Gamma matrices and lorentz algebra

1. Apr 20, 2010

RedX

I'm trying to show that the generators of the spinor representation:

$$M^{\mu \nu}=\frac{1}{2}\sigma^{\mu \nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$$

obey the Lorentz algebra:

$$[M^{\mu \nu},M^{\rho \sigma}]=i(\delta^{\mu \rho}M^{\nu \sigma}-\delta^{\nu \rho}M^{\mu \sigma}+\delta^{\nu \sigma}M^{\mu \rho}-\delta^{\mu \sigma}M^{\nu \rho})$$

However, I'm not getting the right answer, so I was hoping someone could point out where I went wrong:

$$[M^{\mu \nu},M^{\rho \sigma}]= \frac{-1}{16}[[\gamma^{\mu},\gamma^{\nu}],[\gamma^{\rho},\gamma^{\sigma}]]$$

$$=\frac{-1}{16}[2\gamma^{\mu}\gamma^{\nu}-2g^{\mu \nu},2\gamma^{\rho}\gamma^{\sigma}-2g^{\rho \sigma}]=\frac{-1}{8}[\gamma^{\mu}\gamma^{\nu},\gamma^{\rho}\gamma^{\sigma}]$$

Now using these relations:

[AB,CD]=[AB,C]D+C[AB,D]
[AB,C]=A{B,C}-{A,C}B

$$\frac{-1}{8}[\gamma^{\mu}\gamma^{\nu},\gamma^{\rho}\gamma^{\sigma}] =\frac{-1}{8}( [\gamma^{\mu}\gamma^{\nu},\gamma^{\rho}]\gamma^{\sigma}+ \gamma^{\rho}[\gamma^{\mu}\gamma^{\nu},\gamma^{\sigma}] )=\frac{-1}{8}(\gamma^{\mu} \{\gamma^{\nu},\gamma^{\rho} \}\gamma^{\sigma} -\{\gamma^{\mu},\gamma^{\rho} \}\gamma^\nu \gamma^\sigma + \gamma^{\rho}\gamma^{\mu} \{\gamma^{\nu},\gamma^{\sigma} \} - \gamma^\rho \{\gamma^{\mu},\gamma^\sigma \} \gamma^\nu )$$

$$=\frac{-1}{4}(g^{\nu \rho}\gamma^{\mu}\gamma^{\sigma} -g^{\mu \rho}\gamma^{\nu}\gamma^{\sigma} +g^{\nu \sigma}\gamma^{\rho}\gamma^{\mu} -g^{\mu \sigma}\gamma^{\rho}\gamma^{\nu} )$$

This last expression almost looks like the Lorentz algebra, but it is missing the partner in the commutator.

2. Apr 21, 2010

haael

You should put eta here, not delta. It's Lorentz metric, anyway.

3. Apr 21, 2010

dextercioby

I did this computation a few years ago. There's no advice to give, rather than paying attention to the awfully many terms. Use as many as possible blank A4 papers and colored pencils to underline similar terms. Eventually, you should get it.

4. Apr 21, 2010

RedX

Thanks everyone. Using different colored pencils: that's a brilliant idea! I never thought about that before, but that would help on long calculations.

As for using the metric instead of the delta: yes I should! Although you can define spinors on Euclidean space mathematically, physically they should be on Minkowski space and I should have a 'g' instead of a delta.

Anyways, I figured out where I went wrong. Here's the complete proof for anyone interested:

Show that:

$$M^{\mu \nu}=\frac{1}{2}\sigma^{\mu \nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$$

obeys the Lorentz algebra:

$$[M^{\mu \nu},M^{\rho \sigma}]=i(\delta^{\mu \rho}M^{\nu \sigma}-\delta^{\nu \rho}M^{\mu \sigma}+\delta^{\nu \sigma}M^{\mu \rho}-\delta^{\mu \sigma}M^{\nu \rho})$$

$$[M^{\mu \nu},M^{\rho \sigma}]= \frac{-1}{16}[[\gamma^{\mu},\gamma^{\nu}],[\gamma^{\rho},\gamma^{\sigma}]]$$

$$=\frac{-1}{16}[2\gamma^{\mu}\gamma^{\nu}-2g^{\mu \nu},2\gamma^{\rho}\gamma^{\sigma}-2g^{\rho \sigma}]=\frac{-1}{4}[\gamma^{\mu}\gamma^{\nu},\gamma^{\rho}\gamma^{\sigma}]$$

Now using these relations:

[AB,CD]=[AB,C]D+C[AB,D]
[AB,C]=A{B,C}-{A,C}B

$$\frac{-1}{4}[\gamma^{\mu}\gamma^{\nu},\gamma^{\rho}\gamma^{\sigma}] =\frac{-1}{4}( [\gamma^{\mu}\gamma^{\nu},\gamma^{\rho}]\gamma^{\sigma}+ \gamma^{\rho}[\gamma^{\mu}\gamma^{\nu},\gamma^{\sigma}] )=\frac{-1}{4}(\gamma^{\mu} \{\gamma^{\nu},\gamma^{\rho} \}\gamma^{\sigma} -\{\gamma^{\mu},\gamma^{\rho} \}\gamma^\nu \gamma^\sigma + \gamma^{\rho}\gamma^{\mu} \{\gamma^{\nu},\gamma^{\sigma} \} - \gamma^\rho \{\gamma^{\mu},\gamma^\sigma \} \gamma^\nu )$$

$$=\frac{-1}{2}(g^{\nu \rho}\gamma^{\mu}\gamma^{\sigma} -g^{\mu \rho}\gamma^{\nu}\gamma^{\sigma} +g^{\nu \sigma}\gamma^{\rho}\gamma^{\mu} -g^{\mu \sigma}\gamma^{\rho}\gamma^{\nu} )$$

$$=\frac{-1}{4}(g^{\nu \rho}\gamma^{\mu}\gamma^{\sigma}+g^{\nu \rho}\gamma^{\mu}\gamma^{\sigma} \\ -g^{\mu \rho}\gamma^{\nu}\gamma^{\sigma}-g^{\mu \rho}\gamma^{\nu}\gamma^{\sigma} \\ +g^{\nu \sigma}\gamma^{\rho}\gamma^{\mu}+g^{\nu \sigma}\gamma^{\rho}\gamma^{\mu}\\ -g^{\mu \sigma}\gamma^{\rho}\gamma^{\nu}-g^{\mu \sigma}\gamma^{\rho}\gamma^{\nu} )$$
But for example the first two terms:

$$g^{\nu \rho}\gamma^{\mu}\gamma^{\sigma}+g^{\nu \rho}\gamma^{\mu}\gamma^{\sigma}=g^{\nu \rho}[\gamma^\mu,\gamma^\sigma]+2g^{\nu \rho}g^{\mu \sigma}$$ (this comes from the identity $$\gamma^\mu \gamma^\sigma=2g^{\mu \sigma}-\gamma^\sigma \gamma^\mu$$).

All the "gg's" end up cancelling, proving the result. So it's pretty long, but I think it's still short enough that I can put it on my power point presentation I have tomorrow :)