I was looking for Fierz rearrangement for Gamma matrices in the context of Chiral Fermions but couldn't find a simple introductory level lecture note/book ! Here's what I want:

I have this expression: $$ T = [\bar{s}\gamma^\mu\gamma^\sigma \gamma^\nu (1-\gamma^5) d]\otimes [\bar{\nu}\gamma_\mu\gamma_\sigma \gamma_\nu (1-\gamma^5) \nu] $$

I need to see if I can move two of the Gamma matrices from the R.H.S bracket to the L.H.S bracket so that I can have something like: $$ T = [\bar{s}\gamma^\mu\gamma^\sigma \gamma^\nu (1-\gamma^5) d]\otimes [\bar{\nu}\gamma_\mu\gamma_\sigma \gamma_\nu (1-\gamma^5) \nu] $$

$$ T \propto [\bar{s}\gamma^\mu \gamma_\mu \gamma^\sigma \gamma^\nu \gamma_\nu (1-\gamma^5) d]\otimes [\bar{\nu}\gamma_\sigma (1-\gamma^5) \nu] $$

So that I can write:

$$ T \propto [\bar{s} \gamma^\sigma (1-\gamma^5) d]\otimes [\bar{\nu}\gamma_\sigma (1-\gamma^5) \nu] $$
$$ T \propto (\bar{s} d)_{V-A} (\bar{\nu}\nu )_{V-A} $$

I don't need an exact derivation of this particular example, but I need the material where I can learn these techniques. Lecture notes or reviews or books where the author targets the students with a level of understanding of the simple Dirac matrix algebra.

Any help will be really appreciated.

I made some mistakes earlier with the indices but I have corrected them.


2 Answers 2


Ok, after posting this question I have been trying to solve this and finally did, so I am posting the answer.

Just one identity is enough and that is:

$$ \gamma^\mu\gamma^\nu\gamma^\lambda = g^{\mu\nu}\gamma^\lambda + g^{\nu\lambda}\gamma^\mu - g^{\mu\lambda}\gamma^\nu + i \epsilon^{\sigma\mu\nu\lambda} \gamma_\sigma\gamma^5 $$

And a few basic properties of Dirac matrices like: $ (\gamma^5)^2 = 1 $, $ \{ \gamma^\mu, \gamma^5 \} = 0 $, $ \{\gamma^\mu, \gamma^\nu \} = 2 g^{\mu\nu} $ etc.

So we had:

$$ T = [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}\gamma_\mu\gamma_\sigma\gamma_\nu (1-\gamma^5) \nu] $$
We implement the above mentioned identity to the right bracket

$$ T \propto [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}( g_{\mu\sigma}\gamma_\nu + g_{\sigma\nu}\gamma_\mu - g_{\mu\nu}\gamma_\sigma + i \epsilon_{\alpha\mu\sigma\nu} \gamma^\alpha\gamma^5 ) (1-\gamma^5) \nu] $$
Now using using the symmetric property of the metric tensor the we simplify the expression a lot:

$$ T \propto (4+4+4) \{ [\bar{s} \gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}\gamma_\nu(1-\gamma^5) \nu] + i \epsilon_{\alpha\mu\sigma\nu} [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu} \gamma^\alpha\gamma^5 (1-\gamma^5) \nu] $$

Now if we again use the identity of the product of three Gamma matrices in the first bracket of the last term and again the metric tensor-terms will drop because of the anti-symmetric property of the Levi Civita and so the last term will become:

$$ T_{last} = i \epsilon^{\alpha\mu\sigma\nu} [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha\gamma^5 (1-\gamma^5) \nu] = i \epsilon_{\mu\sigma\nu\alpha} i\epsilon^{\mu\sigma\nu\delta} [\bar{s}\gamma_\delta \gamma^5 (1-\gamma^5) d] \otimes [\bar{\nu} \gamma^\alpha\gamma^5 (1-\gamma^5) \nu] $$

Now we can apply the identity: $$ \epsilon_{\mu\sigma\nu\alpha} \epsilon^{\mu\sigma\nu\delta} = 4! \delta^\alpha_\delta $$ and also using the properties of $\gamma^5$ finally we have,
$$ T_{last} = - 24 [\bar{s}\gamma^\alpha(1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha(1-\gamma^5) \nu] $$

$$ T \propto [\bar{s}\gamma^\alpha(1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha(1-\gamma^5) \nu] = (\bar{s} d)_{V-A}(\bar{\nu}\nu)_{V-A} $$

  • $\begingroup$ The contraction of the two Levi-Civita tensors should be 3! instead of 4! $\endgroup$
    – Luca
    Commented Mar 22, 2023 at 4:21

Scrednicki's textbook chapter 50. is about spinor helicity techniques. He uses a different notation though. Fierz identity is one of the problems.

  • $\begingroup$ Hi, I am sorry, I was writing the answer myself but it took so long to post. But anyway thanks. $\endgroup$
    – quanta
    Commented Aug 14, 2015 at 15:52

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