Accidental Degeneracy and the Runge-Lenz Symmetry

The radial solution 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 $\gamma =kZ{e}^2$, conserves energy and angular momentum $\vec{L}=\vec{r}\times \vec{p}$. It conserves one more vector, the Laplace-Runge-Lenz vector

Direct differentiation using Newton's equation $\dot{\vec{p}}=-\gamma \hat{r}/r^2$ gives $\mathbf{d}\vec{A}/\mathbf{d}t=0$ 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

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.

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 is¹²

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

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

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.

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

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:

Each generates its own spin-like representation with eigenvalues $I^2=i(i+1){\hslash }^2$ and $K^2=k(k+1){\hslash }^2$, where $i$ and $k$ run over $0,\frac{1}{2},1,\frac{3}{2},\dots$. 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

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

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 ${\hslash }^2$:¹

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

The left side is fixed by the representation: since ${\hat{I}}^2=\frac{1}{4}({\hat{L}}^2+{\hat{D}}^2)$ when $\hat{\vec{L}}\cdot \hat{\vec{D}}=0$,

Equating the two expressions,

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:

The angular-momentum content of that $n^2$-dimensional space, decomposed back into $SO(3)$ representations, is the direct sum $\ell =0,1,\dots ,n-1$, 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 $2p$.

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.
• 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.
• 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.

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.