The positive-energy condition in quantum field theory for Hamiltonians associated with different timelike Killing vectors
The answer is no, and ironically, the example I used to motivate the question is actually a counterexample: the spectrum of the Rindler Hamiltonian does not have a lower bound.
The Rindler Hamiltonian generates boosts in Minkowski spacetime. An expression in terms of the stress-energy tensor is shown in equation (25) in
- Jacobson, "Black holes and Hawking radiation in spacetime and its analogues", https://arxiv.org/abs/1212.6821
That expression makes it clear that the Rindler Hamiltonian cannot have a lower bound.
In hindsight, this is obvious by symmetry. The inverse of a boost is the same as a boost combined with a spatial reflection. A spatial reflection doesn't change the spectrum, but the inverse flips the sign of the spectrum. The only way these can be the same is if the spectrum is symmetric about zero. Therefore, if the spectrum doesn't have an upper bound, it can't have a lower bound, either.
Notes:
Jacobson's paper (cited above) considers only a partial Hamiltonian obtained by integrating over one "Rindler wedge", but that integration surface is not a Cauchy surface. To see the full Hamiltonian on a Cauchy surface, we need to consider the left and right Rindler wedges together, and then it's evident that the full Hamiltonian cannot have a lower bound.
Beware that some of the Unruh-effect literature tacitly redefines the name "vacuum state" to mean something different than "lowest-energy state."
For a careful analysis of some subtleties, see Requardt, "The Rigorous Relation between Rindler and Minkowski Quantum Field Theory in the Unruh Scenario", https://arxiv.org/abs/1804.09403
In QFT (quantum field theory) the Lagrangian density $\mathcal L$ is constructed to be Lorentz invariant. Based on the Lagrangian you build a Hamiltonian density $\mathcal H$, which is requested to be positive definite.
If you change the reference system, formally the Lagrangian does not change, hence the Hamiltonian will not either. Consequently the positive definiteness of the Hamiltonian will maintain, even if applied to transformed fields.
Suppose that you can start a the Minkowski vacuum $(H-E_{\Omega})|{\Omega}\rangle=0$. Then for any time-like Killing vector (which I’ll think of as specifying a time-like curve or some accelerated observer) we can ask whether there is vacuum. Locally the region in space over which the killing field is defined can be put in the form of Rindler coordinates. In other words, at each instance of proper time we know what the acceleration is and general covariance tells you that locally physics is the same as Minkowski space. So the Minkowski vacuum for this observer should look like a thermal state, maybe with a varying temperature. In other words, an accelerated observer always sees an effective horizon to which one can assign a temperature, so your questions should be answered by the Unruh effect.