# Proof of the Fierz Identity

In p.51 of the Peskin's QFT book, the author derives the Fierz identity as the following:

$$(\bar{u}_{1R}\,\sigma^\mu\, u_{2R})(\bar{u}_{3R}\,\sigma_\mu\, u_{4R}) = 2\epsilon_{\alpha\gamma}\,\bar{u}_{1R,\alpha}\, \bar{u}_{3R,\gamma}\,\epsilon_{\beta\delta}\,u_{2R,\beta}\, u_{4R,\delta} \\= -(\bar{u}_{1R}\,\sigma^\mu\, u_{4R})(\bar{u}_{3R}\,\sigma_\mu\, u_{2R}). \tag{3.78}$$

My question is about the order of the spinors right after the first equality. Since $$(\bar{u}_{1R}\,\sigma^\mu\, u_{2R})$$ and $$(\bar{u}_{3R}\,\sigma_\mu\, u_{4R})$$ are scalars and they are expressed as $$(\bar{u}_{1R,\alpha}\,\sigma^\mu_{\alpha\beta}\, u_{2R,\beta})$$ and $$(\bar{u}_{3R,\gamma}\,\sigma_{\mu,\gamma\delta}\, u_{4R,\delta})$$ with the explicit spinor indices, I think the middle part of the above equation should be $$2\epsilon_{\alpha\gamma}\,\bar{u}_{1R,\alpha}\, u_{2R,\beta}\,\epsilon_{\beta\delta}\,\bar{u}_{3R,\gamma}\, u_{4R,\delta}\,,$$ which leads to an additional minus sign due to anticommuting spinors. Why all the barred spinors ($$\bar{u}$$) must appear first? I appreciate any help.

When you write spinor products without indices, the ordering of the symbols is important: it is implying that neighboring symbols are being contracted with each other. Then commuting two symbols past each other implies a change in the way the symbols are being contracted, which could introduce a sign.

When you are writing out your indices explicitly, all of the information is carried by your choice of how you assigned the indices to the symbols. Then it is no longer relevant which order you write them in: as long as the index assignments are preserved, it's implying the same contractions.

• Thank you for your answer but it seems the order of the spinors really matter. The below comment is my derivation. Could you check where am I making mistakes? I used the relation $(\sigma^\mu)_{\alpha\beta}(\sigma_\mu)_{\gamma\delta} = 2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}$.
– asdf
Dec 18, 2020 at 5:27
• $$(\bar{u}_{1R}\sigma^\mu u_{2R})(\bar{u}_{3R}\sigma_\mu u_{4R}) = \bar{u}_{1R,\alpha}\sigma^\mu_{\alpha\beta}u_{2R, \beta}\bar{u}_{3R,\gamma}\sigma_{\mu,\gamma\delta}u_{4R,\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}\bar{u}_{1R,\alpha}u_{2R, \beta}\bar{u}_{3R,\gamma} u_{4R,\delta} = 2\epsilon_{\alpha\gamma}(-\epsilon_{\delta\beta})(-\bar{u}_{1R,\alpha} u_{4R, \delta}\bar{u}_{3R,\gamma}u_{2R,\beta})=\bar{u}_{1R,\alpha}\sigma^\mu_{\alpha\delta} u_{4R,\delta} \bar{u}_{3R,\gamma}\sigma_{\mu,\gamma\beta}u_{2R,\beta} = (\bar{u}_{1R}\sigma^\mu u_{4R})(\bar{u}_{3R}\sigma_\mu u_{2R})$$
– asdf
Dec 18, 2020 at 5:30
• @asdf why have you added another minus sign in front of $\bar{u}_{1R,\alpha}$? Dec 18, 2020 at 5:53
• @Nihar Karve Because the spinors anticommute. Is it wrong?
– asdf
Dec 18, 2020 at 5:56
• @asdf My point is that the order of the spinors is encoded by the explicit spinor indices that you have assigned to them. So after your 3rd equality above, that set of four spinors is the same as it was before because you left all the same indices on the same spinors. So there is no minus sign. I'll update my answer to make this clearer when I have a minute. Dec 18, 2020 at 6:08