Weak isospin current I cannot understand the product of a Dirac gamma matrix and a Pauli matrix in this formula of the weak isospin current:
$$J_α^i(x)=\frac12\bar \psi_L(x)\gamma_\alpha\tau^i\psi_L(x),$$ where $γ_α$ is a gamma matrix and $\tau^i$ a Pauli matrix.
(From Manda & Shaw, Quantum Field Theory 2nd, page 393.)
Can anybody help me?
 A: Short answer: this is just an (outer) tensor product, not a matrix product. I agree that the notation is a bit confusing though.
To see this, try substituting the left chiral duplet of fermions, i.e. for leptons:
$$ \psi_L = \left( \begin{matrix} e_L \\ \nu_e \end{matrix} \right). $$
The matrix $\tau_i$ is sandwiched between the row $\bar{\psi}_L$ and the column $\psi_L$, thus mixing electrons and neutrinos.
The Dirac gamma-matrix $\gamma_{\alpha}$ is acting on spacetime components of the individual fermions. I.e. to give you some concreteness, let's calculate $J_{\alpha}^1$:
$$ J_{\alpha}^1 = \left( \begin{matrix} \bar{e}_L & \bar{\nu}_e \end{matrix} \right) \gamma_{\alpha} \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) \left( \begin{matrix} e_L \\ \nu_e \end{matrix} \right) =\bar{e}_L \gamma_{\alpha} \nu_e + \bar{\nu}_e \gamma_{\alpha} e_L. $$
Here both summands are of form $\bar{\varphi} \gamma_{\mu} \psi$ with $\varphi, \psi$ – Dirac spinors. These transform as 4-vectors (easy to prove) and are, of course, well known from QED where the electromagnetic current was given by $\bar{e} \gamma_{\mu} e$.
One important difference is that here only the left-handed part of the electron field is nonzero.
