Stability of the vacuum state of interacting quantum fields

Question

"Stability" is generally taken to be the justification for requiring that the spectrum of the Hamiltonian should be bounded below. The spectrum of the Hamiltonian is not bounded below for thermal sectors, however, but thermal states are nonetheless taken to be stable because they satisfy thermodynamic constraints. In classical Physics, we would say that the thermal state has lowest free energy, which is a thermodynamic concept distinct from Hamiltonian operators that generate time-like translations.

The entropy component of free energy, meanwhile, is a nonlinear functional of the quantum state (presuming that the definition of entropy in quantum field theory would be at least this much like von Neumann's definition in terms of density operators), so we can reasonably expect the sum of the energy and entropy components to have a minimum in the state of greatest symmetry, as we see for thermal states. [It seems particularly notable in this context that the entropy is not an observable in the usual quantum mechanical sense of a linear functional of the quantum state.]

The presence of irreducible randomness in quantum mechanics presumably puts quantum field theory as much in the conceptual space of thermodynamics as in the conceptual space of classical mechanics, despite the quasi-functorial relationship of "quantization", so perhaps we should expect there to be some relevance of thermodynamic concepts.

Given this background (assuming, indeed, that no part of it is too tendentious), why should we think that requiring the Hamiltonian to have a spectrum that is bounded below should have anything to do with stability in the case of an interacting field? The fact that we can construct a vacuum sector for free fields in which the spectrum of the Hamiltonian is bounded below does not seem enough justification for interacting fields that introduce nontrivial biases towards statistically more complex states.

This question is partly motivated by John Baez' discussion of "quantropy" on Azimuth. I am also interested in the idea that if we release ourselves from the requirement that the Hamiltonian of interacting fields must have a spectrum that is bounded below, then we will have to look for analogues of the KMS condition for thermal states that restore some kind of analytic structure for interacting fields.

I asked a related Question here, almost a year ago. I don't see an answer to the present Question in the citations given in Tim van Beek's Answer there.

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Comments

You might enjoy reading the work of Hollands & Wald on QFT on curved spacetimes. They suggest an axiomatic alternative to the usual stability condition in curved spacetimes, where one doesn't necessarily have Hamiltonians or preferred vacuum states.

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If a free Hamiltonian spectrum being bounded below is OK with you, then a temporary interaction can only change the occupation numbers in the system. If your interaction is permanent, then your system must be represented as a set of free quasi-particles with a free quasi-particle Hamiltonian to which the previous reasoning is applicable. The latter case may require efforts to obtain a physical Hamiltonian.

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If the Hamiltonian is unbounded from below in the vacuum sector then there is no thermal sector: there in no thermal equilibrium. Another POV is that in finite volume the Hamiltonian has to bounded from below in all sectors. I think it should imply the bound for the vacuum sector in infinite volume since negative energy states survive transition to finite volume

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@Squark, By unbounded below, I mean, loosely, that the energy of every state in the Hilbert space is infinitely positive (as is the case for a thermal sector), that there is no state of lowest energy in the Hilbert space, not that there are states of negative energy. The states in the Hilbert space are constructed by the action of the *-algebra of observables on a given(constructed) Lorentz and translation invariant state (together with closure in the GNS-norm that results from that state). Managing the infinities in some explicit way to construct a thermal sector of course may not be trivial.

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This is not possible in the vacuum sector because of Poincare invariance. The vacuum state has to be Poincare invariant hence its energy is exactly zero. If the Hamiltonian is unbounded from below it means that in the spectral decomposition we have representations that live in the negative light cone

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@Squark, Introduce a number operator $N$, $N\left|0\right>=0$ for which $[N,a_f^\dagger]=a_f^\dagger$, for a smeared creation operator $a_f^\dagger$, $[a_f,a_g^\dagger]=(f,g)$, $a_f\left|0\right>=0$, field $\phi_f=a_{f^*}+a_f^\dagger$. A generating function for the VEVs is $\omega(e^{i\lambda\phi_f})=e^{-\lambda^2(f^*,f)/2}$. For a different state, take $\omega_\mu(e^{i\lambda\phi_f})=\frac{Tr[e^{i\lambda\phi_f}e^{-\mu N}]}{Tr[e^{-\mu N}]}=e^{-\lambda^2(f^*,f)_\mu/2}$, $(f,g)_\mu=\int \tilde f^*(k)2\pi\delta(k^2-m^2)\theta(k_0)\mathrm{coth}(\mu/2)\tilde g(k) d^4k/(2\pi)^4$.

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Sorry, that was a tight fit. The point is to use the number operator to construct a state over the free field that has "extra quantum fluctuations" instead of using the Hamiltonian to construct a thermal state, which has quantum+thermal fluctuations. The energy of every state in such a sector is infinitely positive relative to the energy of the vacuum state. To derive the above expression, introduce a resolution of the identity in terms of coherent states, then turn the handle with BCH identities and $[N,a_f^\dagger]=a_f^\dagger$.

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Peter I'm not sure I followed what you did there but surely there are no Poincare invariant pure states in free field theory except the usual vacuum. Why do you think the functional you constructed yields a positive pure state?

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@Squark, the state constructed is a mixture, not pure. It can be rejected out of hand because it doesn't satisfy the Wightman axioms, however on the same grounds one would have to reject thermal states and the corresponding sectors. Changing $N$ to $H/\mathsf{kT}$ results in a thermal state, so that $\mu/2$ in the expression for the inner product $(f,g)_\mu$ changes to $\hbar k_0/2\mathsf{kT}$. The Poincare invariant fluctuations of the new state are greater than the fluctuations of the conventional vacuum state, but the field (and the annihilation/creation operator) algebra is unchanged.

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Peter my point is that the Hamiltonian only has to be bounded below in vacuum sectors. Such sectors are by definition generated by pure Poincare invariant states

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