# Proof of trace theorems for gamma matrices

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1. Apr 23, 2017

### tb87

Hi,

I'm currently going through Griffith's Particle Physics gamma matrices proofs. There's one that puzzles me, it's very simple but I'm obviously missing something (I'm fairly new to tensor algebra).

1. The problem statement, all variables and given/known data

Prove that $\text{Tr}(\gamma^\mu \gamma^\nu) = 4g^{\mu\nu}$, $\text{Tr}$ being the trace and $g^{\mu\nu}$ the Minkowski metric.

2. Relevant equations

$g_{\mu\nu} \gamma^\mu = \gamma_\nu$
$\text{Tr}(\alpha A) = \alpha\text{Tr}(A)$
$g_\mu g^\mu = 4I_4$

3. The attempt at a solution

I get : $\text{Tr}(\gamma^\mu \gamma^\nu) = \text{Tr}(g^{\mu\nu} \gamma_\nu \gamma^\nu) = g^{\mu\nu} \text{Tr}(4I_4) = 16 g^{\mu\nu}$

What am I missing there?

Edit : I think this is related to the space in which you evaluate $\gamma$ (which I still have a hard time to understand - Minkowski space vs 4-vect space I think? where $\gamma$ is a set of matrices in the first and a vector in the second?). In fact, Griffith says that $\gamma_\mu\gamma^\mu=4$ while on Wikipedia it is rather $\gamma_\mu\gamma^\mu=4I_4$. My proof would work if I considered Giffith's identity, but then how do I distinguish in which context $\gamma_\mu\gamma^\mu$ is equal to 4 rather than $4I_4$ ?

Thanks!
Alex

Last edited: Apr 23, 2017
2. Apr 23, 2017

### vela

Staff Emeritus
You don't have a summation, so the relationship $\gamma_\mu \gamma^\mu = 4$ isn't useful here. Note that $\gamma_\mu\gamma^\mu$ is a matrix, so when you say it's equal to 4, that's a sloppy way of saying it's 4 times the identity matrix.

Try starting with the anticommutation relation the gamma matrices satisfy and take the trace of both sides.

3. Apr 23, 2017

### tb87

vela,

I'm not sure I follow when you say to take the trace on both side of the anticommutation relation because the way Griffith presented it, the relation $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2g^{\mu\nu}$ says that when specifying $\mu$ and $\nu$, you have $\gamma^\mu \gamma^\nu = 1$ for $\mu=\nu=0$, and $\gamma^\mu \gamma^\nu=-1$ for $\mu=\nu \in \{1,2,3\}$, and $\gamma^\mu \gamma^\nu=0$ otherwise. So, according to Griffith, $g^{\mu\nu}$ is actually considered a scalar in this context? (Parenthesis : Griffith says that he uses the convention that sometimes, scalars mean an identity matrix multiplied by that scalar and this is really hard to follow when you've never dealt with tensors). Moreover, $\text{Tr}(g^{\mu\nu}=-2)$, so it doesn't seem to solve the problem?

Actually, I mostly wanted to know why my solution wasn't working, because it highlights a miscomprehension I have with tensors/gamma matrices representation. It's not so much about the exercise per se than why the way I tackled it wasn't working. On this matter, you say that the relation $\gamma_\mu \gamma^\mu = 4I_4$ isn't of any help because I have no summation, but in writing $\text{Tr}(\gamma^\mu\gamma_\nu) = \text{Tr}(g^{\mu\nu}\gamma_\nu \gamma^\nu$ (using the relation $g_{\mu\nu} \gamma^\mu = \gamma_\nu$) am I not getting specifically a summation -- $\gamma_\nu \gamma^\nu$ being the inner product?

Thanks,
Alex

4. Apr 23, 2017

### vela

Staff Emeritus
No. What you're doing is mathematically incorrect. The $\nu$ in $\gamma^\mu = g^{\mu \nu}\gamma_\nu$ is distinct from the one in $\text{Tr}(\gamma^\mu\gamma^\nu).$ What you should write is
$$\text{Tr}(\gamma^\mu\gamma^\nu) = \text{Tr}(g^{\mu\rho}\gamma_\rho\gamma^\nu)$$ but it doesn't seem particularly helpful to do that.

5. Apr 23, 2017

### tb87

Right I get it now : you can't repeat indices on both sides of the equality, thus the contradiction. Thanks vela.

About the $g^{\mu\nu}$ metric though, is my comprehension at least correct? The anticommutation relation is a scalar equality, i.e. the product $\gamma^\mu \gamma^\nu$ gives either 1, -1 or 0 (those are scalars, not identity matrices multiplied by those numbers) ? It really gets confusing that $\gamma$ can be both vectors and matrices!

6. Apr 23, 2017

### vela

Staff Emeritus
$\gamma^\mu$ is a matrix, period. And there are four matrices, $\gamma^0$, $\gamma^1$, $\gamma^2$, and $\gamma^3$. The anticommutator $\{\gamma^\mu,\gamma^\nu\}$ is a matrix, so the other side of the equality also has to be a matrix.

7. Apr 24, 2017

### tb87

Oh okay, thanks for the precision. I might have been mixed up by Griffith's scalar/matrix notation, and by my teacher telling me that $\gamma^\mu$ can be both vectors in the Minkowski space or matrices in spinors space.

8. Apr 24, 2017

### king vitamin

It can sometimes be useful to write gamma matrices as

$$\gamma^{\mu}_{ab}$$

where the latin indices contract with spinors and other gamma matrices, while greek indices contract with spacetime objects like derivatives or the metric. At least I find this useful when trying to keep everything straight. Traces are taken over spinor indices. So the anticommutation relation reads

$$\gamma^{\mu}_{ab}\gamma^{\nu}_{bc} + \gamma^{\nu}_{ab}\gamma^{\mu}_{bc} = 2 g^{\mu \nu} \delta_{ac}$$

The repeated spinor index "b" on the LHS is summed.

9. Apr 24, 2017

### tb87

Right! Thanks king vitamin for the additional info, I do realize that it's even clearer written that way indeed!

Oh by the way, I'm new to PF, I think they tell you to "rate" or "like" the people that take the time to answer your questions, how do I do that? I do see a Like button, is there anything else?