# Lorentz group generator in Srednicki

I'm reading through Srednicki's QFT.

In Chapter 2, the author denotes an infinitesimal transformation by $$U(1+\delta \omega)=I+\frac{i}{2 \hbar}\delta \omega_{\mu \nu} M^{\mu \nu}.\tag{2.11}$$ Then using $$U(\Lambda)^{-1}U(\Lambda')U(\Lambda)=U(\Lambda^{-1}\Lambda'\Lambda) \quad(\Lambda'=1+\delta \omega')$$, the author gets $$\delta \omega_{\mu \nu}U(\Lambda)^{-1}M^{\mu \nu}U(\Lambda)=\delta \omega_{\mu \nu}\Lambda^\mu_{\ \ \rho}\Lambda^\nu_{\ \ \sigma}M^{\rho \sigma}.\tag{2.12}$$ But I don't quite understand how RHS goes to this form. I used $$(\Lambda^{-1}M\Lambda)^{\mu \nu}=(\Lambda^{-1})^\mu_{\ \ \rho}M^{\rho\sigma}\Lambda_\sigma^{\ \ \nu}$$ and $$(\Lambda^{-1})^\mu_{\ \ \rho}=\Lambda_\rho^{\ \ \mu}$$ for calculation, and I just can't get the right form in the textbook.

What's the right way to do the calculation?

## 1 Answer

$$\text{RHS : } U(\Lambda^{-1}\Lambda'\Lambda)=U(\Lambda^{-1}(1+\delta w)\Lambda)=U(1+(\Lambda^{-1}\delta w\Lambda))$$ $$\Rightarrow I \;+ \frac{i}{2\hbar} (\Lambda^{-1}\delta w\Lambda)_{\rho \sigma}M^{\rho \sigma} \tag{1}$$

$$(\Lambda^{-1} \delta w \Lambda)_{\rho \sigma}={(\Lambda^{-1}\delta w \Lambda)^{\alpha}}_{\sigma}g_{\alpha \rho}={(\Lambda^{-1})^{\alpha}}_{\mu}{\delta w^{\mu}}_{\nu}{\Lambda^{\nu}}_{\sigma} g_{\alpha \rho} = {\Lambda_{\mu}}^{\alpha}{\delta w^{\mu}}_{\nu}{\Lambda^{\nu}}_{\sigma} g_{\alpha \rho} = {\Lambda^{\mu}}_{\rho}\delta w_{\mu \nu}{\Lambda^{\nu}}_{\sigma} \tag{2}$$

Substitute $$(2)$$ in $$(1)$$ and you should get the desired result.

$$\underline{\text{Elaboration}}$$

$$\text{ LHS : }U(\Lambda^{-1})U(\Lambda')U(\Lambda)=U(\Lambda^{-1})(I+\frac{i}{2\hbar}\delta w_{\mu \nu} M^{\mu \nu})U(\Lambda)$$ $$\Rightarrow I+\frac{i}{2\hbar}\delta w_{\mu \nu} U(\Lambda^{-1})M^{\mu \nu}U(\Lambda) \tag{3}$$

$$\text{ RHS (After substituting (2) in (1)) : }I + \frac{i}{2\hbar} {\Lambda^{\mu}}_{\rho}\delta w_{\mu \nu}{\Lambda^{\nu}}_{\sigma} M^{\rho \sigma} \tag{4}$$

Equating $$(3)$$ and $$(4)$$, we get $$\delta w_{\mu \nu} U(\Lambda^{-1})M^{\mu \nu}U(\Lambda)=\delta w_{\mu \nu}{\Lambda^{\mu}}_{\rho}{\Lambda^{\nu}}_{\sigma} M^{\rho \sigma}$$

• Thanks a lot! Seems that I have put U^{-1} and U at the wrong place. – Rosetta Nov 3 '19 at 7:25