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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 $\Lambda '_{\mu \nu}=M_{\mu \nu}$ ?

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

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  • $\begingroup$ 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 $U(1+\omega)$. $\endgroup$ – aitfel Mar 22 '20 at 13:47
  • $\begingroup$ 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. $\endgroup$ – MannyC Mar 22 '20 at 13:54

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