Proof of the Fierz Identity

Question

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.

Answer 1

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.