In classical mechanics, we assumed a particle to have a definite momentum and a definite position. Afterwards, with Quantum mechanics, we gave up the concept of a time-dependend position and momentum, and instead have propability distribution stuff and a Hilbert space containing all the information about the state of the particle. Still, we can "retrieve" the concept of a point particle, by stating that the quantum mechanical state gives a propability distribution for the (point-particle)-properties position and momentum, and moreover, the mean-values for position and momentum follow the classical rules.

What I am seeking for now is basically the same, but not for QM, but for QFT. For an abitrary Quantum-Field-Theory, is there a way to "construct" a particle-concept that gives position and momentum for one particle? I know that in QFT, particles are just excitations of the field, but still, is there a way to assign position and momentum to certain types of excitations?

For example, in Theories of free fields (see my other Question here), one can identify the hilbert-space with a fock-space, and by that one can "construct" some wave-package state, that then (in the context of many-body QM) has a localized position and momentum. Is something like that also possible in an interacting theory (despite the fact that the hilbert spaces of interacting theories are not in correspondence to a fock space)?

My thought is that this should be possible in principle, since the QFT is somewhat a generalization of the QM, that intendes to describe Nature better than QM. There should be something like "backward-compability".

  • $\begingroup$ A "particle" in CM is the name for the approximation that reduces the true motion of an extended body to the motion of its center of mass coordinates. A planet in the Kepler problem is a particle, an atom or molecule in kinetic gas theory is not (because rotations and molecular vibrational degrees of freedom are included). A particle is therefor not a thing, but a property and it's not even a property of a thing but a property of the description of the thing. Obviously, none of this applies in QM or QFT, which is why there we are talking about quanta rather than particles. $\endgroup$
    – CuriousOne
    Commented Jun 24, 2016 at 19:58
  • $\begingroup$ still, we have a concept of position and momentum, even in QM, and I also stumbled over the formulation "particle-properties" refering to an electron in the context of QM. $\endgroup$ Commented Jun 24, 2016 at 21:43
  • $\begingroup$ The concepts of physical position and momentum in QM are operators acting on fields. They are not localized to an abstract that wasn't of much use beyond entry level Newtonian mechanics. Whenever someone refers to electrons as particles in anything but a handwaving sense in a quantum mechanical context, you already know that something is either off or they expect you to make the translation to "Yeah, I am really talking about quanta here and I expect you to know that.". Many "problems" with QM occur because folks don't know what the speaker means (except when the speaker is incompetent). $\endgroup$
    – CuriousOne
    Commented Jun 24, 2016 at 23:06
  • $\begingroup$ Well, do you know what I mean here? I guess my usage of the word particle was not wrong, and I should instead have used the word quantum, yet this is not the core of my question. $\endgroup$ Commented Jun 25, 2016 at 11:20
  • $\begingroup$ There are no particle states in interacting QFTs. All particle states you will ever see live in the asymptotic Fock spaces of the free fields. $\endgroup$
    – ACuriousMind
    Commented Jun 25, 2016 at 11:23

2 Answers 2


In QFT, a single particle does not scatter, hence its (renomalized) wave function in an interacting theory is the same as the corresponding asymptotic wave function in the asymptotic Fock space.

However, the multiparticle picture breaks down as the interacting Hilbert space cannot be identified with the asymptotic Fock space, by Haag's theorem. Thus multiparticle states make sense only asymptotically.


If I'm understanding correctly, you're asking whether an arbitrary QFT admits an asymptotic Fock basis. If a QFT does admit a Fock basis, you can talk about particles and do scattering theory in momentum space and construct position operators for single particles.

But it's not guaranteed that a given QFT admits asympotic Fock bases. Quantum field theory in general deals with the quantum behavior of field systems. Sometimes the fields and interactions are such that you get particles, but this is a property that depends on the details of the QFT, not a property of QFTs in general. Conformal QFTs, for example, do not have asymptotic particle states or scattering matrices. Nor do theories which lack a mass gap.


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