# Spin operator: tricky proof using gamma matrices

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?

• also how come I can't use \cancel to draw a slanted strikethrough the $2^{nd}$ and $4^{th}$ terms in the second line? – SuperCiocia Apr 9 '15 at 19:52
• 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. – Constandinos Damalas Apr 9 '15 at 20:01
• Isn't that what I did in the last step? – SuperCiocia Apr 9 '15 at 21:20

• 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.. – SuperCiocia Apr 10 '15 at 14:04