# Does anything guarantee that a field theory will have a lower bound on energy, so that a vacuum exists?

It's not #1, because it's easy to write down QFTs which are unstable against pair production. One simple example is $$\mathcal{L} = \frac12 (\partial_\mu \phi)^2 + \frac12 m^2 \phi^2$$ which corresponds to particles with negative $$m^2$$. Since $$E^2 = p^2 + m^2$$, it is energetically favorable to produce infinitely many particles, since the rest energy is negative. (Note this corresponds to imaginary mass, not negative mass.) There's no vacuum state at all, so #2 doesn't hold either.

For a scalar field theory, as long as we have $$\mathcal{L} = \frac12 (\partial_\mu \phi)^2 - V(\phi)$$ where the potential $$V(\phi)$$ is bounded below, then we will have a vacuum state, simply because the Hamiltonian is bounded below. Heuristically, if we start, say, at a maximum of the potential rather than a minimum, particle production will occur until the vacuum expectation of the field is shifted to the minimum, which is our true vacuum state. Another way to phrase this is that the particles produced interact with each other (noninteracting particles correspond to a quadratic potential, which cannot have both a maximum and minimum), so as this process goes on it becomes less energetically favorable to create more particles, and the process stops at the true vacuum. All of this is a simplified description of what happened to the Higgs field at some point in the early universe.

Since it's essentially #3, I suppose your real question boils down to: how do people enforce the existence of a stable vacuum in the first place? Well, if you work on formal QFT, you simply assume it. That's called the spectrum condition, and it's a reasonable requirement to make of any field theory, like assuming that a spacetime is time orientable in general relativity.

If you're a model builder adding stuff to the Standard Model, there are a couple things you can use:

• if your new particles are weakly coupled, the energies are just $$E = \sqrt{p^2 + m^2}$$ plus small corrections, so you can essentially read off the result from the potential (this covers most papers)
• if your new particles are weakly coupled but the above treatment is too loose, you can try computing the effective potential perturbatively, e.g. the Coleman-Weinberg potential
• if your new particles are strongly coupled but analogous to QCD, we assume it's fine because QCD has a stable vacuum, e.g. whenever any paper says "consider a confining hidden sector"
• if your theory has spontaneously broken supersymmetry, the vacuum energy density is positive

The trickiest part of deciding whether the Standard Model itself has a stable vacuum is QCD. We know the naive vacuum isn't stable, and lattice simulations tell us there is a vacuum containing a so-called chiral condensate of quarks. So in some sense your question about the sign of the pion mass could have gone the other way, and in fact it already has, because the chiral condensate formed and we now define pions about that.

If you are allowed to assume that QCD with massless quarks has a stable vacuum, then it's straightforward to show that pions have positive $$m^2$$ once you account for quark masses. But actually showing that statement is difficult nonperturbative physics. I don't know how to do it, and I don't know if anybody knows.

Option 3 is the closest match, but it's a bit like saying "Nothing guarantees that spacetime has a Lorentzian signature." We normally only consider spacetimes that do, because so much else depends on it. It's a requirement, not a theorem.

Similarly, for relativistic QFT in flat spacetime, we normally only consider QFTs whose total energy has a finite lower bound. The Lorentz-symmetric statement of this condition is that the spectrum of the generators of spacetime translations is restricted to the future light-cone. This is called the spectrum condition. It's one of the basic conditions that we usually require, just like microcausality. Theories that don't satisfy these basic conditions are rejected as unphysical, because so many other things rely on them. For example, the spin-statistics theorem and the CPT theorem both rely on the spectrum condition.

That's for QFT in flat spacetime. For QFT in a generic curved spacetime, we lose translation symmetry, so there are no "generators of spacetime translations," and the spectrum condition becomes undefined. Candidate replacements have been proposed, like the microlocal spectrum condition, but as far as I know this is still an unsettled research topic. The goal, I suppose, is to find a condition that allows things like the spin-statistics theorem to be derived even in curved spacetime. (If I remember right, this has sort of already been done, but the approach that I'm vaguely remembering relies on the flat spacetime proof, and something about it didn't seem quite satisfying to me. If you're interested, I can try to find that paper and post a link.)

Returning to flat spacetime...

is this natural? Do we need fine-tuning?

Depends on what you mean. If we define the Hamiltonian (total energy operator) to be the generator of time-translations, then the constant term can be shifted by an arbitrary finite value with no observable effects. In that sense, there is no fine-tuning problem. But the real world includes gravity even if our favorite QFT doesn't, and gravity does care about that constant term in the Hamiltonian. In that sense, there is a fine-tuning problem, also known as the cosmological constant problem: if we define our QFT with a short-distance cutoff, then the constant term in the total energy (or the cosmological constant) is extremely sensitive to the precise value of the cutoff, even though the cutoff is artificial. It's not a "real" problem in a QFT that doesn't include gravity anyway, but it's a symptom that QFT and gravity probably don't get along with each other in the way we might have naively expected.