I How Peskin & Schroeder simplified this horrible product of bilinears?

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0
Summary
I couldn't understand how to pass from sigma to sigma bar Pauli matrices.
P&S had calculated this expression almost explicitly, except that I didn't find a way to exchange the $$\nu \lambda$$ indices, but I'm sure the below identity is used,

$$
\begin{aligned}\left(\overline{u}_{1 L} \overline{\sigma}^{\mu} \sigma^{\nu} \overline{\sigma}^{\lambda} u_{2 L}\right)\left(\overline{u}_{3 L} \overline{\sigma}_{\mu} \sigma_{\nu} \overline{\sigma}_{\lambda} u_{4 L}\right) &=2 \epsilon_{\alpha \gamma} \overline{u}_{1 L \alpha} \overline{u}_{3 L \gamma} \epsilon_{\beta \delta}\left(\sigma^{\nu} \overline{\sigma}^{\lambda} u_{2 L}\right)_{\beta}\left(\sigma_{\nu} \overline{\sigma}_{\lambda} u_{4 L}\right)_{\delta} \\ &=2 \epsilon_{\alpha \gamma} \overline{u}_{1 L \alpha} \overline{u}_{3 L \gamma} \epsilon_{\beta \delta} u_{2 L \beta}\left(\sigma^{\lambda} \overline{\sigma}^{\nu} \sigma_{\nu} \overline{\sigma}_{\lambda} u_{4 L}\right)_{\delta} \end{aligned}
$$

The identity is,

$$
\epsilon_{\alpha \beta}\left(\sigma^{\mu}\right)_{\beta \gamma}=\left(\overline{\sigma}^{\mu T}\right)_{\alpha \beta} \epsilon_{\beta \gamma}
$$
 

strangerep

Science Advisor
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When asking questions of this kind, you'd get more help more quickly if you state the precise page number(s) and equation number(s), instead of expecting potential helpers to trawl through the book trying to find it.
 
4
0
When asking questions of this kind, you'd get more help more quickly if you state the precise page number(s) and equation number(s), instead of expecting potential helpers to trawl through the book trying to find it.
it's on page 51, Peskin & Schroeder's book.
 

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