What do field operators in QFT act on? I have been self-studying physics and QFT for a quite a while now and there are a couple of basic ideas of QFT that I just can't find the answer to no matter how hard I try. I know this might sound silly to some of you but I'm very eager to understand QFT and I have no opportunity to ask teachers and I feel stuck and would greatly appreciate any help. So my question is this:
What do field operators do?
I have been absolutely unable to find any straightforward answers to this. The only available answers I keep stumbling on (and I take pains to find them) are so heavily technical that I find them hopeless to understand with my humble mathematical background.
I understand that in second quantization we treat each mode of a field as a quantum harmonic oscillator and that we define field operators as integrals over creation and annihilation operators for each mode and we define commutation relations between them to make our theory quantum. I have also read at some places (and this is again extremely obscure in most sources - if they even care to mention it) that the wavefunction in QFT is not a function on spacetime coordinates like in regular QM but a functional on a space of possible field configurations that doesn't have time dependence (because we are in the Heisenberg picture).
What I don't understand is then what is it exactly that these operators operate on? Do they operate on the wavefunction like operators in regular QM? If, so do they have a set of eigenstates with associated eigenvalues that tell us something about some physical observable related to the field? I saw a video in which they seemed to show just this: that a field operator applied to the wavefunction showed the expectation value of the field at that point. But I also read several times that an operator creates a particle at position x (?!).
I would be more than thankful if someone could explain it to me in a non-fancy way so I can just get an idea of what the operators even do and what is the something they act on to start with and thus have a chance to study the technicalities afterwards.
 A: I agree it can be very confusing to appreciate what's going on so hopefully this helps and sync's up with the linked discussions in the comments:
There are (at least) two ways to introduce 'quantum fields' which illustrate what they do. One way is to start with a classical system of $N$ coordinates $q_i(t)$ and $N$ momenta $p_i(t)$ satisfying the Heisenberg equations, which in terms of Poisson brackets read as
$$
\frac{d q_i}{dt} = \{H,q_i\} $$
$$
\frac{dp_i}{dt} = \{H,p_i\} $$
where e.g. $\{p_i,q_j\} = \delta_{ij}$, and then promote the coordinates and momenta to operators
$$
q_i(t) \ \ \to \ \ \hat{q}_i(t) \ \ , \ \ p_i(t) \ \ \to \ \ \hat{p}_i(t)$$
and Poisson brackets to commutators, e.g.
$$
\{p_i,q_j\} = \delta_{ij} \ \ \to \ \ [\hat{p}_i,\hat{q}_j] = - i \delta_{ij}.$$
We then send
$$
N \to \infty.$$
Before going any further, one should note that we are setting up the transition from a discrete to a continuous system, and this is familiar when setting up the classical dynamics of a continuous string starting from a model of a finite chain of springs (ref [1]), note this is modelled as a finite chain of identical particles. In this model one recalls that the coordinates $q_i(t)$ end up being replaced by a field $\varphi(x,t)$ i.e.
$$
q_i(t) \ \ \to \ \ q_{x}(t) = \varphi(x,t)$$
giving the amplitude from it's rest position at $(x,t)$, so by analogy we expect that the classical degrees of freedom $q_i(t)$ that we are quantizing with $i$ going to $\infty$ should end up as operators depending on space and time (if the operators $\hat{q}(t)$ are Heisenberg picture operators, or just space if they are Schrodinger operators $\hat{q}$, let's leave time in here). Thus, modulo a subtlety I will come back to, the transition from classical mechanics of a finite set of positional degrees of freedom to a continuum of quantum 'position' degrees of freedom is
$$
q_i(t) \ \ \to \ \ \hat{q}_{\mathbf{x}}(t) = \hat{\varphi}(\mathbf{x},t)$$
For example, this (in the Schrodinger picture) is roughly how quantum fields are introduced in [3]. Just as your classical mechanics Lagrangian's and Hamiltonian's like $L = \frac{1}{2}m \dot{q}_i^2 - V(q_i)$ involve positions and coordinates, so too does e.g. the free Klein-Gordon Lagrangian density
$$
\mathcal{L} = \frac{1}{2} \partial_{\mu} \hat{\varphi} \partial^{\mu} \hat{\varphi} - \frac{1}{2} m \hat{\varphi}^2.$$
So a quantum field operator is (modulo a subtlety) like a position operator at each point of space/space-time, at this stage of setting things up I could have used completely different notation and then you'd probably never confuse this with the familiar Schrodinger picture Schrodinger wave function in
