Aug 2, 2014

$\beta$ and $\gamma^0$

$\gamma^0$ with the other three gamma matices $\vec\gamma$, furnishes a representation of the Clifford algebra $\mathrm{C\ell}_{1,3}(\mathbb R)$ \[\{ \gamma^\mu, \gamma^\nu \} = 2 g^{\mu\nu}, \] where $\{X, Y\} = XY + YX$, $g^{\mu\nu}$ is the Minkowski space metric.

The parity matrix $\beta$ is a spinor representation of the parity operator: \[
P^{-1} \Psi(x) P = \beta \Psi(\mathcal P\cdot x)
\] where $\mathcal P^\mu_\nu = \mathrm{diag}\{1, -1, -1, -1\}$ is the 4-vector representation of the parity operator. The operator $P$ here, of course, furnishes a field representation, with the help of $\beta$ and $\mathcal P$. Then $\beta$ satisfies the parity relations: $\beta^2 = 1$, $\beta^{-1} \vec\gamma \beta = -\vec\gamma$, $\beta^{-1} \gamma^0 \beta = \gamma^0$.

It is popular to take $\beta = \gamma^0$. But this may not always be the correct one. The reason is the metric tensor g. There are two popular sign conventions of $g^{\mu\nu}$: $\mathrm{diag}\{+1, -1, -1, -1\}$ and $\mathrm{diag}\{-1, +1, +1, +1\}$. Under the first convention, $\gamma^0 \gamma^0 = 1$. $\beta$ is readily to be chosen as $\gamma^0$. It is easy to check all the parity relations are satisfied. Under the second conventions, however, $\gamma^0 \gamma^0 =-1$, which means $\beta$ should be $\pm i\gamma^0$.

Of course, all unitary representations in quantum theory is only determined up to an over all phase factor: \[
P^{-1} \Psi(x) P = \exp(i\eta) \beta \Psi(\mathcal P\cdot x), \quad (\eta \in \mathbb R)
\] It is possible to choose phase factors such that $\beta = \gamma^0$ always holds. For example, for $g_{00} = +1$, we choose $\exp(i\eta) = 1$; for $g_{00}=-1$, we choose $\exp(i\eta) = i$. This is essentially what is done in the literature, such as Peskin & Schroeder, Mark Srednicki etc. But keep in mind that, the phase factors for T, C, and P are not all independent. It is important to be self-consistent at the end of the day. Weinberg's book keeps the phase factors open.