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?


1 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) $$ \phi(x,t)\to - \eta_T^* {\mathcal T}^{-1} \phi^*(x,-t) $$

  • $\begingroup$ Thank you for your answer! I still have a question: QFT books like Peskin&Schroeder or Schwartz use $\mathcal T=\gamma^1\gamma^3$, which does indeed fulfill your second equation. But where does it come from? And why does it work differently for $S(\Lambda_T)$? $\endgroup$
    – ersbygre1
    Jul 16, 2020 at 22:54
  • 1
    $\begingroup$ There is little connection between your def of $S(\Lambda)T)$ and what is called "time reversal".Your def would work for time reflection in Euclidean siganture, but that operation turns particles into antiparticles. What is called "time reversal" today adds an extra charge conjugation to time reflection so as to undo this, and then adds a complex conjugation in changing to Minkowski signature. P&S formula is basis-dependent are most such textbook expressions. That's why I gave the transformation formula. $\endgroup$
    – mike stone
    Jul 17, 2020 at 0:03
  • $\begingroup$ Oh wow, thanks for clearing that up. To be sure, let me try: 1) $\mathcal T = K \exp(\pm i\pi S_y)$? 2) Time reversal ($\mathsf T,\mathcal T$) = time reflection $(\Lambda_T)$ + charge conjugation ($\mathsf C,\mathcal C$) + complex conjugation ($K$)? If so, 3) where does your second equation (transformation formula) come from? and 4) why is $(\Lambda_T)$ time reflection in Euclidean space? $\endgroup$
    – ersbygre1
    Jul 17, 2020 at 3:49

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