Degeneracy in Separable Systems

Degeneracy means that more than one linearly independent state has the same energy. In a separable system, this happens often because the energy is built from several quantum labels, and different label combinations can produce the same number.

If

$$\hat H\lvert E,\alpha\rangle = E\lvert E,\alpha\rangle,$$

where $\alpha$ distinguishes independent states with the same energy, then the degeneracy of $E$ is the dimension of the corresponding energy eigenspace:

$$g_E = \dim\ker(\hat H-EI).$$

The important lesson is that “same energy” does not always mean “same state” or “same physical reason.” Degeneracies can come from exact symmetries, from additive separable spectra, from number-theoretic coincidences, or from hidden structure of a special Hamiltonian.

Helpful background. Separation of Variables to turn a separable wave equation into ordinary differential equations.

Why Separability Creates Repeated Energies

In many separable problems, the Hamiltonian splits into commuting pieces:

$$\hat H = \hat H_1+\hat H_2+\hat H_3, \qquad [\hat H_i,\hat H_j]=0.$$

Product eigenstates then have additive energies:

$$\psi_{abc} = \psi_a^{(1)}\psi_b^{(2)}\psi_c^{(3)}, \qquad E_{abc} = E_a^{(1)}+E_b^{(2)}+E_c^{(3)}.$$

Degeneracy occurs when two or more triples give the same sum:

$$E_{abc}=E_{a'b'c'}.$$

Sometimes the equality is enforced by a symmetry. Sometimes it is a coincidence caused by the arithmetic form of the spectrum. The distinction matters because a symmetry protects degeneracy within each irreducible multiplet. Equal energies of different multiplets, including coincidences caused by a separable model, can split even under perturbations that preserve the visible geometric symmetry.

Energy Labels Are Not Complete State Labels

If an energy is nondegenerate, the energy label may identify the energy eigenstate up to an overall phase. If an energy is degenerate, energy alone labels an eigenspace, not a unique vector.

Inside a degenerate eigenspace, any orthonormal basis is allowed unless additional commuting observables or boundary conditions select a preferred basis. A complete set of commuting observables supplies enough compatible labels to distinguish states within the model. For a box these labels may be $(n_x,n_y,n_z)$. For a central potential they may be $(E,\ell,m)$ or $(n,\ell,m)$ when the spectrum is discrete.

This is why degeneracy is not merely a counting issue. It changes how measurements, perturbations, and state expansions are described.

Cubic Box Degeneracy

For a rectangular hard-wall box, the energies are

$$E_{n_xn_yn_z} = \frac{\pi^2\hbar^2}{2m} \left( \frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2} \right),$$

with

$$n_x,n_y,n_z=1,2,3,\ldots .$$

If the side lengths are all different, most energy levels are nondegenerate. In the cubic case,

$$L_x=L_y=L_z=L,$$

the energy depends only on the sum

$$n_x^2+n_y^2+n_z^2.$$

The triples

$$(1,1,2), \qquad (1,2,1), \qquad (2,1,1)$$

therefore have the same energy. The equal side lengths make these permutations degenerate; this first excited triplet also forms a symmetry-protected cubic multiplet.

Higher box levels can also show arithmetic coincidences: $(1,1,5)$ and $(3,3,3)$ both give a sum of squares of $27$. Moreover, a larger set of permuted product modes need not form one irreducible cubic multiplet; a cubic-symmetric perturbation can split it into smaller multiplets, as Exercise 2 demonstrates. Deforming the side lengths generally removes permutation degeneracies as well.

