For a proper infinitesimal Lorentz transformation
$$
\Lambda_\mu^\nu=\delta_\mu^\nu+\varepsilon_\mu^\nu,
$$
show that an $S$ which satisfies (5.57) is
$$
S_L=1-\frac{i}{4} \sigma_{\mu v} \varepsilon^{\mu \nu} .
$$
Hence, show that
$$
\begin{aligned}
S_L^{-1} & =\gamma^0 S_L^{\dagger} \gamma^0 \\
\gamma^5 S_L & =S_L \gamma^5 .
\end{aligned}
$$
For space inversion, or the parity operation,
$$
\Lambda_\mu^\nu=\left(\begin{array}{cccc}
1 & & & \\
& -1 & & \\
& & -1 & \\
& & & -1
\end{array}\right) .
$$
Then, (5.57) becomes
$$
\begin{aligned}
& S_P^{-1} \gamma^0 S_P=\gamma^0, \\
& S_P^{-1} \gamma^k S_P=-\gamma^k \quad \text { for } k=1,2,3,
\end{aligned}
$$
which is satisfied by
$$
S_P=\gamma^0 \text {. }
$$
In the Dirac-Pauli representation of $\gamma^0,(5.51)$, the behavior of the four components of $\psi$ under parity is therefore
$$
\psi_{1,2}^{\prime}=\psi_{1,2} \quad \text { and } \quad \psi_{3,4}^{\prime}=-\psi_{3,4} \text {. }
$$
The "at rest" states, $(5.22)$, are therefore eigenstates of parity, with the positive and negative energy states (that is, the electron and the positron) having opposite intrinsic parities.
Armed with $S_I$ and $S_P$, we can now check the claimed properties of the bilinear covariants. First, we note that
$$
\begin{aligned}
\bar{\psi}^{\prime} & =\psi^{\dagger} \gamma^0=\psi^{\dagger} S^{\dagger} \gamma^0=\psi^{\dagger} \gamma^0 S^{-1} \\
& =\bar{\psi} S^{-1},
\end{aligned}
$$
where we have used (5.56) and (5.60). As an example, let us establish the character of $\bar{\psi} \gamma^\mu \psi$. Under Lorentz transformations,
$$
\bar{\psi}^{\prime} \gamma^\mu \psi^{\prime}=\bar{\psi} S_L^{-1} \gamma^\mu S_L \psi=\Lambda_\nu^\mu\left(\bar{\psi} \gamma^\nu \psi\right),
$$
using (5.64) and (5.57), while under the parity operation,
$$
\bar{\psi}^{\prime} \gamma^\mu \psi^{\prime}=\bar{\psi} S_P^{-1} \gamma^\mu S_P \psi=\left\{\begin{array}{r}
\bar{\psi} \gamma^0 \psi, \\
-\bar{\psi} \gamma^k \psi .
\end{array}\right.
$$
These transformation properties are precisely what we expect for a Lorentz four-vector.
From (5.56) and (5.64), if follows immediately that $\bar{\psi} \psi$ is a Lorentz scalar. The probability density $\rho=\psi^{\dagger} \psi$ is not a scalar, but is the time-like component of the