# 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?

• Hi Joan and welcome to physics.SE. You need to be a good bit more specific as to what aspect you don't follow, say by showing what your own attempt is (and where you get stuck) and also by spending two minutes formatting your post using math.meta.stackexchange.com/questions/5020/…, would help you get a good reception. Is it an intuitive picture you are after? – user184990 Feb 16 '18 at 0:09
• The spinors have suppressed spinor indices saturated by those of the γ matrix, and likewise suppressed isospin indices saturated with those of the Pauli isospin matrix. One omits such indices to uncluttered the messy bilinear in the fields, but they are simple to insert if you need to understand them. – Cosmas Zachos Feb 16 '18 at 1:14
• Thank you for your suggestions,Countto10. I’m stuck at understanding how a 4x4 gamma matrix could be multiplied by a 2x2 Pauli matrix. It is a Kronecker product? At the end of this five term product between a real, an adjoint 8-component spinor, a 4x4 gamma matrix, a2x2 Pauli matrix and a 8-component spinor one should obtain the alpha component of a four-current... – Joan Feb 16 '18 at 10:36

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.

• @Joan you can upvote and accept it if it clarifies your confusion. – Prof. Legolasov Feb 17 '18 at 1:51