Interpretation of Hilbert space in the Wightman Axioms for QFT

Question

My confusion is about the different Hilbert spaces we meet in QFT.

In a first introduction to QFT, the Hilbert space is often taken to consist of wavefunctionals on classical fields on $\mathbb{R}^3$. In this picture, the state as seen by a given observer contains information about what is going on at all points of space: for example, the wavefunctional might represent a disturbance localised around some faraway point $\mathbf{x}$. Note that the effect on the system of a spatial translation of reference frame is clear: it just shifts the wavefunctional in the obvious way, e.g. a spatial translation by $\mathbf{a}$ will move the disturbance to $\mathbf{x-a}$. So its unitary representation $U(\mathbf{a})$ is simply the map which shifts all arguments by $\mathbf{a}$.

By contrast, in the Wightman Axioms, the Hilbert space is left pretty much arbitrary (bar some technical assumptions). The state as seen by a given observer doesn't look like a bunch of superposed fields: it's just some abstract vector in Hilbert space, and doesn't lend an obvious interpretation like "a disturbance over at $\mathbf{x}$". In this picture, the unitaries $U(\mathbf{a})$ which represent spatial translations are left arbitrary.

The Wightman picture feels more elegant to me, since it assumes less structure. It also puts space and time on a more equal footing, since in the wavefunctional picture the effect of spatial translations is fixed but time translations are arbitrary, whereas in the Wightman picture all spacetime translations are arbitrary. However, the states in the Wightman picture are completely "bare", without the nice interpretation that wavefunctionals have. Moreover, as far as I can tell, in practice the Hilbert spaces are taken to be Fock spaces, which are closer to the wavefunctional picture (they admit a nice interpretation in terms of particles at different locations in space).

So which of these pictures is "correct"? Should I stop thinking about wavefunctionals and just accept the abstract Hilbert space of Wightman? Does this abstract space really give us enough structure to do QFT? Does this all matter in practice?

Apologies if this is a bit vague - I'll be grateful for any wisdom on the topic even if it doesn't directly answer my questions.

Answer 1

"Should I stop thinking about wavefunctionals and just accept the abstract Hilbert space of Wightman?".

Yes.

The wave-function interpretation is only valid in theories with a mass gap, and only in the weakly-interacting regime. Most QFTs are not of this form, so most QFTs must be understood in the abstract sense.

Unfortunately, there isn't much more to say here. The old-school approach to QFT is very limited, it was introduced when people didn't know what a QFT should be in the first place. It is very outdated, it is not useful in general. It is best if we, as a community, move on from such viewpoint.

All QFTs carry a Hilbert space, almost by definition. In most QFTs, this Hilbert space cannot be understood as the space of wavefunctions representing a physical disturbance at $\boldsymbol x$. For example, gauge theories have fields that cannot be localised (as gauge transformations may move the "disturbance" somewhere else), and therefore the very construction of the Hilbert space is surprisingly elaborate (especially in the case where the whole machinery of Batalin-Vilkovisky is required). Conformal theories, or topological theories, are other examples where the old-school interpretation of the Hilbert space is unfit.

So yes, it is best if you forget the textbook definition of the Hilbert space of a QFT (or any quantum theory, for that matter) and accept the abstract definition. A quantum theory assigns a vector space (with positive inner product) to whatever data defines the theory (e.g., coupling constants). This vector space is in principle an abstract space: no interpretation is built-in in the definition of the QFT. Only in some restricted cases can a physical interpretation be given. These are the exception rather than the rule.

Answer 2

Wightman axioms are simply axioms. They state that there is a Hilbert space, but they do not specify what it is, because specifying this would make them not axioms. In a concrete theory with concrete fields, you might be able to define a "Hilbert space of wave functionals on fields", and you might be able to prove that it satisfies Wightman axioms.

We want axioms to be abstract, because we want them to be general. If you said something about "wave functionals" in an axiomatic framework, you would exclude things like most interacting conformal field theories, where there is no obvious (definitely not unique) set of fields on which you can define the functionals.

