QFT question - anti-commutator

[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 { γ^µ,{γ^ργ^σ} } = 2 { γ^µ, I₄ }$$\eta^{\rho \sigma}$$

I understand that { γ^µ,{γ^ργ^σ} } = 2{ γ^µ,$$\eta^{\rho \sigma}I_4$$ } (as {γ^ρ,γ^σ}= $$2\eta^{\rho \sigma}$$) 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?

 

[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).

 

[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.

 

[CompuChip]

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

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

That will answer your first question.

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.

 

[ansgar]

well

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

which is the Id-matrix