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

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.

• I don't understand why $H$ and $\hat H$ must act on the same Hilbert space. I'd imagine the only requirement is that the observables are coordinate invariant. I'd imagine as long as the states and the Hilbert space change together, the spectrum should be unchanged.
– Guy
Aug 16, 2020 at 3:31
• Have you seen Letaw & Pfautsch, 1981 on canonical quantization in stationary coordinates of Minkowski space? Aug 17, 2020 at 20:44
• @A.V.S. Thank you again for the link to the paper. Unfortunately, the paper doesn't address my question. It re-defines "vacuum state" to mean something that is generally different than lowest-energy state (which the paper acknowledges), and then proceeds to derive results about that other type of state. Not sure why they would choose to confuse the language by assigning a new meaning to an established name instead of just choosing a new name... but considering my own username, I guess I shouldn't be pointing fingers. :P Aug 19, 2020 at 0:06
• @ChiralAnomaly If we neglect gravity, then the spectrum should be unchanged up to an overall additive constant. However, gravity is sensitive to the total energy of the system, so a consistent theory of QFT and GR should leave the spectrum entirely invariant in any frame. That said, I don't know whether a consistent picture exists.
– Guy
Aug 19, 2020 at 15:17
• @David It occurred to me that expressing the question in terms of coordinate transformations might have been distracting, because the question is not about expressing the same thing in different coordinates. I rewrote the question in coordinate-independent language to make this clear. The question is about the spectra of two different operators that are not related to each other by any symmetry. Aug 20, 2020 at 12:50

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

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

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.

• @Chiral Anomaly. Please have a look at "S. M. Carroll, Spacetime and geometry - An introduction to general relativity": Chapter 9.4 Quantum field theory in curved spacetime. It can provide some hint. Aug 18, 2020 at 7:27
• Actually, in quantum field theory, the Hamiltonian density is not necessarily positive definite and cannot necessarily be made positive definite. Only the Hamiltonian (density integrated over space) can be positive definite. See Fewster (2005) "Energy Inequalities in Quantum Field Theory", 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." Sep 8, 2020 at 13:05

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.

• My first reaction to the locally-Rindler idea was that it can't be sufficient, because the spectrum condition is non-local (the Hamiltonian density doesn't have a lower bound even if the Hamiltonian does). But then I remembered that timelike Killing vector fields in flat spacetime are not very abundant: each one corresponds to some linear combination of the generators of translations, boosts, and rotations. That's pretty constraining, so maybe the locally-Rindler argument is sufficient when combined with that constraint. I'll need to think about this, but thank you for inspiring the thought! Sep 13, 2020 at 4:12
• I'm awarding the bounty to this answer, because the idea of trying relate the general case to the Rindler case led me to think more carefully about the Rindler case itself, and that in turn led me to realize that the Rindler case is actually a counterexample (as explained in my self-answer). Sep 15, 2020 at 0:06