Harmonic-Oscillator Degeneracy

A three-dimensional isotropic oscillator has Hamiltonian

$$\hat H = \sum_{i=x,y,z} \left( \frac{\hat p_i^2}{2m} + \frac12m\omega^2\hat x_i^2 \right).$$

Because all three frequencies are equal, the energy is

$$E_{n_xn_yn_z} = \hbar\omega \left( n_x+n_y+n_z+\frac32 \right),$$

with

$$n_x,n_y,n_z=0,1,2,\ldots .$$

For a fixed shell

$$N=n_x+n_y+n_z,$$

the degeneracy is the number of nonnegative integer triples with that sum:

$$g_N = \frac{(N+1)(N+2)}{2}.$$

For example, $N=2$ has six states:

$$(2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1).$$

This degeneracy is larger than a simple axis-permutation count because all partitions of the same total excitation number have the same energy. If the frequencies become anisotropic,

$$E_{n_xn_yn_z} = \hbar\omega_x\left(n_x+\frac12\right) + \hbar\omega_y\left(n_y+\frac12\right) + \hbar\omega_z\left(n_z+\frac12\right),$$

the degeneracy usually disappears unless the frequencies have exact rational relations or residual symmetries.

Central-Potential Degeneracy

For a central potential,

$$V(\mathbf r)=V(r),$$

the wavefunction separates as

$$\psi(r,\theta,\phi) = R_{E\ell}(r)Y_\ell^m(\theta,\phi).$$

The radial equation depends on $\ell$ but not on $m$. Therefore, for any rotationally invariant central potential, the $2\ell+1$ states

$$m=-\ell,-\ell+1,\ldots,\ell$$

are degenerate within a fixed $\ell$ sector. This is symmetry-enforced rotational degeneracy: no axis is preferred, so the Hamiltonian cannot distinguish the different orientations.

Degeneracy between different $\ell$ sectors is a separate question. A generic central potential does not have it, because changing $\ell$ changes the centrifugal term in the effective radial potential:

$$\frac{\hbar^2\ell(\ell+1)}{2mr^2}.$$

The nonrelativistic Coulomb potential is special: its bound-state energy depends only on the principal quantum number $n$, producing the familiar $n^2$ spatial degeneracy. That stronger result is explained in Degeneracy of the Hydrogen Atom.

Symmetry, Accident, and Hidden Symmetry

It is useful to classify degeneracies by their source.

Source	Example	What Usually Lifts It
Exact visible symmetry	$m$ degeneracy in a central potential	external fields or symmetry-breaking boundaries
Equal separable parameters	cubic box or isotropic oscillator	unequal side lengths or unequal frequencies
Arithmetic coincidence	distinct box triples with the same sum of squares	small generic deformation
Hidden symmetry	Coulomb $\ell$ degeneracy	relativistic, radiative, or non-Coulomb corrections
Continuous label degeneracy	free-particle directions with fixed $\lvert\mathbf k\rvert$	boundaries, fields, or interactions selecting directions

The word “accidental” is often used for degeneracies not explained by the most obvious symmetry. It should not mean “unimportant” or “numerically approximate.” Some accidental degeneracies reveal a hidden symmetry, as in hydrogen; others are simple arithmetic coincidences.

Lifting Degeneracy

A perturbation lifts a degeneracy when it acts differently on states that previously had the same energy. The classic qualitative example is an external field selecting an axis.

For a central potential, a weak interaction proportional to $\hat L_z$ gives shifts of the form

$$\Delta E_m = \gamma\hbar m,$$

where $\gamma$ is a field-dependent constant in the simplified model. The different $m$ values, previously degenerate under full rotational symmetry, now separate because the field has chosen a preferred $z$ direction.

This is the basic idea behind magnetic-field splitting. Detailed Zeeman physics involves spin, magnetic moments, perturbation theory, and atomic structure, so it belongs in later volumes. The lesson here is more general: once the symmetry or exact equality responsible for a degeneracy is broken, energy levels usually split.

Common Mistakes

• Treating degeneracy as if it always has the same explanation.
• Assuming that energy alone labels an eigenstate in a degenerate spectrum.
• Calling every repeated energy “accidental” without checking for symmetry.
• Assuming cubic-box permutation degeneracy survives in a rectangular box.
• Treating hydrogen’s $\ell$ degeneracy as generic for all central potentials.
• Forgetting that a perturbation must be diagonalized inside a degenerate eigenspace, not just evaluated on one arbitrary basis vector.

