Kramers' theorem

By the definition of an antiunitary operator, ${\textstyle \langle Tu,Tw\rangle =\langle w,u\rangle }$, where ${\textstyle u}$ and ${\textstyle w}$ are vectors in ${\textstyle {\mathcal {H}}}$. Replacing ${\textstyle u=Tv}$ and ${\textstyle w=v}$ and using that ${\textstyle T^{2}=-1}$, we get ${\textstyle -\langle v,Tv\rangle =\langle T^{2}v,Tv\rangle =\langle v,Tv\rangle ,}$which implies that ${\textstyle \langle v,Tv\rangle =0}$.

Consequently, if a Hamiltonian ${\textstyle H}$ is time-reversal symmetric, i.e. it commutes with ${\textstyle T}$, then all its energy eigenspaces have even degeneracy, since applying ${\textstyle T}$ to an arbitrary energy eigenstate ${\textstyle |n\rangle }$ gives another energy eigenstate ${\textstyle T|n\rangle }$ that is orthogonal to the first one. The orthogonality property is crucial, as it means that the two eigenstates ${\textstyle |n\rangle }$ and ${\textstyle T|n\rangle }$ represent different physical states. If, on the contrary, they were the same physical state, then ${\displaystyle T|n\rangle =e^{i\alpha }|n\rangle }$ for an angle ${\displaystyle \alpha \in \mathbb {R} }$, which would imply

${\displaystyle T^{2}|n\rangle =T(e^{i\alpha }|n\rangle )=e^{-i\alpha }e^{i\alpha }|n\rangle =+\,|n\rangle }$

To complete Kramers degeneracy theorem, we just need to prove that the time-reversal operator ${\textstyle T}$ acting on a half-odd-integer spin Hilbert space satisfies ${\textstyle T^{2}=-1}$. This follows from the fact that the spin operator ${\textstyle \mathbf {S} }$ represents a type of angular momentum, and, as such, should reverse direction under ${\displaystyle T}$:

${\displaystyle \mathbf {S} \to T^{-1}\mathbf {S} T=-\mathbf {S} .}$

Concretely, an operator ${\textstyle T}$ that has this property is usually written as

${\displaystyle T=e^{-i\pi S_{y}}K}$

where ${\textstyle S_{y}}$ is the spin operator in the ${\textstyle y}$ direction and ${\textstyle K}$ is the complex conjugation map in the ${\textstyle S_{z}}$ spin basis.[2]

Since ${\textstyle iS_{y}}$ has real matrix components in the ${\displaystyle S_{z}}$ basis, then

${\displaystyle T^{2}=e^{-i\pi S_{y}}Ke^{-i\pi S_{y}}K=e^{-i2\pi S_{y}}K^{2}=(-1)^{2S}.}$

Hence, for half-odd-integer spins ${\textstyle S={\frac {1}{2}},{\frac {3}{2}},\ldots }$, we have ${\textstyle T^{2}=-1}$. This is the same minus sign that appears when one does a full ${\displaystyle 2\pi }$ rotation on systems with half-odd-integer spins, such as fermions.