Degeneracy Lifting

Degeneracy lifting occurs when states with equal energy in a reference Hamiltonian acquire different energies after a perturbation. The useful question is which degeneracy can split and which symmetry still protects part of it. A magnetic field can resolve angular-momentum projections; a rotationally invariant spin–orbit interaction can instead separate total-angular-momentum multiplets while leaving each multiplet internally degenerate.

Helpful background. Exact symmetry identifies conserved labels, explicit symmetry breaking distinguishes a changed Hamiltonian from a changed state, and degenerate perturbation theory supplies the calculation in a degenerate subspace.

The perturbation within a degenerate subspace

Let $H(\lambda)=H_0+\lambda V$ have a finite-dimensional eigenspace $\mathcal D$ of $H_0$ at energy $E_0$, separated from the remaining spectrum. The dimensionless parameter $\lambda$ controls the perturbation; its relevant matrix elements must be small compared with the separation to other levels. Merely writing $|\lambda|\ll1$ is insufficient without an energy comparison.

If $P$ projects onto $\mathcal D$, the first-order shifts are eigenvalues of the restricted operator $W=PVP|_{\mathcal D}$:

$$W|\alpha\rangle=w_\alpha|\alpha\rangle, \qquad E_\alpha(\lambda)=E_0+\lambda w_\alpha+O(\lambda^2).$$

Diagonal elements in an arbitrary old basis do not suffice. For example, $W=g\sigma_x$ has zero diagonal in the $\sigma_z$ basis but shifts the two energies by $\pm\lambda g$. Its eigenstates are equal superpositions of the old basis states, however small nonzero $\lambda$ is. Small energy corrections can accompany order-one changes of the chosen eigenvectors inside a degenerate space.

If $W=cI$ on $\mathcal D$, all states receive the same first-order shift. This establishes only the absence of first-order splitting. Higher orders or a different allowed perturbation can still split them.

Residual symmetry constrains the splitting

Among the original symmetries $U(g)$ of $H_0$, those satisfying $U(g)VU(g)^{-1}=V$ remain symmetries of the perturbed Hamiltonian. Restricting an original multiplet to this smaller group can decompose it into smaller irreducible representations. Their energies need not agree.

For unitary symmetries, Schur’s lemma explains the remaining degeneracy. Within one irreducible representation a symmetric operator acts as a scalar. If an irreducible representation occurs several times, however, a symmetric operator can mix those copies and split their multiplicity labels while preserving the degeneracy within each resulting multiplet. A block decomposition is therefore more informative than simply counting conserved operators. See degeneracy and multiplets.

Kramers degeneracy has a different, antiunitary origin. When the full Hamiltonian preserves a time-reversal operator with $\Theta^2=-I$ in the relevant sector, normalizable energy eigenstates remain paired exactly. This is a theorem about the full symmetric Hamiltonian, not an inference from one scalar first-order matrix. See symmetry-protected structure.

Zeeman, Stark, and spin–orbit examples

In a simplified central orbital problem, $V_Z=\omega_L L_z$ gives shifts $\hbar\omega_L m$. Full rotational invariance is lost but $L_z$ and $L^2$ still commute with this particular Hamiltonian. Thus a reduced symmetry group does not imply that every old Casimir label is lost. In more general axial perturbations only the projection may remain exact. Real Zeeman spectra also depend on spin, hyperfine coupling, and field strength.

For a particle of signed charge $q$ in $\mathbf E=\mathcal E\hat{\mathbf z}$, choose electrostatic potential $\phi=-\mathcal Ez$. Its Stark perturbation is

$$V_S=q\phi=-q\mathcal Ez=-\mathbf d\cdot\mathbf E, \qquad \mathbf d=q\mathbf r.$$

The sign follows from the field direction and signed charge; it is not an independent convention once those are specified. Since $z$ is odd under parity, a nondegenerate parity eigenstate has $\langle z\rangle=0$ and no first-order shift. Within a degenerate manifold containing opposite parities, off-diagonal matrix elements can be nonzero and produce a linear splitting. The Stark effect and hydrogen degeneracy develop the physical examples.

A central spin–orbit term $\xi(r)\mathbf L\cdot\mathbf S$ preserves $L^2,S^2,J^2,J_z$, with $\mathbf J=\mathbf L+\mathbf S$. The separate projections $L_z,S_z$ generally cease to be conserved. Using

$$\mathbf L\cdot\mathbf S=\frac12(J^2-L^2-S^2),$$

one finds different shifts for different $j$, while simultaneous spatial and spin rotations preserve the $m_j$ degeneracy within each $j$ multiplet. It is incorrect to describe this central interaction as breaking all rotational symmetry. The atomic application works out the $\ell=1,s=1/2$ example.

In a crystal field, a rotational multiplet instead decomposes into representations of the local point group. Multidimensional point-group representations can retain degeneracy, and spinful time reversal can add Kramers pairing. The energy ordering and actual splittings require coupling strengths, not just group labels.

Splitting first appearing at second order

Take two states of energy zero and one state at $\Delta>0$, coupled by

$$H=\begin{pmatrix} 0&0&\lambda g\\ 0&0&0\\ \lambda g&0&\Delta \end{pmatrix},\qquad g\in\mathbb R.$$

In the two-dimensional zero-energy subspace, $PVP=0$. Yet the first state couples virtually to the upper state while the second does not. Eliminating that upper state gives, to second order,

$$H_{\rm eff}^{(2)}= -\frac{\lambda^2g^2}{\Delta} \begin{pmatrix}1&0\\0&0\end{pmatrix}.$$

The two low energies thus split by $\lambda^2g^2/\Delta$ to leading order. The approximation requires $|\lambda g|\ll\Delta$; when another level approaches this scale it must be included in a larger quasi-degenerate subspace.

Crossings and avoided crossings

Near a crossing, the same reasoning gives a two-state Hamiltonian with diagonal entries $E_1,E_2$ and off-diagonal coupling $v$. Its gap is

$$E_+-E_-=\sqrt{(E_1-E_2)^2+4|v|^2}.$$

At equality of the diagonal energies, a nonzero $v$ opens a gap $2|v|$. Different exact symmetry sectors can forbid $v$, allowing a crossing. The absence of a selection rule does not logically force $v\ne0$; accidental zeros and further tuning can still occur. Generic allowed mixing produces an avoided crossing, as illustrated in good quantum numbers.

Exercises

1. For $W=\begin{pmatrix}a&b\\b^*&c\end{pmatrix}$, with real $a,c$, find the first-order shifts and the condition for no first-order splitting.

Solution

The eigenvalues are

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

They agree exactly when $a=c$ and $b=0$. The shifts are $\lambda w_\pm$; a scalar restricted matrix does not alone settle higher-order splitting.

2. Diagonalize the three-state example exactly and verify the second-order low-energy splitting.

Solution

The uncoupled second state stays at zero. The other two energies are

$$E_\pm=\frac{\Delta\pm\sqrt{\Delta^2+4\lambda^2g^2}}2.$$

For $|\lambda g|\ll\Delta$, $E_-=-\lambda^2g^2/\Delta+O(\lambda^4g^4/\Delta^3)$. It differs from zero despite $PVP=0$. Its negative sign is the level repulsion from a higher-energy state.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977. See stationary perturbation theory and angular momentum.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003; originally McGraw-Hill, 1964. See symmetry constraints on degenerate subspaces.