---
title: The Isotropic Oscillator and Hidden Symmetry
module: Central Potentials
moduleNumber: 7
lessonNumber: 3
order: 703
summary: >
  The three-dimensional isotropic harmonic oscillator solves in both Cartesian
  and spherical bases, and the two solutions must agree on the degeneracy of
  every level. That agreement, and the accidental degeneracy of hydrogen, both
  come from a symmetry larger than rotation: the oscillator carries an SU(3)
  invariance built from a conserved quadrupole tensor, and the Coulomb problem
  carries an SO(4) invariance built from the conserved Runge–Lenz vector. These
  hidden symmetries pin the degeneracies that rotational invariance alone leaves
  unexplained.
topics: [Central Potentials]
sources:
  - book: Shankar
    ref: "Ch. 12 — Rotational Invariance and Angular Momentum; Ch. 13 §13.4 The Degeneracy of the Hydrogen Spectrum"
  - book: Sakurai & Napolitano
    ref: "Ch. 3 §3.9 Spin Correlations; Ch. 4 §4.1 Symmetries, Conservation Laws, and Degeneracies"
  - book: Cohen-Tannoudji
    ref: "Ch. VII Complement B — Stationary States of the Three-Dimensional Harmonic Oscillator"
draft: false
---

Two central potentials stand apart from the rest: the Coulomb attraction $-k/r$
and the isotropic oscillator $\tfrac12\mu\omega^2 r^2$. 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](/quantum-mechanics/central-potentials/the-hydrogen-atom)
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 $V(r) = \tfrac12\mu\omega^2 r^2 = \tfrac12\mu\omega^2(x^2+y^2+z^2)$
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,

$$
H = H_x + H_y + H_z,
\qquad
H_i = \frac{p_i^2}{2\mu} + \frac12\mu\omega^2 x_i^2,
$$

and the eigenstates are products of one-dimensional
[oscillator states](/quantum-mechanics/oscillator-and-symmetry/ladder-operators-and-the-number-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,

$$
E = \left(n_x + \tfrac12\right)\hbar\omega
  + \left(n_y + \tfrac12\right)\hbar\omega
  + \left(n_z + \tfrac12\right)\hbar\omega
  = \left(N + \tfrac32\right)\hbar\omega,
$$

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,

> **Theorem (Oscillator degeneracy).** The number of Cartesian states with
> $n_x + n_y + n_z = N$ is
> $$
> g_N = \binom{N+2}{2} = \frac{(N+1)(N+2)}{2}.
> $$
> The ground state $N = 0$ is unique; the first excited level $N = 1$ is
> threefold degenerate; $N = 2$ is sixfold, and so on.

The count follows from a standard stars-and-bars argument: distributing $N$
identical quanta into $3$ labelled boxes has $\binom{N+3-1}{3-1}$ 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.

$$
% caption: Cartesian states of the isotropic oscillator as lattice points
% $(n_x, n_y, n_z)$ on planes of constant $N = n_x + n_y + n_z$; each plane
% holds $(N+1)(N+2)/2$ states, drawn here for the first three levels.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% simple triangular arrays: level N has a triangle of dots (N+1 rows)
% N=0
\node[anchor=south, black, font=\scriptsize] at (0.0,2.7) {$N=0$};
\fill[acc] (0,2) circle (2.6pt);
\node[anchor=north, black, font=\scriptsize] at (0,1.7) {1 state};
% N=1: 3 dots
\node[anchor=south, black, font=\scriptsize] at (3.2,2.7) {$N=1$};
\fill[acc] (2.8,2) circle (2.6pt);
\fill[acc] (3.2,2) circle (2.6pt);
\fill[acc] (3.6,2) circle (2.6pt);
\node[anchor=north, black, font=\scriptsize] at (3.2,1.7) {3 states};
% N=2: 6 dots (triangle 3+2+1)
\node[anchor=south, black, font=\scriptsize] at (6.6,2.7) {$N=2$};
\fill[acc] (6.0,2.15) circle (2.6pt);
\fill[acc] (6.4,2.15) circle (2.6pt);
\fill[acc] (6.8,2.15) circle (2.6pt);
\fill[acc] (7.2,2.15) circle (2.6pt);
\fill[acc] (6.2,1.75) circle (2.6pt);
\fill[acc] (6.6,1.75) circle (2.6pt);
\node[anchor=north, black, font=\scriptsize] at (6.6,1.5) {6 states};
\end{tikzpicture}
$$

## 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 $\tfrac12\mu\omega^2 r^2 + \hbar^2\ell(\ell+1)/2\mu r^2$. Extracting the
$r \to 0$ behavior $r^{\ell}$ and the $r \to \infty$ behavior
$e^{-\mu\omega r^2/2\hbar}$ and solving the residual series gives associated
Laguerre polynomials in $r^2$, with energies

$$
E = \left(2n_r + \ell + \tfrac32\right)\hbar\omega,
\qquad
n_r = 0, 1, 2, \dots,
$$

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

$$
N = 2n_r + \ell.
$$

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

$$
\ell = N, N-2, N-4, \dots, \ 1 \ \text{or}\ 0,
$$

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.

