Degeneracy Lifting

Degeneracy lifting is the splitting of states that had the same energy in a reference Hamiltonian. It is the spectral signature of the fact that a perturbation has distinguished states that the unperturbed problem treated as equivalent.

The standard setup is

$$H(\lambda) = H_0+\lambda V, \qquad \lvert \lambda\rvert \ll 1,$$

where $H_0$ has a degenerate eigenspace $\mathcal D$ at energy $E^{(0)}$:

$$H_0|\psi_a\rangle = E^{(0)}|\psi_a\rangle, \qquad a=1,\ldots,d.$$

The degeneracy is lifted if the exact energies near $E^{(0)}$ are no longer all equal when $\lambda\ne0$. In first-order perturbation theory the relevant object is not the diagonal matrix element in an arbitrarily chosen old basis. It is the operator

$$W = PVP, \qquad P = \sum_{a=1}^d |\psi_a\rangle\langle\psi_a|,$$

restricted to $\mathcal D$. Its eigenvalues $w_\alpha$ give

$$E_\alpha = E^{(0)} + \lambda w_\alpha + O(\lambda^2).$$

If the $w_\alpha$ differ, the degeneracy splits at first order. If $W$ is proportional to the identity on all of $\mathcal D$, the degeneracy survives at first order, either because it is symmetry-protected or because the first splitting appears at higher order.

The computational method is covered in Degenerate Perturbation Theory. This page explains the symmetry meaning of the result.

What Gets Lifted

Degeneracy is always degeneracy of a specified Hamiltonian. Before asking whether it is lifted, identify:

• the reference Hamiltonian $H_0$;
• the degenerate subspace $\mathcal D$;
• the symmetry, hidden structure, or fine-tuning that produced the degeneracy;
• the perturbation $V$;
• the residual symmetries of $H_0+\lambda V$.

For a group $G$ represented by $U(g)$, an exact symmetry of $H_0$ means

$$U(g)H_0U(g)^{-1}=H_0.$$

An added perturbation usually leaves only the subgroup

$$G_V = \{g\in G:U(g)VU(g)^{-1}=V\}.$$

The full perturbed Hamiltonian has the common symmetry of $H_0$ and $V$. Degeneracy lifting often says that a large representation of $G$ has decomposed into smaller representations of $G_V$.

This is the reason the phrase “symmetry breaking splits a multiplet” is useful but incomplete. It must be supplemented by the residual symmetry. The residual symmetry determines which states can still share an energy and which labels remain good.

Protected and Unprotected Degeneracy

A protected degeneracy cannot be split by perturbations that preserve the protecting symmetry. An unprotected degeneracy can be split by some symmetry-allowed perturbation.

Inside the degenerate subspace, protection is encoded by the allowed form of $PVP$. If symmetry forces

$$PVP = cP$$

for every allowed perturbation $V$, then all states in $\mathcal D$ receive the same first-order shift. If the symmetry allows a more general matrix on $\mathcal D$, generic perturbations will split the subspace.

This is not only a first-order statement. If a symmetry theorem requires a degeneracy, such as Kramers degeneracy for appropriate time-reversal-invariant systems, the degeneracy is protected to all orders by that symmetry. If the degeneracy is only an accidental symmetry of the ideal model, small corrections often reveal that the original equality of energies was not robust.

Residual Quantum Numbers

Degeneracy lifting usually changes the best labels for states. Suppose $H_0$ has a large symmetry and $V$ reduces it. The old quantum numbers may no longer all label exact eigenstates, while the quantum numbers of the residual symmetry remain exact.

For example, if a rotationally invariant Hamiltonian is perturbed by a fixed field along $\hat{\mathbf z}$, full rotational symmetry is reduced to rotations about the $z$ axis. The exact conserved angular momentum label is then the projection along the field, not the full spatial multiplet structure.

In equations, if

$$[H_0,\mathbf J^2]=0, \qquad [H_0,J_z]=0,$$

but the perturbation satisfies

$$[V,J_z]=0, \qquad [V,J_x]\ne0$$

or $[V,J_y]\ne0$, then $m$ can remain a good label while the degeneracy among different $m$ values need not survive.

Approximate labels remain useful when the splitting is small compared with the resolution, linewidth, temperature scale, or dynamical timescale of interest. That regime is discussed in Approximate Symmetry.

