Degeneracy and Multiplets

Degeneracy means that an eigenspace has dimension greater than one. A multiplet is a collection of states that transform together under a symmetry. Symmetry often explains degeneracy because a Hamiltonian that commutes with a symmetry cannot distinguish states inside an irreducible symmetry multiplet.

The slogan is useful but incomplete:

$$\text{symmetry} \quad\Rightarrow\quad \text{organization into multiplets},$$

and sometimes

$$\text{multiplet dimension} \quad\Rightarrow\quad \text{degeneracy}.$$

The word “sometimes” matters. Abelian symmetries may have one-dimensional irreducible representations, degeneracies may be accidental, and perturbations can split a multiplet when they break the protecting symmetry.

Degenerate Eigenspaces

Let $H$ be a Hamiltonian. An energy $E$ is degenerate if the eigenspace

$$\mathcal H_E = \{\lvert\psi\rangle:H\lvert\psi\rangle=E\lvert\psi\rangle\}$$

has dimension

$$\dim\mathcal H_E>1.$$

Equivalently, there is more than one linearly independent eigenstate with the same energy. Degeneracy is always relative to a specified Hamiltonian and Hilbert-space sector. A spinless Hamiltonian, a spinful Hamiltonian, and a Hamiltonian with external fields may have different degeneracies even when they describe related physics.

Symmetry Leaves Energy Eigenspaces Invariant

Suppose a group $G$ is represented by unitary operators $U(g)$ and the Hamiltonian is invariant:

$$U(g)HU(g)^\dagger = H$$

for every $g\in G$. Then

$$HU(g) = U(g)H.$$

If $\lvert\psi\rangle$ has energy $E$, then

$$\begin{aligned} H\,U(g)\lvert\psi\rangle &= U(g)H\lvert\psi\rangle \\ &= E\,U(g)\lvert\psi\rangle. \end{aligned}$$

Thus $U(g)\lvert\psi\rangle$ lies in the same energy eigenspace. The eigenspace $\mathcal H_E$ carries a representation of the symmetry group.

This is the basic reason symmetry and degeneracy are linked. Symmetry does not merely add labels after a spectrum is found; it constrains the possible structure of the eigenspaces.

Irreducible Multiplets

When an energy eigenspace contains an irreducible representation of dimension $d$, the states in that irrep form a multiplet of $d$ states. For a compact symmetry represented unitarily in finite dimension, a Hamiltonian that commutes with the symmetry acts as a scalar on each irreducible sector:

$$H\big|_{\mathcal H_\lambda} = E_\lambda I$$

on a single irreducible representation space $\mathcal H_\lambda$.

The representation-theoretic reason is that an operator commuting with every symmetry operation cannot distinguish directions inside an irreducible representation. This is the physics content of the Schur-lemma argument used throughout angular-momentum theory.

If the irrep has dimension $d>1$, this forces a $d$-fold degeneracy within that symmetry sector. If the irrep is one-dimensional, symmetry may label states without forcing degeneracy.

Rotation Multiplets

The most familiar example is rotational symmetry. A rotationally invariant Hamiltonian satisfies

$$[H,J_i]=0, \qquad i=x,y,z.$$

Angular-momentum eigenstates are labeled

$$\lvert j,m\rangle, \qquad m=-j,-j+1,\ldots,j.$$

For fixed $j$, there are

$$2j+1$$

states. If the Hamiltonian is fully rotationally invariant and the relevant states form a single spin-$j$ multiplet, all $m$ values in that multiplet have the same energy.

For a spinless central potential, this becomes the familiar orbital result:

$$m=-\ell,\ldots,\ell, \qquad \text{degeneracy }2\ell+1.$$

The states differ by orientation relative to a chosen axis. In the absence of a physical axis, the Hamiltonian cannot assign different energies to different $m$ values in the same $\ell$ multiplet.

Degeneracy Is Not Always Forced

Symmetry does not automatically imply degeneracy. A Hamiltonian can have a symmetry group whose relevant irreducible representations are one-dimensional. Parity is the simplest example: even and odd states carry different one-dimensional irreps of the two-element parity group.

For a parity-symmetric one-dimensional potential, states may be labeled even or odd, but parity alone does not require an even state and an odd state to have the same energy. In fact, generic one-dimensional bound-state spectra are nondegenerate even when parity is present.

The better statement is:

symmetry forces degeneracy when the relevant symmetry representation
has dimension greater than one, or when an antiunitary theorem such as
Kramers degeneracy applies.

Accidental and Hidden Degeneracy

Degeneracy can exceed what a visible symmetry requires. The ideal nonrelativistic hydrogen atom is the standard example.

For a generic central potential, rotational symmetry explains the $2\ell+1$ degeneracy among $m$ values at fixed $\ell$. It does not require different $\ell$ values to share an energy. The Coulomb problem is special: in the ideal spinless model, the bound-state energy depends only on the principal quantum number $n$, giving

$$g_n = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2$$

spatial states at fixed $n$.