$$
i \frac{\partial \psi}{\partial t} = \hat{H} \psi$$
It might be better to think of the Dirac Lagrangian
$$
\mathcal{L} = \hat{\overline{\psi}} (i \gamma^{\mu} \partial_{\mu} - m) \hat{\psi}$$
since this can directly be written in the form of a Schrodinger equation, but now with $\hat{\psi}$ an operator
$$
i \frac{\partial \hat{\psi}}{\partial t} = \hat{H} \hat{\psi}$$
In other words, it's like we took the regular Schrodinger equation and replaced $\psi$ with $\hat{q}_k$
$$
i \frac{\partial \hat{q}_k}{\partial t} =^? \hat{H} \hat{q}_k$$
but now we have to have $k = (\mathbf{x},t)$ i.e. the Schrodinger operator act on the labels that the 'position operator' depends on, but one even could say the same thing about the wave equation one derives in the continuum model of a string so it's not that bizarre. Maybe the bizarre thing is how this is natural even starting from the Schrodinger picture Schrodinger equation as we'll see below.
Before going further I will note most of this is the setup in ref [2]. Also, I will note that a common way of introducing the above is via the example of setting up the electromagnetic field Lagrangian/Hamiltonian (e.g. Schwartz ch. 2 or [5]) and reducing it to the analysis of Harmonic Oscillators, thus suggesting promoting the classical modes to quantum creation and annihilation operators.
This makes us face up to a subtlety. Above I said a quantum field was basically just a position operator but now with a continuous label, $\hat{q}_{\mathbf{x}}(t) = \hat{\varphi}(\mathbf{x},t)$, and the electromagnetic field example gives things which act like position and momentum operators in the sense that they combine like position and momentum operators do in the Harmonic Oscillator  example when we define creation and annihilation operators. But the quantities in the electromagnetic case definitely aren't position and momentum operators, they are just 'conjugate variables' (sometimes, such as in [5], one literally uses the position/momentum notation in the electromagnetic field case due to this).
Recall in the classical mechanics Hamiltonian formalism that the very meaning of what is a position variable and what is a momentum variable is actually completely arbitrary, all we need are canonically conjugate variables, we can even do a canonical transformation and interchange what we call position and what we call momentum...
So a quantum field is a quantity which classically behaves like one of the 'coordinates' in a canonically conjugate pair of degree of freedom which then gets quantized by promoting Poisson brackets to commutators. For all intents and purposes it behaves similarly to how a position operator acts, which is is why it's introduced as above e.g. in [2] or [3].
So, now that you see all this formalism arose by promoting position $q_i$ to a operator $\hat{q}_i$ and sending $N \to \infty$ so that we get $\hat{q}_{\mathbf{x}}(t) = \hat{\varphi}(\mathbf{x},t)$, or more generally thinking in terms of a canonically conjugate pair of degrees of freedom, it hopefully makes sense that quantum fields act on state vectors the same way that a position operator $\hat{q}$ in quantum mechanics also acts on state vectors.
One can ask (as your post does), why didn't I just say $\hat{\varphi}$ acts on a Schrodinger wave function $\psi(\mathbf{x},t)$ the way that $\hat{q}$ acts on a Schrodinger wave function $\psi(\mathbf{x},t)$ as usual in introductory quantum mechanics, indeed $\psi$ is sometimes loosely called as a state vector.
The 'cheat' answer is that in Harmonic oscillator example we apply creation and annihilation operators to state vectors that look roughly like $|n> = \hat{a}^{\dagger}|0>$ so that's why we do it for the $\hat{\varphi}$'s (recall $\hat{q} \approx (\hat{a} + \hat{a}^{\dagger})$ in the Harmonic oscillator case), but one can ask why can't we also blindly just apply it to $\psi(\mathbf{x},t)$, when we examine the Schrodinger wave function below we'll see why the above is the natural thing to do.
We can then as usual solve the free Lagrangian or Hamiltonian equations of motion related to a given model using e.g. Fourier methods (e.g. sec. 2.3 and 2.4 of [3]) and we get expansions like
$$
\hat{\varphi}(t,\mathbf{x}) = \int \frac{d^3 \mathbf{p}}{(2 \pi)^3 \sqrt{2E_{\mathbf{p}}}} (\hat{a}_{\mathbf{p}} e^{-ipx} + \hat{a}_{\mathbf{p}}^{\dagger} e^{+ipx})|_{p_0 = E_{\mathbf{p}}}$$
where the time evolution is always contained in these scalars $e^{\pm ipx}$. In other words, here we just casually note for all intents and purposes if we didn't plug an operator wave function into e.g. Klein-Gordon and instead plugged in scalar wave functions, our solutions would look the exact same just without the hat's. Indeed note with these mode expansions that things like the Hamiltonian reduce to sums like
