1
$\begingroup$

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

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

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

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

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

$$(1^{\alpha\dot{\alpha}})^{\ast}=-1^{\dot{\alpha}\alpha}.$$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.