> **Worked example.** Take $N = 2$. The spherical ladder gives $\ell = 2$
> ($n_r = 0$) and $\ell = 0$ ($n_r = 1$), contributing $(2\cdot 2+1) + (2\cdot 0+1)
> = 5 + 1 = 6$ states, matching the Cartesian $g_2 = \tfrac{(3)(4)}{2} = 6$. The
> two bases describe the same six-dimensional level: one sorts the states by
> Cartesian occupation, the other by angular momentum, and a unitary change of
> basis connects them.

$$
% caption: The same oscillator level decomposed two ways. The Cartesian basis
% sorts the six $N = 2$ states by $(n_x, n_y, n_z)$; the spherical basis sorts
% them by angular momentum into a $d$ multiplet (5 states) and an $s$ state.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0,
  bx/.style={draw, minimum width=24mm, minimum height=11mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[bx, draw=acc, text=acc, thick] (L) at (0,0) {$N = 2$ level\\(6 states)};
\node[bx] (C) at (5.4,1.5) {Cartesian:\\6 triples $(n_x\,n_y\,n_z)$};
\node[bx] (S) at (5.4,-1.5) {spherical:\\$d$ (5) plus $s$ (1)};
\draw[->, acc, thick] (L) -- (C);
\draw[->, acc, thick] (L) -- (S);
\node[anchor=west, black, font=\scriptsize] at (7.7,1.5) {sorted by axis};
\node[anchor=west, black, font=\scriptsize] at (7.7,-1.5) {sorted by $L^2$};
\end{tikzpicture}
$$

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

$$
% caption: Equally spaced oscillator levels $E = (N + 3/2)\hbar\omega$, adjacent
% levels separated by $\hbar\omega$; each level's degeneracy $g_N$ and its
% angular-momentum content (all of one parity) are listed at right.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% vertical energy axis
\draw[->, black] (0,0) -- (0,6.1) node[anchor=south, black] {$E$};
% equally spaced levels
\def\olev#1#2#3{\draw[acc, very thick] (0.9,#1) -- (3.4,#1); \node[anchor=east, black!70, font=\scriptsize] at (0.8,#1) {#2}; \node[anchor=west, black, font=\scriptsize] at (3.55,#1) {#3};}
\olev{0.8}{$N=0$}{$g = 1$ \ (s)}
\olev{2.1}{$N=1$}{$g = 3$ \ (p)}
\olev{3.4}{$N=2$}{$g = 6$ \ (s, d)}
\olev{4.7}{$N=3$}{$g = 10$ \ (p, f)}
% spacing brace-free indicator
\draw[<->, black] (4.9,0.8) -- (4.9,2.1);
\node[anchor=west, black, font=\scriptsize] at (5.0,1.45) {equal gap};
\end{tikzpicture}
$$

## 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](/quantum-mechanics/oscillator-and-symmetry/ladder-operators-and-the-number-states)
$a_i$ with $[a_i, a_j^\dagger] = \delta_{ij}$, one per Cartesian axis. The
Hamiltonian is

$$
H = \hbar\omega\left(\sum_i a_i^\dagger a_i + \tfrac32\right).
$$

Consider the nine bilinear operators $T_{ij} = a_i^\dagger a_j$. 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,

$$
[H, T_{ij}] = 0.
$$

The trace $\sum_i T_{ii} = \sum_i a_i^\dagger a_i$ is the number operator, itself
$H/\hbar\omega - \tfrac32$; 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.

> **Definition (Hidden or dynamical symmetry).** A symmetry of a specific
> Hamiltonian generated by operators that commute with $H$ but are not the
> geometric symmetries (rotations, translations, reflections) of space. Its
> multiplets have larger dimension than the rotational $(2\ell+1)$, and they
> account for accidental degeneracies. For the isotropic oscillator the group is
> $\mathrm{SU}(3)$; for the Coulomb problem it is $\mathrm{SO}(4)$.

## 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**

$$
\vec A_{\text{cl}} = \vec p \times \vec L - \mu k\,\hat r,
$$

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.

$$
% caption: The Runge–Lenz vector points from the force center along the major
% axis to the perihelion of a Kepler ellipse. Its constancy fixes the orbit's
% orientation, so a pure inverse-square orbit closes without precessing.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% ellipse centered so focus at origin: semi-major 3, semi-minor 2.2, center at (-1.4,0) so focus ~ (0,0)
\draw[black, very thick] (-1.4,0) ellipse (3 and 2.2);
% focus (force center)
\fill[black] (0,0) circle (2.2pt);
\node[anchor=north, font=\scriptsize] at (-0.1,-0.18) {center of force};
% perihelion is nearest point on major axis: right vertex at x = -1.4+3 = 1.6
\fill[acc] (1.6,0) circle (2.0pt);
\node[acc, anchor=west, font=\scriptsize] at (1.74,0) {perihelion};
% Runge-Lenz vector from focus toward perihelion
\draw[->, acc, very thick] (0,0) -- (1.5,0);
\node[acc, anchor=south, font=\scriptsize] at (0.75,0.06) {$A$};
% a sample orbit point with position vector
\fill[black] (-3.70,1.41) circle (1.8pt);
\draw[->, black, thick] (0,0) -- (-3.58,1.36);
\node[black, anchor=south, font=\scriptsize] at (-1.85,0.75) {$r$};
\end{tikzpicture}
$$

