I have not dealt with the gamma matrices extensively so I am having a bit of trouble here.

Basically I want to show that the spin operator defined by $$ \mathbf{\hat{S}} = \frac{1}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}$$

saitisfies the commutation relation $ [H,\mathbf{S}] = \gamma^0 \boldsymbol{\gamma} \times \nabla$ with the Hamiltonian: $$ H = \gamma^0(-i\boldsymbol{\gamma}\cdot\nabla + m) .$$

My work so far:

$$ [H,\mathbf{S} ]\color{blue}{\psi} = \\ H\mathbf{S}\color{blue}{\psi} - \mathbf{S} H\color{blue}{\psi} = \\ \gamma^0(-i\boldsymbol{\gamma}\cdot\nabla + m)*\frac{1}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}\color{blue}{\psi} - \frac{1}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}* \gamma^0(-i\boldsymbol{\gamma}\cdot\nabla + m)\color{blue}{\psi} = \\ -i\gamma^0\boldsymbol{\gamma}\cdot {\nabla \left ( \frac{1}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}\color{blue}{\psi} \right )} + {\frac{m}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}}\color{blue}{\psi} + \frac{i}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma} \gamma^0 \boldsymbol{\gamma}\cdot\nabla\color{blue}{\psi} - \frac{m}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}\color{blue}{\psi} = \\ -i\gamma^0\boldsymbol{\gamma}\cdot {\nabla \left ( \frac{1}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma}\color{blue}{\psi} \right )} + \frac{i}{2}\gamma^5 \gamma^0 \boldsymbol{\gamma} \gamma^0 \boldsymbol{\gamma}\cdot\nabla\color{blue}{\psi} $$

switching to index notation now $[H, S^i]$ :

$$ \frac{-i}{2}\gamma^0 \gamma^5\gamma^0 \gamma^i \gamma^k\partial^k + \frac{i}{2}\gamma^5\gamma^0\gamma^i\gamma^0 \gamma^j\partial^j, $$ rearranging: $$ -i\gamma^5\gamma^i\gamma^j\partial^j $$

Now, the answer is $ \gamma^0 \boldsymbol{\gamma} \times\nabla$, and to get the $ \times $ in there I need a Levi-Civita symbol. Which I guess comes from the definition of $$ \gamma^5 = \frac{i}{4!}\epsilon_{\mu\nu\alpha\beta}\gamma^{\mu}\gamma^{\nu}\gamma^{\alpha}\gamma^{\beta}, $$ from which I would have $$[H, S^i] = \frac{1}{4!}\epsilon_{\mu\nu\alpha\beta}\gamma^{\mu}\gamma^{\nu}\gamma^{\alpha}\gamma^{\beta} \gamma^i \gamma^j \partial^j $$ where the greek letters run from $0$ to $4$ whereas the latin ones only from $1$ to $3$.

How do I proceed?

  • $\begingroup$ also how come I can't use \cancel to draw a slanted strikethrough the $2^{nd}$ and $4^{th}$ terms in the second line? $\endgroup$
    – SuperCiocia
    Apr 9, 2015 at 19:52
  • $\begingroup$ Have you tried it using components of $\mathbf{S}$, i.e $S^i$ instead of the vector $\mathbf{S}$? It would probably be easier, and it can be generalised very easily. $\endgroup$
    – PhotonBoom
    Apr 9, 2015 at 20:01
  • $\begingroup$ Isn't that what I did in the last step? $\endgroup$
    – SuperCiocia
    Apr 9, 2015 at 21:20

1 Answer 1


\begin{align*} -i\gamma^5 \gamma^i \gamma^j \partial_j &= \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^i \gamma^j \partial_j \\ &= \tfrac{1}{3!}\gamma^0 \epsilon_{klm}\gamma^k\gamma^l\gamma^m\gamma^i\gamma^j\partial_j\\ &= \tfrac{1}{2}\gamma^0 \epsilon_{kli}\gamma^k\gamma^l\gamma^j\partial_j\\ &= \gamma^0 \epsilon_{kji}\gamma^k\partial_j\\ &=\gamma^0 \vec{\gamma}\times\vec{\nabla} \end{align*} There are two steps that need justification, but this is left for you to do.

  • $\begingroup$ OK, thanks. I have tried to work through it, but obviously I am missing a trick or a key identity. I follow the step between the first and the second line. But between the second and the third, how do you get rid of the $\gamma^m \gamma^i$? I have tried to write it as $-\gamma^i \gamma^m + 2g^{im}$ but I don't get the same answer.. $\endgroup$
    – SuperCiocia
    Apr 10, 2015 at 14:04
  • $\begingroup$ ACTUALLY sorry no, I don't follow. Can you please elaborate? $\endgroup$
    – SuperCiocia
    Apr 10, 2015 at 23:03

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