Zeeman Splitting

A magnetic field provides a clean example because it selects a spatial direction. For a field

$$\mathbf B = B\hat{\mathbf z},$$

a simple magnetic perturbation has the form

$$V_Z = -\boldsymbol{\mu}\cdot\mathbf B.$$

In a spinless orbital model with charge convention fixed separately, this often reduces to a term proportional to $L_z$:

$$V_Z = \omega_L L_z$$

up to the sign and coefficient appropriate to the particle. Since $L_z|\ell m\rangle=\hbar m|\ell m\rangle$, different $m$ values receive different shifts:

$$\Delta E_m^{(1)} = \omega_L\hbar m.$$

The field has broken full rotational symmetry down to axial symmetry. The projection $m$ remains a good quantum number, while the original degeneracy among the $m=-\ell,\ldots,\ell$ states is generally lifted.

Real atomic Zeeman patterns involve spin, spin–orbit coupling, fine structure, hyperfine structure, and the strength of the applied field. The historical and spectroscopic context is in Zeeman Effect Revisited, with the compact named-effect entry at Zeeman Effect.

Stark Splitting

An electric field selects a polar direction and commonly adds an electric-dipole perturbation

$$V_S = -\mathbf d\cdot\mathbf E.$$

For a particle of charge $q$ in a uniform field $\mathbf E=E\hat{\mathbf z}$, one often writes

$$V_S = qEz,$$

with the sign depending on the charge convention. The coordinate $z$ is odd under parity, so this perturbation breaks inversion symmetry:

$$\mathcal P z \mathcal P^{-1} = -z.$$

For a nondegenerate parity eigenstate, the first-order Stark shift from $qEz$ vanishes because

$$\langle n|z|n\rangle = 0$$

when $|n\rangle$ has definite parity. The leading shift is then often second order. In a degenerate subspace containing opposite-parity states, however, $z$ can have nonzero off-diagonal matrix elements. One must diagonalize $PzP$, and a linear Stark splitting can appear.

Hydrogen is the classic case: the ideal Coulomb problem has degeneracies between states of different $\ell$ at fixed principal quantum number $n$. The electric field mixes allowed opposite-parity states inside that degenerate manifold. The named-effect overview is Stark Effect, and the Coulomb degeneracy is explained in Degeneracy of the Hydrogen Atom.

Spin–Orbit Splitting

Spin–orbit coupling does not simply “break rotational symmetry.” In a central problem with spin, a term

$$H_{\mathrm{SO}} = \xi(r)\mathbf L\cdot\mathbf S$$

breaks the separate conservation of orbital and spin angular momentum, but preserves simultaneous rotations generated by

$$\mathbf J = \mathbf L+\mathbf S.$$

The useful labels change from separate $\ell,m_\ell,s,m_s$ labels to total angular momentum labels $j,m_j$. Since

$$\mathbf L\cdot\mathbf S = \frac{1}{2} \left( \mathbf J^2-\mathbf L^2-\mathbf S^2 \right),$$

the perturbation can split states with different $j$ while leaving the $m_j$ degeneracy within a given $j$ multiplet intact, as long as no additional field breaks rotational symmetry.

Thus spin–orbit coupling is a good example of partial lifting. The original product-space degeneracy is reorganized into multiplets of the residual exact symmetry. The detailed angular-momentum algebra is developed in Spin–Orbit Coupling.

Crystal-Field Splitting

In atoms, the ideal central potential has rotational symmetry. In a molecule or crystal, neighboring ions and ligands produce an environment with only a finite point-group symmetry. Full rotational multiplets then split into irreducible representations of the point group.

Schematically, a rotational multiplet may branch as

$$\mathcal H_\ell \downarrow_G = \Gamma_1 \oplus \Gamma_2 \oplus \cdots,$$

where $G$ is the point group of the local environment. The perturbation is not arbitrary; it is constrained by $G$. States belonging to inequivalent point-group irreducible representations need not remain degenerate, while degeneracies required by multidimensional irreducible representations of $G$ remain protected by that point-group symmetry.

This is the symmetry core of crystal-field splitting. Detailed ligand-field theory belongs to molecular and quantum-matter pages, but the logic is the same as in any degenerate perturbation problem: restrict the perturbation to the degenerate subspace and diagonalize it subject to the residual symmetry.

First Order, Higher Order, and No Splitting

If $PVP$ has distinct eigenvalues, splitting appears at first order. If $PVP$ is proportional to the identity, there are three common possibilities.

First, a symmetry may protect the degeneracy exactly. No symmetry-preserving perturbative correction can split it.

Second, first-order splitting may vanish but higher-order splitting may appear through virtual coupling to states outside $\mathcal D$. The second-order effective operator has the schematic form

$$W_{\mathrm{eff}}^{(2)} = PVQ \frac{1}{E^{(0)}-QH_0Q} QVP,$$

where $Q=1-P$, with the inverse understood on the separated subspace. If this operator is not proportional to $P$, the degeneracy can split at second order.

