# What does it actually mean “to define a field” in QFT?

First of all: how does one define one operator in a Hilbert space? This is just a mathematics question and the answer is simple: we have a Hilbert space at hand $\mathcal{H}$, then we define a function $A : D(A)\subset \mathcal{H}\to \mathcal{H}$ that is linear, $D(A)$ being its domain.

Then we must define the function. In other words, we must say how $A$ acts on $D(A)$. This means that we must say what $A|\psi\rangle$ is for each $|\psi\rangle \in D(A)$. Usually we do so establishing a rule in terms of a general $|\psi\rangle$, perhaps using a basis, or sometimes we can do so in an indirect manner.

As for definining a function, this happens in all mathematics: to define a function, we need a set, and then we define the function on some subset of this set. So it is not possible to define a function, unless we know beforehand: (i) the set $\mathcal{H}$, (ii) the domain $D(f)\subset \mathcal{H}$ and (iii) the range $\mathcal{H}'$.

In the case of Hilbert spaces, $\mathcal{H}$ is a known Hilbert space to the problem, $D(f)$ is the domain of the operator and $\mathcal{H}'=\mathcal{H}$. Let us call $\mathcal{L}(\mathcal{H})$ the set of all operators on the Hilbert space $\mathcal{H}$.

This is just mathematics. Now, if we want to define a field of operators on spacetime, what do we need? Well, following this logic (which is the standard math, not anything fancy) we need a function $\phi : M\to \mathcal{L}(\mathcal{H})$. But, wait a minute, to build this function we need to say how it acts. In other words, for each event $x\in M$ we must say what $\phi(x)$ is.

Fine, so how do we say what is $\phi(x)$? It is an operator in $\mathcal{H}$. Hence to define $\phi(x)$ we need to say how it acts on $\mathcal{H}$, in its domain $D(\phi(x))$. In other words, we need to specify for each $x\in M$ what is the action $\phi(x)|\psi\rangle$ for each $|\psi\rangle \in \mathcal{H}$, otherwise we haven't defined $\phi$ at all.

One can argue that quantum fields must be viewed as operator-valued distributions, although I'm still unsure if this is the standard approach, but anyway, the problem remains and it is the same thing. To define the quantum field $\phi(f)$ we must say how it acts $\phi(f)|\psi\rangle$ for each test function $f$ and each $|\psi\rangle\in \mathcal{H}$.

And of course, we need $\mathcal{H}$. Although this is less important since all Hilbert spaces of same dimension are isomorphic. On the other hand, truly defining $\phi$ is vital.

Now comes the question: what do physicists do in QFT? They pick one scalar field $\phi(x)$ and say: "ok, now we just make $\phi(x)$ become operators obeying $[\phi(x),\pi(y)]=i\delta(x-y)$ and we are done". Even more is done! One writes $\phi(x)$ in terms of other operators $a(p)$, which are also not known and relate to the commutation relation. Then you imagine: "fine, the next step is naturally to define these operators" and nothing is done, all operators are left undefined with the commutation relations proposed. How can one have commutation relations and not have the operators?

In quantum mechanics I can even accept that. The state space $\mathcal{E}$ is there by postulate, the observables are there by postulate, and we assume all of them (we can even invoke Stone-von Neumann theorem if we want to be more rigorous). In QFT we have a field, so we need the functional dependency $\phi(x)$, and neither $x\mapsto \phi(x)$ nor $|\psi\rangle \mapsto \phi(x)|\psi\rangle$ are ever defined.

In that sense I'm really confused. What does actually mean for physicists in the context of QFT to define a field? How can one work with operator-valued fields (distributions) when one just says that commutation relations are obeyed without ever defining the fields and operators? What is actually going on here?

• The abstract definition of the ccr and car C*-algebras and of their irreducible representations would probably clarify you many things. If I have the time later I may write a detailed answer – yuggib Mar 31 '17 at 14:31

You don't need to define objects if you assume they exist. I know that sounds silly but that's what axioms do for us - they hand us things without the needs to construct or justify those things, in mathematics and physics alike. Basically, in quantum field theory we assume that we are handed the operator-valued fields acting on some Hilbert space. That's the first two Wightman axioms: States are rays in Hilbert space $H$ and a "field" is a distribution with values in the space of operators on $H$. Nothing more is known in a generic QFT - we cannot describe $H$ explicitly, and the action of the fields is likewise mysterious to us in the general case.

