- #1
ChrisVer
Gold Member
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I am not looking for a solution, just a "starting point"/guidance for calculating the expression:
[itex][M^{\mu \nu} , \phi_a][/itex]
with [itex]M^{\mu \nu}[/itex] being the angular-momentum operators and [itex]\phi_a[/itex] being the field's component, which happens to transform under Lorentz Transformations:
[itex]x'^\mu = x^\mu + \delta \omega^{\rho \sigma} X ^\mu_{\rho \sigma}~~ X^{\mu}_{\rho \sigma} = \eta^{\mu}_\rho x_\sigma - \eta^\mu_\sigma x_\rho[/itex]
like:
[itex]\phi_a'(x') = \phi_a(x) + \frac{1}{2} \delta \omega^{\mu \nu} (\Sigma_{\mu \nu})_a^b \phi_b(x)[/itex]
I have a tendency of writing [itex]M^{\mu \nu} =i( x^\mu \partial^\nu - x^\nu \partial^\mu )[/itex] but I am not sure that this can help.
[itex][M^{\mu \nu} , \phi_a][/itex]
with [itex]M^{\mu \nu}[/itex] being the angular-momentum operators and [itex]\phi_a[/itex] being the field's component, which happens to transform under Lorentz Transformations:
[itex]x'^\mu = x^\mu + \delta \omega^{\rho \sigma} X ^\mu_{\rho \sigma}~~ X^{\mu}_{\rho \sigma} = \eta^{\mu}_\rho x_\sigma - \eta^\mu_\sigma x_\rho[/itex]
like:
[itex]\phi_a'(x') = \phi_a(x) + \frac{1}{2} \delta \omega^{\mu \nu} (\Sigma_{\mu \nu})_a^b \phi_b(x)[/itex]
I have a tendency of writing [itex]M^{\mu \nu} =i( x^\mu \partial^\nu - x^\nu \partial^\mu )[/itex] but I am not sure that this can help.
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