Commutation Relations and Unitary Operators

[abode_x]

I have a problem with deriving another result. Sorry I am new to this field. Please see the attached PDF - everything is there.

 

[dextercioby]

What you're trying to prove is wrong. Srednicki is wrong. If you make the unitary transformation of the 4-momentum operator the way it should be done, namely

\bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)U^{\dagger} (\Lambda)

then you get the correct RHS

\bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)PU^{\dagger} (\Lambda)= \Lambda^{\mu}{}_{\nu} P^{\nu}(x)

which leads to the commutation relation

\left[P^{\mu}(x),M^{\nu\lambda}(x)\right]_{-}= i\hbar \ g^{\mu[\nu}P^{\lambda]}(x)

 

[Avodyne]

abode_x: you got
{i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}] <br /> =\delta\omega^\mu{}_\nu P^\nu
Next,
\delta\omega^\mu{}_\nu P^\nu<br /> =\delta\omega_{\rho\nu}g^{\rho\mu} P^\nu<br /> =\delta\omega_{\rho\sigma}g^{\rho\mu} P^\sigma
Next, use that \delta\omega_{\rho\sigma} is antisymmetric, and so we can replace g^{\rho\mu} P^\sigma with its antisymmetric part,
{1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)
This gives
{i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]={1\over2}\delta\omega_{\rho\sigma}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)
For this to be true for any \delta\omega_{\rho\sigma}, it must be true of each coefficient. So,
{i\over2\hbar}[P^\mu,M^{\rho\sigma}]<br /> ={1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)
Multiply by -2i\hbar and switch the two terms on the right to get
[P^\mu,M^{\rho\sigma}] =i\hbar(g^{\sigma\mu}P^\rho-g^{\rho\mu} P^\sigma)
Using g^{\sigma\mu}=g^{\mu\sigma}, this matches Srednicki's eq.(2.18).