The Isotropic Oscillator and Hidden Symmetry

Two central potentials stand apart from the rest: the Coulomb attraction $-k/r$ and the isotropic oscillator $\frac{1}{2}\mu {\omega }^2{r}^2$. Both produce energy levels that depend on a single combined quantum number rather than on the radial count and the angular momentum separately, an accidental degeneracy that a generic $V(r)$ such as the spherical box does not share. Classically the same two potentials are the only central forces whose bound orbits close on themselves. The two statements are the same statement: a closed orbit is fixed in space by a conserved vector beyond angular momentum, and the quantum image of that extra conserved quantity is the extra symmetry that forces the degeneracy.

The isotropic oscillator in Cartesian coordinates

The potential $V(r)=\frac{1}{2}\mu {\omega }^2{r}^2=\frac{1}{2}\mu {\omega }^2(x^2+y^2+z^2)$ splits into three independent one-dimensional oscillators, since it is a sum of a term in each Cartesian coordinate and the kinetic energy is likewise a sum. The Hamiltonian separates,

and the eigenstates are products of one-dimensional oscillator states, labelled by three occupation numbers $(n_x,n_y,n_z)$ with each $n_i=0,1,2,\dots$. The energy is the sum of three ladders,

where $N=n_x+n_y+n_z$ is the total quantum number. The energy depends only on the sum $N$, not on how the quanta are distributed among the three axes. The degeneracy of level $N$ is the number of ways to write $N$ as an ordered sum of three non-negative integers,

The count follows from a standard stars-and-bars argument: distributing $N$ identical quanta into $3$ labelled boxes has $\left(\genfrac{}{}{0ex}{}{N+3-1}{3-1}\right)$ arrangements. This degeneracy grows quadratically in $N$, faster than the $(2\ell +1)$ that a single $\ell$ contributes, which already signals that a single value of $N$ gathers several angular momenta.

The isotropic oscillator in the spherical basis

The same operator is central, so it also solves in the spherical basis with $\psi =R_{n_r\ell }(r)Y_{\ell }^m$. The radial equation carries the effective potential $\frac{1}{2}\mu {\omega }^2{r}^2+{\hslash }^2\ell (\ell +1)/2\mu r^2$. Extracting the $r\to 0$ behavior $r^{\ell }$ and the $r\to \infty$ behavior $e^{-\mu \omega r^2/2\hslash }$ and solving the residual series gives associated Laguerre polynomials in $r^2$, with energies

where $n_r$ is the number of radial nodes. Comparing with the Cartesian result identifies the combined quantum number

For a given $N$, the allowed $\ell$ are those with $\ell =N-2n_r\ge 0$, so

all of the same parity as $N$, since parity is $(-1{)}^{\ell }=(-1{)}^N$. The oscillator levels contain only even $\ell$ (for even $N$) or only odd $\ell$ (for odd $N$), never both — a selection the Coulomb problem does not share. Summing the $2\ell +1$ over this ladder reproduces the Cartesian count.

The energy levels are equally spaced by $\hslash \omega$, and each successive level adds one more angular-momentum value of the alternating parity while its degeneracy climbs as $(N+1)(N+2)/2$.

The hidden symmetry of the oscillator

That the energy depends on $N$ alone, gathering several $\ell$ into one level, is the oscillator's accidental degeneracy. Its source is a symmetry group larger than the rotation group $\mathrm{SO}(3)$. Introduce the three ladder operators$a_i$ with $[a_i,a_j^{†}]={\delta }_{ij}$, one per Cartesian axis. The Hamiltonian is

Consider the nine bilinear operators $T_{ij}=a_i^{†}{a}_j$. Each commutes with $H$, because $H$ counts total quanta and $T_{ij}$ moves one quantum from axis $j$ to axis $i$ without changing the total,

The trace ${\sum }_i{T}_{ii}={\sum }_i{a}_i^{†}{a}_i$ is the number operator, itself $H/\hslash \omega -\frac{3}{2}$; the remaining eight independent traceless combinations generate the group $\mathrm{SU}(3)$. Three of them are the angular-momentum components (the antisymmetric part $T_{ij}-T_{ji}$), and the other five form a symmetric traceless quadrupole tensor. The states of a fixed $N$ fill exactly one symmetric irreducible representation of $\mathrm{SU}(3)$, whose dimension is $(N+1)(N+2)/2$ — the degeneracy computed above, now explained as the size of a single symmetry multiplet.

