In any relativistic quantum field theory, we require that the spectrum is bounded from below. The typical explanation is that this condition enforces the stability of the theory. However, to me this intuition necessarily requires the existence of an external system coupled to the field theory in order for a state to collapse down to arbitrarily low energies. For a theory that is intended to be "complete" in the sense that there are no other external systems it interacts with, I don't see how positivity enforces this notion of stability. So what exactly do people mean when they say this? Are there other physical reasons for this axiom?
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It turns out this issue is actually much simpler than I thought. We first have the following theorem:
Theorem: In a Wightman field theory, the energy-momentum spectrum is additive. In particular, if $p_1, p_2\in \sigma(P)$, then so is $p_1+p_2$.
Corrolary: For a Field theory that satisfies the Wightman axioms except perhaps for the spectral condition, the following are the only possible kinds of spectra:
- The energy-momentum spectrum is inside the closed forward light-cone.
- The energy-momentum spectrum is inside the backward closed light-cone.
- The energy-momentum spectrum is all of $\mathbb{R}^4$
Given the theorem, it is easy to see why this Corollary is true. If you have vectors both in the forward and backward lightcones, you can construct spacelike vectors and eventually every other vector by linear combination.
The theorem follows for any Wightman theory with a unique state that is invariant under spacetime translations. Clustering and covariance allows you to explicitly construct states with the summed momenta that have positive norm.
The consequence is that for such a field theory, the spectrum must either lie in a cone (which one doesn't really matter, it's a conventional choice), or causality must be violated in the sense that spacelike momenta are present in the theory.
This result is proven in Haag's Local Quantum Physics as Theorem 5.4.1.
Prox
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The Energy of the system should be positive definite so that the system could remain stable. If that bound is not in the theory then system would go to low and low energy forever and no any stability would be achieved. One of the reason for this bound is stability. There could be other reasons also but I am not aware about that currently.
Sambhav Antriksh
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You are assuming, as with plenty of the pioneers of quantum theory, that a closed quantum system in an excited state cannot radiate energy away.
That is an assumption. It might turn out to be true.
However, standard QFT does not obey this. i.e. when you treat QED properly, any excited state can decay and radiate photons away. i.e. when the EM field is properly quantised, the result is that photons can be generated by excited systems at any time. Only ground states have any chance to be stable. Even there, I am not totally sure. I only am sure that QED would cause all excited states to have an imaginary part to the "energy eigenvalues", leading to decay.
If it is true that closed quantum systems do not radiate, then modification needs to be done to the standard QFT, even though we would have no way to know how to do that. After all, there is no system possible in such an idealisation, so we may well not ever know what particular modifications we need to use to enforce that.
What is more physically sensible to consider is that both the physics (no closed system) and the mathematics (QED) are telling us that there would be spontaneous radiation of excited states, leading to decay, and that should be a good reason to require energy positivity.
naturallyInconsistent
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You are right: in a theory which describes a system not interacting with others, it does not matter if the energy is unbounded from below. However, that is a boring universe with no dynamics or interactions (between distinct systems) whatsoever. In such a universe, no energy or momentum is exchanged between (distinct) systems. Our physical universe is not like this. We do have different systems that interact with each other every moment. If one of the systems had negative energy, it would interact with a positive-energy system continuously and produce more systems (of both positive and negative energy) indefinitely (since conservation of energy is not violated). This is what people mean when they talk about instability in this context. Obviously, our universe is not blowing up all around us, proliferating new earths and galaxies out of nowhere. For more discussions on the same topic, see my answers to following similar questions:
What does it mean for the Hamiltonian to not be bounded from below?
Avantgarde
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