Kramers Degeneracy

Kramers degeneracy is the degeneracy forced by an antiunitary time-reversal symmetry whose square is $-I$. The theorem combines three facts:

• time reversal is antiunitary;
• the Hamiltonian is invariant under time reversal;
• the time-reversal operator squares to $-I$ on the Hilbert-space sector being studied.

For a single spin-$1/2$ degree of freedom, the last condition comes from

$$\Theta=-i\sigma_yK, \qquad \Theta^2=-I,$$

as shown in Time Reversal for Spin-1/2 Particles. Kramers degeneracy is what remains of that algebra in more complicated systems: atoms, molecules, quantum dots, spin–orbit-coupled bands, and other systems with an odd half-integer time-reversal structure.

Kramers Theorem

Let $H$ be a self-adjoint Hamiltonian and let $\Theta$ be an antiunitary time-reversal operator such that

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

on the domain of $H$. Suppose also that, on the sector of interest,

$$\Theta^2=-I.$$

If

$$H\lvert\psi\rangle = E\lvert\psi\rangle$$

for a normalizable energy eigenstate, then $\Theta\lvert\psi\rangle$ is a distinct orthogonal eigenstate with the same energy:

$$H\,\Theta\lvert\psi\rangle = E\,\Theta\lvert\psi\rangle, \qquad \langle\psi|\Theta\psi\rangle=0.$$

Every finite-dimensional energy eigenspace therefore has even dimension. After selecting one orthonormal pair, its orthogonal complement in the eigenspace is also invariant under $\Theta$, so the pairing can be repeated. The pair

$$\lvert\psi\rangle, \qquad \Theta\lvert\psi\rangle$$

is called a Kramers pair.

Both partners must belong to the same physical sector and Hamiltonian. Reversing an external magnetic field can relate two different spectra, but does not supply a symmetry of either fixed-field problem. Continuous spectra require a spectral-subspace formulation instead of treating unnormalizable plane waves as discrete eigenvectors.

Proof

First show that the partner has the same energy. From $\Theta H\Theta^{-1}=H$ one has $H\Theta=\Theta H$. Therefore

$$\begin{aligned} H\Theta\lvert\psi\rangle &= \Theta H\lvert\psi\rangle \\ &= \Theta\left(E\lvert\psi\rangle\right) \\ &= E^*\Theta\lvert\psi\rangle. \end{aligned}$$

Because $H$ is self-adjoint, $E$ is real, so $E^*=E$. Hence $\Theta\lvert\psi\rangle$ has the same energy.

Now show orthogonality. Define

$$c = \langle\psi|\Theta\psi\rangle.$$

Antiunitarity gives

$$\langle\Theta\psi|\Theta^2\psi\rangle = \langle\psi|\Theta\psi\rangle^* = c^*.$$

Using $\Theta^2=-I$, the same left-hand side is

$$\langle\Theta\psi|-\psi\rangle = -\langle\Theta\psi|\psi\rangle = -c^*.$$

Thus $c^*=-c^*$, so $c=0$. The state and its time-reversed partner are orthogonal.

Why Antiunitarity Is Essential

A unitary symmetry can commute with the Hamiltonian without forcing a degeneracy. If $U$ is unitary and $U\lvert\psi\rangle=e^{i\alpha}\lvert\psi\rangle$, then a one-dimensional energy eigenspace can carry that symmetry by a phase.

For an antiunitary operator with $\Theta^2=-I$, a one-dimensional invariant subspace is impossible. If one tried

$$\Theta\lvert\psi\rangle = c\lvert\psi\rangle,$$

then antiunitarity would imply

$$\Theta^2\lvert\psi\rangle = c^*\Theta\lvert\psi\rangle = |c|^2\lvert\psi\rangle,$$

which cannot equal $-\lvert\psi\rangle$. This is the short proof that Kramers degeneracy is fundamentally antiunitary.

Which Systems Have the Minus Sign?

For a single angular-momentum multiplet with quantum number $j$, time reversal satisfies

$$\Theta^2=(-1)^{2j}I.$$

Thus half-integer $j$ has $\Theta^2=-I$, while integer $j$ has $\Theta^2=+I$.

