Making precise the statement "particles are excitations in a quantum field"

I've been trying to self teach QFT lately. I find that the basic physical idea makes sense, and I can keep up with the mathematical formalism without too much trouble, but I'm having trouble connecting the two.

It is usually said that QFT is necessary because particles can appear and disappear, and so rather than treat a particle as a basic entity as in non-relativistic QM, we instead have a field, and particles are excitations or quanta in that field. In the canonical formalism, the field is a field of operators which creates and destroys particles. For example, if we have a real scalar field $\phi$, then we know that its Fourier components $a_\mathbf{p}$ and $a_\mathbf{p}^\dagger$ act on the vacuum as $a_\mathbf{p}^\dagger |0\rangle = |\mathbf{p}\rangle$. Also, if in the Schrödinger picture $\phi$ is independent of time, that is, $\phi=\phi(\mathbf{x})$, then in the Heisenberg picture we can get time dependence as $\phi(x) = e^{-iHt}\phi(\mathbf{x})e^{iHt} = e^{-ipx}\phi(0)e^{ipx}$.

Looking at the last expression, I don't see how the particles come into play. It seems to me that the current state of the world is "encoded" in the kets, so for example if we have two particles with momentum $\mathbf{p_1}$ and $\mathbf{p_2}$, the state is $|\mathbf{p_1}\ \mathbf{p_2}\rangle$. But it seems that the time evolution of the field depends only on the Hamiltonian and not on (for example) the number of particles. Is this true? If it is, then how does that statement "particles are excitations in a quantum field" translate into the formalism?

Perhaps more simply: is the field any different if we have one particle or if we have two particles?

Answer

The talk about "particles" being "excitations" is handwavy, and comes about precisely because of the analogy of the usage of Fourier components that you talk about.¹

Perhaps more simply, is the field any different if we have one particle or if we have two particles?

"The field" is inaccessible. Some fields aren't even observables (because they may not be (gauge) invariant in and of themselves), and even of these that are, you can only measure the expectation values w.r.t. some states. The statement "particles are excitations" is perhaps most "precise" in the LSZ formalism, where you take some definite-momentum particle-in/out states and obtain that you can calculate

$$\langle p_1,\dots,p_m \vert q_1,\dots,q_n \rangle$$

by looking at some integral whose integrand has $\langle \Omega \rvert T \phi(x_1)\dots\phi(x_m)\phi(y_1)\dots\phi(y_n)\lvert \Omega \rangle$ as a factor, so you could say that this "looks like the field creates the particles".

In QFT, you have to abandon the notion that there is a real value for the field at every point - that would be the classical picture of the field just being the solution of the classical equation of motion. All you can have are expectation values, and the idea that there is some kind of field actually being "excited" is...ill-defined. It isn't clear at all what one would want to say with that. It's just a cute picture that many like, not a mathematically rigorous statement.

Of course, the time evolution will act differently on many-particle states than it will on one-particle states. But these states are states in the Hilbert space of the theory, and only by analogy (since they are created by Fourier components of free fields) identified with "excitations", and analogy that breaks down if we would not consider the asymptotic Fock space but instead the full (but sady, mostly unknown) Hilbert space of the interacting QFT.

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^{1I wrote about this some time ago. The analogy is not very helpful in my view. The mathematics are of the same form as that for an oscillator, but to say that particles are excitations is endowing the concept of field with an ontological weight that is unnecessary, and just tends to confuse people about non-perturbative, interacting QFTs, where it is not helpful at all to think in such terms.}