# Commutators of Lorentz generators

Homework Helper
Gold Member
Equation 2.14 of Srednicki is:

$$U(\Lambda)^{-1} M^{\mu \nu} U(\Lambda) = \Lambda^{\mu}_{\,\,\rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma}$$

He says that writing $\Lambda = 1 + \delta \omega$, one obtains the usual commutation relation of the $M_{\mu \nu}$:

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

I get, starting from the first equation

$$[ M^{\mu \nu},M^{\rho \sigma} ] \delta \omega_{\rho \sigma} = i (\eta^\mu_\rho \delta \omega^\nu_\sigma M^{\rho \sigma} + \eta^\nu_\sigma \delta \omega^\mu_\rho M^{\rho \sigma} )$$

I don't see how to extract the commutator because the omega on the right side contain one of the indices that is not summed over. Its' not clear to me how to proceed. Any suggestion?

StatusX
Homework Helper
I think it should work if you stick in a kronecker delta.

Check the right hand side of the equation, you can write

$$\delta\omega^\nu_\sigma = \delta\omega_{\rho\sigma} \eta^{\rho \nu}$$

Edit:

Also M is anti-symmetric, which means you can write it as

$$M^{uv} = \frac{1}{2} (M^{uv} - M^{vu})$$

Last edited:
what level of quantum physics is this? grad?

Homework Helper
Gold Member
Check the right hand side of the equation, you can write

$$\delta\omega^\nu_\sigma = \delta\omega_{\rho\sigma} \eta^{\rho \nu}$$
Of course! (Slapping myself right now...I should have seen it). Thanks!!

Fredrik
Staff Emeritus