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.


4 Answers 4


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


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.

  • $\begingroup$ Can I just clarify that when you write "The wave-function interpretation is only valid in theories with a mass gap, and only in the weakly-interacting regime", you are referring to the wave-function$\textit{al}$ interpretation? If so, is it right to say that the unitaries corresponding to spatial translations need not be (unitarily equivalent to) the obvious action of translations on wavefunctionals (where we just translate arguments), but rather can just be any old unitaries? Which representation does nature actually use? $\endgroup$ Jul 24, 2020 at 19:51
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    $\begingroup$ @JacobDrori: Since you seem to have trouble because of two kinds of translation, you should perhaps recall the formulas defining these two operations in your post. Please don't use verbal descriptions which will only increase any existing confusion. $\endgroup$ Jul 24, 2020 at 21:12
  • $\begingroup$ @JacobDrori I'm going to have to echo Abdelmalek and ask you to be more precise about these two notions you are referring to. While I'm not entirely sure what you mean, I guess the answer is going to be: a QFT is said to be Poincare invariant if its Hilbert space realizes some representation of the Poincare group. Which representation depends on the specific theory, there is no general answer. Given such representation, say, $U$, a field is said to be covariant if it satisfies $U(g)\phi U(g)^\dagger=g\cdot\phi$, where $\cdot$ denotes some action of Poincare (that is also case-dependent). $\endgroup$ Jul 25, 2020 at 19:21
  • $\begingroup$ I just wanted to clarify whether "The wave-function interpretation is only valid in theories with a mass gap" should instead read "The wave-function$\textit{al}$ interpretation..." $\endgroup$ Jul 26, 2020 at 13:37

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.

  • $\begingroup$ I suppose my question is then: What is the minimal amount of colouring we need to do in order to get a complete picture of nature? Does the Hilbert space need to have some interpretation in terms of space, with translation operators acting in the expected way, or can we leave it abstract? $\endgroup$ Jul 24, 2020 at 19:20

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.

  • $\begingroup$ Shouldn't we almost never expect the wavefunctional (WF) representation to hold? In the WF rep, the spatial translation operators $U(\mathbf{a})$ act in the obvious way: if $\Psi$ is a WF and $T_\mathbf{a}$ translates classical fields by $\mathbf{a}$, then $U(\mathbf{a})$ sends $\Psi\mapsto \Psi\circ T_\mathbf{a}$. So if we let $\mathcal{B}$ be the basis of WFs which are supported on a single classical field, then each $U(\mathbf{a})$ sends $\mathcal{B}$ to itself, i.e. it just permutes a basis. But this is a very non-generic feature for a rep to have. So does nature actually use this rep? $\endgroup$ Jul 24, 2020 at 19:42
  • $\begingroup$ Hard to understand the comment: is "rep" the representation in terms of WFs or is it the group representation $U$? Also, there is no such thing as a basis $\mathcal{B}$. $\endgroup$ Jul 24, 2020 at 21:05
  • $\begingroup$ By "the WF rep" I mean the group representation $U$ which maps a translation $\mathbf{a}$ to a unitary map $U(\mathbf{a})$ on the space of WFs. And perhaps $\mathcal{B}$ fails to be a basis for analytic reasons - this is not the point I'm trying to make. My point is that the WF rep is extremely non-generic, and I want to know whether this is the rep that actually occurs in nature. Sorry for any confusion. $\endgroup$ Jul 24, 2020 at 21:21

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.

  • $\begingroup$ The various Hilbert spaces you mentioned are indeed isomorphic as Hilbert spaces, but I'm more worried about representations. The representation of the QFT translation operators $U(a)$ is, in general, not equivalent to the "obvious" representation on wavefunctionals, where spatial translations act in the natural way by translating arguments. Does nature use the wavefunctional representation, or should I just forget all about it? $\endgroup$ Jul 24, 2020 at 19:18
  • $\begingroup$ @JacobDrori I can't answer the metaphysical question about what nature uses, but if you're really asking for a general perspective that works well in all quantum theories, then this one does: Forget about Hilbert space entirely until after the theory's observables have been specified as an abstract algebra. Then construct the Hilbert space from the algebra of observables (or the algebra of field operators, which is typically bigger). The QFT translation operators act in the natural way by translating arguments of the operators. The Hilbert space is constructed from them, so there you go. $\endgroup$ Jul 24, 2020 at 20:52
  • $\begingroup$ @JacobDrori The observables-first perspective also exposes the fact that there can be unitarily inequivalent Hilbert-space representations of the algebra of observables (or field operators), even though the Hilbert spaces are isomorphic, and even though the algebra is the same. This is what the positive-energy requirement is about (which underlies the spin-statistics theorem): it selects a specific representation. By the way, the path-integral formulation implicitly selects that special positive-energy representation, at least if we start in "Euclidean" signature and then Wick-rotate. $\endgroup$ Jul 24, 2020 at 20:55
  • $\begingroup$ @JacobDrori For example, consider the QFT of the free Dirac spinor field. To construct a Hilbert-space rep'n from the algebra of field operators, we assert that one state is annihilated by some set of field operators, and then we construct the rest of the Hilbert space by applying other field operators to that one state. We can get unitarily inequivalent rep'ns by choosing different sets of field operators to annihilate the one state. Most of them are physically "bad" because the energy has no lower bound. One of them is a "good" rep'n. This is what the "filling the Dirac sea" poetry is about. $\endgroup$ Jul 24, 2020 at 21:08
  • $\begingroup$ Thank you for the in-depth reply. Suppose we represent our algebra of observables by letting them act on a space of wavefunctionals. We both agree that "The QFT translation operators act in the natural way by translating arguments of the operators". Is it right to say that, in general, the QFT translation unitaries will $\textit{not}$ act on wavefunctionals in the natural way (i.e. by translation of arguments)? $\endgroup$ Jul 24, 2020 at 22:04

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