Q:

Can I define a unitary representation of the Lorentz group on the Hilbert space of a theory which breaks Lorentz invariance?

Let's say I have some Lorentz-violating theory. For concreteness we could imagine a scalar field
\begin{equation}
S=\int {\rm d}^4 x \left(-\frac{1}{2} (\partial \phi)^2 + V(x)^\mu \partial_\mu \phi\right)
\end{equation}
where $V(x)^\mu$ is some space-time dependent vector field (ie a non-trivial external current sourcing $\partial_\mu \phi$).
Classically, I can define a Lorentz transformation that acts in the normal way, $x^\mu \rightarrow \Lambda^{\mu}_{\ \ \nu}x^\nu$, $\phi(x) \rightarrow \phi(\Lambda^{-1}x)$. Of course, this transformation won't be a symmetry because the action will not be invariant.
What I want to understand is what the analog of this logic is in quantum field theory. Ie: I would expect that I could define a Lorentz transformation, but I should find there is some condition defining a symmetry that is violated by this transformation.
Now let's say I quantize the theory, and work in the canonical formalism. Surely I can define the action of Lorentz transformations $\Lambda$ on the Hilbert space. After all, given a state $|\Psi\rangle$, I should be able to ask what the state looks like under a boost. Furthermore, I would expect the set of transformations to form a representation on Hilbert space, $R(\Lambda)$, since the Lorentz transformations themselves form a group.
However, it's not clear to me if the transformations can be taken to be unitary. I basically see two possibilities... (1) it is possible to construct a unitary representation of the Lorentz group on the Hilbert space of the theory, but it won't have any physical relevance because the operators controlling the dynamics (4-momentum operator, angular momentum operator) won't transform as tensors under these transformations, or (2) it is not possible to construct a unitary representation.
In favor of possibility (1): in non-relativistic quantum mechanics, I could always define angular momentum operators $L_i = \epsilon_{ijk}x_j p_k$ which obey the algebra for $SU(2)$, even if the Hamiltonian is not Lorentz invariant, so it seems I could always construct a unitary representation of rotation operators by exponentiating these.
My questions are:

Is constructing $R(\Lambda)$ possible in a Lorentz-violating theory? (If not, what stops you from constructing it?)
Can we choose a unitary representation for $R(\Lambda)$ (possibility (1)), or not (possibility (2))?

A:

If I understand the question correctly, then your question is more general than relativity. For example, you can ask the same question about rotations in a non-relativistic theory. In the spirit of trying to address the problem in the simplest situation possible, allow me to even downgrade to regular particle quantum mechanics, not a QFT. The Hamiltonian is of course,
$$H = \frac{\textbf{p}^2}{2m} + V(\textbf{x}).$$
Let's look at the rotation $R_{\hat{n}}(\theta)$, which is the rotation around the unit vector $\hat{n}$ by an angle $\theta$. The action of rotations on the Hilbert space is defined regardless of what is the Hamiltonian. For example we might simply define on the transformation operation on the Hilbert space as,
$$U(R_{\hat{n}}(\theta))|\textbf{x}\rangle  = |R_{\hat{n}}(\theta) \  \textbf{x}\rangle.$$
This is true whether the rotation is a symmetry of the Hamiltonian or not.
Whenever you meet a new operation, you need to figure out how it acts on the Hilbert space. Even though all symmetries of the Hamiltonian are unitary, not all unitary operations are symmetry of the Hamiltonian/Lagrangian. I hope this helps.

