Representation of the Lorentz group

Question

1. Is there any representation of the Lorentz group where $$U^{-1} f(x) U = f(\lambda^{-1}x)$$ other than the (0,0) representation?

2. If not then is it possible for a field (with a well defined polynomial basis) to behave like a scalar field under the Lorentz group?

3. Will such fields still be called the (0,0) representation of the Lorentz group?

Answer 1

It precisely one of the Wightman axioms that the infinite-dimensional unitary representation ¹ $U : \mathrm{SO}(1,3)\to\mathrm{U}(\mathcal{H})$ on the space of states $\mathcal{H}$ of the theory upon which the field acts as operator is compatible with the field transformation law under the finite-dimensional representation $\rho_\text{fin}: \mathrm{SO}(1,3)\to\mathrm{GL}(V)$ where $V$ is the target space of the field. For a real scalar field, $V=\mathbb{R}$ and $\rho_\text{fin}$ is the trivial representation. Being "compatible" means that $$ U(\Lambda)^\dagger\phi_i(x)U(\Lambda) = \sum_j\rho_\text{fin}(\Lambda)_{ij}\phi_j(\Lambda^{-1}(x))$$ holds as an operator equation on the space of states.

Now, if $\phi$ is scalar, then $\rho_\text{fin}$ is trivial. However, this does not mean, in any way, that $U$ is trivial. The infinite-dimensional unitary representations of the Poincare group $\mathrm{SO}(1,3)\ltimes \mathbb{R}^4$ are given by Wigner's classification, and the scalar field creates particles with mass and momentum, so the unitary representation is not trivial - the trivial unitary representation is just the vacuum.

---

^{1No finite-dimensional representation can be unitary.}