As I understand it, a naive interpretation of the state space of a quantum field theory is an infinite tensor product
$$\otimes_{x\in M} H_x,$$
where $x$ runs over the points of space. This corresponds to the fact that a field $\phi$ and the conjugate momentum $\pi$ can be viewed as a composite system of the array of $\phi(x)$ and $ \pi(x)$. Thus, again naively, the amplitude assigned by a quantum state $\Psi(\phi, \pi)$ to a classical initial condition $(\phi, \pi)$ is a tensor product of the amplitudes $\Psi(\phi(x), \pi(x))\in H_x$.
Of course, this doesn't quite make sense, for many reasons including the fact that the infinite tensor product is quite badly behaved. Instead, the standard way of quantising, say a scalar field satisfying the Klein-Gordon equation, is to write it in terms of Fourier modes
$$\phi(x,t)=(2\pi)^{-3/2}\int [a(p)e^{i(px-\epsilon_pt)}+a^*(p)e^{-i(px-\epsilon_pt)}]\frac{d^3p}{2\epsilon_p}$$
with $\epsilon_p=\sqrt{p^2+m^2}$ (this being the KG equation). The canonical commutation relation for $\phi$ and $\pi$ become
$$[a(p), a^*(p')]=2\epsilon_p\delta(p-p'); \ [a(p), a(p')]=[a^*(p), a^*(p')]=0,$$
which can individually be quantised in a Segal-Bargmann fashion to act on a Hilbert space $H_p$. To quantise all these operators as we run over all momenta, we would again require an infinite tensor product
$$\otimes_p H_p.$$
This is avoided by imposing an additional condition, the existence of a vector $\Psi_0$ (interpreted as the vacuum), satisfying $$a(p)\Psi_0=0$$ for all $p$. After this, it all works out and we have a nicely quantised free field by putting the operators into the integral above. I think I sort of understand this procedure, with the level of uncertainty that I'm normally stuck with when thinking about physics.
However, I've come upon the following passage in the book by Streater and Wightman, page 86-87.
When, in fact, do non-separable Hilbert spaces appear in quantum mechanics? There are two cases which deserve mention. The first arises when one takes an infinite tensor product of Hilbert spaces… Infinite tensor products of Hilbert spaces are always non-separable. Since a Bose field can be thought of as a system composed of an infinitely of oscillators, one might think that such an infinite tensor product is the natural state space. However, it is characteristic of field theory that some of its observables involve all the oscillators at once, and it turns out that such observables can be naturally defined only on vectors belonging to a tiny separable subset of the infinite tensor product. It is the subspace spanned by such a subset that is the natural state space rather than the whole infinite tensor product itself. Thus, while it may be a matter of convenience to regard the state space as part of the infinite tensor product, it is not necessary.
My question is, how does one relate this passage to the usual quantisation procedure described above. In particular, what is the 'tiny separable subset' alluded to by Streater and Wightman?
Because the infinite tensor product picture is so intuitively compelling (this is emphasised, I think, by all authors on QFT), it would be nice to spell out the relation between it and standard quantisation with at least some level of mathematical clarity.
Best Answer
For free quantum fields I think this issue can be dealt with using this theory:
We describe a well-behaved notion of grounded tensor product for a possibly infinite collection of grounded Hilbert spaces: that is, Hilbert spaces $(K_\lambda)_{\lambda \in \Lambda}$ equipped with unit vectors $z_\lambda \in K_\lambda$. If each $K_\lambda$ is separable and the index set $\Lambda$ is countable, this tensor product is separable!
This doesn't help you for a tensor product of uncountably many Hilbert spaces, but it still helps you a little with your question. There is a way to reduce the Hilbert space of a free quantum field to an infinite but countable tensor product of grounded Hilbert spaces.
Namely:
When you have a free bosonic quantum field, the single-particle Hilbert space $H$ is a countable direct sum of 1-dimensional spaces $H_\lambda$. Quantizing each $H_\lambda$ is just like quantizing a harmonic oscillator: the Fock space of $H_\lambda$, say $K_\lambda$, is a Hilbert space completion of the polynomial algebra on $H_\lambda$.
Moreover, each $K_\lambda$ is grounded: there's an obvious 'vacuum vector' $z_\lambda \in H_\lambda$, namely the element 1 in the polynomial algebra. And here's the best part: the Fock space of $H$, say $K$, is the grounded tensor product of the $K_\lambda$:
$$ H = \bigoplus_{\lambda} H_\lambda \implies K = \bigotimes_{\lambda} K_\lambda $$
where, just to emphasize, the tensor product here is the grounded tensor product.
If we're dealing with a free quantum field on a spacetime $\mathbb{R} \times S$ where the spatial manifold is compact, we can do the decomposition
$$ H = \bigoplus_{\lambda} H_\lambda $$
using momentum or energy eigenstates, since the Laplacian and other elliptic operators on $S$ will have discrete spectrum.
If we're working on Minkowski spacetime, as you are, this doesn't work: your momentum $p$ takes a continuum of values. So you're trying to write $H$ not as a direct sum but as a direct integral of 1-dimensional Hilbert spaces.
So, what seem to be asking for is a generalization of the grounded tensor product to a kind of 'grounded continuous tensor product' operation that sets up an analogy
direct sum : grounded tensor product :: direct integral : grounded continuous tensor product
My hunch is that this should be doable. For one thing, physicists are implicitly using a nonrigorous version of this idea in their daily work on quantum field theory---as you pointed out. For another, it's one of those situations where the final answer you're shooting for has been made rigorous, and you're just looking for a new way of getting there.
However, I am happy enough knowing that countable tensor products of grounded Hilbert spaces work as they should. In the book, we use them to investigate the question of when a linear symplectic transformation of $H$ can be quantized to obtain a unitary operator on $K$.