Fierz identity for chiral fermions [closed]

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?

• 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? May 6 '16 at 14:42
• @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. May 6 '16 at 14:51
• As long as the question contains the sentence "I would like to know if this result is right or not", I'm not convinced. May 6 '16 at 15:03
• If you finish to read all the post (until the end I mean), maybe you will convince yourself. Maybe. May 6 '16 at 15:04
• 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. May 6 '16 at 15:13