PT transformation of a spinor

Question

By demanding the the Dirac equation be invariant under general Lorentz transformations, we get an equation for the transformation matrix of a Dirac spinor, $$ S^{-1}(\Lambda) \gamma^\mu S(\Lambda) = {\Lambda^\mu}_\nu \gamma^\nu, $$ such that the spinor transforms as $$ \psi \quad\stackrel{{\Lambda^\mu}_\nu}\to\quad S(\Lambda)\psi $$ If we go through this three times for $$ {(\Lambda_P)^\mu}_\nu = {diag(+1,-1,-1,-1)^\mu}_\nu \\ {(\Lambda_T)^\mu}_\nu = {diag(-1,+1,+1,+1)^\mu}_\nu \\ {(\Lambda_{PT})^\mu}_\nu = {diag(-1,-1,-1,-1)^\mu}_\nu $$ the relevant equations can be solved by $$\begin{align} S(\Lambda_P) &= \pm\gamma^0 \\ S(\Lambda_T) &= \pm\gamma^1\gamma^3 \qquad\leftarrow\text{(wrong! see answer below!)} \\ S(\Lambda_{PT}) &= \pm\gamma^5 \end{align}$$ Naively, I would expect that since $P\circ T=PT$, we would also have $S(\Lambda_P)S(\Lambda_T)=S(\Lambda_{PT})$. This is obviously not the case. Why?

Answer 1

I find that your $S(\Lambda_T)$ should be $\gamma^1\gamma^2\gamma^3$ as that flips the $\gamma^0$ three times, but flips the other three gammas twice.

But be careful the actual action of $T$ on spinor fields and states works differently: On fields $$ {\mathsf T}^{-1} \hat \psi(x,t) {\mathsf T}= \eta_T {\mathcal T} \hat \psi(x,-t) $$ where $\mathsf T$ is antiunitary and ${\mathcal T}$ is a unitary matrix obeying $$ {\mathcal T} \gamma^\mu {\mathcal T}^{-1} =(\gamma^\mu)^T $$ The form of ${\mathcal T}$ in terms of the gamma matrices depeneds on the representation because if $\gamma^\mu\to S\gamma^\mu S^{-1}$ then ${\mathcal T}\to S^T{\mathcal T} S$.

On the wavefunction $\phi(x,t)= \langle 0|\hat \psi(x,t)|\phi\rangle$ of a single-particle state $|\phi\rangle$ we have (in four dimensions where ${\mathcal T}$ is skew symmetric) $$ \phi(x,t)\to - \eta_T^* {\mathcal T}^{-1} \phi^*(x,-t) $$