The Runge–Lenz vector and the Coulomb symmetry

The Coulomb problem carries its own conserved vector. Classically, a particle in a $-k/r$ potential conserves the Runge–Lenz vector

which points along the major axis of the elliptical orbit from the focus to the perihelion. To say it is conserved is to say the ellipse does not precess: the orbit closes because its orientation is fixed by a constant of motion. Any perturbation away from the pure $1/r$ law makes ${\vec{A}}_{\text{cl}}$ drift and the perihelion precess, as it does for Mercury.

The quantum Runge–Lenz vector must be Hermitized, since $\vec{p}\times \vec{L}$ and $\vec{L}\times \vec{p}$ differ as operators. The symmetric choice is

and a direct computation confirms $[\vec{A},H]=0$: it is conserved. Two further identities close the algebra. First, $\vec{A}$ is perpendicular to $\vec{L}$, $\vec{A}\cdot \vec{L}=\vec{L}\cdot \vec{A}=0$. Second, its square relates the two Casimir-like quantities to the energy,

On a bound-state subspace of fixed negative energy $E$, rescale $\vec{K}=\sqrt{-\mu /2E}\vec{A}$ to give it the dimensions of angular momentum. The commutators of $\vec{L}$ and $\vec{K}$ then close,

which is the Lie algebra of $\mathrm{SO}(4)$, the rotation group of four dimensions. The Coulomb bound states carry a symmetry as if they lived on the surface of a sphere in four-dimensional space.

The n-squared degeneracy from SO(4)

The $\mathrm{SO}(4)$ algebra decouples into two independent $\mathrm{SU}(2)$ angular momenta by taking the combinations

each obeying its own angular-momentum algebra and commuting with the other. Since $\vec{A}\cdot \vec{L}=0$ forces $\vec{K}\cdot \vec{L}=0$, the two Casimirs are equal, $({\vec{J}}^{(1)}{)}^2=({\vec{J}}^{(2)}{)}^2=j(j+1){\hslash }^2$ with a common $j$. The relation for $A^2$ becomes, in operator form on an energy eigenspace,

Writing $n\equiv 2j+1$, with $j=0,\frac{1}{2},1,\dots$ giving $n=1,2,3,\dots$, this is precisely the Bohr spectrum

derived here from symmetry alone, with no differential equation solved. The degeneracy is the dimension of the joint $\mathrm{SU}(2)\times \mathrm{SU}(2)$ multiplet, $(2j+1)(2j+1)=n^2$, reproducing the hydrogen count. The extra $\ell$-degeneracy that rotational invariance could not explain is exactly the statement that $\vec{K}$ moves between different $\ell$ within a fixed-$n$$\mathrm{SO}(4)$ multiplet.

Closed orbits and the two special potentials

The classical parallel makes the pattern memorable. Bertrand's theorem states that of all central potentials, only the inverse-square force ($\propto 1/r^2$, from $-k/r$) and the linear force ($\propto r$, from $\frac{1}{2}\mu {\omega }^2{r}^2$) produce bound orbits that close after a single revolution. For every other $V(r)$, a bound orbit is a rosette whose perihelion precesses and never exactly repeats.

The equivalence runs in both directions. A closed orbit is one whose shape is frozen by a conserved vector — the Runge–Lenz vector for Kepler, the analogous symmetric tensor for the oscillator — and the quantized version of that conserved quantity is the generator of a symmetry beyond rotation. The accidental degeneracy in the spectrum, the closure of the classical orbit, and the existence of a hidden conserved quantity are three faces of one fact. Everything past these two potentials — screened Coulomb fields, anharmonic wells, multi-electron atoms — loses the extra symmetry, and with it the exact degeneracy, which is why the fine-structure corrections and screening in real atoms split the hydrogenic $n^2$ levels apart.¹²³