How Peskin & Schroeder simplified this horrible product of bilinears?

[hamad12a]

TL;DR 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]

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.

 

[hamad12a]

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.

 

[mdb71]

Hey good question. This comes down to a few basic observations about sigma matrices.
First of all, we note
$$
(i\sigma ^2)_{\alpha\beta} = \epsilon_{\alpha\beta}.
$$
Next we rephrase the statement of the problem in a slightly more illuminating form:
$$
(i\sigma^2 \sigma^{\mu})_{\alpha\beta}= ((\bar{\sigma^{\mu}})^T)i\sigma^2)_{\alpha\beta}
$$
and of course two matrices are equal if and only if all their entries are equal one by one, so it is easy to see how the indices nomenclature of the statement of the problem follows (note $\beta$ is summed over).
Now, notice from the particular form of the Pauli matrices that they are all symmetric but $\sigma^2$. This means that only $\sigma^2$ changes sign under transposition.
Moreover, consider the identity
$$
\{\sigma^i, \sigma^j\} = \delta^{ij}
$$.
Thus commuting $\sigma^2$ past all sigmas from the left (or if you wish you can reverse left and right of the identity) - and the identity matrix in the first position - in the four vector $\sigma^{\mu}$, we obtain the desired identity, since $\sigma^2$ anti-commutes with all components except itself; but the change of sign here is taken care of by the transposition.

I hope this was clear enough.
mdb71.

 

[PeterDonis]

@mdb71 the [CODE] BB code tag is for displaying program source code, not for LaTeX. LaTeX only needs to be delimited by $$ for separate equations or ## for inline (these replace the old [tex] and [itex] delimiters which no longer work).

I have used magic moderator powers to edit your post accordingly.

 

[mdb71]

Thanks a lot, appreciate. Got it now, I was still getting used to it.