1
$\begingroup$

We know that $S^{-1}=\gamma^0S^\dagger\gamma^0$ and not equal to $S^\dagger$. How to explicitly prove this? I have searched the books but could not find this and am at a loss on how to proceed. Even some pointers would be helpful.

$\endgroup$
2
  • $\begingroup$ May you provide the definition of S? $\endgroup$
    – Quillo
    Oct 14, 2022 at 3:22
  • 1
    $\begingroup$ S=exp$(−\frac{i}{4}ω_{μν}σ^{μν})$ $\endgroup$
    – Lelouch
    Oct 14, 2022 at 15:29

1 Answer 1

5
$\begingroup$

We have $S = \exp\left(-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}\right)$, where $\sigma^{\mu\nu} = \frac i2[\gamma^\mu, \gamma^\nu]$. This gives $S^\dagger = \exp\left(\frac i4\omega_{\mu\nu}\sigma^{\mu\nu\dagger}\right)$. Now, try to prove $\sigma^{\mu\nu\dagger} = \gamma^0\sigma^{\mu\nu}\gamma^0$ which will give the desired result.

$\endgroup$
1
  • $\begingroup$ The $ω_{μν}$ is supposed to be antisymmetric then on taking a hermitian conjugate why it does not introduce a negative sign? $\endgroup$
    – Lelouch
    Oct 18, 2022 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.