Graduate Vector and Axial vector currents in QFT

[RicardoMP]

I'm currently working out quantities that include the vector and axialvector currents $j^\mu_B(x)=\bar{\psi}(x)\Gamma^\mu_{B,0}\psi(x)$ where B stands for V (vector) or A (axialvector). The gamma in the middle is a product of gamma matrices and the psi's are dirac spinors. Therefore on the left I have a 1x4 matrix, on the left a 4x1 matrix and in the middle a 4x4 matrix, thus this current is just a number. Am I correct?

 

[vanhees71]

Usually one write $\psi(x)$ as a column of four numbers. It's a socalled bispinor or Dirac spinor. Then by definition $\overline{\psi}(x)=\psi^{\dagger}(x) \gamma^0$, and your $\Gamma^{\mu}$'s are some product of $\gamma$-matrices. For the vector current it's $\gamma^{\mu}$ for the axial vector current $\gamma^{\mu} \gamma_5=-\gamma_5 \gamma^{\mu}$ (where I use the usual commutation relations of $\gamma_5$ in (1+3)-dimensional Minkowski space; if you do dimensional regularization, sometimes there are extra rules for $\gamma_5$ making it easier to deal with anomalies, but that's more advanced stuff).

So what you multiply in the sense of matrix multiplication is a "row bi-spinor" $\times$ "$4 \times 4$ matrix" $\times$ a "column bi-spinor", which gives a number.

Of course, in QFT the $\psi$ are operators, and thus also the $j^{\mu}$'s become operators (transforming as a four-vector or an axial-four-vector operator, respectively).