# How to trace over spinor indices?

1. Nov 21, 2014

### gentsagree

I would like to take the trace over spinorial indices of the following expression:

$$(\gamma_{\mu}\gamma^{0})_{\alpha}^{\beta}=(\gamma_{\mu})_{\alpha}^{\gamma}(\gamma^{0})_{\gamma}^{\beta}$$.

How do I go about doing this? I reckon I could expand the trace out (let's say I want to do this in 4D) and use a particular representation of the algebra of the gammas, but is there a representation-independent way of doing it?

Also, it confuses me that in the equation above, if one traces over it, it becomes $(\gamma_{\mu})_{\alpha}^{\beta}(\gamma^{0})_{\beta}^{\alpha}$, where the beta index is summed over properly according to the spinor northwest-southeast convention, while the alphas aren't. Is this a problem?

I guess my main question is, how do you take a trace over spinor indices?

Thanks

2. Nov 21, 2014

### Orodruin

Staff Emeritus
You can start from the anti-commutation relation $\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu\nu}$ and use the cyclic property of the trace.

3. Nov 21, 2014

### gentsagree

So you mean:

$$Tr(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu})=2Tr(g^{\mu\nu})$$

and the LHS can be expanded as

$$Tr(\gamma^{\mu}\gamma^{\nu})+Tr(\gamma^{\nu}\gamma^{\mu})=Tr(\gamma^{\mu}\gamma^{\nu})+Tr(\gamma^{\mu}\gamma^{\nu})=2Tr(\gamma^{\mu}\gamma^{\nu})$$

So $Tr(\gamma^{\mu}\gamma^{\nu})=Tr(g^{\mu\nu})=4I$

4. Nov 21, 2014

### Orodruin

Staff Emeritus
Exactly, apart from the last step. The last step is really a trace of the identity matrix multiplied by the metric ($\mu$ and $\nu$ are Lorentz indices, not spinor indices) and hence $tr(g^{\mu\nu}) = g^{\mu\nu} tr(\mathbb 1) = 4g^{\mu\nu}$.

5. Nov 21, 2014

### gentsagree

Of course, you are right. Thanks!