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- 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}

$$

$$

\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}

$$