Kramers' theorem

Remarks

Since $E$ is real and $T$ is antilinear, $HT\psi = TH\psi = T(E\psi) = E\, T\psi$. Apply the antiunitary property to $\varphi = T\psi$: it gives $\ip{T^2\psi}{T\psi} = \overline{\ip{T\psi}{\psi}} = \ip{\psi}{T\psi}$. The left side equals $-\ip{\psi}{T\psi}$ because $T^2 = -I$, so $\ip{\psi}{T\psi} = 0$. On a finite-dimensional eigenspace $V$, the map $T$ is an antilinear map with $T^2 = -I$, so it makes $V$ a vector space over the quaternions, and $\dim_\C V$ is even.

Example: for one particle of spin $1/2$ the time reversal operator on $\C^2$ can be taken as $T = \sigma_y K$. Here $K$ is complex conjugation of the coordinates and $\sigma_y$ is the Pauli matrix with rows $(0, -i)$ and $(i, 0)$. Then $T^2 = \sigma_y \overline{\sigma_y} = -\sigma_y^2 = -I$. For $N$ such particles $T^2 = (-1)^N I$, so the theorem applies when $N$ is odd.

The hypothesis $T^2 = -I$ is necessary. For a spinless particle on $L^2(\R)$ time reversal is $T\psi = \overline{\psi}$, with $T^2 = I$, and the harmonic oscillator Hamiltonian commutes with it while all its eigenvalues are simple.

Going deeper

For a student

The proof only uses that an antiunitary $T$ with $T^2 = -I$ cannot fix the line through any vector: $T\psi$ is always orthogonal to $\psi$. The pairs $\{\psi, T\psi\}$ are the Kramers doublets. A magnetic field breaks time reversal symmetry, and then the doublets can split, as in the Zeeman effect. The degeneracy can be larger than two, but each eigenspace splits into orthogonal Kramers doublets, so its dimension is even.