I found the following identities about Pauli matrices from the lecture notes of Supersymmetry.

$$((\sigma^{\mu})^{\alpha\dot{\alpha}})^{\ast}=(\bar{\sigma}^{\mu})^{\dot{\alpha}\alpha} \tag{1}$$

$$((\sigma_{\mu\nu})^{\alpha}_{\,\,\,\beta})^{\ast}=-(\bar{\sigma}_{\mu\nu})_{\dot{\beta}}^{\,\,\,\dot{\alpha}} \tag{2}$$

$$((\sigma_{\mu\nu})^{\alpha\beta})^{\ast}=(\bar{\sigma}_{\mu\nu})^{\dot{\alpha}\dot{\beta}} \tag{3}$$

Please give me some hints on how to derive the above identities.

For the first identity, if we take $\mu=0$, then $\sigma^{0}\equiv 1$, which is real. Then, the identity says


Why does the complex conjugation acting on this identity matrix $1_{2\times 2}$ exchange the two spinorial indices? Where does the minus sign come from?


2 Answers 2


Remember that $\bar{\sigma}^{\mu \dot{a}b}=(1,-\vec{\sigma})^{\dot{a}b}$


$(\sigma^{\mu b \dot{a}})^*=\bar{\sigma}^{\mu \dot{a}b}$

Taking $\mu=0$, would simply yield $1=1$ from the definition.


In any convention only one of $\sigma^\mu$ and $\bar\sigma_\mu$ can have both its spinor indices up. But in the original post above, in eq. (1) (which is as given on page 81 of the lecture notes) both $\sigma^\mu$ and $\bar\sigma_\mu$ have been shown with their spinor indices up. There are some typos. Eq. (1) should read as



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