The positive-energy condition in quantum field theory for Hamiltonians associated with different timelike Killing vectors

Question

The Unruh effect is a well-known example in which two Hamiltonians $H$ and $\hat H$ associated with different timelike Killing vector fields both have a lower bound, in the same Hilbert-space representation, even though they are not related to each other by any spacetime isometry. This question asks about a generalization.

Consider a quantum field theory in flat spacetime, expressed in terms of field operators acting on a Hilbert space. Let $K$ and $\hat K$ be two different timelike Killing vector fields, not necessarily related to each other by any isometry, and not necessarily covering the whole spacetime. (As an example, think of Rindler coordinates.) Let $R$ be the region of spacetime in which both Killing vector fields are defined, and consider the algebra of observables in $R$. Let $H$ and $\hat H$ be the operators (Hamiltonians) that generate translations of these observables along $K$ and $\hat K$, respectively.

Question: Suppose that the algebra is represented on a Hilbert space in such a way that the spectrum of one of the Hamiltonians $H$ has a lower bound. Does this imply that the spectrum of the other Hamiltonian $\hat H$ also has a lower bound (in the same Hilbert-space representation)?$^\dagger$

I'm not looking for a watertight proof, just a compelling argument — something clear enough that I could check each step in a free field theory.

By the way, in case this isn't familiar: the Hamiltonian density is not necessarily positive definite in quantum field theory, not even in a representation where the Hamiltonian itself is positive definite. See Fewster (2005) "Energy Inequalities in Quantum Field Theory", https://arxiv.org/abs/math-ph/0501073, which says (page 2):

quantum fields have long been known to violate all such pointwise energy conditions [4] and, in many models, the energy density is in fact unbounded from below on the class of physically reasonable states.

$^\dagger$ ^{The question refers to how the operators are represented on a Hilbert space. That's important because $H$ typically does not have a lower bound in most Hilbert-space representations even if it does in one of them. The spectrum condition is a property of a specific Hilbert-space representation, not just a property of the abstract algebra of observables.}

Answer 1

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:

1. 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.

2. Beware that some of the Unruh-effect literature tacitly redefines the name "vacuum state" to mean something different than "lowest-energy state."

3. 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

Answer 2

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.

Answer 3

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.