# Proving a representation of the Lorentz Algebra from Clifford Algebra/Gamma matrices.

1. Dec 1, 2011

### Onamor

Paraphrasing Peskin and Schroeder:

By repeated use of
$\left\{ \gamma^{\mu} , \gamma^{\nu} \right\}= 2 g^{\mu\nu} \times \textbf{1}_{n \times n}$ (Clifford/Dirac algebra),
verify that the n-dimensional representation of the Lorentz algebra,
$S^{\mu \nu}=\frac{i}{4}\left[\gamma^{\mu},\gamma^{\nu}\right]$,
satisfies the commutation relation
$\left[J^{\mu \nu},J^{\rho \sigma}\right]=i\left(g^{\nu \rho}J^{\mu \sigma}-g^{\mu \rho}J^{\nu \sigma}-g^{\nu \sigma}J^{\mu \rho}+g^{\mu \sigma}J^{\nu \rho}\right)$.

I've tried many lengthy computations and always seem to be missing something.
Most obvious thing to try is just
$\left[S^{\mu \nu},S^{\rho \sigma}\right]=S^{\mu \nu}S^{\rho \sigma}-S^{\rho \sigma}S^{\mu \nu}=\frac{-1}{16}\left(\left[\gamma^{\mu},\gamma^{\nu}\right]\left[\gamma^{\rho},\gamma^{\sigma}\right]-\left[\gamma^{\rho},\gamma^{\sigma}\right]\left[\gamma^{\mu},\gamma^{\nu}\right]\right)$
$=\frac{-1}{16}\left(\left(\gamma^{\mu}\gamma^{\nu}-\gamma^{\nu}\gamma^{\mu}\right)\left(\gamma^{\rho}\gamma^{\sigma}-\gamma^{\sigma}\gamma^{\rho}\right)-\left(\gamma^{\rho}\gamma^{\sigma}-\gamma^{\sigma}\gamma^{\rho}\right)\left(\gamma^{\mu}\gamma^{\nu}-\gamma^{\nu}\gamma^{\mu}\right)\right)$
$=\frac{-1}{16}\left( \gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma} - \gamma^{\mu} \gamma^{\nu} \gamma^{\sigma} \gamma^{\rho} - \gamma^{\nu} \gamma^{\mu} \gamma^{\rho} \gamma^{\sigma} + \gamma^{\nu} \gamma^{\mu} \gamma^{\sigma} \gamma^{\rho} - \gamma^{\rho} \gamma^{\sigma} \gamma^{\mu} \gamma^{\nu} + \gamma^{\rho} \gamma^{\sigma} \gamma^{\mu} \gamma^{\nu} + \gamma^{\sigma} \gamma^{\rho} \gamma^{\mu} \gamma^{\nu} - \gamma^{\sigma} \gamma^{\rho} \gamma^{\nu} \gamma^{\mu} \right)$

and then I've tried a few different commutation relations but to no avail.
Would be very grateful for any help in finishing this off.

Last edited: Dec 1, 2011
2. Dec 1, 2011

### dextercioby

Re: Proving a representation of the Lorentz Algebra from Clifford Algebra/Gamma matri

This is indeed a difficult computation. You must cheat in a way, in the sense that you already know what the final answer looks like. So the the LHS with those 8 terms must lead to the RHS which has also 8 terms (4 times J, but each J has 2 times gammas). The g's will appear when you use the clifford algebra as

$$\gamma^{\mu}\gamma^{\nu} = 2g^{\mu\nu}-\gamma^{\nu}\gamma^{\mu}$$

So try to group the 8 terms of 4 gammas into the desired form according to how the J's indices occur in the RHS of what you're trying to prove.