Representation of $SO^{+}(3,1)$ for scalar fields

Question

As far as i know, the generators of the representation of the group of the orthochronous Lorentz transformations $SO^{+}(3,1)$ can bewritten in the following form: $$J^{\mu \nu} = i(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})$$ So this would mean that we could write for the Lorentz transformation of the scalar field $\phi(x)$: $$\phi'(x')=\exp(i\omega_{\mu \nu}J^{\mu \nu})\phi(x)$$ But i know that the scalar field is invariant under Lorentz transformations, so $$\phi'(x')=\phi(x)$$ How can these two expressions not contradict each other? If one expands the exponent, i.e. $$\exp(i\omega_{\mu \nu}J^{\mu \nu})=1+i\omega_{\mu \nu}J^{\mu \nu}+O(\omega^2)$$ then one has terms like $i\omega_{12}i(x\partial_y-y\partial_x)+\ ...$

Why does this exponential acting on the scalar field leave the scalar field invariant?

Answer 1

The scalar field transforms in the trivial representation of the Lorentz group, so $J^{\mu\nu} = 0$ for scalar fields.

The equation $J^{\mu\nu} = \mathrm{i}\left(x^\mu\partial^\nu - x^\nu \partial^\mu\right)$ is only true when momentum is represented as $p^\mu = \partial^\mu$, i.e. this expression is for $J^{\mu\nu}$ acting on a wavefunction, not on a quantum field.

Note that there are two different representations you can think of in the context of a quantum field - the finite-dimensional one on the classical target space of the field, and the infinite-dimensional unitary one on the space of states of the quantum theory, see also this answer of mine. The statement $J^{\mu\nu} = 0$ for a scalar field is in the context of the finite-dimensional representation.