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The QM spin operator can be expressed in terms of gamma matrices and I am trying to do an exercise where I prove an identity which uses $\gamma^5$ and ${\mathbf{\alpha}}$:

$$\mathbf{S}=\frac{1}{2}\gamma^5\mathbf{\alpha}$$

In my first attempt I did this directly in the Dirac representation but the exercise states that I cannot do this, can anyone advise? Is there some identity or trick which would enable me to do this?

To clarify, $\alpha$ is the following matrix where the non-zero elements are the Pauli matrices:

$ \alpha^i= \left[ {\begin{array}{cc} 0 & {\sigma^i} \\ {\sigma^i} & 0 \\ \end{array} } \right] $

$\textbf{S}=\frac{1}{2}\Sigma$

where

$ \Sigma= \left[ {\begin{array}{cc} {\sigma^i} & 0 \\ 0 & {\sigma^i} \\ \end{array} } \right]=-i\alpha_{1}\alpha_{2}\alpha_{3}\mathbf{\alpha} $

Tom
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  • What is $\alpha$ and ${\bf S}$ explicitly? – Prahar Mar 19 '18 at 11:08
  • Alpha is the matrix whose entries not on the leading diagonal are Pauli matrices, but not sure how that helps. – Tom Mar 19 '18 at 12:18
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    How do you expect us to help you prove an identity without a clear definition of all the symbols involved? – Prahar Mar 19 '18 at 12:35
  • The trouble is that the only way that I know of writing $\alpha$ is to pin it down in the Dirac representation and the exercise asks you not to do this, for that reason I feel like someone who knows this area well is really the only one who can assist. – Tom Mar 19 '18 at 14:06
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    @Hollis Surely you can at least say what $\alpha$ is supposed to mean. It's not a standard notation like the gamma matrices are. – knzhou Mar 19 '18 at 14:11
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    $\mathbf{\alpha}$ is as standard as the $\gamma$ matrices. Most standard physics books introduce $\mathbf{\alpha}$ even before the $\gamma$ matrices. – David Bar Moshe Mar 19 '18 at 14:36
  • @DavidBarMoshe - That might be true, but once you're out of the initial stage of learning the Dirac equation, one drops the $\alpha$ and $\beta$ notation and deals exclusively with $\gamma^\mu$. Thus, while I am in a good position to help OP here (having had a decent amount of experience with gamma matrix algebra), I am impeded by notation I have not seen in a long time. It would help if all notation is simply clarified in the question itself instead of me having to dig up textbooks to see what it all means. Especially in this case, where it is not at all difficult to write it down. – Prahar Mar 19 '18 at 14:55
  • @Hollis - Please also define ${\bf S}$. – Prahar Mar 19 '18 at 14:55
  • @DavidBarMoshe @ Hollis - Additionally, spinor conventions are by far the most confusing of all. Every text uses their own favorite convention and therefore especially in these sort of questions, it is most imporatant to lay down the details clearly. – Prahar Mar 19 '18 at 15:13

1 Answers1

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I'm following the conventions of Wikipedia with the following definitions $$ \Sigma^{\mu\nu} = \frac{i}{4} [ \gamma^\mu , \gamma^\nu ] , \qquad S^i = \frac{1}{2} \epsilon^{ijk} \Sigma^{jk}, \qquad \alpha^i = \gamma^0 \gamma^i , \qquad \gamma^5 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 . $$ where $$ \{ \gamma^\mu , \gamma^\nu \} = 2 \eta^{\mu\nu} , \qquad \eta^{\mu\nu} = \text{diag}(1,-1,-1,-1). $$ Having said this, we now note $$ S^i = \frac{i}{4} \epsilon^{ijk}\gamma^j\gamma^k $$ Explicitly, $$ S^1 = \frac{i}{2} \gamma^2 \gamma^3 , \qquad S^2 = \frac{i}{2} \gamma^3 \gamma^1, \qquad S^3 = \frac{i}{2} \gamma^1 \gamma^2 $$ Then, $$ \frac{1}{2} \gamma^5 \alpha^1 = \frac{1}{2} i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^0 \gamma^1 = \frac{i}{2} \gamma^2 \gamma^3 = S^1 , \\ \frac{1}{2} \gamma^5 \alpha^2 = \frac{1}{2} i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^0 \gamma^2 = - \frac{i}{2} \gamma^1 \gamma^3 = S^2 , \\ \frac{1}{2} \gamma^5 \alpha^3 = \frac{1}{2} i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^0 \gamma^3 = - \frac{i}{2} \gamma^1 \gamma^2 = S^3 , \\ $$ Thus, $$ S^i = \frac{1}{2} \gamma^5 \alpha^i. $$

Prahar
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