[In view of the discussion let me mention that the context is that of Poincare-covariant quantum field field theories . It is clear that giving up covariance makes many things possible that are not possible otherwise, and allow to make rogorous sense of renormalization in simpler situations such as for free covariant fields interacting with classical external fields, or for the Lee model.]
The fact that naive perturbation theory produces infinite corrections, no matter which finite counterterms are used, proves that no Fock space supports an interacting quantum field theory. Hence there is no Hilbert space featuring at every time physical particles.
Here is my non-rigorous proof:
If there were a Fock space defining the interacting theory at every coupling parameter $g$, it would represent the particles by annihilation fields $a_g(x)$ corresponding to some mass $m(g)$. Taking the limit $g\to 0$ (assuming it exists) we see that the Fock spaces at $g$ have the same structure as the limiting Fock space. By continuity, only continuous labels of Poincare representations can change; the other will be fixed for small enough $g$. The only continuous label is the mass, so the Fock spaces differ only by the mass $m_g$. All other structure is rigid and hence preserved. In particular, if we assume the existence of a power series in $g$, the fields are given by
\[\Phi_g(x)=\frac12(a_g^*(x)+a_g(x))+O(g).\]
Now consider the operator field equations at coupling $g$. For simplicity take $\Phi^4$ theory, where they take (the limit term guarantees a correct the form
\[ \nabla^2 \Phi_g(x)+ m(g)^2 \Phi_g(x) + g \lim_{\epsilon\to 0} \Phi_g(x+\epsilon u)\Phi_g(x)\Phi_g(x-\epsilon u):=0.\]
(This is called a Yang-Feldman equation.) Multiplying with the negative propagator $(\nabla^2 +m_g^2)^{-1}$, we find a fixed point equation for $\Phi_g(x)$, which can be expanded into powers of $g$, and all coefficients will be finite because the $\Phi_g(x)$ and hence their Taylor coefficients are (after smearing) well-defined operators on the corresponding Fock space. Going to the Fourier domain and taking vacuum expectation values, one find a perturbative expansion with finite coefficients which is essentially the textbook expansion of vacuum expectation values corresponding to perturbation around the solution with the correct mass.
This proof is not rigorous, but suggestive, surely at the same level as Ron's arguments. I don't know whether it can be made rigorous by more careful arguments. But it only makes the sort of assumptions physicists always make when solving their problems. If these assumptions are not satisfied but an interacting theory is still Fock, it at least means that there is no mathematically natural way of constructing the operators, even perturbatively. Thus for practical purposes, i.e., to use it for physical computations, it amounts to the same thing as if the Fock representation didn't exist.
Q&A (4871)
