- #1

LAHLH

- 409

- 1

If [tex] M^{\mu\nu} [/tex] are the generators of the Lorentz group, i.e. they obey

[tex] [M^{\mu\nu}, M^{\rho\sigma}]=i\hbar(g^{\mu\rho}M^{\nu\sigma}-g^{\nu\rho}M^{\mu\sigma})+.....(1)[/tex]

and [tex] L^{\mu\nu} [/tex] is defined by, [tex] L^{\mu\nu} := \frac{\hbar}{i} (x^{\mu} \partial^{\nu}-x^{\nu} \partial^{\mu})[/tex]

I have found that L also obeys a similar commutation relation to the generators, namely:

[tex] [L^{\mu\nu}, L^{\rho\sigma}]\phi=i\hbar(g^{\mu\rho}L^{\nu\sigma}-g^{\nu\rho}L^{\mu\sigma})+......(2)[/tex]

I also know that [tex] [\phi(x),[M^{\mu\nu}, M^{\rho\sigma}]]=[L^{\mu\nu}, L^{\rho\sigma}]\phi(x) (3) [/tex]

I am now trying to solve a problem that asks to verify equation (1) upto a term on the RHS that commutes with [tex] \phi(x) [/tex] and its derivatives, by using equations 2 and 3. I have already proved equation 1 by other means, but no idea how to go about it this way. Apparently the term that might arise on RHS is called the central charge.

Just plugging things in takes me to:

[tex]\phi(x)[M^{\mu\nu}, M^{\rho\sigma}]-[M^{\mu\nu}, M^{\rho\sigma}]\phi(x)=i\hbar(g^{\mu\rho}L^{\nu\sigma}-g^{\nu\rho}L^{\mu\sigma})+...... [/tex]

I'm not sure where one would go from here to verify the generator equation, equation (1).

Thanks for any suggestions