Are there translation-invariant hamiltonians that are not parity symmetric? Are there translation-invariant hamiltonians that are not parity symmetric? I am primarily thinking in terms of the state space of a single massive particle in one or more dimensions, but I would like to deliberately keep the question slightly vague to see just how pathological an example you'd have to bring up to have a system with that kind of symmetry.
Moreover, I'm mostly interested in systems that do not have any symmetry that could be reasonably interpreted as a parity transformation. By this I mean that you can make a 1D hamiltonian that "breaks parity" by having
$$
H=\frac12p^2 + p_0 p
$$
where $p_0$ is a constant, but this is clearly not that interesting since you can come up with a deformed 'parity' operator of the form $P'=e^{-ip_0x}Pe^{ip_0x}$ (i.e. a boost by $p_0$, parity, and a boost back by $p_0$). In these terms, the clearest marker of success would be a translationally invariant hamiltonian with large chunks of nondegenerate spectrum.
Is this possible? How far do you need to bend the normal examples to get there?

Edit: To explain a bit the motivation for this question, this related thread circles around claims of the form

if $H$ is translation-invariant and $|\psi\rangle$ is an eigenfunction of $H$, then $|\psi\rangle$ also needs to be translation invariant

which are normally scuppered by the fact that a translation-invariant $H$ is usually parity symmetric in that direction, which introduces a degeneracy into almost all of the spectrum, and therefore makes the usual non-degeneracy argument useless. 
Now, translation invariance and inversion symmetry normally come together in real-world hamiltonians, but they are formally independent and there's no reason the former cannot come without the latter for a 'pathological enough' hamiltonian. The question here is, then, what does 'enough' mean after that pathological? How far off the beaten track do you need to go? And how many of the desirable properties of a hamiltonian (such as e.g. boundedness from below, or the existence of a ground state) can you preserve in the process?
 A: What about $H=\frac12(p_1^2+p_2^2)+V(p,x_1-x_2)$, where $V(p,r)$ is an arbitrary odd function, i.e., $V(-p,-r)=-V(p,r)$. It should be possible to choose $V$ such that no gauge transformation can simplify the Hamiltonian to a parity invariant one.
There are plenty of such $V$. For example all linear combinations of products of any odd number of variables $p_1,p_2,r$ work. These depend on infinitely many parameters. Whereas the physically natural simplifying transformations that would allow calling the transformed parity to be still sensibly viewed as parity have few parameters only. Thus for most of the $V$ there is no natural parity that would be conserved. 
If you want translation invariance in 3D, the same works with vector-valued $p_1,p_2,x_1,x_2$ of diemsnion 3.
A: OK, so thinking a bit more about this, and suggested in terms of working with the momentum eigenfunctions and where to put their eigenvalues in a way that will avoid both degeneracies and unboundedness-from-below, here is one example. Working in $L_2(\mathbb R)$, consider the hamiltonian
$$
\hat{H} = h_0 \exp(a\hat{p}/\hbar) = h_0\exp\left(-i a\frac{\mathrm d}{\mathrm dx}\right),
$$
where $a$ and $h_0$ are constants with dimensions of length and energy, respectively, and $\hat{p}$ is the usual momentum operator. This operator is translation-invariant but it does not seem to have any easy relationship with its mirror version (and, more importantly for the linked motivation, it does not have any degeneracies).
Moreover, the spectrum is bounded from below, but unfortunately it does not seem to have any clear ground state (since the sequence $\psi_n(x)=e^{-inx/a}$ has eigenenergies $h_0e^{-n}$ which asymptotically approach zero but never reach it), and thinking in terms of hamiltonians of the form $\hat{H} = h_0 f(a\hat{p}/\hbar)$ does not suggest any obvious ways to get a clear ground state without introducing degeneracies into the spectrum.
Thus, I'll consider this a partial answer - hopefully a similar example can come along which does have a ground state.
