Lorentz transformation of field with components

[ChrisVer]

I am not looking for a solution, just a "starting point"/guidance for calculating the expression:
[M^{\mu \nu} , \phi_a]
with M^{\mu \nu} being the angular-momentum operators and \phi_a being the field's component, which happens to transform under Lorentz Transformations:
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
like:
\phi_a'(x') = \phi_a(x) + \frac{1}{2} \delta \omega^{\mu \nu} (\Sigma_{\mu \nu})_a^b \phi_b(x)

I have a tendency of writing M^{\mu \nu} =i( x^\mu \partial^\nu - x^\nu \partial^\mu ) but I am not sure that this can help.

 

[MathematicalPhysicist]

In case you want a solution, I believe there's one in Atkinson's textbook, you should check it out.

 

[samalkhaiat]

under Lorentz Transformations:
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

1) You have a factor of half missing in the Lorentz transformation of the coordinates: \bar{x}^{\tau} = x^{\tau} + \delta x^{\tau} , \delta x^{\tau} = \frac{1}{2}\omega_{\mu\nu} (\eta^{\tau \mu} x^{\nu} - \eta^{\tau \nu} x^{\mu}) . \ \ \ \ (1)
2) The commutator [M , \varphi ] means that you are subjecting the field operator to an infinitesimal Lorentz transformation.
3) As an operator, \varphi transforms by (infinite dimensional) unitary representation of the Lorentz group \bar{\varphi}_{a}(x) = e^{-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} \ \varphi_{a}(x) \ e^{\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} . Expanding this to first order in \omega, you obtain \delta \varphi_{a}(x) = - \frac{i}{2}\omega_{\mu\nu}[M^{\mu\nu}, \varphi_{a}(x)] , \ \ \ \ \ \ \ \ \ (2) \delta \varphi_{a}(x) \equiv \bar{\varphi}_{a}(x) - \varphi_{a}(x) .
4) Since \varphi_{a}(x) is a finite-component field on Minkowski space, Lorentz group mixes its components by (finite-dimensional representation) matrix D(\omega), and transform its argument by \Lambda^{-1} \bar{\varphi}_{a}(x) = D_{a}{}^{b} (\omega) \varphi_{b}(\Lambda^{-1}x) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) Infinitesimally, we may write D_{a}{}^{b} (\omega) = \delta_{a}{}^{b} - \frac{i}{2} \omega_{\mu\nu} \left( \Sigma^{\mu\nu}\right)_{a}{}^{b} , \Lambda^{-1}x = x - \delta x . Using these expressions and keeping only the terms linear in \omega, Eq(3) becomes \delta \varphi_{a}(x) = - \delta x^{\tau} \partial_{\tau}\varphi_{a} - \frac{i}{2}\omega_{\mu\nu} \left(\Sigma^{\mu\nu}\right)_{a}{}^{b} \varphi_{b} .
Now, in this expression, substitute (1) and (2) to obtain [M^{\mu\nu} , \varphi_{a}(x)] = i \left(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}\right) \varphi_{a} + \left(\Sigma^{\mu\nu}\right)_{a}{}^{b} \varphi_{b} .

M^{\mu \nu} =i( x^\mu \partial^\nu - x^\nu \partial^\mu )

This expression has no meaning in field theory: on the left hand side, you have M^{\mu\nu} which is an operator constructed out of \varphi_{a} and its conjugate \pi^{a}, but the differential "operator" on the RHS, x^{[\mu} \partial^{\nu]} is not an operator in field theory. If the generator M^{\mu\nu} acts on the field \varphi according to [M^{\mu\nu} , \varphi (x)] = i (x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}) \varphi (x) , this means that \varphi is a scalar field, i.e., the spin matrix \Sigma = 0.