# Deriving an identity of Lorentz group representation

I have a representation of Lorentz group on Hilbert space by following rule: $$|\alpha\rangle_{F'}=U(\Lambda)|\alpha\rangle_{F}$$ where $$\Lambda$$ is Lorentz transformation satisfying $$x^{\mu'}=\Lambda^{\mu}_{\nu}x^{\nu}$$ Also $$U(\Lambda \Lambda')=U(\Lambda)U(\Lambda')$$

For infinitesimal transformation, we have $$U(1+\omega)=1+\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}+O(\omega^2)$$ I have to prove the following identity $$U^{-1}(\Lambda)M^{\mu\nu}U(\Lambda)=\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}M^{\rho\sigma}$$

Then hint is given that us the relation $$U(\Lambda \Lambda')=U(\Lambda)U(\Lambda')$$ and the transformation $$\Lambda^{-1}\Lambda ' \Lambda$$.

I don't understand even how to proceed. Am I suppose to assume $$\Lambda '_{\mu \nu}=M_{\mu \nu}$$ ?

• I guess you mean to say $U^{-1}(\Lambda)M^{\mu\nu}U(\Lambda)=\Lambda^{\mu}_{\rho}\Lambda^{\color{red}\nu}_{\sigma}M^{\rho\sigma}$. – Abhay Hegde Mar 22 '20 at 8:45
• @expikx yeah it was a typo – aitfel Mar 22 '20 at 9:06

It is easier to assume that the finite transformation is correct and derive the infinitesimal one from it. The infinitesimal transformation is given by the commutator $$[M^{\rho\sigma},M^{\mu\nu}] = \eta^{\rho\mu}M^{\sigma\nu} + \eta^{\sigma\nu}M^{\rho\mu}-\eta^{\sigma\mu}M^{\rho\nu} - \eta^{\rho\nu}M^{\sigma\mu}\,. \tag{1}\label{comm}$$ The transformation is infinitesimal if $$\Lambda^\mu_{\phantom{\mu}\nu} =\delta^\mu_\nu + i\omega^\mu_{\phantom{\mu}\nu}\,,$$ with $$\omega$$ small (i.e. only linear orders will be kept). So the transformation for $$M$$ reads \begin{align} \Lambda^\mu_{\phantom\mu\rho}\Lambda^\nu_{\phantom\nu\sigma} M^{\rho\sigma} &= (\delta^\mu_\rho + i\omega^\mu_{\phantom{\mu}\rho})(\delta^\nu_\sigma+ i\omega^\nu_{\phantom{\nu}\sigma}) M^{\rho\sigma} \\&= M^{\mu\nu} + i\omega^\mu_{\phantom\mu\rho}M^{\rho\nu} + i\omega^\nu_{\phantom\nu\sigma}M^{\mu\sigma} + O(\omega^2) \\&\overset{?}{=}M^{\mu\nu} + \left[\tfrac{i}2\omega_{\rho\sigma}M^{\rho\sigma}, M^{\mu\nu}\right] + O(\omega^2)\tag{2}\label{qm} \,. \end{align} The last line (the one with the question mark) is the one that we have to prove, namely that $$U(1+\omega)$$ written as you did in your post is actually generating the infinitesimal version of the transformation of $$M$$. I remind you why we are interested in the commutator: $$U(1+\omega) M \,U(1+\omega)^{-1} = e^{\tfrac{i}2 \omega\cdot M} M e^{-\tfrac{i}2 \omega\cdot M} = \tfrac{i}2(\omega\cdot M) M - \tfrac{i}2 M (\omega \cdot M) + O(\omega^2)\,.$$ The last equality is the commutator in \eqref{qm}. So we now use \eqref{comm} to show that it works. $$\eqref{qm} -M^{\mu\nu}= \tfrac{i}2\left(\omega^\mu_{\phantom\mu\sigma} M^{\sigma\nu} + \omega_\rho^{\phantom\rho\nu} M^{\rho\mu} - \omega_\rho^{\phantom\rho\mu}M^{\rho\nu} - \omega^\nu_{\phantom\nu\sigma}M^{\sigma\mu}\right)\,.$$ Now you see why I put the indices staggered: the antisymmetry property of $$\omega$$ states $$\omega^\mu_{\phantom\mu\nu} = - \omega^{\phantom\nu\mu}_{\nu}$$. The reason is that the only object which is actually defined is $$\omega_{\mu\nu}$$. Whereas the ones with mixed indices simply mean $$\omega^\mu_{\phantom\mu\rho}=\eta^{\mu\nu}\omega_{\nu\rho}$$ and $$\omega^{\phantom\rho\mu}_{\rho}=\omega_{\rho\nu}\eta^{\nu\mu}$$. So tidying up the formula above one finds $$\eqref{qm} - M^{\mu\nu} = i \omega^\mu_{\phantom\mu\sigma} M^{\sigma\nu} + i \omega^\nu_{\phantom\nu\rho} M^{\mu\rho}\,,$$ as you see I combined the first and third term together, and also the second and last after swapping the indices in $$M$$. Then by renaming $$\rho\leftrightarrow \sigma$$ we obtain the line immediately above \eqref{qm} as we wanted to prove.
• I think my main confusion is coming from the fact that I don't know what $M^{\mu \nu}$ actually is? Is it a Linear operator in Hilbert space or is it a c-valued number, for example, $M^{23}=1+i$? I have taken a look at the wiki and without doubt it's an operator but then I don't under the expansion of $U(1+\omega)$. – aitfel Mar 22 '20 at 13:47
• No no, $M^{\mu\nu}$ is a matrix, $(M^{\mu\nu})_{\rho\sigma}$ for fixed $\mu,\nu,\rho,\sigma$ is a number. You can think of it as acting on the infinite dimensional Hilbert space of states, but in the end it only rotates the Lorentz indices among themselves, so it is a finite dimensional linear operator. – MannyC Mar 22 '20 at 13:54