First of all I define the convention I use.

The matrices $\bar{\sigma}^\mu$ I will use are $\{ Id, \sigma^i \}$ where $\sigma^i$ are the Pauli matrices and $Id$ is the 2x2 identity matrix. I will use the Chiral Fierz Identity $$(\bar{\sigma}^\mu)[\bar{\sigma}^\nu] = (\bar{\sigma}^\mu][\bar{\sigma}^\nu) + (\bar{\sigma}^\nu][\bar{\sigma}^\mu) - \eta^{\mu\nu}(\bar{\sigma}^\lambda][\bar{\sigma}_\lambda) + i\epsilon^{\mu\nu\rho\lambda}(\bar{\sigma}_\lambda][\bar{\sigma}_\rho)$$ where I used the Takashi notation.

Let us consider the left-handed component $\chi$ of a massless fermion field $\psi$ and the operator defined as $$\mathcal{O} = \chi^\dagger \bar{\sigma}^\mu\chi (\partial_\mu\partial_\nu\chi^\dagger)\bar{\sigma}^\nu\chi.$$

If I have use the Chiral Fierz identity I get $\mathcal{O} = 2\mathcal{O}$ where I used $\partial_\mu\partial^\mu \chi = 0$. So, I get $\mathcal{O}=0$.

This equality, if true, suggests me there is another way to show that this operator is null for massless fermions. Is there any way? Do you suggest anything?


closed as off-topic by David Z May 6 '16 at 14:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Questions that just ask if a result is right or not aren't really suitable for this site. Could you edit it to focus on a conceptual issue? $\endgroup$ – David Z May 6 '16 at 14:42
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    $\begingroup$ @DavidZ I am not asking an answer like "yes" or "no". I didn't "just" ask if the result is right. I look for an alternative way to demonstrate that $\mathcal{O} = 0 $. I already explained it, so please do not tag my question as off-topic. If you want I can post all my calculations but I think they are not useful. $\endgroup$ – apt45 May 6 '16 at 14:51
  • $\begingroup$ As long as the question contains the sentence "I would like to know if this result is right or not", I'm not convinced. $\endgroup$ – David Z May 6 '16 at 15:03
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    $\begingroup$ If you finish to read all the post (until the end I mean), maybe you will convince yourself. Maybe. $\endgroup$ – apt45 May 6 '16 at 15:04
  • $\begingroup$ I hope you don't really believe I skipped reading the end of the post ;-) Yes, I did read the whole thing prior to putting it on hold. Let me ask you this: if you were to simply remove the sentence I quoted, would it materially change the question? If not, you could make that edit and I would be happy to remove the hold. $\endgroup$ – David Z May 6 '16 at 15:13