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As far as I understand, the Hamiltonian of QED and QCD are positive definite. The QED Hamiltonian in Coulomb Gauge is given by the following (credits David Tong QFT Notes):

$$ H=\int d^{3} x\left[\frac{1}{2} \vec{A}^{2}+\frac{1}{2} \vec{B}^{2}+\bar{\psi}\left(-i \gamma^{i} \partial_{i}+m\right) \psi-e \vec{j} \cdot \vec{A}+\frac{e^{2}}{2} \int d^{3} x^{\prime} \frac{j^{0}(\vec{x}) j^{0}\left(\vec{x}^{\prime}\right)}{4 \pi\left|\vec{x}-\vec{x}^{\prime}\right|}\right] $$

where $ \vec{j}=\bar{\psi} \vec{\gamma} \psi \text { and } j^{0}=\bar{\psi} \gamma^{0} \psi $

Looking at the $e\vec{j}.\vec{A}$ term, the Hamiltonian Density could be made arbitrarily negative or positive, at least naively. Of course, the coupling between the current and the potential may conspire to suppress or possibly enhance the "negativity" of this term. My question is, could this term be made arbitrarily negative in a localised region of space to give a sub-vacuum energy density? I know this term appears due to the covariant derivative term in the Lagrangian, so this must also apply to QCD as well. However, aren't there Quantum Energy Inequalities which forbid the existence of negative energies or are such restrictions only global, and that there are no locally negative energies?

Would such a negative energy behave negatively in the gravitational sense? This would be interesting from a phenomenological point of view.

I am new to QFT, please be gentle.

Joeseph123
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1 Answers1

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Negative energy density in QFT

The energy density is not necessarily positive definite in quantum field theory (refs 1 and 2), not even in a representation where the total energy is positive definite. Reference 1 says this on page 2:

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

This occurs even in the simplest model of all, the free scalar field. In this case, the result might seem surprising at first, because the naive expression for the Hamiltonian (total energy operator) is the spatial integral over a manifestly positive-definite integrand. The catch is that in order to make the total energy finite, we need to subtract a constant term that depends on the high-energy cutoff used to define the theory. (We can do this without any mathematical ambiguity by treating space as a very fine discrete lattice, for example.) This constant diverges as a function of the cutoff, so even though the naive energy density is manifestly positive definite, the energy density with the required constant term is not.

What about gravity?

What are the implications for gravity? I don't know. That's a tricky issue for more than one reason. But since the question asks specifically about the Standard Model instead of about a nonperturbative quantum theory of gravity, I'll mention this: The usual ways to account for gravity in the Standard Model are

  • Treat the metric field as a prescribed background field, in which case it doesn't care about the energy density (or total energy) because it's prescribed.

  • Treat the metric field as a quantum field, but only perturbatively (small-coupling expnansion), in which case it still doesn't even care if the total energy has a lower bound. Perturbation theory works fine for models like the $\phi^3$ model, whose total energy cannot have a lower bound.

Nonperturbatively, one of the basic principles of quantum field theory in flat spacetime is that the total energy must have a finite lower bound, so models like the $\phi^3$ model are not allowed. The appropriate way to generalize this total-energy condition to a curved spacetime background is a topic of current research ("total energy" is ambiguous in a generic curved spacetime), and identifying a good set of general principles for nonperturbative quantum gravity is also a topic of current research.

Non-negative total energy in QED/QCD

By the way, to prove that the total energy in QED/QCD is nonnegative, one approach is to use the path-integral formulation. After Wick rotation to Euclidean signature, we can show that the action has a property called reflection positivity, and then we can appeal to a general theorem about the existence of a representation on a Hilbert space with a Hamiltonian having a nonnegative spectrum. This is explained in ref 3.

  1. Fewster (2005) "Energy Inequalities in Quantum Field Theory" (https://arxiv.org/abs/math-ph/0501073)

  2. Fewster (2005) "Quantum Energy Inequalities and Stability Conditions in Quantum Field Theory" (http://arxiv.org/abs/math-ph/0502002)

  3. Montvay and Münster (1994), Quantum Fields on a Lattice, Cambridge University Press

Chiral Anomaly
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  • Excellent answer! Could you comment on this particular $e\vec{j}.\vec{A}$ term, and how there might exist configurations of current and vector potential producing a dominant negative contribution? From an experimental standpoint, I am interested in controllable methods of producing such "negative energies," not just those linked to vacuum fluctuations. I know about the Casimir effect, but I am looking for an experiment with a more pronounced negative energy. – Joeseph123 Jan 27 '21 at 01:08
  • @Joeseph123 The relationship between the vacuum state and the field operators is surprisingly non-obvious in relativistic QED, so I'll need to think about that carefully. Before you posted this question, I was already thinking about similar things anyway for a different reason (related to chiral anomalies -- my username!), so it is on my mind and probably will be for quite a while. I'll post another comment if I come up with something worth sharing, but I'm not there yet. – Chiral Anomaly Jan 27 '21 at 01:53