"stability" is invoked as the justification for the axiomatic requirement that the spectrum of the generators of the translation group must be confined to the forward light-cone. The spectrum condition has pervasive, significant effects in axiomatic QFT. There seems to be no proof, however, that the spectrum condition actually ensures that a quantum field will be stable, partly because there is, AFAIK, no mathematical specification of what stability consists of in QFT. IF there were, I suppose the stability axiom would be central to axiomatic QFT instead of the positive spectrum condition.

Stability is intimately related with positive energy in classical physics, of course, but the concept of energy is rather different in classical relativistic field physics than in quantum mechanics, being the 00 component of the stress-energy tensor instead of being the 0 component of the 4-vector of generators of translations. The relationship between positive energy and stability in classical physics does not seem enough to justify an uncritical adoption of the spectrum condition in quantum theory as an axiom, which is supposed to be obvious enough that it is almost beyond question. Negative frequencies are certainly not ruled out for classical field theories, because the energy is not a linear functional of the frequency of the Fourier components of the field.

An axiomatic definition of stability would presumably have to specify what deformations would or should not affect the stability of a given construction. A building is only stable, for example, provided a strong enough earthquake does not occur, it is not stable sine die. Given that the deformations that are possible in quantum field theory are more varied than the deformations that are possible in classical field theory, the spectrum condition seems to require a more substantial justification.

Less axiomatically, Feynman integrals include negative frequency/energy components in intermediate calculations, though not in observables, which seems to bring the spectrum condition into at least some question.

Haag discusses the relationship of stability with the spectrum condition only extremely perfunctorily (p.29 of the 2nd edition of Local Quantum Physics), and I am not aware of an elaborate discussion by other authors. Is there one?

EDIT: Streater & Wightman, in PCT, Spin & Statistics, and all that, discuss collision states. Their discussion is entirely in terms of perturbation theory, which seems not adequate enough for an axiomatic discussion. However, because of asking this question I'm starting to see slightly more clearly why a conventional Physicist might be entirely satisfied with what there is on this.

EDIT(after acceptance of Tim van Beek's Answer): The other aspect of this is that the restriction to positive frequency is apparently not enough to ensure “confinement”, at least not in an elementary way. That seems to me to be more what “stability” ought to mean.

  • $\begingroup$ I dont have access to Haag so I am not sure what "stability" means here informally. The building example almost suggests a link with Chaos theory and dynamical systems theory, but I cannot imagine that Haag went that far. Is stability a reference to preventing QFT perturbations (as in pertubation theory) from introducing negative frequencies? Is it a reference to lagrangian modifications - $\phi^4th$ theory is good, but its deformation $\phi^4th + a$ is not "stable"? $\endgroup$ Feb 24 '11 at 19:37
  • $\begingroup$ Haag says, on p.29, only that "One of the most important principles of quantum field theory, ensuring the stability, demands that the energy should have a lower bound." There is also a discussion of stability in the context of equilibrium states on pp.227-9. I will add an EDIT above on what Streater & Wightman have to say on pp.24-27 of PCT, Spin & Statistics, and all that. $\endgroup$ Feb 24 '11 at 21:18
  • $\begingroup$ I have found some formalisations of "stability" as used by Haag in the context of dynamical systems theory (ergodic theory) as applied to quantum statistical mechanics. The idea is that any observables are invariant under small perturbations of the underlying Hamiltonian. I cannot post an answer because I cannot yet work out whether: spectrum condition implies stability is a requirement and how one gets from Quantum Statistical mechanics to Axiomatic QFT. $\endgroup$ Feb 24 '11 at 23:20

Stability in this context is usually defined as "there is a vacuum state and there is no state with energy below that of the vacuum state". For this to be true, you need the energy to be bounded from below. All in all the spectrum condition can be expressed in words as

"The spectrum condition is a relativistically invariant way of requiring that the total energy in the theory be nonnegative with respect to every inertial frame of reference and that the quantum system is stable in the sense that it cannot decay to energies below that of the vacuum state."

You'll find a nice exposition of both this statement and a lot of other interesting results about the vacuum state in AQFT in this paper:

  • Stephen Summers: "Yet More Ado About Nothing: The Remarkable Relativistic Vacuum State" (arxiv).

It's maybe worth noting that the spectrum condition can only be formulated in Minkowski spacetime, because it needs the Poincare group for its very formulation. About 17 years ago an alternative formulation was found which is entirely local and can therefore be used to define the "positivity of energy" on any suitable Lorentzian manifold. You can find all necessary references in this paper:

  • Alexander Strohmaier, Rainer Verch, Manfred Wollenberg: "Microlocal analysis of quantum fields on curved spacetimes: Analytic wavefront sets and Reeh-Schlieder theorems" (arxiv)
  • $\begingroup$ I'm not sure why I hadn't seen the Stephen Summers paper before, because I make a point of reading everything of his that I see go by. This is no different from his usual standard. Very helpful. Many thanks. I have so far not managed to encompass microlocal analysis in my world view; I'll have to look to see whether I can complete my current constructions without getting my hands dirty with it. Many thanks for that pointer also. $\endgroup$ Feb 25 '11 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.