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

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

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,

$$
A^2 = k^2 + \frac{2H}{\mu}\big(L^2 + \hbar^2\big).
$$

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,

$$
[L_i, L_j] = i\hbar\,\varepsilon_{ijk}L_k,
\qquad
[L_i, K_j] = i\hbar\,\varepsilon_{ijk}K_k,
\qquad
[K_i, K_j] = i\hbar\,\varepsilon_{ijk}L_k,
$$

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

$$
\vec J^{(1)} = \tfrac12(\vec L + \vec K),
\qquad
\vec J^{(2)} = \tfrac12(\vec L - \vec K),
$$

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, $(\vec J^{(1)})^2 = (\vec J^{(2)})^2 = j(j+1)\hbar^2$ with a common $j$. The
relation for $A^2$ becomes, in operator form on an energy eigenspace,

$$
-\frac{\mu k^2}{2H} = \big(L^2 + K^2 + \hbar^2\big)
  = \hbar^2\big(4j(j+1) + 1\big) = \hbar^2(2j+1)^2.
$$

Writing $n \equiv 2j + 1$, with $j = 0, \tfrac12, 1, \dots$ giving
$n = 1, 2, 3, \dots$, this is precisely the Bohr spectrum

$$
E_n = -\frac{\mu k^2}{2\hbar^2 n^2},
\qquad
k = \frac{e^2}{4\pi\epsilon_0},
$$

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](/quantum-mechanics/central-potentials/the-hydrogen-atom). 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.

> **Theorem (Symmetry origin of the degeneracies).** The accidental degeneracies
> of the two special central potentials are the dimensions of hidden-symmetry
> multiplets. The isotropic oscillator carries $\mathrm{SU}(3)$ with multiplet
> dimension $\tfrac12(N+1)(N+2)$; the Coulomb bound states carry $\mathrm{SO}(4)
> \cong \mathrm{SU}(2)\times\mathrm{SU}(2)$ with multiplet dimension $n^2$. A
> generic central potential has only $\mathrm{SO}(3)$, whose multiplets have the
> mandatory dimension $2\ell+1$ and no more.

## 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 $\tfrac12\mu\omega^2 r^2$)
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.

$$
% caption: A pure inverse-square orbit closes into a fixed ellipse (left); any
% deviation makes the perihelion advance each revolution, tracing a
% non-closing rosette (right), the classical face of the lost extra symmetry.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% left: closed ellipse
\begin{scope}[xshift=0cm]
  \fill[black] (0,0) circle (1.8pt);
  \draw[acc, very thick] (-0.9,0) ellipse (2.0 and 1.35);
  \node[anchor=north, font=\scriptsize] at (-0.9,-1.7) {closed orbit};
  \node[anchor=south, font=\scriptsize, acc] at (-0.9,1.45) {$\frac{1}{r}$ or $r^2$};
\end{scope}
% right: precessing rosette (several rotated ellipses)
\begin{scope}[xshift=6.4cm]
  \fill[black] (0,0) circle (1.8pt);
  \foreach \a in {0,40,80,120,160} {
    \draw[black, thick, rotate=\a] (-0.75,0) ellipse (1.7 and 1.1);
  }
  \node[anchor=north, font=\scriptsize] at (0,-1.7) {precessing orbit};
  \node[anchor=south, font=\scriptsize, black] at (0,1.55) {generic $V(r)$};
\end{scope}
\end{tikzpicture}
$$

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](/quantum-mechanics/approximation-methods/fine-structure-and-the-real-hydrogen-atom)
and screening in real atoms split the hydrogenic $n^2$ levels apart.[^shankar-degeneracy][^sakurai-sym][^ct-oscillator]

[^shankar-degeneracy]: Shankar, _Principles of Quantum Mechanics_, 2nd ed. (Springer, 1994), Ch. 13 §13.4 — the degeneracy of the hydrogen spectrum, the Runge–Lenz vector, and the $\mathrm{SO}(4)$ symmetry of the Coulomb bound states. Publisher: https://doi.org/10.1007/978-1-4757-0576-8
[^sakurai-sym]: Sakurai & Napolitano, _Modern Quantum Mechanics_, 3rd ed. (Cambridge, 2021), Ch. 4 §4.1 — symmetries, conservation laws, and degeneracies, including the connection between a conserved operator commuting with $H$ and symmetry-forced degeneracy. Publisher: https://doi.org/10.1017/9781108587280
[^ct-oscillator]: Cohen-Tannoudji, Diu & Laloë, _Quantum Mechanics_, Vol. I (Wiley, 1977), Ch. VII Complement B — the three-dimensional isotropic harmonic oscillator in Cartesian and spherical bases and the degeneracy of its levels.