimum width=13mm, minimum height=8mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, thick] (n1) at (0,0) {$n=1$};
\node[box, thick] (n2) at (0,-1.4) {$n=2$};
\node[box, thick] (n3) at (0,-2.8) {$n=3$};
% n=1: 1 box
\node[box, minimum width=9mm] at (2.4,0) {s};
\node[black, anchor=west] at (5.6,0) {$1^2=1$};
% n=2: s p(3)
\node[box, minimum width=9mm] at (2.4,-1.4) {s};
\node[box, minimum width=9mm] at (3.5,-1.4) {p};
\node[box, minimum width=9mm] at (4.6,-1.4) {p};
\node[box, minimum width=9mm] at (5.7,-1.4) {p};
\node[black, anchor=west] at (7.0,-1.4) {$2^2=4$};
% n=3: s p(3) d(5)
\foreach \x/\lab in {2.4/s, 3.5/p, 4.6/p, 5.7/p, 6.8/d, 7.9/d, 9.0/d}
  \node[box, minimum width=9mm] at (\x,-2.8) {\lab};
\node[black, anchor=west] at (10.3,-2.8) {$3^2=9$};
\end{tikzpicture}
$$

## Lifting the degeneracy

The accidental degeneracy is fragile because $\hat{\vec A}$ is conserved only for
the exact $1/r$ potential. Every real correction spoils it and separates the
$\ell$ values that hydrogen had fused:

- **Core screening.** In an alkali atom the valence electron sees $-1/r$ far out
  but a deeper potential where it penetrates the ion core. The deviation from
  $1/r$ makes low-$\ell$ (penetrating) states more bound than high-$\ell$ ones,
  the [quantum defect](/atomic-physics/quantum-hydrogen-atom/quantum-defects-alkali-spectra).
- **Relativistic and spin-orbit terms.** The $p^4$ and $\vec L\cdot\vec S$
  corrections depend on $\ell$ and $j$, splitting the level into
  [fine structure](/atomic-physics/fine-structure-and-the-dirac-atom/relativistic-kinetic-correction).
- **External fields.** A uniform field breaks rotational symmetry too; the
  linear Stark effect in hydrogen is large precisely because the unperturbed
  $\ell$-states are already degenerate and mix at first order.

$$
% caption: The exact 1/r potential fuses 3s, 3p, 3d into one level; a deviation
% from 1/r (screening, relativity, a field) lifts the ℓ-degeneracy, and the more
% penetrating low-ℓ states drop lowest.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% degenerate level (left)
\draw[black, very thick] (0,1.4) -- (2.4,1.4);
\node[black, anchor=east] at (0,1.4) {$n=3$};
\node[black!70, anchor=south] at (1.2,1.45) {threefold};
\node[align=center, black!70] at (1.2,-0.3) {exact Coulomb\\ degenerate};
% split levels (right)
\begin{scope}[xshift=5.0cm]
\draw[acc, thick] (0,0.5) -- (2.4,0.5); \node[acc, anchor=west] at (2.5,0.5) {3s};
\draw[acc, thick] (0,1.2) -- (2.4,1.2); \node[acc, anchor=west] at (2.5,1.2) {3p};
\draw[acc, thick] (0,1.8) -- (2.4,1.8); \node[acc, anchor=west] at (2.5,1.8) {3d};
\node[align=center, black!70] at (1.2,-0.3) {symmetry\\ broken};
\end{scope}
% connectors
\draw[black, dashed] (2.4,1.4) -- (5.0,0.5);
\draw[black, dashed] (2.4,1.4) -- (5.0,1.2);
\draw[black, dashed] (2.4,1.4) -- (5.0,1.8);
\end{tikzpicture}
$$

> **Worked example.** Count the $n=3$ states and their orbital content. With
> $i = (n-1)/2 = 1$, each $SO(4)$ multiplet is a spin-$1$ triplet, and the level
> is $(1,1)$ of dimension $3\times 3 = 9$. Decomposing the product of two spin-1
> representations into total-$\ell$ pieces gives $\ell = 0$ (1 state), $\ell = 1$
> (3 states), and $\ell = 2$ (5 states), summing to $1+3+5 = 9$. The count
> matches $\sum_{\ell=0}^{2}(2\ell+1) = n^2 = 9$ obtained by direct enumeration.

The symmetry viewpoint explains what the differential equation only computes: the
degeneracy is $n^2$ because a hidden four-dimensional rotational symmetry
organizes each level into an irreducible representation, and the number $n$ is
$2i+1$ for the spin of that representation. When the symmetry is exact the levels
are fused; when it is broken, the spectrum fans out into the fine and hyperfine
structure that makes atomic physics quantitative.

[^bj-3]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., Ch. 3 — the degeneracy of the hydrogen levels and its connection to the conserved Runge-Lenz vector and the enlarged symmetry of the Coulomb problem. <https://www.pearson.com/en-gb/subject-catalog/p/physics-of-atoms-and-molecules/P200000005386>
[^gs-rl]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_, 3rd ed., §4.2 and end-of-chapter problems — the Hermitian Runge-Lenz operator, its commutator with $\hat H$, and the algebraic route to the hydrogen spectrum.