Kramers Degeneracy

Kramers degeneracy is the degeneracy forced by an antiunitary time-reversal symmetry whose square is $-I$. It is not an ordinary symmetry degeneracy from a commuting unitary operator. Its power comes from combining 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.

Statement

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:

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

Thus every such energy eigenspace has even dimension. The pair

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

is called a Kramers pair.

Assumptions Matter

The compact theorem is often stated as “time reversal plus $T^2=-1$ implies double degeneracy.” The useful version is a little more precise:

• $\Theta$ must be antiunitary, not unitary.
• $\Theta^2=-I$ must hold on the relevant sector.
• $H$ must be time-reversal invariant: $\Theta H\Theta^{-1}=H$.
• The transformed state must remain in the same physical Hilbert-space sector.
• The eigenstate statement is cleanest for discrete normalizable eigenstates; continuous spectra require a spectral-subspace formulation.
• Fixed magnetic fields, magnetic order, and other time-reversal-breaking backgrounds invalidate the theorem unless they are transformed as part of the comparison.

The last point is not technical fussiness. It is the difference between a protected doublet and an ordinary Zeeman splitting.

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 several independent spin-$1/2$ factors, the product time-reversal operator has the schematic square

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

on the spin part. An odd number of spin-$1/2$ factors gives a Kramers structure; an even number does not automatically do so. In many-electron language, sectors with an odd number of electrons carry the characteristic $\Theta^2=-I$ behavior, while even-electron sectors can have $\Theta^2=+I$.

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

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

If $\mathbf B$ is a fixed external background, $H(\mathbf B)$ is not invariant unless $\mathbf B=0$. 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 can occur within the same crystal momentum sector. Away from those momenta, the partner is generally at $-\mathbf k$. Symmetry of Bloch States owns this crystalline specialization, including time-reversal-invariant momenta, $\mathcal P\mathcal T$-protected doublets, and the distinction from valley or unitary little-group degeneracy.

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 $-\mathbf k$.

Cross-Links

• Time Reversal
• Antiunitary Time Reversal
• Time Reversal for Spinless Particles
• Time Reversal for Spin-1/2 Particles
• Symmetry Classification Preview
• Symmetry-Protected Structure Preview
• Topological Insulators
• Degeneracy and Multiplets
• Antiunitary Symmetries
• Spin in Magnetic Fields
• Spin–Orbit Coupling
• Symmetry Constraints on Hamiltonians
• Kramers Degeneracy Reference Card
• Time-Reversal Operator

References

• H. A. Kramers, “Theorie generale de la rotation paramagnetique dans les cristaux,” Proceedings of the Royal Netherlands Academy of Arts and Sciences 33, 959-972, 1930.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• 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.

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. For a single spin-$1/2$ Hamiltonian $H=aI+\mathbf b\cdot\boldsymbol\sigma$, use time reversal to find the allowed $\mathbf b$ when no external time-odd parameter is present.

Solution

Spin-$1/2$ time reversal sends

$$\Theta\boldsymbol\sigma\Theta^{-1} = -\boldsymbol\sigma.$$

Therefore

$$\Theta H\Theta^{-1} = aI-\mathbf b\cdot\boldsymbol\sigma.$$

The condition $\Theta H\Theta^{-1}=H$ requires

$$\mathbf b=0.$$

So the Hamiltonian is $H=aI$ and the two spin states are degenerate.

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

Solution

Let each spin have time reversal $\Theta_i$ with $\Theta_i^2=-I$. On the product spin space, the schematic product operator is

$$\Theta_{12} = \Theta_1\otimes\Theta_2.$$

Its square is

$$\Theta_{12}^2 = \Theta_1^2\otimes\Theta_2^2 = (-I)\otimes(-I) = I.$$

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 the doublet is split unless $B=0$ or another symmetry enforces a separate degeneracy.