In representation (5.4) and (5.8) of the γ-matrices, show that a possible choice of C is C γ^0=i

Step 1: **Verify the choice of ( C ) and ( C gamma^0 )** - Given ( C gamma^0 = i gamma^2 ), w

In representation (5.4) and (5.8) of the γ-matrices, show...

In representation (5.4) and (5.8) of the $\gamma$-matrices, show that a possible choice of $C$ is $$ C \gamma^0=i \gamma^2=\left(\begin{array}{llll} & & & 1 \\ & -1 & & \end{array}\right) . $$ Try this operation out on a particular spinor, and show, for instance, that $$ \psi_C^{(1)}=i \gamma^2\left[u^{(1)}(\mathbf{p}) e^{-i p \cdot x}\right]^*=u^{(4)}(-\mathbf{p}) e^{i p \cdot x}=v^{(1)}(\mathbf{p}) e^{i p \cdot x} . $$ Further, show that in this representation $$ \begin{aligned} C^{-1} \gamma^\mu C & =\left(-\gamma^\mu\right)^T, \\ C & =-C^{-1}=-C^{\dagger}=-C^T, \\ \bar{\psi}_C & =-\psi^T C^{-1} . \end{aligned} $$ We have seen that the electron current is $$ j^\mu=-e \bar{\psi} \gamma^\mu \psi \text {. } $$ The current associated with the charge conjugate field is therefore $$ \begin{aligned} j_C^\mu & =-e \bar{\psi}_C \gamma^\mu \psi_C \\ & =+e \psi^T C^{-1} \gamma^\mu C \bar{\psi}^T \\ & =-e \psi^T\left(\gamma^\mu\right)^T \bar{\psi}^T \\ & =-(-) e \bar{\psi} \gamma^\mu \psi \end{aligned} $$ [see (5.39)]. The origin of the extra minus sign introduced in the last line is subtle but important. It is clearly necessary for a physically meaningful result, that is, if $j_C^\mu$ is to be the positron current. The minus sign is related to the connection between spin and statistics; in field theory, it occurs because of the antisymmetric nature of the fermion fields. In field theory, the charge conjugation operator $C$ changes a positive-energy electron into a positive-energy positron, and the formalism is completely $\mathrm{e}^{+} \leftrightarrow \mathrm{e}^{-}$symmetric. However, in a single-particle (electron) theory, positron states are not allowed; rather, $C$ changes a positive-energy electron state into a negative-energy electron state. As a result, we can show that we must add to our Feynman rules the requirement that we insert by hand an extra minus sign for every negative-energy electron in the final state of the process. The $C$ invariance of electromagnetic interactions then follows: $$ j_\mu^C\left(A^\mu\right)^C=\left(-j_\mu\right)\left(-A^\mu\right)=j_\mu A^\mu . $$

In representation (5.4) and (5.8) of the $\gamma$-matrices, show that a possible choice of $C$ is
$$
C \gamma^0=i \gamma^2=\left(\begin{array}{llll}
& & & 1 \\
& -1 & &
\end{array}\right) .
$$

Try this operation out on a particular spinor, and show, for instance, that
$$
\psi_C^{(1)}=i \gamma^2\left[u^{(1)}(\mathbf{p}) e^{-i p \cdot x}\right]^*=u^{(4)}(-\mathbf{p}) e^{i p \cdot x}=v^{(1)}(\mathbf{p}) e^{i p \cdot x} .
$$

Further, show that in this representation
$$
\begin{aligned}
C^{-1} \gamma^\mu C & =\left(-\gamma^\mu\right)^T, \\
C & =-C^{-1}=-C^{\dagger}=-C^T, \\
\bar{\psi}_C & =-\psi^T C^{-1} .
\end{aligned}
$$

We have seen that the electron current is
$$
j^\mu=-e \bar{\psi} \gamma^\mu \psi \text {. }
$$

The current associated with the charge conjugate field is therefore
$$
\begin{aligned}
j_C^\mu & =-e \bar{\psi}_C \gamma^\mu \psi_C \\
& =+e \psi^T C^{-1} \gamma^\mu C \bar{\psi}^T \\
& =-e \psi^T\left(\gamma^\mu\right)^T \bar{\psi}^T \\
& =-(-) e \bar{\psi} \gamma^\mu \psi
\end{aligned}
$$
[see (5.39)]. The origin of the extra minus sign introduced in the last line is subtle but important. It is clearly necessary for a physically meaningful result, that is, if $j_C^\mu$ is to be the positron current. The minus sign is related to the connection between spin and statistics; in field theory, it occurs because of the antisymmetric nature of the fermion fields. In field theory, the charge conjugation operator $C$ changes a positive-energy electron into a positive-energy positron, and the formalism is completely $\mathrm{e}^{+} \leftrightarrow \mathrm{e}^{-}$symmetric. However, in a single-particle (electron) theory, positron states are not allowed; rather, $C$ changes a positive-energy electron state into a negative-energy electron state. As a result, we can show that we must add to our Feynman rules the requirement that we insert by hand an extra minus sign for every negative-energy electron in the final state of the process. The $C$ invariance of electromagnetic interactions then follows:
$$
j_\mu^C\left(A^\mu\right)^C=\left(-j_\mu\right)\left(-A^\mu\right)=j_\mu A^\mu .
$$

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