Exercises

1. In a cubic hard-wall box, find the degeneracy of the shell with $n_x^2+n_y^2+n_z^2=14$.

Solution

The positive integer triple

$$(1,2,3)$$

has

$$1^2+2^2+3^2=14.$$

All three entries are distinct, so there are

$$3!=6$$

permutations. No entry can exceed $3$. If all entries were at most $2$, the sum would be at most $12$. One entry must therefore be $3$; the other two squares must sum to $5$, giving $1$ and $2$. The degeneracy is exactly $6$.

2. Can cubic symmetry survive while the sixfold shell in Exercise 1 splits? Let $0<x,y,z<L$, define centered dimensionless coordinates $\xi_x=x/L-1/2$ and similarly for $y,z$, and add

$$W=g(\xi_x^2\xi_y^2+\xi_y^2\xi_z^2+\xi_z^2\xi_x^2), \qquad |g|\ll E_0=\frac{\pi^2\hbar^2}{2mL^2}.$$

Here $g$ has units of energy. This potential preserves all signed permutations of the centered axes, hence the cube’s geometric symmetry. Use the normalized one-dimensional box modes to calculate its first-order shifts in the shell $E=14E_0$. The general prescription is Degenerate Perturbation Theory: diagonalize $W$ within the six-dimensional eigenspace.

Solution

Write $\phi_{n_xn_yn_z}$ for the normalized product of sine modes. The one-dimensional integrals give

$$a_n=\langle n|\xi^2|n\rangle=\frac{1}{12}-\frac{1}{2\pi^2n^2}, \qquad b=\langle1|\xi^2|3\rangle=\frac{3}{8\pi^2}.$$

For example, $b$ follows by integrating $(u-1/2)^2[\cos(2\pi u)-\cos(4\pi u)]$ from $0$ to $1$. Reflection about the center gives mode $n$ parity $(-1)^{n+1}$, so the even operator $\xi^2$ cannot connect an even $n$ to an odd $n$.

Every permutation of $(1,2,3)$ has one center-odd factor, the mode $n=2$. Its axis cannot change under $W$. The six states therefore form three independent pairs. In the pair $(\phi_{123},\phi_{321})$, and likewise in its cyclic partners, the perturbation matrix is

$$W_{\rm pair}=g\begin{pmatrix}A&B\\B&A\end{pmatrix}, \qquad A=a_1a_2+a_2a_3+a_3a_1, \qquad B=b^2=\frac{9}{64\pi^4}.$$

Only $\xi_x^2\xi_z^2$ contributes to this pair’s off-diagonal element; the other terms contain an orthogonal spectator overlap. The normalized sum and difference $(\phi_{123}\pm\phi_{321})/\sqrt2$ have shifts $g(A\pm B)$. There are three states of each kind, giving two triplets:

$$E_\pm=14E_0+g(A\pm B)+O(g^2/E_0).$$

Their first-order separation is $2|g|B=9|g|/(32\pi^4)$. The perturbation preserves cubic symmetry but couples coordinates and removes the original additive separation. The calculation makes explicit why the sixfold level need not stay sixfold. Fernández analyzes the geometric and dynamical symmetries behind this example.

3. Explain why a perturbation proportional to $\hat L_z$ splits an $m$ multiplet.

Solution

The states $Y_\ell^m$ obey

$$\hat L_zY_\ell^m = \hbar mY_\ell^m.$$

If the perturbation is $\gamma\hat L_z$, then the first-order shift is

$$\Delta E_m = \gamma\hbar m.$$

Different $m$ values generally receive different shifts. Physically, the perturbation has selected a $z$ axis, so the full rotational symmetry that protected the $m$ degeneracy is broken.

References

• F. M. Fernández, “On the symmetry of the quantum-mechanical particle in a cubic box,” arXiv:1310.5136v2 (2014), §§3–4.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.