# False vacuum in axiomatic QFT

There is an elegant way to define the concept of an unstable particle in axiomatic QFT (let's use the Haag-Kastler axioms for definiteness), namely as complex poles in scattering amplitudes. Stable particles are much simpler from this point of view, since they correspond to the discrete part of the Poincare group spectrum of the theory (of course they also correspond to real poles).

The concept of a vacuum state is rather straightforward to define in the axiomatic framework. But what about false (unstable) vacua?

What is the definition of "QFT false vacuum" in the Haag-Kastler axiomatric approach to QFT?

EDIT: I have a wild guess. Perhaps a false vacuum sector corresponds to an irreducible Poincare-invariant continuous representation of the observable algebra which is non-Hermitean, i.e. the representation space is a Banach, or maybe a Hilbertian Banach space (regarded as a topological vector space, without preferred norm or inner product) and no condition involving the *-structure is satisfied. This representation is supposed to have a unique Poincare invariant vector corresponding to the false vacuum itself. It should be possible to define "expectation value" in this setting if some kind of a spectral decomposition exists, and the energy-momentum tensor has expectation value $\epsilon \eta_{\mu\nu}$ where $\epsilon$ is a complex number, the imaginary part signifying the decay rate (as Lubos suggested below). Btw, is it possible to prove the existence of the energy-momentum tensor in Haag-Kastler? Anyway, this is a purely intuitive guess and I don't see how to connect it to the actual physics

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Dear @Squark, the first sentence of your text is loaded because it suggests that the elegance of the complex poles – or even the discovery itself – came from axiomatic QFT. It has nothing to do with this particular research program. This program only "borrowed" a well-known physics fact. I don't see why you're talking about a failed research direction at all. Why don't you ask the same question in the context of proper QFT rather than "axiomatic" QFT? –  Luboš Motl Nov 13 '11 at 8:04
Obviously, an unstable vacuum is at least formally a vacuum with a complex energy density where the imaginary part is equal to the decay probability density per spacetime. However, any such non-real Hamiltonian eigenvalue requires one to discuss a broader picture (like scattering amplitudes of stable particles in the first case) and it's true here, too. All unstable vacua (their Hilbert spaces) have to be embedded into lower-C.C. (AdS or Minkowski) stable vacua. –  Luboš Motl Nov 13 '11 at 8:05
I haven't intended to suggest that. The reason I'm using axiomatic QFT as a context is that I want an answer which can be made mathematically precise, at least in principle. –  Squark Nov 19 '11 at 18:31
If you want that, axiomatic QFT is the worst way--- it's just a political rewriting to QFT to fit better into mathematics politics, it has no useful results, and the ordinary QFT is more mathematically well defined, through path integration. –  Ron Maimon Sep 8 '12 at 15:52
"ordinary QFT is more mathematically well defined, through path integration" nice sense of humor. –  MBN Sep 8 '12 at 16:32

## 1 Answer

The exponential decay dynamics of an unstable particle (defined in terms of a complex pole) is dissipative, since the decay products are ignored in the description. Therefore its symmetries are described by the Poincare semigroup only, where momenta are restricted to be timelike, resulting in a dynamics forward in time without reversibility.

Then the classification of irreducible subunitary representations (characterized by $U(g)^*U(g)\le 1$, which replaces unitarity in the dissipative case) allows further possibilities, among which one finds those for unstable particles. See Schulman, Annals of Physics 59 (1970), 201-218.

The space where the semigroup acts is not a Hilbert space, so this doesn't quite fit the $C^*$-algebra framework of algebraic QFT. Instead one needs the rigged Hilbert space extension of quantum mechanics to accomodate unstable particles. See, e.g., Bohm et al. hep-th/9911059. The rigged Hilbert space can accomodate a deformed inner product in which the continuous spectum is moved far enough into the nonphysical sheet that the pole there becomes visible.

See also my answer to http://physics.stackexchange.com/a/29765/7924

Edit: On the other hand, a false vacuum is given by a tachyonic state with negative mass square, not by an unstable particle, whose complex mass has a positive real part. Thus it corresponds to a unitary but unphysical representation of the Poincare group. It is known that these sectors cannot satisfy causal commutation rules, hence they are excluded in algebraic QFT.

Indeed, in the algebraic treatment of gauge theories by the Epstein-Glaser method (see Scharf's true ghost story), broken symmetries don't arise as tachyons but lead directly to a unitary representation with generated masses.

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But this doesn't answer the main question: the rigged Hilbert space can't include the unstable vacuum can it? –  Ron Maimon Jun 9 '12 at 2:22
@RonMaimon: see the addition to my answer –  Arnold Neumaier Jun 10 '12 at 10:21
The unstable vacuum dioesn't necessarily have tachyons--- that's only if it's perturbatively unstable. It could have all positive mass particles and still be unstable to large deformations (this can naturally happen with a Coleman Weinberg potential from loops stabilizing a phi-4 potential at the symmetric point, but with a true stable vacuum at a distant location). The algebraic treatment doesn't allow for unstable vacua at all, as far as I can see. –  Ron Maimon Jul 10 '12 at 20:04