 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 $({\gamma }^2{)}^{†}=-{\gamma }^2$ and $v=i{\gamma }^2{u}^{\ast }$,Transpose this scalar and use $({\gamma }^2{)}^T={\gamma }^2$, $B^T=A$ and ${\bar{u}}_B=u_B^{†}{\gamma }^0$:Here $A^2=1$ and $A({\gamma }^5{)}^{T}A={\gamma }^5$. The supplied identity therefore yieldsHenceThe 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.