# Calculating the Invariant Matrix

1. May 29, 2014

### ChrisVer

I am having one question... If we know the form of the effective Lagrangian, let's say the form:

$L= g (\bar{\psi}_{e} \gamma^{\mu} P_{L} \psi_{\nu})(\bar{\psi}_{p} \gamma_{\mu} P_{L} \psi_{n})$

How can someone calculate the spin averaged invariant matrix $\large M$?
I mean I can do the whole calculation if it's to have the $u,v$ in place of $\psi$. I am having also a problem of seeing when this is done in QED scatterings too, since we know that $M= j^{\mu}j_{\mu}$ with $j^{\mu}= \bar{\psi} \gamma^{\mu} \psi$...
but I don't know if I have the whole Dirac spinor $\psi$ what someone is supposed to do?

In most cases for the weak interaction, I'm seeing $M$ given by $u,v$ (like in Halzen & Martin)...

Thanks

2. May 29, 2014

### Einj

It's exactly the calculation you are used to in terms of the u and v spinors. What appears in the Lagrangian (for example the Fermi Lagrangian that you wrote) is the Dirac field which can be expressed as a sum of Dirac spinors, u and v.

So suppose you want to describe the effective muon decay, $\mu^-\to e^-+ \nu_\mu +\bar\nu_e$. Then your effective Lagrangian is going to be:
$$L=\frac{G_F}{\sqrt{2}}\left(\bar\psi_\mu\gamma^\alpha P_L\psi_{\nu_\mu}\right)\left(\bar \psi_{\nu_e}\gamma_\alpha P_L\psi_e\right).$$
However, when computing the matrix element using the Feynman diagram this Lagrangian leads to:
$$\mathcal{M}=\frac{G_F}{\sqrt{2}}\left(\bar \nu_\mu\gamma_\alpha P_L\mu\right)\left(\bar e\gamma^\alpha P_L\nu_e\right).$$
Now in the matrix element you have the spinors appropriate for each particle and thus you can use all the well-known trace technology to compute the square amplitude averaged over the spins.