A QFT with vanishing vacuum expectation value and perturbation theory

[The Tortoise-Man]

In This wikipedia article is said:

"If the quantum field theory can be accurately described through perturbation theory, then the properties of the vacuum are analogous to the properties of the ground state of a quantum mechanical harmonic oscillator, or more accurately, the ground state of a measurement problem. In this case the vacuum expectation value (VEV) of any field operator vanishes. For quantum field theories in which perturbation theory breaks down at low energies (for example, Quantum chromodynamics or the BCS theory of superconductivity) field operators may have non-vanishing vacuum expectation values called condensates."

Question 1: Why if perturbation theory is applicable on a quantum field theory, then necessary the VEV of every field operator in this theory must vanish? Can it be directly proved?

Question 2: Can this also be "reversed" and used as a criterion when a QTF is approachable by techniques from perturbation theory? Namely if and only if all VEVs of every field operator in this theory vanish?

Meta question: Can it be summarized that that's exactly THE reason that whenever it is possible the field theories with vanishing VEV's are preferred BECAUSE these allow to use techniques from PT?

 

[vanhees71]

Ad 1: That's of course wrong. The electroweak part of the standard model is perfectly suited for perturbative calculations, and the VEV of the Higgs field is non-zero, which is crucial for the entire phenomenology of the weak interaction, described by a Higgsed local gauge theory.

The same holds true for QCD and effective hadronic models, where we have a gluon as well as a quark condensate, the former being responsible for almost all of the mass of the light hadrons and the latter for the spontaneous breaking of the approximate chiral symmetry of the light-quark sector with the pions as the pseudo-scalar pseudo-Goldstone modes.

 

[The Tortoise-Man]

Thank you, so the statement from wikipedia is wrong. Nevertheless are there in general any fundamental advantages to work with field theories with vanishing vacuum expectation values in sense of that only in case of fields with vanishing vev certain calculational techniques are available? Or in other words where exactly for theories with non-vanishing vev arise the characteristic obstructions which we in other case not have?

 

[vanhees71]

I don't know, which "obstructions" you mean. Perturbation theory for scattering processes, i.e., considering a few (usually 2) particles in the initial state and asking for the cross section for all kinds of reactions in the collision needs stable ground state. In the case that the VEV of a field (due to Poincare invariance that's usually a constant expectation value of some scalar field) is non-zero there is nothing special concerning perturbation theory except that in the Feynman rules usually this VEV occurs.