The extra degeneracy across different $\ell$ sectors is accidental from the viewpoint of ordinary rotations. It points to a larger hidden structure associated with the Coulomb problem. The canonical counting is in Degeneracy of the Hydrogen Atom, while the conceptual taxonomy is in Accidental Symmetry and Hidden Symmetry.

Degeneracy Lifting

A perturbation can split a degenerate subspace if it breaks the symmetry or if the original degeneracy was not protected by the remaining symmetry.

Let $\mathcal D$ be a degenerate eigenspace of $H_0$, with projector $P$. For a perturbed Hamiltonian

$$H = H_0+\lambda V,$$

first-order splitting inside $\mathcal D$ is governed by

$$PVP.$$

If symmetry forces $PVP$ to be proportional to the identity on the whole multiplet, the degeneracy survives at first order. If the perturbation distinguishes states inside the multiplet, the degeneracy can split.

The Zeeman effect is the standard picture. A field along $z$ reduces full rotational symmetry to rotations about $z$. The magnetic quantum number $m$ can remain good, but the different $m$ states usually acquire different energies. The full perturbative and spectroscopic story belongs to Degeneracy Lifting.

Kramers Degeneracy

Kramers degeneracy is different from ordinary unitary multiplet degeneracy. It comes from an antiunitary time-reversal symmetry $\Theta$ satisfying

$$\Theta^2=-I$$

on the relevant sector, together with

$$\Theta H\Theta^{-1} = H.$$

If $H\lvert\psi\rangle=E\lvert\psi\rangle$, then $\Theta\lvert\psi\rangle$ is an orthogonal state with the same energy. This forces at least double degeneracy in the applicable sector.

The proof and assumptions are given in Kramers Degeneracy. The important point here is taxonomic: not every protected degeneracy is a multiplet of an ordinary unitary representation.

Diagnostic Table

Pattern	Typical cause	Robust against
$2\ell+1$ degeneracy in a central potential	rotational symmetry	rotationally invariant perturbations
even or odd labels without paired energies	parity symmetry	parity-preserving labels, not degeneracy
hydrogen $n^2$ spatial degeneracy	special Coulomb hidden structure	not generic central-potential perturbations
Zeeman splitting	symmetry reduced by external magnetic field	residual axial symmetry only
Kramers doublets	antiunitary time reversal with $\Theta^2=-I$	time-reversal-preserving perturbations in the same sector
isolated level crossing	fine tuning or incompatible symmetry sectors	usually not generic perturbations

Use the table as a diagnostic, not a proof. The proof always comes from the actual symmetry action, Hamiltonian, and allowed perturbations.

Common Mistakes

• Assuming every degeneracy is explained by the first symmetry group you notice.
• Assuming every symmetry forces degeneracy.
• Confusing a multiplet subspace with a single invariant state.
• Forgetting that degeneracy can be lifted by symmetry-breaking perturbations.
• Treating accidental degeneracy as robust without testing allowed perturbations.
• Applying Kramers degeneracy without checking antiunitarity, $\Theta^2=-I$, and time-reversal invariance.
• Counting spin degeneracy, orbital degeneracy, and hidden degeneracy without stating which Hamiltonian includes which degrees of freedom.

Cross-Links

• Symmetry Groups and Representations
• States, Observables, and Hamiltonians
• Symmetry Constraints on Hamiltonians
• Angular Momentum Algebra
• Central Potentials and Rotational Symmetry
• Accidental Symmetry
• Hidden Symmetry
• Degeneracy Lifting
• Kramers Degeneracy
• Degeneracy of the Hydrogen Atom
• Degenerate Perturbation Theory

References

• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• 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.

Exercises

1. A central-potential Hamiltonian has an $\ell=2$ orbital multiplet. How many $m$ states are degenerate by rotational symmetry?

Solution

For fixed $\ell$, the allowed values are

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

For $\ell=2$,

$$m=-2,-1,0,1,2,$$

so there are $2\ell+1=5$ states in the rotational multiplet.

2. Why does parity symmetry in a one-dimensional potential not by itself imply double degeneracy?

Solution

Parity has one-dimensional irreducible representations, even and odd. A state can carry either label without needing a partner at the same energy. Parity organizes states and selection rules, but it does not force an even state and an odd state to have the same energy in a generic one-dimensional bound-state problem.

3. A magnetic field along $z$ is added to a rotationally invariant atom. Which part of the original rotational degeneracy is most likely to be lifted first, and which label can remain good?

Solution

The field selects the $z$ axis and reduces full rotational symmetry to axial symmetry. Degeneracy among different $m$ values in a multiplet can split because the perturbation may depend on $J_z$ or related magnetic moments. The projection label $m$ can remain good when the perturbation commutes with $J_z$.

4. Why is Kramers degeneracy not just ordinary two-dimensional unitary representation degeneracy?

Solution

Kramers degeneracy relies on an antiunitary operator $\Theta$ with $\Theta^2=-I$. Antiunitarity prevents a one-dimensional invariant subspace in the relevant sector and forces an orthogonal partner $\Theta|\psi\rangle$ at the same energy when $H$ is time-reversal invariant. This mechanism is different from a unitary nonabelian symmetry forcing a multiplet of dimension greater than one.