For $N$ spin-$1/2$ factors in a product $S_z$ basis, write $\Theta_N=(-i\sigma_y)^{\otimes N}K$, with one conjugation of the full coefficient array. Its square is

$$\Theta_N^2=(-1)^N I$$

on the spin part. An odd number of spin-$1/2$ factors gives a Kramers structure; an even number does not automatically do so. On the electronic Hilbert space, an odd electron number gives $\Theta^2=-I$ and an even number gives $+I$. If nuclear spins or other internal degrees of freedom are included, their time-reversal squares must also be included; counting electrons alone no longer specifies the full sector.

This is a sector statement. One should not say simply that “the system has Kramers degeneracy” without specifying the Hilbert-space sector, particle-number sector, or effective doublet being discussed.

Magnetic Fields Break the Protection

For spin-$1/2$,

$$H(\mathbf B) = -\gamma\mathbf S\cdot\mathbf B.$$

Time reversal flips spin:

$$\Theta\mathbf S\Theta^{-1} = -\mathbf S.$$

Therefore

$$\Theta H(\mathbf B)\Theta^{-1} = H(-\mathbf B).$$

For $\gamma\ne0$ and fixed $\mathbf B\ne0$, this spin-$1/2$ Hamiltonian is not invariant. The Zeeman term then lifts the Kramers degeneracy. This is why a magnetic field splits a spin doublet, while a time-reversal-preserving perturbation cannot split a Kramers pair by itself.

The sign and splitting conventions for this Hamiltonian are discussed in Spin in Magnetic Fields.

Spin–Orbit Coupling Does Not Automatically Break It

Spin–orbit coupling is often present in Kramers-degenerate systems. A central spin–orbit term has angular structure

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

Under time reversal,

$$\Theta\mathbf L\Theta^{-1} = -\mathbf L, \qquad \Theta\mathbf S\Theta^{-1} = -\mathbf S.$$

The scalar product is therefore even:

$$\Theta(\mathbf L\cdot\mathbf S)\Theta^{-1} = \mathbf L\cdot\mathbf S.$$

Spin–orbit coupling can split levels according to total angular momentum, but if time reversal remains intact and the sector has $\Theta^2=-I$, each allowed level still contains Kramers partners. The angular-momentum organization is developed in Spin–Orbit Coupling. Spin–Orbit Coupling in Solids applies the theorem to Bloch bands and distinguishes Kramers degeneracy at invariant momenta from $\mathcal P\mathcal T$ degeneracy at every momentum.

Where the Degeneracy Appears

The theorem has different faces in different settings:

• In a single spin-$1/2$ with no magnetic field, any time-reversal-invariant Hamiltonian is proportional to $I$, so the two spin states form a protected doublet.
• In atoms or molecules with an odd number of electrons, crystal-field or molecular-field perturbations can split multiplets, but time-reversal-preserving perturbations leave Kramers doublets.
• In a quantum dot with an odd electron number, spin–orbit coupling and confinement can reshape the doublet, while a magnetic field can split it.
• In crystals, time reversal maps Bloch momentum $\mathbf k$ to $-\mathbf k$. At momenta equivalent to their negatives modulo a reciprocal lattice vector, Kramers degeneracy is forced within the same crystal momentum sector. Away from those momenta, the partner is generally at $-\mathbf k$. Symmetry of Bloch States explains these invariant momenta and the additional conditions under which combined inversion and time reversal protect a doublet at every momentum.

These examples use the same theorem but different Hilbert-space labels. The theorem protects a pair only when both partners belong to the same symmetry setting being compared.

What It Does Not Say

Kramers degeneracy does not say that every time-reversal-invariant system is degenerate. Spinless systems commonly have $\Theta^2=+I$ and need not have paired eigenstates.

It also does not say that every twofold degeneracy is a Kramers degeneracy. Degeneracies can arise from rotations, translations, parity plus other symmetries, accidental parameter values, or approximate model assumptions.

Finally, it does not protect a pair against all perturbations. It protects against perturbations that preserve the same antiunitary time-reversal symmetry and remain in the same $\Theta^2=-I$ sector.

Common Mistakes

• Applying the theorem when $\Theta^2=+I$.
• Forgetting that $\Theta$ is antiunitary and therefore conjugates coefficients.
• Saying “time reversal symmetry” while keeping a fixed magnetic field unchanged.
• Treating spin–orbit coupling as time-reversal breaking by itself.
• Assuming a Kramers pair must be the naive $|\uparrow\rangle,|\downarrow\rangle$ basis. Interactions can mix the pair.
• Confusing Kramers degeneracy with rotational degeneracy from $m_j=-j,\ldots,j$.
• Forgetting that in a crystal the time-reversed partner of a state at $\mathbf k$ is generally at $-\mathbf k$.

