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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.

For ordinary complex unitary representations, a multiplet of dimension greater than one forces degeneracy. A one-dimensional representation, such as even or odd parity, supplies a label without requiring a partner at the same energy. Antiunitary symmetries can protect degeneracies by a different argument.

Let HH be a Hamiltonian. An energy EE is degenerate if the eigenspace

HE={∣ψ⟩:H∣ψ⟩=E∣ψ⟩}\mathcal H_E = \{\lvert\psi\rangle:H\lvert\psi\rangle=E\lvert\psi\rangle\}

has dimension

dim⁡HE>1.\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

Section titled “Symmetry Leaves Energy Eigenspaces Invariant”

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

U(g)HU(g)†=HU(g)HU(g)^\dagger = H

for every g∈Gg\in G. Then

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

If ∣ψ⟩\lvert\psi\rangle has energy EE, then

H U(g)∣ψ⟩=U(g)H∣ψ⟩=E U(g)∣ψ⟩.\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)∣ψ⟩U(g)\lvert\psi\rangle lies in the same energy eigenspace. The eigenspace HE\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.

When an energy eigenspace contains an irreducible representation of dimension dd, the states in that irrep form a multiplet of dd states. If this single irreducible subspace is also invariant under HH, Schur’s lemma says that HH acts on it as a scalar:

H∣Hλ=EλIH\big|_{\mathcal H_\lambda} = E_\lambda I

on a single irreducible representation space Hλ\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.

Repeated copies need care. In the finite-dimensional decomposition H=⨁λ(Mλ⊗Vλ)\mathcal H=\bigoplus_\lambda(M_\lambda\otimes V_\lambda), the most general invariant Hamiltonian is

H=⨁λ(hλ⊗IVλ).H=\bigoplus_\lambda (h_\lambda\otimes I_{V_\lambda}).

It may mix equivalent copies through the matrix hλh_\lambda. After diagonalizing that matrix, each eigenvalue repeats at least dλ=dim⁡Vλd_\lambda=\dim V_\lambda times. Symmetry does not force distinct copies to share an energy. If dλ=1d_\lambda=1, no degeneracy follows from this argument.

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

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

Angular-momentum eigenstates are labeled

∣j,m⟩,m=−j,−j+1,…,j.\lvert j,m\rangle, \qquad m=-j,-j+1,\ldots,j.

For fixed jj, there are

2j+12j+1

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

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

m=−ℓ,…,ℓ,degeneracy 2ℓ+1.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 mm values in the same ℓ\ell multiplet.

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. For a regular one-dimensional scalar Schrödinger problem on a connected line with ordinary bound-state boundary conditions, the bound eigenvalues are nondegenerate even when parity is present. Disconnected wells, singular domains, periodic geometries, or internal spin degrees of freedom require separate analysis.

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 2ℓ+12\ell+1 degeneracy among mm values at fixed ℓ\ell. It does not require different ℓ\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 nn, giving

gn=∑ℓ=0n−1(2ℓ+1)=n2g_n = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2

spatial states at fixed nn.

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

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 D\mathcal D be a degenerate eigenspace of H0H_0, with projector PP. For a perturbed Hamiltonian

H=H0+λV,H = H_0+\lambda V,

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

PVP.PVP.

If symmetry forces PVPPVP 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 zz reduces full rotational symmetry to rotations about zz. The magnetic quantum number mm can remain good, but the different mm states usually acquire different energies. The full perturbative and spectroscopic story belongs to Degeneracy Lifting.

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

Θ2=−I\Theta^2=-I

on the relevant sector, together with

ΘHΘ−1=H.\Theta H\Theta^{-1} = H.

If H∣ψ⟩=E∣ψ⟩H\lvert\psi\rangle=E\lvert\psi\rangle, then Θ∣ψ⟩\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. Thus a protected doublet need not be an irreducible multiplet of an ordinary unitary group.

Pattern Typical cause Robust against
2ℓ+12\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 n2n^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 Θ2=−I\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.

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

For fixed ℓ\ell, the allowed values are

m=−ℓ,−ℓ+1,…,ℓ.m=-\ell,-\ell+1,\ldots,\ell.

For ℓ=2\ell=2,

m=−2,−1,0,1,2,m=-2,-1,0,1,2,

so there are 2ℓ+1=52\ell+1=5 states in the rotational multiplet.

  1. 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.

  1. A magnetic field along zz 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 zz axis and reduces full rotational symmetry to axial symmetry. Degeneracy among different mm values in a multiplet can split because the perturbation may depend on JzJ_z or related magnetic moments. The projection label mm can remain good when the perturbation commutes with JzJ_z.

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

Kramers degeneracy relies on an antiunitary operator Θ\Theta with Θ2=−I\Theta^2=-I. Antiunitarity prevents a one-dimensional invariant subspace in the relevant sector and forces an orthogonal partner Θ∣ψ⟩\Theta|\psi\rangle at the same energy when HH is time-reversal invariant. This mechanism is different from a unitary nonabelian symmetry forcing a multiplet of dimension greater than one.

  1. Two copies of a spin-1/21/2 irrep have an invariant Hamiltonian H=h⊗I2H=h\otimes I_2, where h=(avv∗b)h=\begin{pmatrix}a&v\\v^*&b\end{pmatrix} and a,ba,b are real. Find the energies and their symmetry-enforced multiplicity.
Solution

Diagonalizing the multiplicity matrix gives

E±=a+b2±(a−b2)2+∣v∣2.E_\pm=\frac{a+b}{2}\pm\sqrt{\left(\frac{a-b}{2}\right)^2+|v|^2}.

Each energy is a doublet because of the factor I2I_2. They coincide only if a=ba=b and v=0v=0, an additional condition not required by rotation symmetry. This is why one must distinguish repeated irreps from magnetic states within one irrep.

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