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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".

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  • 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. – CuriousOne Jun 24 '16 at 19:58
  • 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. – Quantumwhisp Jun 24 '16 at 21:43
  • 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). – CuriousOne Jun 24 '16 at 23:06
  • 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. – Quantumwhisp Jun 25 '16 at 11:20
  • 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. – ACuriousMind Jun 25 '16 at 11:23
  • What I don't understand about this is that Fields are never free, yet the description of electrons in in manybody-QM (which is somewhat like the next approximation for me) works out great and knows "particle-states". – Quantumwhisp Jun 25 '16 at 11:34
  • I honestly don't know what you are asking. QFT is QM, it's just a particular application of QM and in the low energy case of electrons it reproduces the same phenomenology as the ad-hoc Schroedinger equation with suitable potentials (some numerical values will be a little different). I don't know why you seem to think that there are particles in the Schroedinger equation. There never were. It's an equation for quanta. – CuriousOne Jun 25 '16 at 17:58
  • @CuriousOne: I'm pretty sure that the question is about the fact that in not-field QM, we can usually write down the Hilbert space pretty explicitly, and describe how the states in it are created and related to what we measure. In interactive QFTs, such a description of the space of states is unknown - the question is simply currently unanswerable, nobody knows what the state of space of interacting QFTs is - all the particle states and almost all the computations rely on the notion of particle states in the asymptotic free theories. – ACuriousMind Jun 26 '16 at 11:09
  • Slightly. @CuriousOne: In the moment you write down Schrödinger equation, you choose a physical System (for example, two electrons - two particles), you choose a hamiltonian that describes these two electrons, and you choose a Hilbert space that describes the states of these two particles, in that case a tensor product of two one-electron (or one particle)-hilbert-spaces. That is what I mean, when I say that Schrödingers Equation works with particles. Maybe we can discuss this with private messages? I think we don't understand each other. – Quantumwhisp Jun 26 '16 at 19:17
  • @ACuriousMind: The physical phenomenology we are describing is exactly the same in both descriptions. The Schroedinger equation covers the non-relativistic single excitation behavior of a massive field. That's as close to the classical case of "particles" as one can get and even for that case we can't make any sense of the particle picture, but we can self-consistently interpret it as a prescription for the exchange of quanta between sub-systems. The same interpretation holds for QFT. I don't see any problems with that and it has absolutely nothing to do with perturbation theory. – CuriousOne Jun 26 '16 at 20:04
  • @Quantumwhisp: You are already doing that in CM. What you chose to do on paper is completely irrelevant for nature, though. Earth still has earthquakes and its poles still nutate, even when you are reducing it to the Kepler problem's particle description. Your problem, however, doesn't live on that level. You still seem to be confused about the difference between a particle and a quantum. I can only suggest that you study the actual phenomenology of quanta and compare them with the phenomenology of point particles. They are nothing alike, not even in the Schroedinger equation. – CuriousOne Jun 26 '16 at 20:09
  • @CuriousOne: I am well aware what a particle in cm is and what the excitation of a Field is. Schrödingers equation makes a statement about one electron. About the quantum-object electron. Maybe it's not a particle in sense of a pointlike particle having position, momentum and mass. I don't mean those particles when I I say particle. To me it seems we're just talking about language here. – Quantumwhisp Jun 26 '16 at 20:30
  • Even one electron is a quantum, even at the Schroedinger level. It does not have an identity, it does not have a position or momentum until measured. The "information" in the Schroedinger wave function is information about a quantum field with exactly one excited state, which is a good approximation for excitations that are far below the rest mass energy of the minimal excitation. That's why I don't understand your problem. Nature has given us a much more simple description for the non-relativistic "single-particle" case, but that doesn't change the nature of the phenomenon. – CuriousOne Jun 26 '16 at 20:37

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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.

Arnold Neumaier
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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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