In reality the small fraction of the air you hadn't managed to remove.
Even if you could make a perfect vacuum pump (you can't) then there would be molecules from the material making up the walls of the container which will boil off into the vacuum.
Ignoring engineering details and with a perfect vacuum there would still be a sea of infrared photons emitted from the walls and of course there is an ever present stream of neutrinos from the sun and space sources which are constantly streaming through everything.
But even if you had a perfect vacuum, at absolute zero, in space, an infinite distance from any stars (!) there would be quantum effects which caused virtual particle pairs to be created out of nothing
First of all, classically (neglecting loop corrections), we obviously want to expand around the true minima that are found in the vacuum, which means around one of the states $|0_\pm\rangle$ – these two are really equivalent to one another due to the gauge symmetry. The state around $\phi=0$ is a maximum of energy, not a minimum, so Nature doesn't spend much time over there before it rolls down to the true minimum.
If we expanded around a shifted scalar field, there would be, assuming that the Lagrangian is Taylor-expanded, also first-order terms in the scalar field. They would produce Feynman "vertices" with one external line connected to a "cross" (in which a Feynman diagram ends like an external line but isn't associated with an actual external particle). These simple decorations of Feynman diagrams could be resummed and their effect would be simple – effective shift the scalar field back to the minimum.
When loop corrections are taken into account, the true minimum isn't exactly given by the classical approximation and the vev isn't exactly as the location of the minimum, too. All these things get quantum corrections – suppressed formally by positive powers of $\hbar$ and more quantitatively by the small dimensionless values of the coupling constants.
The calculations in a quantum field theory shift the scalar field to the new, true minimum (incorporating quantum corrections) automatically. Do you remember I took about one-external-leg vertices of Feynman diagrams that we could eliminate by a clever shift? Well, when you consider loop diagrams, there is a similar subtlety – tadpole diagrams such as this one:
It's a one-loop diagram and the loop at the end plays the same role as the small "cross" indicating the direct Feynman diagram vertex from the beginning of the discussion. But even if we eliminate the linear terms from the action we start with, quantum loops effectively generate their own tadpoles that may be attached through other vertices to any Feynman diagram and that effectively shift the scalar field to the right minimum corrected by the corrections suppressed by powers of $\hbar$.
The LSZ formula doesn't break at all. If you correctly add all the Feynman diagrams, there will also be the Feynman diagrams with the tadpoles attached. They fixed the "internal" part of the Feynman diagram so that it knows about the true minimum. The external part of the LSZ formula is also fixed automatically because the particles in LSZ are connected with the creation action by a near-mass-shall mode of the scalar fields. And it doesn't matter whether you pick a Fourier component of $\phi$ or $\phi-c$ for a constant $c$ – only the actual field $c$ will get magnified near $k^2=m^2$ while the constant term is annihilated by $(k^2-m^2)$, anyway.
So:
- Yes, there is a linear term generated by quantum effects.
- All such quantum corrections are accounted for by summing over all the Feynman diagrams automatically.
- There is always some freedom about how you parameterize the fields etc. Aside from the simple linear shifts above, you may consider nonlinear redefinitions of the scalar fields or redefinitions that depend on derivatives of $\phi$ (and, therefore, indirectly on the energy-momentum vector of the quanta). These different approaches are all possible and they give you different "renormalization schemes", if I use the most general buzzword for such choices. All the measurable physical predictions will ultimately be independent of the renormalization scheme if you adjust the parameters correctly in a given scheme (the correct way depends on the scheme).
So you shouldn't worry about any of these things. The calculation has intermediate steps that are not unique but if you carefully follow your conventions, you don't have to do anything special and the summation over all the Feynman diagrams produces the right physical answers without any extra "cures".
Best Answer
Freeze it in liquid helium. Any gas inside will condense out.
Spin it quickly then stop it. The internal turbulence of the spinning gas will be visible with a sensitive detector.
Apply a short sharp impact to one side. If there is gas inside, the sound energy peak from the sound transiting the gas will be temporally distinct from the spectrum of the sound transiting through the glass.