-1
$\begingroup$

I am trying to write product of Pauli matrices in terms of its indicies. I am trying to find a proof of it.
$\sigma^{z}_{\mu \nu}\sigma^{z}_{\alpha \beta}=\delta_{\mu \beta}\delta_{\nu \alpha}$

$\sigma^{x}_{\mu \nu}\sigma^{x}_{\alpha \beta}=-\epsilon_{\mu \beta}\epsilon_{\nu \alpha}$

And the product of unlike Pauli matrices are

$\sigma^{z}_{\mu \nu}\sigma^{x}_{\alpha \beta}=i\delta_{\mu \beta}\epsilon_{\alpha \nu}$

Here $\sigma$'s are Pauli matrices and $\epsilon_{\alpha \beta}$ are Levi civita symbols.

Any help will be highly appreciated.

$\endgroup$
  • $\begingroup$ How about just verifying them using the explicit Pauli matrices? Aren’t these relations true only for the well-known matrices, not for other matrices equivalent through conjugation? $\endgroup$ – G. Smith Jun 26 at 2:07
  • $\begingroup$ Checked versus the completeness relations? $\endgroup$ – Cosmas Zachos Jun 26 at 2:55
  • $\begingroup$ $\sigma^{2}_{i} = I.$ You can test the others with $\sigma_{i}\sigma_{j}+\sigma_{i}\sigma_{j}=0.$ Note, $\sigma_{x}\sigma_{y} \neq \sigma_{y}\sigma_{x}$ so you're missing $3$ products. For instance, $\sigma_{x}\sigma_{y}=-\sigma_{y},\sigma_{x}\sigma_{z}=-i\sigma_{y},$ and $\sigma_{y}\sigma_{z}=i\sigma_{x}$ so it you know $3$ products where $i\neq j$ you know them all. $\endgroup$ – Cinaed Simson Jun 26 at 4:26
  • $\begingroup$ @ Cosmas Zachos Yes i have checked the completeness relation, but it speaks of sum on product of all $\sigma$s not individual products like $\sigma^{z}_{\mu \nu} \sigma^{z}_{\alpha \beta}$. I have $\endgroup$ – Hazoor Imran Jun 26 at 5:02
1
$\begingroup$

If you want to do it using only indices, maybe you can write the Pauli matrices in this way (making the indices vary in $\{0,1\}$ and $\epsilon_{01} = 1$) $$ \sigma^x_{\alpha\beta} = \epsilon_{\alpha\beta} (-1)^\alpha\,,\qquad \sigma^y_{\alpha\beta} = -i\epsilon_{\alpha\beta} \,,\qquad \sigma^z_{\alpha\beta} = \delta_{\alpha\beta} (-1)^\beta\,. $$ Then you'll need only the identities of the $\delta$ and $\epsilon$ tensors.

Anyway your formulas appear to be wrong. I checked two at random and I found $$ \begin{aligned} 1)&&\sigma^z_{01} \,\sigma^z_{10} &= 0 \neq \delta_{11}\delta_{00}\,. \\ \\ 4)&&\sigma^x_{01} \,\sigma^y_{01} &= -i \neq \delta_{10}\epsilon_{01}\,. \end{aligned} $$

$\endgroup$
  • 1
    $\begingroup$ Thanks MannyC. Yes there is mistake. $\endgroup$ – Hazoor Imran Jun 26 at 3:11
  • $\begingroup$ Did you figure out the correct relations you are supposed to prove now? $\endgroup$ – MannyC Jun 26 at 3:14
  • $\begingroup$ This is used in this article "journals.aps.org/prb/pdf/10.1103/PhysRevB.86.245309" Eq (5.18) and Eq. (5.24). It may seem some how obscure though. $\endgroup$ – Hazoor Imran Jun 26 at 3:24
  • 1
    $\begingroup$ You are trying verify the product $T_{\mu\nu} T_{\alpha\beta}$, right? I believe they are just using the completeness relations that @Cosmas Zachos linked in your post. $\endgroup$ – MannyC Jun 26 at 3:31

Your Answer

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

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