One might say that that is precisely the thing that makes QFT fundamentally more difficult from quantum mechanics with finitely-many degrees of freedom. Thanks to the Stone-von Neumann theorem, just assuming that there exist operators $x,p$ fulfilling the canonical commutation relations on some space allows us to know that this space is unitarily equivalent to $L^2(\mathbb{R}^n)$ and $x$ and $p$ act my multiplication and differentiation, meaning we can view states as wavefunctions, etc. QM in finitely many degrees of freedom is concrete in the sense that we can explicitly write down the space of states and the operators acting on them. But we don't define $x$ and $p$ to act in this way a priori, it is the SvN theorem that grants us the power to do so.

The proper axiomatization of this attitude, that is assuming the existence of disembodied "operators" with commutation relations without any definite space they act on, is to axiomatize quantum theories as the theory of certain linear functionals (states, expectation values) on $C^\ast$-algebras. an abstract $C^\ast$-algebra does not act on anything, and states are merely linear functionals on it that correspond to taking expectation values. This is an "intrinsic" view of quantum mechanics, assuming nothing more than the structure of the operators as an algebra - there's no Hilbert space, no operators acting on anything, nothing, so from this point of view, the question of how to "define" a quantum field just looks silly - you write down the generators of the algebra and their relations, and that's it (modulo defining the Banach structure on it). The contact to the more familiar world of Hilbert spaces is made through the notion of $C^\ast$-representations and in particular the GNS construction.

In QFT, that is to say quantum mechanics with infinitely many degrees of freedom, we lack the power of Stone and von Neumann. There are uncountably many unitarily inequivalent representations of the CCR (this is one aspect of Haag's theorem), so we cannot write down any specific action of the field on some specific space just by assuming the space and the fields exist. So we construct explicitly the free field (which you should view as first defining the $a_p,a_p^\dagger$ as the ladder operators on a Fock space, then putting together the field), and do clever tricks to somehow gain knowledge about the interacting theories out of this, e.g through the LSZ formula, which really is the cornerstone of the canonical formalism of QFT.

Now, if you're worried how physicists "define" the fields associated to certain Lagrangians, then the LSZ formula is the closest you get to an answer - through LSZ, you get all the vacuum expectation values or "Wightman functions", and the point of the Wightman axioms is precisely that they allow the Wightman reconstruction theorem to hold that states that the n-point functions are sufficient to reconstruct the fields. Now, physical theories are unfortunately rarely known to be Wightman theories, but this is the roadmap for how you'd rigorously hope to define quantum fields in the Wightman approach.

In the abstract setting, though, what you need to define is not the fields as functions themselves, but their $C^\ast$-algebra. And given a collection of classical fields and a Lagrangian, that gives you a notion of commutation relations by taking the CCR/CAR between the fields and their canonical momenta, and therefore an algebra. So in the abstract setting there's not that much to define, after all, even in the case of quantum fields.

• Thanks for your very complete answer @ACuriousMind ! It certainly clarifies a lot. But since I'm new to QFT I still have some doubts. For example: in Classical Field Theory, what we seek is the field itself $\phi(x)$ so we don't assume we already have it. That is the objective of solving the field equations after all , like the Maxwell's equations to find $\mathbf{E}$ and $\mathbf{B}$. But now in QFT, we are just assuming the field is already there? I thought in QFT the objective was the same as in CFT: find what is the field and how it evolves with time. Am I missing the point in QFT? – user1620696 Apr 1 '17 at 5:38
• Maxwell's equations already contain E and B. So you don't seek any field and later find them. – Helen Apr 1 '17 at 6:06
• @user1620696 QFT is just quantum mechanics with more degrees of freedom. We're not solving for $\phi(x)$ any more than we are solving for $x$ in quantum mechanics. In both cases, $x$ and $\phi(\vec x)$ are givens - in the Heisenberg picture, these operators evolve, but they are not the primary object of interest, states and their time evolution are, instead. – ACuriousMind Apr 1 '17 at 7:48
• @ACuriousMind: LSZ does not tell you anything about getting interacting fields. It assumes you already have your interacting field, say through its VEVs or Minkowski correlation functions, and then it spits out the S-matrix. That's a different question from what the OP is asking. – Abdelmalek Abdesselam Apr 9 '17 at 20:41