(adsbygoogle = window.adsbygoogle || []).push({}); Given [M[itex]^{\mu \nu}[/itex],M[itex]^{\rho\sigma}[/itex]] = -i([itex]\eta^{\mu\rho}M^{\nu\sigma}+\eta^{\nu\sigma}M^{\mu\rho}-\eta^{\mu\sigma}M^{\nu\rho}-\eta^{\nu\rho}M^{\mu\sigma}[/itex])

and [[itex] P^{\mu},P^{\nu}[/itex]]=0

I need to show that

[M[itex]^{\mu\nu},P^{\mu}[/itex]] = i[itex]\eta^{\mu\rho}P^{\nu} - i\eta^{\nu\rho}P^{\mu}[/itex]

We've been given the following as a 'hint':

Lorentz transformation [itex]x'^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}[/itex],

[itex]\Lambda[/itex] is a 4x4 matrix such that xy is Lorentz invariant.

xy = [itex]\eta_{\mu\nu}x^{\mu}y^{\nu} = x'y' = \eta_{\mu\nu}\Lambda^{\mu}_{\rho}x^{\rho}\Lambda^{\nu}_{\sigma}y^{\sigma}[/itex]

[itex]\eta_{\mu\nu} = \eta_{\rho\sigma}\lambda^{\rho}_{\mu}\Lambda^{\sigma}_{\nu}[/itex] (every Lambda solving this is a Lorentz transformation),

[itex]\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu}+\eta^{\mu\rho}\omega_{\rho\nu}[/itex]

[itex]\delta x^{\mu} = \eta^{\mu\rho}\omega_{\rho\nu} x^{\nu}[/itex]

--> [itex]\omega_{\mu\nu} = - \omega_{\nu\mu}[/itex]

Then [itex]\delta x^{\mu} = i \omega_{\rho\sigma}(M^{\rho\sigma})^{\mu}_{\nu} x^{\nu}[/itex]

Then have two expressions for [itex]\delta x[/itex].

My problem is basically that I do not see the relation between the question and the 'hint'. Presumably we can substitute something from the hint into the commutator we want to calculate, right? But what?

Then we are also asked tofind differential operators M such that

[itex]\delta x^{\mu} = \eta^{\mu\rho}\omega_{\rho\nu} x^{\nu}.[/itex]

We are given the answer:

[itex](M^{\rho\sigma})^{\mu}_{\nu} = L^{\rho\sigma}\delta^{\mu}_{\nu}[/itex]

where

[itex] L^{\rho\sigma} = i (x^{\rho}ā^{\sigma} - x^{\sigma}ā^{\rho})[/itex] and [itex]P^{\mu} = -iā^{\mu}[/itex]

but I do not know how to get there.

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# Lie algebra of the Poincare group

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