So no picture is "the correct one", they are just different things. Wightman axioms are like a coloring book, and "Hilbert space of wave functionals on fields" is a way of coloring it.

Answer 3

For variety, let me offer a dissenting opinion.

First, "in practice the Hilbert spaces are taken to be Fock spaces" is incorrect. If the QFT is not free, then the Fock space representation is lost.

For QFTs which are Nelson-Symanzik (NS) positive, the wave functional representation should still hold. Let $S_n(t_1,x_1;\ldots;t_n,x_n)$ denote the Euclidean correlations on $\mathbb{R}^{1+d}$ for say a scalar field. NS positivity is the property $$ \sum_{m,n\ge 0}\int_{(\mathbb{R}^{1+d})^{m+n}} S_n(t_1,x_1;\ldots;t_{m+n},x_{m+n}) \overline{f_m(t_1,x_1;\ldots;t_n,x_n)}f_n(t_{m+1},x_{m+1};\ldots;t_{m+n},x_{m+n}) \ge 0 $$ for all collection of nice decaying functions $f_0,f_1,\ldots$ of an increasing number of Euclidean spacetime points.

If this holds then there should be a probability measure $\mu$ on classical fields $\phi(t,x)$ in $\mathscr{S}'(\mathbb{R}^{1+d})$ such that $$ S_n(t_1,x_1;\ldots;t_n,x_n)=\int d\mu(\phi)\ \phi(t_1,x_1)\cdots\phi(t_n,x_n)\ . $$ Suppose you can define a sharp time restriction map $\mathscr{S}'(\mathbb{R}^{1+d})\rightarrow \mathscr{S}'(\mathbb{R}^{d})$ sending the field $\phi(t,x)$ to its time zero slice $\phi(0,x)$. You then get a (push-forward or here marginal) probability measure $\nu$ on $\mathscr{S}'(\mathbb{R}^{d})$ which is the law of the time zero field $\phi(0,x)$ or $\phi_0(x)$. The physical Hilbert space should be $\mathcal{H}=L^2(\mathscr{S}'(\mathbb{R}^{d}),d\nu)$. Its elements are wave functionals $F(\phi_0)$ which are square-integrable with respect to the measure $d\nu$, just like in basic quantum mechanics, i.e., when $d=0$.

For the free field, $d\nu$ is Gaussian and $\mathcal{H}$ has a Fock space representation (called Wiener chaos by probabilists). For theories at the intersection of quantum field theory and statistical field theory, NS positivity should hold. For example CFTs like Ising in 2D and 3D coming from an honest lattice random field, should have that property.

Of course, I am not claiming this probabilistic picture with $L^2$ wave functionals should always hold. I am just disagreeing with the opposite belief that this picture never holds, in particular for CFTs.

Answer 4

Good answers have already been posted, so I'll just add one other piece.

When we take the Hilbert space "to consist of wavefunctionals on classical fields," we're just expressing the Hilbert space in a way that streamlines the construction of some particular set of observables. However we express it, the Hilbert space itself is a pretty featureless thing: it's a vector space with an inner product satisfying certain conditions. It's still the same Hilbert space that we use in single-particle QM in $1$-dimensional space, or single-particle QM in $27$-dimensional space, or the Ising model, or lattice Yang-Mills theory, or absolutely any other quantum theory in which the number of mutually orthogonal states is not finite. All of these Hilbert spaces are isomorphic to each other (as Hilbert spaces).$^\dagger$ Different theories are distinguished from each other by their observables, and different-looking ways of constructing that same Hilbert space are used only to facilitate constructing those different observables.

$^\dagger$ ^{I'm assuming that the Hilbert space is separable, which is usually a requirement in quantum theory. In hindsight, the origin of the name "quantum" can be traced to the separability of the Hilbert space (the existence of a countable orthogonal basis).}

So the featureless appearance of the Hilbert space in the Wightman Axioms isn't some strange quirk of the Wightman Axioms. Instead, it is the same featureless appearance that Hilbert space always has whenever we're careful to distinguish between the Hilbert space and the observables. The Wightman Axioms make that distinction more clear.