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- Thread starter abode_x
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- #2

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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

[tex] \bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)U^{\dagger} (\Lambda) [/tex]

then you get the correct RHS

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

which leads to the commutation relation

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

[tex] \bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)U^{\dagger} (\Lambda) [/tex]

then you get the correct RHS

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

which leads to the commutation relation

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

Last edited:

- #3

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[tex] {i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]

=\delta\omega^\mu{}_\nu P^\nu [/tex]

Next,

[tex]\delta\omega^\mu{}_\nu P^\nu

=\delta\omega_{\rho\nu}g^{\rho\mu} P^\nu

=\delta\omega_{\rho\sigma}g^{\rho\mu} P^\sigma[/tex]

Next, use that [tex]\delta\omega_{\rho\sigma}[/tex] is antisymmetric, and so we can replace [tex] g^{\rho\mu} P^\sigma [/tex] with its antisymmetric part,

[tex]{1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]

This gives

[tex]{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)[/tex]

For this to be true for any [tex]\delta\omega_{\rho\sigma}[/tex], it must be true of each coefficient. So,

[tex]{i\over2\hbar}[P^\mu,M^{\rho\sigma}]

={1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]

Multiply by [tex]-2i\hbar[/tex] and switch the two terms on the right to get

[tex] [P^\mu,M^{\rho\sigma}] =i\hbar(g^{\sigma\mu}P^\rho-g^{\rho\mu} P^\sigma)[/tex]

Using [tex]g^{\sigma\mu}=g^{\mu\sigma}[/tex], this matches Srednicki's eq.(2.18).

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