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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} $

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  • $\begingroup$ What is $\alpha$ and ${\bf S}$ explicitly? $\endgroup$
    – Prahar
    Mar 19, 2018 at 11:08
  • $\begingroup$ Alpha is the matrix whose entries not on the leading diagonal are Pauli matrices, but not sure how that helps. $\endgroup$
    – Tom
    Mar 19, 2018 at 12:18
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    $\begingroup$ How do you expect us to help you prove an identity without a clear definition of all the symbols involved? $\endgroup$
    – Prahar
    Mar 19, 2018 at 12:35
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    $\begingroup$ @Hollis Surely you can at least say what $\alpha$ is supposed to mean. It's not a standard notation like the gamma matrices are. $\endgroup$
    – knzhou
    Mar 19, 2018 at 14:11
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    $\begingroup$ $\mathbf{\alpha}$ is as standard as the $\gamma$ matrices. Most standard physics books introduce $\mathbf{\alpha}$ even before the $\gamma$ matrices. $\endgroup$ Mar 19, 2018 at 14:36

1 Answer 1

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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, $$ S^i = \frac{1}{2} \gamma^5 \alpha^i. $$

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