Covariant Bilinears: Fierz Expansion of Dirac gamma matrices products

1. Apr 4, 2012

PLuz

1. The problem statement, all variables and given/known data

So my question is related somehow to the Fierz Identities.

I'm taking a course on QFT. My teacher explained in class that instead of using the traces method one could use another, more intuitive, method. He said that we could use the fact that if we garante that we have the same number os indexes at each side of the expression and only use the base matrices (scalar, vector, pseudoscalar, tensor and axial) one would get the same results as using the traces method.

He then gave an example and advised for us to try with some other example.

I then tried to write $\gamma_5\gamma^{\alpha}\gamma^{\mu}$ using that method.
2. Relevant equations

$\sigma^{\alpha\mu}=\frac{i}{2}[\gamma^{\alpha},\gamma^{\mu}]$
$\eta^{\alpha \mu}$ is the minkowski metric and $I_{4}$ is the identity matrix in 4-spacetime.

3. The attempt at a solution

The attempt of a solution goes as:
$\gamma_{5}\gamma^{\alpha}\gamma^{\mu}= a*\eta^{\alpha\mu}I_{4}+b*\eta^{\alpha\mu}\gamma_{5} +c*\sigma^{\alpha\mu}$

Is this correct?

If I contract $\gamma_5\gamma^\alpha\gamma^\mu$ with $\eta_{\alpha \mu}$ I get a=0 and b=1. But if the above expression is correct, how can I get $c$?

2. Apr 4, 2012

fzero

From the definition of $\sigma^{\alpha\mu}$ and the anticommutation relations, you can see that there must be a term $\gamma_5\sigma^{\alpha\mu}$ in the expansion of $\gamma_5\gamma^\alpha\gamma^\mu$. I guess we'd call this a pseudotensor term.

3. Apr 5, 2012

PLuz

I thought so, but I have read that such term would not be independent of the other 16 matrices.

$\sigma^{\alpha \mu}\gamma_{5}=\frac{i}{2}\epsilon^{\alpha \mu \nu \beta}\sigma_{\nu \beta}$

,where $\epsilon^{\alpha \mu \nu \beta}$ is the Levi-Civita symbol

So the above expression for $\gamma_{5}\gamma^{\alpha} \gamma^{\mu}$ is incomplete and I should add a extra term
$d* \epsilon^{\alpha \mu \nu \beta}\sigma_{\nu \beta}$ ?

Last edited: Apr 5, 2012
4. Apr 5, 2012

PLuz

I got it now.

One as to add such term and then get the correct answer. For other readers with a similar doubt I got:

$\gamma_{5}\gamma^{\alpha}\gamma^{\mu}=\eta^{\alpha\mu}I_{4}+\frac{1}{2}\epsilon^{\alpha\mu\nu\beta}σ_{\nu\beta}$

Which I believe, it's the correct answer.