Exercises

1. Let $\Theta$ be antiunitary, $\Theta^2=-I$, and $\Theta H\Theta^{-1}=H$. Prove that a nondegenerate energy eigenstate is impossible.

Solution

If $H\lvert\psi\rangle=E\lvert\psi\rangle$, then

$$H\Theta\lvert\psi\rangle = \Theta H\lvert\psi\rangle = \Theta(E\lvert\psi\rangle) = E^*\Theta\lvert\psi\rangle,$$

where $E$ is real. Thus $\Theta\lvert\psi\rangle$ has the same energy. Orthogonality follows from

$$\langle\Theta\psi|\Theta^2\psi\rangle = \langle\psi|\Theta\psi\rangle^*$$

and $\Theta^2=-I$, giving

$$\langle\psi|\Theta\psi\rangle=0.$$

So the same eigenspace contains at least two orthogonal states.

2. Consider a spin-$3/2$ with $H=D(S_z/\hbar)^2$, where $D$ has energy units and is nonzero. Does the anisotropy break time reversal? Find the energies and their degeneracies. Then add a transverse anisotropy,

$$H=\frac{D S_z^2+C(S_x^2-S_y^2)}{\hbar^2},$$

where $C$ is a real energy. Find the new energies and explain why mixing the $S_z$ eigenstates does not split each Kramers pair.

Solution

Time reversal sends $S_z$ to $-S_z$, so $S_z^2$ is invariant. The $m=\pm1/2$ states have energy $D/4$ and the $m=\pm3/2$ states have energy $9D/4$. Rotational symmetry is reduced, yet each energy is a Kramers doublet. The ordering depends on the sign of $D$.

The transverse term is also time-reversal even. Use $S_x^2-S_y^2=(S_+^2+S_-^2)/2$: it couples states whose $m$ values differ by two. In the ordered subspaces $(|3/2\rangle,|-1/2\rangle)$ and $(|-3/2\rangle,|1/2\rangle)$, the Hamiltonian has the same block,

$$H_{\mathrm{block}}= \begin{pmatrix} 9D/4&\sqrt3\,C\\ \sqrt3\,C&D/4 \end{pmatrix}.$$

The ladder coefficients supply the off-diagonal entry $\sqrt3\,C$. Diagonalizing gives

$$E_\pm=\frac{5D}{4}\pm\sqrt{D^2+3C^2},$$

each twice. With $\Theta|j,m\rangle=(-1)^{j-m}|j,-m\rangle$, time reversal maps the first subspace into the second, so a mixed eigenvector in one block has an orthogonal partner in the other. For $C\ne0$, $S_z$ is no longer conserved, but the doublets persist. Taking $C\to0$ recovers the two axial-anisotropy energies, including their ordering for either sign of $D$.

3. Show that two independent spin-$1/2$ degrees of freedom do not automatically have $\Theta^2=-I$.

Solution

Let $U=-i\sigma_y$ and let $K$ conjugate coefficients in the full product basis. Then the joint time reversal is

$$\Theta_{12} = (U\otimes U)K.$$

Its square is

$$\Theta_{12}^2 = (UU^*)\otimes(UU^*) = (-I)\otimes(-I) = I.$$

Thus the two-spin sector need not have Kramers degeneracy from time reversal alone.

4. A Kramers doublet is perturbed by $H_Z=-(\gamma\hbar B/2)\sigma_z$. Does this perturbation preserve the Kramers degeneracy when $B$ is fixed?

Solution

No. Since

$$\Theta\sigma_z\Theta^{-1} = -\sigma_z,$$

one has

$$\Theta H_Z(B)\Theta^{-1} = -H_Z(B) = H_Z(-B).$$

For fixed nonzero $B$, this is not equal to $H_Z(B)$. The eigenvalues are

$$E_\pm = \mp\frac{\gamma\hbar B}{2},$$

so this two-state model has a nonzero gap $|\gamma\hbar B|$ whenever $\gamma B\ne0$. Breaking time reversal removes the guarantee; in a more general model it does not by itself prove that every level must split.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• M. S. Dresselhaus, G. Dresselhaus, and A. Jorio, Group Theory: Application to the Physics of Condensed Matter, Springer, 2008.
• H. A. Kramers, “Théorie générale de la rotation paramagnétique dans les cristaux,” Proceedings Koninklijke Akademie van Wetenschappen 33, 959–972, 1930. Author bibliography and original paper.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.