# Average fermion current

1. Jan 24, 2013

### parton

Hi!

I have a little problem.

Consider a 4-fermion interaction (neglecting constant factors) of the form $\overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}}$ .
I want to average this interaction over a background consisting of fermions (so it corresponds to the situation where fermions propagate in a background consisting of fermions).

To this purpose, the left-handed current $\overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}}$ is approximated by the average value $\langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \rangle$

There is the following relation for the averaged value produced by this interaction:
\begin{align} \begin{split} \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} \to & \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \rangle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} + \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} \rangle \\ & \quad + \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{d \mathrm{L}} \rangle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}} + \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{d \mathrm{L}} \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}} \rangle \\ & \quad - \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{b \mathrm{L}} \rangle \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} \rangle - \langle \overline{\psi_{a \mathrm{L}}} \gamma^{\lambda} \psi_{d \mathrm{L}} \rangle \langle \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}} \rangle. \end{split} \end{align}

But I don't really understand how to obtain this relation. The first two terms look reasonable, but I don't understand the remaining ones.

Ok, the 3rd and 4th terms might correspond to an "exchange term", where fermion c and d are interchanged, because in the case where c is equal to d, we cannot distinguish between the propagating and the background fermion (on the other hand, if c is not equal to d, this exchange term should be 0). But why is there no additional minus sign, because of Fermi-Dirac statistics?

Maybe it has something to do with a Fierz transformation, where the minus sign cancels out, i.e.,
$$\overline{\psi}_{a \mathrm{L}} \gamma^{\lambda} \psi_{b \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{d \mathrm{L}} = \overline{\psi}_{a \mathrm{L}} \gamma^{\lambda} \psi_{d \mathrm{L}} \overline{\psi_{c \mathrm{L}}} \gamma_{\lambda} \psi_{b \mathrm{L}}.$$

And how could the last two terms be interpreted?

I hope somebody could help me understanding the relation above.

2. Jan 26, 2013

### andrien

can you provide a reference for that averaging procedure.How is it defined.Also it is not Fierz reshuffling otherwise you will also get pseudoscalar and pseudovector part(apart from identity)