Kramers degeneracy

The definition of $K$ given above may be generalized like

\[\begin{aligned} T^2\left|\phi\right\rangle &= UKUK\left|\phi\right\rangle =UU^*\left|\phi\right\rangle =\left(\text i\sigma_y\right)^{2n}\left|\phi\right\rangle =(-)^n\left|\phi\right\rangle, \\ T\left(a\left|\phi\right\rangle+b\left|\psi\right\rangle\right) &= UK\left(a\left|\phi\right\rangle+b\left|\psi\right\rangle\right) =a^*UK\left|\phi\right\rangle+b^*UK\left|\psi\right\rangle =a^*T\left|\phi\right\rangle+b^*T\left|\psi\right\rangle \\ \left(\left\langle\phi\right|T\right)\left|\psi\right\rangle &= \left(\left\langle\phi\right|UK\right)\left|\psi\right\rangle =\left[\left\langle\phi\right| \left(UK\left|\psi\right\rangle\right)\right]^* = \left(\left\langle\psi\right|UK\right)\left|\phi\right\rangle = \left(\left\langle\psi\right|T\right)\left|\phi\right\rangle.\end{aligned}\]

For even $n$, nothing changes compared to the spinless case, but for odd $n$, $T^2=-1$.

Total moment $J$

Equivalent to the transformation rules for the spins above, we require

\[J_x = -UJ_xU^{-1}, \quad J_y=UJ_yU^{-1}, \quad J_z=-UJ_zU^{-1}\]

for a state with total moment $J$. The operator for this rotation of $J$ by an angle $\pi$ around the $y$ axis is

\[U=\exp\left(\text i\pi J_y\right),\]

generalizing the single-spin equation for $U$ above. In the standard $(j,m)$ basis we are using here, $U$ is a $2j+1$-dimensional antidiagonal matrix of the form

\[U= \begin{pmatrix} 0 & \cdots & 0 & 0 & u_j \\ 0 & \cdots & 0 & u_{j-1} & 0 \\ \vdots & 0 & \cdot & 0 & \vdots\\ 0 & u_{-(j-1)} & 0 & \cdots & 0\\ u_{-j} & 0 & 0 & \cdots & 0 \end{pmatrix}\]

with matrix elements $u_m=(-)^{j-m}$ on the antidiagonal, $m=j,j-1\ldots,-j$.

Kramers degeneracy

Already for an even number of electrons, $T\left|\phi\right\rangle$ is not the usual complex congugation, but rather $T\left|\phi\right\rangle=UK\left|\phi\right\rangle=U\left|\phi\right\rangle^*$. For an odd number of electrons, $T\left|\phi\right\rangle$ is orthogonal to $\left|\phi\right\rangle$:

\[\left(\left\langle\phi\right|T\right)\left|\phi\right\rangle = \left(\left\langle\phi\right|T\right) \left(TT\left|\phi\right\rangle\right) = -\left(\left\langle\phi\right|T\right)\left|\phi\right\rangle =0.\]

That means, even without any spatial symmetry, energy levels for a system with total momentum $j=n+1/2$, $n$ nonnegative integer, are at least doubly degenerate, protected by symmetry.

Example: take the ground state of the Yb$^\text{3+}$ ion from above,

\[\begin{aligned} \left|+\right\rangle &= -\alpha{\rm e}^{\text i\phi}\left|\frac72,\frac72\right\rangle +\beta\left|\frac72,\frac12\right\rangle +\gamma{\rm e}^{-\text i\phi}\left|\frac72,-\frac52\right\rangle, \\ \left|-\right\rangle &= T\left|+\right\rangle = U\left(K\left|+\right\rangle\right) \\ &= U\left(-\alpha{\rm e}^{-\text i\phi}\left|\frac72,\frac72\right\rangle^* +\beta\left|\frac72,\frac12\right\rangle^* +\gamma{\rm e}^{\text i\phi}\left|\frac72,-\frac52\right\rangle^*\right) \\ &= \alpha{\rm e}^{-\text i\phi}\left|\frac72,-\frac72\right\rangle +\beta\left|\frac72,-\frac12\right\rangle -\gamma{\rm e}^{\text i\phi}\left|\frac72,\frac52\right\rangle.\end{aligned}\]

We note that, perhaps counter-intuitive, the sign of $\beta$ in the above remains unchanged under time reversal, in contrast to $j=s=1/2$ where

\[T\left|\frac12,\frac12\right\rangle = -\left|\frac12,-\frac12\right\rangle.\]

In general, we obtain from the form of $U$

\[T\left|j,\pm\frac12\right\rangle = (-)^{j\pm1/2}\left|j,\mp\frac12\right\rangle\]

for half-integer angular momentum $j$.