%% ```ad-abstract title:Kramers Theorem For every energy eigenstate of a time-reversal symmetric system with half-integer total spin, there is at least one more eigenstate with the same energy. That is, every energy level is atl east doubly degenerate if it has half-integer spin. [Source](https://en.wikipedia.org/wiki/Kramers_theorem) [Paper](https://www.dwc.knaw.nl/DL/publications/PU00014621.pdf) ``` %% > **Kramers Theorem** > For every energy eigenstate of a time-reversal symmetric system with half-integer total spin, there is at least one more eigenstate with the same energy. That is, every energy level is atl east doubly degenerate if it has half-integer spin. > > [Source](https://en.wikipedia.org/wiki/Kramers_theorem) [Paper](https://www.dwc.knaw.nl/DL/publications/PU00014621.pdf) # Proof of the Theorem Recall that, for spin-1/2 systems, we have $\Theta^2 = -1$ where $\Theta$ is the [[Time-Reversal Operator|time-reversal operator]]. In such a time-reversal invariant system, $\ket{\psi}$ and $\Theta \ket{\psi}$ have the same energy (since $[H, \Theta] = 0$). Thus, if we can prove that these two states are different, then we will have proved Kramer's theorem since there will be two different states with the same energy! %% ```ad-note title: Proof Let's assume the two states are the same, i.e, $ \Theta \ket{\psi} = e^{i\phi} \ket{\psi},\quad \phi \in \mathbb{R}. $ Then $ \Theta^2 \ket{\psi} = \Theta e^{i \phi} \ket{\psi} = e^{-i \phi} \Theta \ket{\psi} = e^{-i \phi}e^{i \phi}\ket{\psi} = \ket{\psi} \implies \Theta^2= 1 \Rightarrow\Leftarrow $ (proof by contradiction) ``` %% > ** Proof** > Let's assume the two states are the same, i.e, > $ > \Theta \ket{\psi} = e^{i\phi} \ket{\psi},\quad \phi \in \mathbb{R}. > $ > Then > $ > \Theta^2 \ket{\psi} = \Theta e^{i \phi} \ket{\psi} = e^{-i \phi} \Theta \ket{\psi} = e^{-i \phi}e^{i \phi}\ket{\psi} = \ket{\psi} \implies \Theta^2= 1 \Rightarrow\Leftarrow > $ > (proof by contradiction) # Solid-State Physics Consider a system with [[Bloch's Theorem|Bloch states]] $\ket{n,\pmb{k},\sigma}$ where $n$ is the band index, $\pmb{k}$ is the crystal momentum and $\sigma$ is the spin. For any such state, there is another state $\Theta \ket{n, \pmb{k}, \sigma}$ with the same energy. $ \Theta \ket{n,\pmb{k},\sigma} = \ket{n,-\pmb{k},-\sigma} $ (Note that we use a notation where there are two bands labeled by the same $n$, one with spin-up and one with spin-down). In general, such a Kramers doublet are located at different momenta $\pmb{k}$ and $-\pmb{k}$. For example, consider the system shown in Fig. 1 below. --- ![[Kramers.png]] **Fig 1:** The band diagram in the first Brillouin zone of a spin-1/2 particle with time-reversal symmetry in 1-dimension. The yellow line shows a constant energy cut, with four states having the same energy. There are two Kramers pairs here, denoted by yellow dots. There is one pair at $\alpha_{\pm}$ and another at $\beta_{pm}$. $\alpha_{\pm}$ and $\beta_{\pm}$ are known as time-reversal conjugate momenta. Moreover, note the special green points at the center and the boundary of the zone (denoted by green dots), which are their own time-reversal conjugates. These points are known as the time-reversal invariant moment (TRIM). --- Note that, in Fig. 1, there are 4 degenerate eigenstates at a constant energy cut because of the shape of the bands, which will always look something like this. ==Does this indicate an extra symmetry?== On the other hand, observe that at the zone boundary and center, we have some special points which are their own time-reversal conjugates. These points are the [[Time-Reversal Invariant Momenta|time-reversal invariant moment]].