# What do the quantum fields represent, mathematically?

I am looking for insight on quantum field theory, and more precisely, I am interested in having a low-detailed idea of what a quantum field theory is about; moreover, I should say hat I am a mathematician with little physical background.

I found this question: What are quantum fields mathematically? but am not fully satisfied with the answers.

I expect that any (mathematical formulation of a) quantum theory should describe three things: what the states are (that is, what mathematical objects represent "all the knowledge of a physical object"), what the observables are (that is, what mathematical objects represent the quantities or properties obtainable through experiments on physical objects). Moreover, it should also describe what symmetries are, but this is not my question today.

According to the answer linked above, I think that quantum fields, i.e. operator-valued distributions on the spacetime manifold, are supposed to be the objects representing an observable, in comparison to quantum theory where observable are represented by self-adjoint operators.

So what are the states? Are they unit vectors (or more accurately, rays) in the Hilbert space on which the operators act?

But I am now puzzled: in quantum theory, the wavefunctions representing the quantum states have a "local content" (since they are maps defined on a space), whereas observables do not. In this case, the observables have a "local content" (since they are "generalized" maps defined on a space) whereas states do not.

Am I thinking of everything the wrong way?

• Regarding your own question about "local content," an valuable exercise would be deriving the non-relativistic quantum mechanics of a single spinless particle from the non-relativistic QFT of a single scalar field. Then you can see how the two models' observables are related to each other, how their states are related to each other, and how their different-looking ways of encoding "local content" are related to each other. Jul 2, 2020 at 1:23
• By the way, the answers to the other question that you linked consider only quantum fields in continuous spacetime, even though the question did not impose any such constraint. The answer is much more straightforward if we allow spacetime (or even just space) to be a finite but extremely-large and extremely-fine lattice, much larger and finer than we could ever resolve experimentally -- a reasonable compromise for QFTs that aren't meant to be theories of everything anyway. If that's of interest, you could search for info about lattice QFT. Jul 2, 2020 at 1:24
• Also, if you're interested in further exploration, if you do a search for "Quantum Field Theory for Mathematicians", you'll find a number of books with some variation of that title, as well as some freely available lecture notes, and other resources like this series of lectures hosted on Perimeter Institute's website (click on the Videos tab; the order is jumbled up, but it seems they are all there). Jul 2, 2020 at 10:14

A little caveat: The quantum fields are a way of organizing the generators of the algebra of observables. They might not be observables themselves. The actual observables are things like $$\phi(f)$$, rather than $$\phi$$.

Regarding your main question: You are correct. States are unit vectors in the Hilbert space upon which the observables act. They are not local in space. Indeed, they encode global information, like total charge and entanglements between widely separated things. Consider the vacuum state: It knows that there's nothing anywhere.

So how to reconcile this picture with quantum mechanics? The answer is that in quantum mechanics, the spacetime is simply the 1-dimensional time manifold. As far as the formalism is concerned, the observables aren't parametrized by a space-manifold. (Think about the quantum mechanics of an electron's internal spin. The spin observable is just an up or down; it doesn't refer at all to the location in space.)

• The vacuum state does not know that there's nothing everywhere; it knows that it's the lowest possible energy state. That is not the same as having nothing, though it does imply that nothing is present under our current understanding of physics. Provided you can rigorously define "nothing", which I'm not sure I can. Jul 2, 2020 at 13:27

First off, I'd want to point out that most to all the complicated mathematical difficulties that go into "in practice" quantum field theory are really more to do with trying to deal with fields that are interacting with each other, and thus don't really have so much to do with how and what I believe you're asking about, which is to get to know the conceptual core of what a quantum field is. And that is, fortunately, much simpler.

A quantum field is the quantum-mechanical version of a classical field, which is a system in which we assign some quantity - in fundamental physics this might be electric and magnetic field vectors, but in more higher-up applications could be, say, the sound-wave field within a solid medium, the value representing the relative compression of the material - to each point in space. That is, a classical field is just a function

$$\phi(P)$$

of a point $$P$$ in space. In terms of coordinates, in three-dimensional space, we'd write this as a function of three arguments:

$$\phi(x, y, z)$$

if we're using Cartesian coordinates for each point, so that $$P = (x, y, z)$$, for example. The value returned by this function is the value of the field quantity at that particular point, e.g. an electric field of 5 V/m (ignoring the vectorial stuff for simplicity), or an increase in pressure of 5 Pa (again, ignoring more technical complexity).

