How do Peskin/Schroeder derive 2-component Fierz identities?

[Theage]

On page 51 Peskin and Schroeder are beginning to derive basic Fierz interchange relations using two-component right-handed spinors. They start by stating the trivial (but tedious) Pauli sigma identity (\sigma^\mu)_{\alpha\beta}(\sigma_{\mu})_{\gamma\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}. They then claim to "sandwich" this identity in between the right-handed part of four Dirac spinors u_1,u_2,u_3,u_4: (\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}. I understand the first identity with elements of the contraction of the Pauli vector perfectly fine, but this one completely mystifies me. The next step in their calculation swaps indices in the Levi-Citiva symbols and uses essentially the same equation in the other direction to get the expected Fierz identity, so if I understood the first equality I would also know the second. However, I just can't see how it easily follows from the Pauli matrix equation. I thus have two questions.

a) Is there an elegant way to actually "sandwich" the identity into four right-handed Weyl spinors, or do I have to manually expand the bilinears?

b) Does right-handedness actually play a role here? That is, it seems to me like this derivation would work just as well with any left-handed spinors, but is this true?

 

[The_Duck]

a) Is there an elegant way to actually "sandwich" the identity into four right-handed Weyl spinors, or do I have to manually expand the bilinears?

So you're asking how to get from the first equation you wrote down to the second equation you wrote down? Just write the matrix multiplication out with explicit indices like this:

(\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
= (\sigma^\mu)_{\alpha\beta} (\sigma_{\mu})_{\gamma\delta}(\bar u_{1R})_\alpha (u_{2R})_\beta (\bar u_{3R})_\gamma (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 (\bar u_{1R})_\alpha \epsilon_{\alpha\gamma} (\bar u_{3R})_\gamma (u_{2R})_\beta \epsilon_{\beta\delta} (u_{4R})_\delta

b) Does right-handedness actually play a role here? That is, it seems to me like this derivation would work just as well with any left-handed spinors, but is this true?

I think between left-handed spinors you are supposed to use the $\bar\sigma$ matrices. That is, I think $u_{1L}\bar\sigma^\mu u_{2L}$ transforms like a proper Lorentz vector, but $u_{1L}\sigma^\mu u_{2L}$ probably doesn't. I like Srednicki's textbook for some of this stuff, because its use of dotted and undotted indices helps to understand the difference between $\sigma^\mu$ and $\bar\sigma^\mu$.

 

[Theage]

Thank you! I was just getting lost in the indices, but now I understand.