$$
\hat{H} = \int \frac{d^3 \mathbf{p}}{(2 \pi)^3} \omega_{\mathbf{p}} \hat{a}^{\dagger}_{\mathbf{p}} \hat{a}_{\mathbf{p}} (+ \text{zero point energy})$$
So you should at this stage ask yourself, what is actually going on with the regular Schrodinger equation through all of this, i.e. what is the Schrodinger picture setup of this model. Well, the whole time the Schrodinger picture situation is that we are starting from a multi-particle system of $N$ identical particles and then sending $N \to \infty$ in
$$
i \frac{\partial }{\partial t}\psi(\mathbf{x}_1,...,\mathbf{x}_N,t) = \hat{H}(\mathbf{x}_1,...,\mathbf{x}_N) \psi(\mathbf{x}_1,...,\mathbf{x}_N,t).$$
But the second we realize we are working with a multi-particle system of identical particles, we immediately go back to the very heart of quantum mechanics [4] (ch. 1), the Heisenberg uncertainty principle, which tells us that working in this position-space wave functions $\psi(\mathbf{x}_1,...,\mathbf{x}_N,t)$ with identical particles is beyond redundant, since the particles are identical what matters is the number of particles in each of the allowable stationary states. Indeed this leads to the (non-relativistic) boson/fermion classification of identical particle wave functions. This is the philosophy behind going to the 'occupation number formalism', or 'second quantization' picture (the term makes sense, but not yet). This is clearly set up from first principles in ref [4].
So, we see there are multiple reasons why the naive Schrodinger picture needs to be modified in QFT, first there is the issue of working with identical particles, something we can even deal with in non-relativistic quantum mechanics, and second that we are sending $N \to \infty$. On top of this, there is the new issue in QFT of dealing with relativity.
If we are happy to just admit that we should work with these occupation numbers from the beginning, and more or less just ignore the above Schrodinger equation but still work as if we're starting from the Schrodinger picture, then ref [4] sets it up nicely. It turns out that this process results in the same quantum field operators that were independently introduced above.
It's useful, however, to blindly start with the above equation and try to end up with this formalism. This is done for example in reference [6]. First assume everything is non-relativistic for simplicity, and then we'll see where things have to change.
If we assume that each of the identical particle stationary states are for example symbolically labelled by energy levels $(1,2,...,\infty)$ and a given energy eigenvalue $E_k$ is one of these allowable values, then the single particle stationary state wave functions are denoted $\psi_{E_k}$ and we can expand $\psi(\mathbf{x}_1,...,\mathbf{x}_N,t)$ as
$$
\psi(\mathbf{x}_1,...,\mathbf{x}_N,t) = \sum_{E_1 .. E_N} C(E_1,...,E_N,t) \psi_{E_1}(\mathbf{x}_1) .. \psi_{E_N}(\mathbf{x}_n)$$
Thus we see that this 'Fourier space' wave function $C(E_1,...,E_N,t)$ more accurately describes the system in the sense that it at least depends on the stationary state energy levels, rather than the space/space-time coordinates and the inherent redundancy (see also [4] for more on this). But it's still not enough, one would like to re-express this in terms of the number of particles in each of the energy levels $(1,2,...)$, e.g. we would like to re-express $C(E_1,...,E_N,t)$ using an infinite set of variables $(n_1,n_2,...)$ where $n_1$ is the number of particles in the stationary state symbolically denoted $1$, etc... Without continuing in detail, which ref [6] does, one then just re-expresses the above Schrodinger equation as a Schrodinger equation for $C(E_1,...,E_N,t)$ (you can see one just plugs in the $\sum_{E_1 .. E_N}$ expression into the Schrodinger equation and then removes the $\psi_{E_k}$'s) and then, e.g. for a boson wave function notes we can use the wave function symmetry to re-arrange the wave functions as, to make up some random example
$$
C(1,2,1,3,2,..,26,33,t) = C(11..;22..;3;44..;..,t) = \tilde{C}(n_1;n_2;1;n_4,..,t)$$
so that e.g. there are $n_1$ particles with energy level $1$, $n_2$ particles with energy level $2$, one particle with energy level $3$ (in this example), etc... and so instead work with (a suitable normalized version of) a new function $\tilde{C}(n_1;n_2;1;n_4,...,t)$ which depends on an infinite number of variables since there are an infinite number of stationary states, even though there are only a finite number of particle (recall $C$ depends on a finite number of variables $N$).