So what happens in quantum mechanics? Well, in quantum mechanics, just as when we develop a quantum theory of a moving particle, i.e. one with position and velocity, we must convert this quantity into a quantum operator: we don't know which one yet, but we first just declare it. Now the return value $$\phi(P)$$ no longer has type "real", but has some sort of operator type, and thus it gets a hat:

$$\hat{\phi}(x, y, z)$$

and in effect we have a field of operators, one at each point. Each operator then should operate on some quantum vector $$|\psi\rangle$$, representing an agent's knowledge about the entire field, so that it is possible to derive from it a field-value wave function

$$[\psi(x, y, z)](\phi)$$

which gives a probability distribution describing what is known about the field value $$\phi$$ at the point $$(x, y, z)$$. (Note this is a "curried function"; I like these because they cast it into a form that makes it more clean what is going on - we're getting a probability density function (pdf) over field values $$\phi$$ specific to the point $$(x, y, z)$$.).

So that's part 1: $$\hat{\phi}(x, y, z)$$ here is what constitute "observables" for the quantum field. But this doesn't really give us much insight now into the next part of the question, which is how do we build the Hilbert space part. I first want to make a note, though: while we are going to do that, technically it really is only the operators above that count and everything can be done in terms of them, the Hilbert space is simply mathematical fluff to make things easier to work with. So you could say "we are done here", but we can do it nonetheless.

To see how to do it, you should note that, in a way, you can consider the field to be as though it were an ordinary multi particle quantum system with an uncountable collection of separate "particles" (these are NOT the usual particles like electrons, photons, etc. but something more mathematical), each one corresponding to a different space-time point $$(x, y, z)$$ and whose "position" is the field value $$\phi(x, y, z)$$. Hence, just as, say, for 2 particles, you have a two-particle wave function

$$\psi(\mathbf{r}_1, \mathbf{r}_2)$$

with the two position components, here you have a $$\psi$$ that takes uncountably many position components, all indexed by coordinates $$(x, y, z)$$ or, in effect, a wave function that takes in a function which is a classical field configuration $$\phi(x, y, z)$$. Such a "wave function" is thus also called a wave functional, and written

$$\psi[\phi]$$

for that quantum field. Then the action of the field operator $$\hat{\phi}(x, y, z)$$ upon such a $$\psi$$ is given by

$$[\hat{\phi}(x, y, z)\psi][\phi] = \phi(x, y, z) \cdot \psi[\phi]$$

just as

$$[\hat{\mathbf{r}}_1 \psi](\mathbf{r}_1, \mathbf{r}_2) = \mathbf{r}_1 \cdot \psi(\mathbf{r}_1, \mathbf{r}_2)$$

. Hence the Hilbert space is suitably-defined equivalence classes of these wave functionals (the whole "same up to a set of measure zero" stuff), and states are the attendant rays.

In short: a quantum field represents a field quantity that is quantum-mechanically fuzzy.

According to the answer linked above, I think that quantum fields, i.e. operator-valued distributions on the spacetime manifold, are supposed to be the objects representing an observable, in comparison to quantum theory where observable are represented by self-adjoint operators.

This is a misconception. quantum fields in particle physics , they are used also in other disciplines, are a type of "coordinate system" on which interactions can be mapped using Feynman diagrams to get to calculable quantities as crossections and decays that can be checked with measurements. For the particles in the standard model of particle physics , fields are defined using the quantum mechanical plane wave wavefunctions of the corresponding free particle equations, on which differential creation and annihilation operators work . The Feynman diagrams are a representation of the integrals necessary to compute many body interactions.

The fields do not represent any observables, it is the interactions that can predict observable quantities.

But I am now puzzled: in quantum theory, the wavefunctions representing the quantum states have a "local content" (since they are maps defined on a space),

The wavefunctions, either simple or using QFT, are not measurable. The $$Ψ^*Ψ$$ entering the Feynman diagrams are the measurable quantities.

whereas observables do not.

The only observables predicted by quantum mechanics, either simple or as a QFT , are connected with probability distributions. Quantum mechanics predicts probabilities through the mathematics of the wavefunctions.

All QFTs obey the postulates of quantum mechanics.