Is this possible and then how? $$((\gamma \otimes \mathbf\sigma)\bullet\mathbf p)(\gamma^\prime\otimes\mathbf 1_2) = \gamma\gamma^\prime\otimes\sigma \bullet \mathbf p $$ where $\gamma$ and $\gamma^\prime$ are using to factorize later we see $\gamma^0=\gamma^\prime\otimes1_2$,$\gamma^1=\gamma\otimes\sigma_1$,$\gamma^2=\gamma\otimes\sigma_2$ $\gamma^3=\gamma\otimes\sigma_3$ where $\gamma^1,\gamma^2,\gamma3$ are Dirac Matrices, $ \mathbf\sigma $ is Pauli Spin Matrix, p is momentum four vector $p=(E/c,-p_1,-p_2,-p_3)$ and $1_2$ is 2x2 unit matrix i.e$$1_2 = \left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)$$ $$\sigma_1 = \left(\begin{matrix} 0 & 1 \\ 1 & 0\end{matrix}\right)$$ $$\sigma_2 = \left(\begin{matrix} 0 & -i \\ i & 0\end{matrix}\right)$$$$\sigma_3 = \left(\begin{matrix} 1 & 0 \\ 0 & -1\end{matrix}\right)$$ I know about matrix mixed product( ($A\otimes B)(C \otimes D)=AC\otimes BD$ ) but I can't understand how this possible I shall be happy to get any answer. Here I put this Image from Where I have got this query. Thanks.Here I put this Image Where I have got this query


1 Answer 1


I think this may be an answer.I am not quite sure about that. $$(\gamma^\prime\otimes 1_2)\{(\gamma\otimes\pmb\sigma)\}\bullet \pmb p\} =\{(\gamma^\prime\otimes 1_2)(\gamma\otimes\pmb\sigma)\}\bullet \pmb p$$ [Here I assume $(\gamma^\prime\otimes 1_2)$ just like scaler] $$=\{\gamma^\prime\gamma\otimes1_2\pmb\sigma\}\bullet \pmb p$$ [By using mixed product theorem $(A\otimes B).(C\otimes D)=AC\otimes BD$]$$=(\gamma^\prime\gamma\otimes\pmb \sigma)\bullet p$$ Similarly $$\{(\gamma\otimes\pmb \sigma)\bullet p\}.\{\gamma^\prime\otimes 1_2\}=\{(\gamma\otimes\pmb \sigma).(\gamma^\prime\otimes 1_2)\}\bullet \pmb p=(\gamma\gamma^\prime\otimes\pmb\sigma 1_2)\bullet p= (\gamma\gamma^\prime\otimes\pmb\sigma)\bullet p$$ Note:I am not sure that there exist any such type rule to exchange inner product with direct product.

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  • $\begingroup$ Hi baponkar, please don't paste screenshots of math. We have MathJax built into the site for this reason (and you even used it in your question!). $\endgroup$
    – Kyle Kanos
    Commented May 17, 2019 at 20:36
  • 1
    $\begingroup$ Well I'm glad you found what you were looking. You might want to see if you can start with Klein-Gordon relativistic wave equation and generate the Dirac equation - which is how Dirac found the $\gamma$ matrices - or the Dirac Clifford algebra $\endgroup$ Commented May 18, 2019 at 9:15
  • $\begingroup$ @CinaedSimson I'm currently trying to reconstruct his original reasoning, but It's still unclear how he deduced the chiral representation of the $\gamma$s (or any other equivalent representation, for that matter). All I have obtained so far - starting from "factoring" the Klein-Gordon equation - is the anti-commutation relation $\{\gamma^\mu, \gamma^\nu\}=2g^{\mu\nu}\mathbb{I}_4$. It's hard to believe that he just guessed correctly, by "extending" the Pauli matrices. There are quite a few ways of doing that $\endgroup$
    – ric.san
    Commented Nov 14, 2023 at 13:24

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