- #1
CAF123
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Homework Statement
Let us study the subgroup of the Poincare group that leaves the point ##x=0## invariant, that is the Lorentz group. The action of an infinitesimal Lorentz transformation on a field ##\Phi(0)## is ##L_{\mu \nu}\Phi(0) = S_{\mu \nu}\Phi(0)##. By use of the commutation relations of the Poincare group, we translate the generator ##L_{\mu \nu}## to a nonzero value of ##x##: $$e^{ix^{\rho}P_{\rho}} L_{\mu \nu} e^{-ix^{\sigma}P_{\sigma}} = S_{\mu \nu} - x_{\mu}P_{\nu} + x_{\nu}P_{\mu}\,\,\,\,\,\,\,\,\,(1),$$ where the RHS is computed using the Baker-Campbell-Hausdorff formula to the second term. Then we can write the action of the generators $$P_{\mu} \Phi(x) = -i\partial_{\mu}\Phi(x) \,\,\,\,\text{and}\,\,\,\,L_{\mu \nu}\Phi(x) = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})\Phi(x) + S_{\mu \nu}\Phi(x)\,\,\,\,\,(2)$$
I understand that the LHS of eqn (1) describes the transformation of ##L_{\mu \nu}## under a space time translation. What I want to understand is how they obtained eqn (2).
Homework Equations
##\Phi'(x') = (\text{Id} - i\omega_{g}T_{g})\Phi(x)##
The Attempt at a Solution
I am trying to understand how we are able to write the action of the generators like that using eqn (1). If we consider the transformation of the fields under a Lorentz transformation then $$\Phi'(x') = (\text{Id} - iL_{\mu \nu}\omega^{\mu \nu})\Phi(0+x)$$ using the equation in relevant equations. Now expand ##\Phi(0+x) \approx \Phi(0) + x^{\mu}\partial_{\mu}\Phi(0)##.
For a scalar field, ##\Phi'(x') = \Phi(x)## and so we have ##\Phi(x) = \Phi(0) + x^{\mu}\partial_{\mu}\Phi(0) + iL_{\mu \nu}\omega^{\mu \nu}\Phi(0) + iL_{\mu \nu}\omega^{\mu \nu}x^{\sigma}\partial_{\sigma}\Phi(0)##. I can't see a way to connect this to eqn (2) exactly.
If I now use the fact that the coordinates change ##x^{\rho} = \omega^{\rho}_{\,\,\,\nu}x^{\nu}## then the above becomes, ignoring the last term quadratic in the parameter, $$\Phi(x) = \Phi(0) - \omega^{\rho}_{\,\,\,\nu}x^{\nu}\partial_{\rho}\Phi(0) + iL_{\mu \nu}\omega^{\mu \nu}\Phi(0)\,\,\,\,(3)$$ Now antisymmetrise over the components of ##\omega##, add the result to (3) gives $$2\Phi(x) = 2\Phi(0) + \omega^{\rho \nu}(x_{\rho}\partial_{\nu} - x_{\nu}\partial_{\rho})\Phi(0) + 2i\omega^{\mu \nu}S_{\mu \nu}\Phi(0)$$
So I am making contact with the orbital and intrinsic generators, but I can't quite get an exact result to match with eqn (1) in order to read off the explicit forms for the generators ##P_{\mu}## and ##L_{\mu \nu}##
Many thanks.