# QFT question - anti-commutator

1. Apr 21, 2010

### vertices

For this question, note that curly brackets {..} is an anti-commutator eg. {AB} = AB+BA where A and B are matrices.

Also note that I4 is the identity 4x4 matrix.

I would like to understand why { γµ,{γργσ} } = 2 { γµ, I4 }$$\eta^{\rho \sigma}$$

I understand that { γµ,{γργσ} } = 2{ γµ,$$\eta^{\rho \sigma}I_4$$ } (as {γρσ}= $$2\eta^{\rho \sigma}$$) but why can we take $$\eta^{\rho \sigma}$$ out of the anti commutator, like in the expression above?

If we can, this means $$\gamma^\mu \eta^{\rho \sigma} = \eta^{\rho \sigma}\gamma^\mu$$. Why is this necessarily true?

2. Apr 21, 2010

### CompuChip

Note that whereas $\gamma^\mu$ is a (4 x 4) matrix with entries $(\gamma^\mu)^\nu{}_\rho$, $\eta^{\rho\sigma}$ is just a number. (For standard Lorentzian metrics, it's 1, 0 or -1).

3. Apr 21, 2010

### vertices

Thanks CompuChip, so the anti-commutator of two gamma matrices is just a number times the identity matrix. So $$\eta_{\mu\rho}$$ is just the entry (a number, which as you say is just 1, 0 or -1) on the µ'th row and ρ'th column of $$\eta$$?

Sorry to do this, but can I please ask another related question:

I'd like to understand why:

$$P_1^\mu P_2^\nu tr[ \gamma_\mu \gamma^0 (\gamma_\nu \gamma^0 + \gamma^0 \gamma_\nu)] - tr[\gamma_\mu \gamma^0 \gamma^0 \gamma_\nu]$$

$$=P_1^\mu P_2^\nu [2tr(\gamma_\mu \gamma^0) \eta_\nu ^0 - \eta^{00}tr(\gamma_\mu\gamma_\nu)]=$$

$$=P_1^\mu P_2^\nu [8\eta_\mu ^0 \eta_\nu ^0 - 4 \eta^{00}\eta_{\mu\nu}]$$

Firstly, why can we write $$\gamma_\mu \gamma^0 (\gamma_\nu \gamma^0 + \gamma^0 \gamma_\nu)=2(\gamma_\nu \gamma^0) \eta_\nu ^0$$ and why is this in turn $$8\eta_\mu ^0 \eta_\nu ^0$$?

Also what does this object mean anyway, physically:

$$\eta_\nu^\mu$$ with one raised and one lowered index?

Any thoughts would be appreciated!

Thanks.

4. Apr 21, 2010

### CompuChip

Note that you can raise and lower indices on the gamma matrices using the standard metric, so

$$\gamma_\nu = \eta_{\nu\rho} \gamma^\rho$$

As for the second one, I'll leave that to someone with some physical intuition about these matters. If I'm not mistaken, $\eta^\mu{}_\nu$ is a complicated way to write the unit matrix, as you can check by using the symmety and the fact that eta also raised and lowers indices on itself.

5. Apr 23, 2010

### ansgar

well

$\eta^\mu{}_\nu \eta^\nu{}_\rho = \delta^\mu{}_\rho$

which is the Id-matrix