Once you agree that we should care only about the number of particles in a given stationary state, we see that the time evolution of a system where things change amounts to the number of particles in a given stationary state changing with time, in other words, we create or annihilate particles in a each of the stationary states at each time. So, without knowing anything, we are unavoidably led to defining operators called creation and annihilation operators which create or annihilate a particle in a given stationary state, and so it's completely natural to think even of the state of a given system with a given number of particles in some given stationary states in terms of these creation operators acting on a 'vacuum' $|0>$.
Thus if you go through the discussion in reference [6] (for example, equation 1.24 and 1.25 which are too big to type) you will see exactly how even if you never heard of a creation operator or quantum field, you would end up defining them yourself - you can see that even a non-relativistic Hamiltonian can be expressed in terms of these creation and annihilation operators, reducing to familiar expressions like
$$
\hat{H} = \sum_i \omega_i \hat{a}^{\dagger}_{i} \hat{a}_{i} $$
which is similar to the above expansion. Note I've assumed a discrete spectrum in all this, and my number of particles $N$ is still fixed, so if you send $N \to \infty$ the labels become $\mathbf{x}$ and $\mathbf{p}$ etc... and it almost looks exactly like the previous discussion.
It's not hard to see from e.g. $\omega_i = \sum_k \int \psi_k^* \hat{H} \psi_i d^3 \mathbf{x}$ that inserting this into this last Hamiltonian makes it natural to define the combinations
$$
\hat{\psi}(\mathbf{x},t) = \sum_i \hat{a}_i \psi_i(\mathbf{x},t)$$
$$
\hat{\psi}^{\dagger}(\mathbf{x},t) = \sum_i \hat{a}_i^{\dagger} \psi_i^*(\mathbf{x},t)$$
which now very explicitly can be seen to involve the ability of creating or annihilating a particle at a given position $(\mathbf{x},t)$ at time $t$, it depends on the state that they act on, exactly like a position operator does (it also really depends on the state it acts on), i.e. syncing up with the previous formalism above, so that the Hamiltonian reads as
$$
\hat{H} = \int d^3 \mathbf{x} \hat{\psi}^{\dagger} \hat{H} \hat{\psi}$$
i.e. the mysterious 'quantum field operator' has again fallen out of our non-relativistic formalism. Note it's literally just a linear combination of stationary state solutions to the Schrodinger equation, but now with time-independent operator coefficients, so of course it still satisfies the non-relativistic Schrodinger equation. So we're again lead to this operator Schrodinger equation
$$
i \frac{\partial \hat{\psi}}{\partial t} = \hat{H} \hat{\psi}$$
Note the Hamiltonian here has the same kind of form that e.g. the Dirac Hamiltonian (which I didn't write explicitly above), an integral of a quantity built out of quantum field operators.
So a big chunk of the formalism you see in a QFT book, can equivalently be applied to non-relativistic quantum field theory, equivalent to the familiar Schrodinger picture Schrodinger equation formalism, but it can be introduced directly via the first method thus skipping over all these intermediate steps rationalizing why it's very natural to reformulate even the usual Schrodinger equation picture in this manner due to working with identical particles.
There is still this extra behemoth of relativity. Based on everything written above, it looks like the only difference is the choice of Hamiltonian you use, and in a sense that's all it is, almost, but even this has big consequences, and it relates to this point you made about the very meaning of a wave function.
In non-relativistic quantum mechanics, the choice of Hamiltonian is fixed by Galilean symmetry. In relativistic quantum mechanics, it is fixed by Lorentz invariance. Thus the fields involved in building up a Hamiltonian that describes identical particles should transform under Lorentz transformations in such a way that the overall theory is Lorentz covariant, i.e. the fields must be representations of the Lorentz group. It turns out that the Hamiltonian describing free electromagnetism can be related to vector representations of the Lorentz group, the Dirac equation (which was historically derived by demanding linear Lorentz covariant equation, thus encoding representation theory from it's inception, that reproduces the mass-shell Klein-Gordon condition) can be expressed using a Hamiltonian built up using spin $1/2$ representations of the Lorentz group (note again that linearity assumption), etc...
