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Operator valued distributions in quantum field theory refer to free fields. The creation operators act in Fock space. They create particle states of various energies and momenta. Every state has a well- defined energy and momentum. Free particles correspond to excitations of fields. One particle can be in a superposition of different energy and momentum states.

Why can't this formalism be applied to interacting fields of particles? Why can't there be fluctuations (virtual particles) in this approach. Virtual particles are necessary for interaction. I don't see why an interaction cannot be viewed in the same approach. Because the operators excite only real, free particles?

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    $\begingroup$ The immediate problem is that the interacting Hamiltonians do not make mathematical sense since they involve products of field operators (which are really distributions), and one cannot generally take the product of distributions. $\endgroup$
    – Pedro
    Commented Jul 16, 2021 at 16:38
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    $\begingroup$ @Pedro But the free Hamiltonian already includes squares of the fields? So you should be more explicit here. $\endgroup$ Commented Jul 16, 2021 at 16:40
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    $\begingroup$ True -- and even in the free case you do end up needing to throw away an "infinite constant" to get the right formula for the Hamiltonian if you derive it as usual from the heuristic procedure via the Lagrangian. But one can also give a direct definition of the Hamiltonian operator in Fock space already that avoids this issue and is completely rigorous. I suppose the issue is that this just breaks down in the interacting case; as I'm not an expert I'll limit myself to this comment. $\endgroup$
    – Pedro
    Commented Jul 16, 2021 at 16:45
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    $\begingroup$ For the Fock space definition see chapter 8 here statweb.stanford.edu/~souravc/qft-lectures-combined.pdf and also 12.4 $\endgroup$
    – Pedro
    Commented Jul 16, 2021 at 16:47
  • $\begingroup$ We must guess (construct) the interaction in a way to only modify occupation number amplitudes. It is still under construction. The ususal way is wrong even in CED. See my popular papers on arXiv. $\endgroup$ Commented Jul 16, 2021 at 17:31

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Of course there is a more mathematically rigorous way of explaining this, but I'll offer a less rigorous and hopefully more intuitive explanation.

In perturbation theory, when the interactions are weak, the formalism of virtual particles is used and is in fact very useful! This is essentially what Feynman diagrams describe.

You can do this when interactions are weak because you can assume that the particles that appear in the theory are the same particles which appear when the interactions are turned off. So the fields that you used to construct the QFT still correspond to the particles in the theory, and have a standard interpretation in terms of creation and annihilation operators.

When interactions become strong, there is no guarantee that the particle spectrum is the same as when the interactions are turned off. The fields you used to construct the QFT therefore no longer necessarily correspond to particles in the theory. A description of the interaction in terms of the 'free' virtual particles no longer makes much sense.

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    $\begingroup$ The fields you used to construct the QFT therefore no longer necessarily correspond to particles in the theory. Unless the fields are elementary. So the particles cant change identity. $\endgroup$
    – user307025
    Commented Jul 16, 2021 at 18:04
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    $\begingroup$ @Vielloosoof even in that circumstance, it is likely that there are more particles in the spectrum than just the elementary ones (for strongly interacting theories). The elementary fields cannot take into account of those particles in the standard way. $\endgroup$
    – fewfew4
    Commented Jul 16, 2021 at 19:55
  • $\begingroup$ But then the extra particles are composed (say that there are more elementsry ones than quarks and leptons). $\endgroup$
    – user307025
    Commented Jul 16, 2021 at 19:58
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    $\begingroup$ @Vielloosoof not necessarily, there are plenty of theories for which this isn't the case. Some theories exhibit magnetic monopoles which are impossible to see at a perturbative level, and these are certainly not composite. Other theories exhibit massive spin $1$ particles which cannot be described in a unitary way using an elementary gauge field. $\endgroup$
    – fewfew4
    Commented Jul 16, 2021 at 20:11
  • $\begingroup$ If real then they can indeed show up at high energies. $\endgroup$
    – user307025
    Commented Jul 16, 2021 at 20:14

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