# Gamma matrices

1. Jun 6, 2005

### Kalimaa23

Greetings,

I've been asked to prove the following identity

$$tr(\gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma}) = 4 (\eta ^{\mu \nu} \eta ^{\rho \sigma} - \eta ^{\mu \rho} \eta ^{\nu \sigma} + \eta ^{\mu \sigma} \eta ^{\nu \rho})$$

I know that

$$tr(\gamma^{\mu} \gamma^{\nu}) = 4 \eta^{\mu \nu}$$

which means I would expect something of the form

$$tr(\gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma}) = 4 \eta^{\mu \nu} \eta^{\rho \sigma}$$

Any suggestions?

2. Jun 6, 2005

### dextercioby

This is standard stuff.It's in every book.

$$Trace\left(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\right)=Trace\left[\left(2\eta^{\mu\nu}-\gamma^{\nu}\gamma^{\nu}\right)\gamma^{\rho}\gamma^{\sigma}\right]=8\eta^{\mu\nu}\eta^{\rho\sigma}-Trace\left[\gamma^{\nu}\left(2\eta^{\mu\rho}-\gamma^{\rho}\gamma^{\mu}\right)\gamma^{\sigma}\right] =...$$

I hope u can carry on.

Daniel.

3. Jun 11, 2005

### CarlB

Dear Dimitri;

Re $$tr(\gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma}) = 4 (\eta ^{\mu \nu} \eta ^{\rho \sigma} - \eta ^{\mu \rho} \eta ^{\nu \sigma} + \eta ^{\mu \sigma} \eta ^{\nu \rho})$$

While it's possible to solve this problem without understanding a damned thing about what you did, it is also possible to learn something about gamma matrices and Clifford algebras.

Here are some useful facts about gamma matrices that will allow the above calculation to be made without pain:

(1) Given any product of gamma matrices, it is possible to get it into a form where each different gamma matrix appears only once or not at all. You do this by anticommutation, with each anticommutation bringing out a factor of -1 to multiply the product. If you have more than one gamma matrix of a given type, then you can cancel it in pairs, with the squares of spatial gamma matrices giving +1 and the square of the temporal gamma matrix giving -1, unless you are on the opposite coast signature.

(2) There are therefore 16 possible products of gamma matrices, i.e.
$$2\times 2\times 2 \times 2$$, where the nth 2 determines whether or not the nth gamma matrix is present or not. Of these sixteen gamma matrix products (which some call bilinear forms), only one has a nonzero trace, and that is the unity matrix, $$\hat{1}$$, which of course has trace 4.

(3) Therefore, the only possible answers you can get for your trace problem is +4, 0 or -4.

(4) You will only get 4 or -4 if the product includes each gamma matrix an even number of times, so that the product reduces to $$\pm \hat{1}$$. In all other cases, the product will be nonzero, but your trace will be zero.

(5) To determine whether you get +4 or -4, you need to count the number of times you have to anticommute, and the number of times you end up with a factor of -1 from squaring a gamma matrix.

I hope that this was not too simple to be useful to you. It is stuff that I was not really aware of when I was a grad student.

Anyway, using the above comments you may be able to derive your equation without having to make use of anything that is not completely obvious in and of itself.

Carl

By the way, a Clifford algebra may be thought of as starting with a given number of generalized gamma matrices (more or less than the usual 4), with the usual anticommutation rules and with each of these squaring to either 1 or -1.

4. May 6, 2011

### earth2

Hi guys,

i know this is a very old topic...so, sorry to reactivate it. But i would like to know if there is an example somewhere to prove this statement. I tried to write down the product of three different gamma matrices in this way, but i can't bring it into a form where each gamma matrix appears only once or not at all. So, i'd welcome some illumination on this :)

Cheers,
earth2