What happens when one "observes' a quantum field, and how do particles get involved?

I've recently begun my journey to understand QFT. I apologize in advance for the length of the post, but there are gaps in my understanding of how I, as an experiementalist, interact with fields to recover observables.

1) In RQM, the probability of observing some value in a multi-particle system is the square of the projection of the system's wave function onto the relevant observable's eigenvector corresponding to that value.

In any of the introductory books I've read, I haven't seen the appropriate analogy for QFT.

Analogously, for example, is it the case that, given a field (replacing the QM particle by analogy), if I'm interested in observing the field's momentum or spin, I can project the field onto the relevant Hermitian operator's eigenbasis, and the square of the resulting coefficient is the probability that I'll observe the given eigenvalue? Moreover, so long as this observable doesn't commute with Newton-Wigner position, is this eigenvalue completely non-localized, and one can only speak of the, say, momentum of a multi-particle field as belonging to the field as a whole and not to any particular location.

2) Once I observe an eigenvalue, how can I extract information about the particles that constitute the resulting field. Do I project the eigenvector associated with the observation, for example, onto a basis of creation/ annihilation operators, and the resulting coefficients encode the probability of observing a given number of particles and their momenta, i.e., the occupation number configuration is just another observable, with its own operators that don't commute with other operators I might be interested in?

3) And, more generally, if I want to "observe" something in QFT, does the field play the role of the wavefunction insofar as, it evolves deterministically through time and observing it collapses it to an eigenvector component, at which point it again begins to behave deterministically?

4) Finally, when I observe a particle, which is completely quotidian in QM and experiments, what does QFT say I'm observing?

In this act I know simultaneously 1) the number of particles, i.e., just 1, and 2) the position of the particle. I've seen explanations in QFT using wave packets, much like QM, but since I'm making an observation, is it the case that these wave packets are simultaneoulsy eigenvectors of both (commuting) position and creation/ number operators?

I apologize if these questions are not well posed. Fundamentally, though, in reading QFT, I feel that my understanding is still disconnected from any meaningful interaction with particles, which, regardless of their reality or lack-of, ultimately what I interact with in a lab. Ultimately, I want this theory to tell me "if you do X, you'll observe Y", and while the texts I've read have given me interesting information how to interact with fields mathematically with operators, I'm ultimately looking for a deeper explanation of everyday phenomena than what normal RQM allows.

While the scattering matrix attempts to do this, it's a very very specific scenario. Ideally, QFT is supposed to explain everything QM does and more, recovering classical mechanics and "normal" QM in a certain limit, right?

Thanks!

• Besides scattering, there are shifts in energy levels (or masses) and perhaps in decay rates due to QFT effects. What other experimental scenarios are you thinking of? – Keith McClary Jul 16 '17 at 2:18
• I think right now I'm just trying to grasp how I can recover non-QFT effects from QFT. E.g., creation operators apply to momentum eigenstates, and with addition of the right momentum eigenstates we can create a wavepacket that "looks like" a particle. Now suppose we use creation operators such that we have 2 wave packets/ particles in our system. Now, if I want to observe some observable of this particle, can I hit the resultant field constructed from the creation operators with my usual QM matrices and get my observables, treating the field as though it were a wave function? – Dragonsheep Jul 16 '17 at 2:38
• I suppose you could say in QFT you are measuring the number operator for each momentum mode whereas in QM you are measuring the momentum operator. But these are far removed from what we actually do in the lab. – Keith McClary Jul 16 '17 at 3:58
• Thanks for the response! That makes a lot of sense. For 2 particles distinguished by different, say, spin or momenta, are there operators we can use to observe 1 particle while keeping the other particle untouched, or do our operators only give us information about the system as a whole? I ask because most second-quantized operators I've seen involve some kind of sum over all momentum states. Despite this, though, if my particles are distinguishable, I intuitively should be know the position of one and momenta of the other. How this translates to second-quantized operators is unclear. – Dragonsheep Jul 16 '17 at 4:05
• Think of a (say, 1 space +1 time dimensional) free (bosonic) QFT "in a box" with momentum cutoff. This is just a multidimensional harmonic oscillator, with each dimension corresponding to a momentum state. The "particles" are the eigenstates of these oscillators. These "particles" are not distinguishable however - if you want that you need a tensor product of two different such theories. As for measuring the position, a very vague statement, you can't really define position states in relativistic QFT because the expression for $\phi(x)$ in terms of $a(k)$ is not a Fourier transform, it has ... – Keith McClary Jul 16 '17 at 5:11