# Algebraic operation on a bilinear expression of Weyl-Spinors

In the book of Srednicki in equation (35.29) a rather peculiar algebraic operation is carried out on spinors that I am not able to understand. It's

$$[\psi_\dot{a}^{\dagger} \overline{\sigma}^{\mu\dot{a}c}\chi_c]^{\dagger} = \chi^{\dagger}_\dot{c} (\overline{\sigma}^{\mu a \dot{c}})^{\ast}\psi_a$$

Actually, I would expect

$$[\psi_\dot{a}^{\dagger} \overline{\sigma}^{\mu\dot{a}c}\chi_c]^{\dagger} =\chi^{\dagger}_\dot{c} (\overline{\sigma}^{\mu \dot{a} c})^{\dagger}\psi_a= \chi^{\dagger}_\dot{c} (\overline{\sigma}^{\mu c\dot{a}})^{\ast}\psi_a$$

as $x^\dagger \equiv (x^{T})^{\ast}$ and "transpose" usually exchanges indices. Furthermore, Srednicki mentions that the hermiticity of $\overline{\sigma}$ is only used in a following step of (35.29) so apparently it has not be used here. If somebody could give me some insight I would be grateful.

You are thinking about the operation $\dagger$ incorrectly.

For $\psi$, $\chi$ being a fermionic fields taking values in some representation (I will gloss over the subtleties corresponding to dotted and undotted indices), define a map $$\dagger: \psi \mapsto \psi^\dagger$$ with the (non-exhaustive list of) properties \begin{align} (\alpha \psi)^\dagger &= \alpha^\ast \psi^\dagger,\\ (\psi^\dagger)^\dagger &= \psi,\\ (\psi^\dagger\chi)^\dagger &= \chi^\dagger\psi. \end{align} Now consider a linear combination $\chi'{}^i = \bar{\sigma}^{ij} \chi_j$, then clearly by the first property $$(\chi'{}^i)^\dagger = (\bar{\sigma}^{ij})^\ast \chi^\dagger_j .$$ With this you will understand what is going on: $$[\psi_i^\dagger \bar{\sigma}^{ij} \chi_j]^\dagger = [\psi_i^\dagger \chi'{}^i]^\dagger = (\chi'{}^i)^\dagger \psi_i = (\bar{\sigma}^{ij})^\ast \chi^\dagger_j \psi_i = \chi^\dagger_j (\bar{\sigma}^\dagger)^{ji} \psi_i ,$$ where now confusingly the symbol $\dagger$ is also used to denote the composition of complex and conjugation of the coefficient matrix $\bar{\sigma}$.

An algebra with such a map is known as a C*-algebra.

• I still don't get it. The statement in Dirac's notation seems to refer on the second equality, but according as it is written it seems to be that $[\overline{\sigma}^{\mu\dot{a}c}]^{\dagger} =(\overline{\sigma}^{\mu\dot{c}a})^{\dagger}$ and I actually don't see why. – Frederic Thomas Sep 1 '18 at 15:46
• In my previous comment I suggested you consider the following: Let $a,b \in \mathbb{R}^n$ and $M$ be an $n\times n$ real matrix. Then you can check easily that $a^T M b = b^T M^T a$. I wanted to call to your attention that it does not make sense to ask what $(a^T M b)^T$ is, since it is a scalar. You can go along these arguments to understand the formula, but I suggest that you try to understand my updated answer. – yalda Sep 7 '18 at 5:22
$$\bar\sigma^{\mu\dot a c}$$ is a component of matrix $$\bar\sigma^\mu$$. The component is a number rather than a matrix. So we have $$(\bar\sigma^{\mu\dot a c})^\dagger=(\bar\sigma^{\mu\dot a c})^*$$ which is just a complex conjugation of a number.
For matrix $$\bar\sigma^\mu$$, we should take the transposition and complex conjugation so that the $$\dot a$$-th row $$c$$-th column of matrix $$(\bar\sigma^{\mu})^\dagger$$ is $$(\bar\sigma^{\mu})^{\dagger\dot a c}=(\bar\sigma^{\mu c\dot a})^*$$.
$$[\psi^\dagger_\dot a \bar\sigma^{\mu\dot a c}\chi_c]^\dagger=[\psi^\dagger\bar\sigma^{\mu}\chi]^\dagger=\chi^\dagger [\bar\sigma^\mu]^\dagger\psi$$
$$=\chi^\dagger_\dot c[\bar\sigma^\mu]^{\dagger \dot c a}\psi_a=\chi^\dagger_\dot c(\bar\sigma^{\mu a\dot c})^*\psi _a$$