# Fierz Identity Substitution Into QED Lagrangian

1. Oct 6, 2013

### welcomeblack

Hi all, I've been playing around with spin 1/2 Lagrangians, and found the very interesting
Fierz identities. In particular for the S x S product,

$(\bar{\chi}\psi)(\bar{\psi}\chi)=\frac{1}{4}(\bar{\chi} \chi)(\bar{\psi} \psi)+\frac{1}{4}(\bar{\chi}\gamma^{\mu}\chi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{4}(\bar{\chi}\sigma^{\mu\nu}\chi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{4}(\bar{\chi}\gamma^{\mu}\gamma^{5}\chi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{4}(\bar{\chi}\gamma^{5}\chi)(\bar{\psi}\gamma_{5} \psi)$

I assume this is valid for any spinors chi and psi. If I then set chi equal to psi,

$(\bar{\psi}\psi)(\bar{\psi}\psi)=\frac{1}{4}(\bar{\psi} \psi)(\bar{\psi} \psi)+\frac{1}{4}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{4}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{4}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{4}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)$

Since adjoint*spinor is a scalar, I can divide by psibar*psi and rearrange to get

$\bar{\psi}\psi=\frac{1}{\bar{\psi}\psi}[\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)]$

Plugging this in to the Dirac Lagrangian,

$\mathcal{L}=i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - m\bar{\psi}{\psi} \\ \mathcal{L}=i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - \frac{m}{\bar{\psi}\psi}[\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)]$

Is everything I've done mathematically allowed? How about physically allowed? If the two representations of psibar*psi are equivalent, shouldn't they give back the same equations of motion?

Last edited: Oct 6, 2013
2. Oct 6, 2013

### fzero

A Fierz identity is a mathematical identity. It means that if we expand both sides of the equation into spinor components, we will find exactly the same combination on the left side as we do on the right. So if you were to expand this transformed Lagrangian in terms of the spinor components, you would find that it was the same as the original one, leading to the same equations of motion.

Now dividing by $\bar{\psi}{\psi}$ is a legitimate process if we are talking about ordinary QM of spinors whose components are wavefunctions. The multiplicative inverse of a c-function exists whenever the function is nonzero. If $\psi$ is considered as a quantum field, then you should consider it as an operator and will need to use some formal definition of division in terms of an inverse operator, which won't always exist.

3. Oct 7, 2013

### andrien

For one thing you should care for how those spinors are distinguished.You should put a subscript for example to do it.Then you will have (ψ4-ψ3)(ψ2-ψ1).you apply fierz reshuffling now to make linear combinations in terms of Ci4-(γ's)ψ1)(ψ2-(y's)ψ3) etc.But you should not use it when 1=2=3=4,because it's trivial and equal to the original one.So you will not get changed anything.

4. Oct 7, 2013

### welcomeblack

Okay so the spinor components on the LHS are the same as those on the RHS. Avoiding any spinor division (since fzero says it isn't generally well defined), if we take the partial of (psibar*psi)(psibar*psi) with respect to psibar, we get from the LHS

$\frac{\partial}{\partial\bar{\psi}} (\bar{\psi}\psi)(\bar{\psi}\psi)=2(\bar{\psi}\psi)\psi$

and from the RHS

$\frac{\partial}{\partial\bar{\psi}} (\bar{\psi}\psi)(\bar{\psi}\psi)=\frac{2}{3}(\bar{\psi}\gamma^{\mu}\psi)\gamma_{\mu}\psi-\frac{2}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)\sigma_{\mu\nu}\psi-\frac{2}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)\gamma_{\mu}\gamma_{5}\psi+\frac{2}{3}(\bar{\psi}\gamma^{5}\psi)\gamma_{5}\psi$

Both LHS and RHS can be interpreted as some operator acting on the spinor psi. Since both sides are equal, the operators themselves must be equal, so

$\bar{\psi}\psi=\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)\gamma_{\mu}-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)\sigma_{\mu\nu}-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)\gamma_{\mu}\gamma_{5}+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)\gamma_{5}$

If I plug this into the Dirac Lagrangian for psibar*psi, then use the Euler-Lagrange equations for psi or psibar, the equations of motion reduce down to the usual ones. Woo! That's what I'd hoped for (ish).

Thanks guys for indulging me in my hobby QFT meanderings.