Third, the perturbation may shift every state in the subspace equally for reasons that are not a deep symmetry theorem in the full problem. A different allowed perturbation, or a correction omitted from the model, may still split the levels.

Avoided Crossings

Degeneracy lifting is closely related to avoided crossings. Consider two levels depending on a parameter $\lambda$. Near a putative crossing, a two-state effective Hamiltonian can be written as

$$H_{\mathrm{eff}}(\lambda) = \begin{pmatrix} E_1(\lambda) & v(\lambda) \\ v(\lambda)^* & E_2(\lambda) \end{pmatrix}.$$

The eigenvalues are

$$E_\pm = \frac{E_1+E_2}{2} \pm \sqrt{ \left(\frac{E_1-E_2}{2}\right)^2 + |v|^2 }.$$

If no symmetry forces $v=0$, the levels repel rather than cross. If a symmetry places the two states in different sectors and forbids mixing, a true crossing can remain. This is the parameter-dependent version of the same lesson: degeneracy is stable only when some structure protects it.

Common Mistakes

• Saying “the degeneracy is lifted” without specifying the reference Hamiltonian and perturbation.
• Treating an arbitrary basis in a degenerate subspace as the perturbed eigenbasis.
• Assuming every perturbation lifts every degeneracy. Symmetry can protect part or all of a multiplet.
• Assuming a small perturbation causes only a small rotation of eigenvectors inside a degenerate subspace.
• Confusing residual symmetry with no symmetry. A magnetic field destroys full rotational symmetry but preserves rotations about the field axis.
• Treating spin–orbit coupling as an external-field splitting. In a central problem it preserves total rotational symmetry.
• Using a named effect, such as Zeeman or Stark, without stating the coupling regime and good quantum numbers.

Related Pages

• Explicit Symmetry Breaking
• Broken Symmetry Preview
• Approximate Symmetry
• Degeneracy and Multiplets
• Accidental Symmetry
• Hidden Symmetry
• Symmetry-Protected Structure Preview
• Degenerate Perturbation Theory
• Quasi-Degenerate Perturbation Theory
• Spin–Orbit Coupling
• Selection Rules
• Zeeman Effect
• Stark Effect

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloe, Quantum Mechanics, Wiley, 1977.
• E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.

Exercises

Exercise 1

Let $\mathcal D$ be a two-dimensional degenerate subspace spanned by $|1\rangle$ and $|2\rangle$. The perturbation restricted to the subspace is

$$W = \begin{pmatrix} a & b \\ b^* & c \end{pmatrix}.$$

Find the first-order energy shifts and state when the degeneracy is not lifted at first order.

Solution

The first-order shifts are the eigenvalues of $W$:

$$w_\pm = \frac{a+c}{2} \pm \sqrt{ \left(\frac{a-c}{2}\right)^2 + |b|^2 }.$$

The degeneracy is not lifted at first order when $w_+=w_-$. This requires $a=c$ and $b=0$, so $W=aI$ on the degenerate subspace.

Exercise 2

A spinless central-potential eigenstate has angular momentum $\ell$ and is degenerate in $m$. Add a weak perturbation proportional to $L_z$. Which degeneracy is lifted, and which quantum number remains exact?

Solution

Since

$$L_z|\ell m\rangle = \hbar m|\ell m\rangle,$$

the shift is proportional to $m$. The degeneracy among different $m$ values is generally lifted. The perturbation still commutes with $L_z$, so $m$ remains an exact label. It also commutes with $\mathbf L^2$, so $\ell$ remains a good label in this simplified model.

Exercise 3

Explain why a nondegenerate parity eigenstate has no first-order shift from a perturbation proportional to $z$, but a degenerate manifold containing opposite-parity states may have a linear Stark splitting.

Solution

The operator $z$ is odd under parity:

$$\mathcal P z\mathcal P^{-1} = -z.$$

For a parity eigenstate $|n\rangle$,

$$\langle n|z|n\rangle = -\langle n|z|n\rangle,$$

so the expectation value vanishes. In a degenerate subspace, however, the perturbation matrix has off-diagonal elements between opposite-parity states. Diagonalizing $PzP$ can produce eigenvalues linear in the applied field.

Exercise 4

Spin–orbit coupling splits an $\ell=1$, $s=1/2$ subspace into which total-angular-momentum multiplets? How many states are in each multiplet?

Solution

Adding $\ell=1$ and $s=1/2$ gives

$$j = \ell\pm\frac{1}{2} = \frac{3}{2},\frac{1}{2}.$$

The $j=3/2$ multiplet has $2j+1=4$ states, and the $j=1/2$ multiplet has $2j+1=2$ states. The six original product states are reorganized as $4+2$. If no external field is present, rotational symmetry preserves the degeneracy among $m_j$ values inside each multiplet.