The way this is usually done is to build up Lorentz invariant Lagrangian's from fields which are representations of the Lorentz group. From this perspective we just note that the free particles that the fields act on can be labelled by their momentum and spin, something we think of as an experimental fact by thinking of the non-relativistic case (e.g. the hydrogen atom). These particle labels, and even the above field representation theory, can be understood as arising in a uniform manner from the representation theory of the Poincare group (which contains the Lorentz group), which is sometimes taken as the absolute starting point of QFT (e.g. [8]).
So you can ask, why doesn't everything in relativistic QFT have the same interpretation as that of the non-relativistic case? Consider only free particles. In the non-relativistic and relativistic case, one can define the notion of a number operator $\hat{N}$. In the non-relativistic case free particle Hamiltonian $\hat{H}$ commutes with the number operator $\hat{N}$. In the relativistic case free particle Hamiltonian $\hat{H}$ does not commute with the number operator $\hat{N}$. The reason traces back to the fact that special relativity allows for those negative energy solutions, which means more terms get added to the non-relativistic number operator thus preventing $\hat{N}$ from commuting with $\hat{H}$. This issue is sketched for example in [7].
This means that any measurement process involving measuring the position of any free particle will unavoidably lead to the creation and annihilation of particles which cannot be detected by the measurement process, making meaningless the position measurement process itself. Indeed this nearly destroyed the whole subject of QFT as it was being set up, and one should ask why anything at all is measurable in QFT, and if so what is it that can be measured. Even worse, what does this imply in terms of describing an interacting system, if we can't even measure some things about free particles.
The transition from non-relativistic to relativistic mechanics involves a fundamental shift in the formalism of classical mechanics, yet here we haven't seen that fundamental shift in the foundations, we've only really seen consequences of quantities like the Hamiltonian and number operator etc... changing in the relativistic case, but even in classical memchanics that happens too in that we replace $S = \int \frac{1}{2} m v^2 dt$ by $S = - mc \int ds$, but special relativity is deeper than this simple replacement, so on this alone one can expect more than this.
If, as in [4] (ch. 1), the Heisenberg Uncertainty Principle is the very core of quantum mechanics, and as seen above things get super complicated in transitioning from one particle to a system of identical particles once we invoke the Uncertainty Principle, you can expect that combining special relativity with the uncertainty principle is the key to the weirdness of (the) relativistic quantum mechanics (of systems of identical particles, which is unavoidable due to the number operator issue mentioned above). I'll just recommend one read [5] to find out what this is.
So the final point to make relevant to your post is the Schrodinger functional picture. You see we had to send $N \to \infty$ in the naive Schrodinger equation, and we ended up reformulating it in a way that bypasses all the coordinate space identical particle redundancy by going to the occupation number formalism and working with (e.g. in the non-relativistic case) an operator analogue $\hat{\psi} = \sum_n \hat{a}_n \psi_n$ of the stationary state mode expansion $\psi = \sum_n a_n \psi_n$ of a single particle wave function (this is obviously why it's called 'second quantization', if we apply all this formalism to a single particle it's literally a second way of working with an expansion in terms of the $\psi_n$'s that describes a particle and gets plugged into the Schrodinger equation). So the question is, why can't we just set up a Schrodinger equation using this new Hamiltonian, where e.g. the momentum in this picture acts like a derivative operator the same way it does in the first quantized picture etc... Obviously you're now differentiating with respect to (the eigenvalues of, on suitable eigenvectors) field variables rather than the underlying coordinates, so it becomes a functional formalism, see ref [9]. So hopefully you can see there's a deeper reason why we use functionals.
References:

*

*Goldstein - Classical Mechanics, Ch. 13.

*Bjorken, Drell - Relativistic Quantum Fields, Ch. 11.

*Peskin, Schroeder - Introduction to Quantum Field Theory, Ch 2.

*Landau, Lifshitz - Quantum Mechanics, Ch. IX.

*Berestetskii, Lifshitz, Pitaevskii - Quantum Electrodynamics, Ch. I.

*Fetter, Walecka - Quantum Theory of Many-Particle Systems, Ch. 1.

*Srednicki - Quantum Field Theory, ch. 1.

*Weinberg, Quantum Theory of Fields, Ch. 2.

*Hatfield - Quantum Field Theory of Point Particles and Strings, Ch. 10.

