1
$\begingroup$

How to prove the identity $$ \tilde {\sigma}_{\alpha}\sigma_{\beta}\tilde {\sigma}_{\gamma} = g_{\alpha \beta}\tilde {\sigma}_{\gamma} + g_{\alpha \gamma}\tilde {\sigma}_{\beta} - g_{\beta \gamma} \tilde {\sigma}_{\alpha} + i\varepsilon_{\alpha \beta \gamma \delta}\tilde {\sigma}^{\delta}, \qquad (1) $$ where $$ \sigma_{\alpha} = (\hat {\mathbf E}, \sigma ), \quad \tilde {\sigma}_{\beta} = (\hat {\mathbf E}, -\sigma ), \quad \varepsilon_{\alpha \beta \gamma \delta } = -\varepsilon_{\beta \alpha \gamma \delta} = ..., \quad \varepsilon_{0123} = 1, \quad g_{\alpha \beta} = diag(1, -1, -1, -1)? $$ I tried to do this by constructing the relation $$ \sigma_{\alpha}\tilde {\sigma}_{\beta} = g_{\alpha \beta}\sigma_{0} + i\varepsilon_{0\alpha \beta \gamma}\sigma^{\gamma} - \delta^{0}_{\alpha}(\delta^{0}_{\beta}\sigma_{0} - \tilde {\sigma}_{\beta}) - \delta^{0}_{\beta}(\delta^{0}_{\alpha}\sigma_{0} - \sigma_{\alpha}), $$ where first summand corresponds to $\alpha = \beta$, the second - to $\alpha , \beta \neq 0$, the third - to $\alpha = 0, \beta \neq 0$, the last - to $\beta = 0, \alpha \neq 0$, and $\sigma^{\mu} = g^{\mu \nu}\sigma_{\nu}$.

But I can't get the answer in a compact form of $(1)$ by using my construction. In particular, I don't understand how $\varepsilon_{\alpha \beta \gamma \delta}$ is appeared in the final answer.

An edit.

I proved it, but the proof is terrible.

$\endgroup$

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.