= Solution
Use the <Dirac representation of the gamma matrices> and commuting external-wavefunction components, rather than anticommuting quantum fields. Let $A=\gamma^0\gamma^2$, $B=\gamma^2\gamma^0=-A$, and $\Gamma^\alpha=\gamma^\alpha(g_V-g_A\gamma^5)$. Since $(\gamma^2)^\dagger=-\gamma^2$ and $v=i\gamma^2u^*$,
$$
\bar v_A\Gamma^\alpha v_B=-u_A^TB\Gamma^\alpha\gamma^2u_B^*.
$$
Transpose this scalar and use $(\gamma^2)^T=\gamma^2$, $B^T=A$ and $\bar u_B=u_B^\dagger\gamma^0$:
$$
\bar v_A\Gamma^\alpha v_B=\bar u_B A(\Gamma^\alpha)^TB u_A.
$$
Here $A^2=1$ and $A(\gamma^5)^TA=\gamma^5$. The supplied identity therefore yields
$$
\begin{aligned}
A(\Gamma^\alpha)^TB&=A(g_V-g_A(\gamma^5)^T)(\gamma^\alpha)^TB\\
&=(g_V-g_A\gamma^5)\gamma^\alpha=\gamma^\alpha(g_V+g_A\gamma^5).
\end{aligned}
$$
Hence
$$
\boxed{\bar v_A(\mathbf p)\gamma^\alpha(g_V-g_A\gamma^5)v_B(\mathbf p')=\bar u_B(\mathbf p')\gamma^\alpha(g_V+g_A\gamma^5)u_A(\mathbf p)}.
$$
The external labels and <momenta> are exchanged; the scalar constants are transposed, not complex-conjugated. At the level of these numerical <fermion bilinears>, charge conjugation reverses the relative sign of the vector and axial terms, relating particle and <antiparticle> weak amplitudes. For $g_V=g_A$, the two displayed spinor chains have opposite <chiral projectors>. A physical charge-conjugation transformation of an operator current also includes the minus sign from reordering <Dirac fields>. In particular this numerical identity does not imply equal electric charges for a particle and its <antiparticle>.
Back to article page
