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This is something I know of but I'm not quite sure I understand the details. Particularly, when it comes to interacting RQFTs, such as even QED, where some posts here have pointed out that it cannot make sense even of a hydrogen atom without some further assumptions. To me that isn't a particularly nice situation, especially for a theory that one would believe from all the hype and the casual talk is supposed to be a "well-trodden area" and not something at the frontier like quantum gravity. And something I think makes sense to want to remedy.

But what I am curious about is, why exactly is this so difficult to do? I've heard of Haag's theorem, but only have access to the Wikipedia terse description so I don't quite get 100% what it's after. I also know of how to make the simple and ubiquitous free-field (R)QFT via the Fock space construction and how you can solve for phonons on a crystal lattice, and so what I don't get is this:

Why can't you just tensor together the two Hilbert (Fock) spaces of the free EM (photon) and free charge (electron/positron) fields, and then write a suitable interaction Hamiltonian?

That is, after all, the way you do interacting systems in NRQM. What makes this naive approach fail? Almost surely this was the first thing tried many decades ago, but can a fairly decent summary of the arguments against it be given?

Qmechanic
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The_Sympathizer
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    I would say that one of the aspects is that quantum fields are operator-valued distributions and the proper handling of their distributional nature is challenging (although not impossible as far as I know). You may want to look up this thread https://physics.stackexchange.com/q/330536/. – Gold Jul 06 '22 at 03:32
  • Why can't you just tensor together the two Hilbert (Fock) spaces of the free EM (photon) and free charge (electron/positron) fields, and then write a suitable interaction Hamiltonian? One of the primary obstructions is that if you try to calculate what look like they should be matrix elements of observables, you invariably get divergent results. – Buzz Jul 06 '22 at 03:33
  • In addition to the other comments, which point to renormalization, another issue is that many interesting QFTs like QCD are strongly coupled (at least at low energies), which means that naive perturbation theory is doomed to fail, and directly trying to construct a Fock space with quarks and gluons is not "close" to a good description of the physics. There is also an analog of this problem in QED in the form of the Landau pole. – Andrew Jul 06 '22 at 03:43
  • @Gold: That's a very nice and accessible statement of the problem. Looking it up, there is apparently such a thing as Colombeau algebra which attempts to define a type of algebras where that you can create a product of distributions. Has this been employed toward RQFT to make sense of those interactive powers? Also, (at least some) distributions become true functions in non-standard analysis, e.g. the Dirac delta becomes a (family of equivalent) spikes of a concrete "infinite" height and infinitesimal thickness. Does it provide any help, or else, how does it fail? What work has been done? – The_Sympathizer Jul 06 '22 at 08:32
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    I don't know much about the approaches to deal rigorously wih these problems. I have been interested in this approximately four years ago, but then decided to just work with standard QFT and holography. So someone more familiarized with the topic will probably have more to say. Still, I have heard that books by Scharf mentioned in the thread I linked deals with the distributions appropriately. Also you might be interested in searching for the field known as constructive QFT: that is the subject of the great answer in that thread by Abdelmalek Abdesselam. – Gold Jul 06 '22 at 13:46
  • Another question: why couldn't P(O)VMs suffice? You can get rid of the need for distributions in continuous single-particle QM by replacing position with a PVM, i.e. the family of projectors that correspond to affirmative answers to questions of the form "Is the particle's position $x$ in set $S$?". Why can't you do the same in RQFT to avoid the need to tangle with distributions? – The_Sympathizer Jul 07 '22 at 05:13

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Why can't you just tensor together the two Hilbert (Fock) spaces of the free EM (photon) and free charge (electron/positron) fields, and then write a suitable interaction Hamiltonian?

The quick mathematical answer to this is that when you do that, and try to write the relativistic interaction term as an operator, it does not make sense.

It does not even make sense weakly as a quadratic form or on some subset of vectors.

To overcome this problem, one should first regularize the interaction. In doing that however, often the relativistic covariance is broken. The aim is then to remove the regularizations, restoring covariance, and obtaining a well-defined operator. In order to do that, some infinities in the wavefunctions and energy shall be taken care of.

There are several ways to do that, that have been successful to some extent (scalar $\phi^4$ theory can be defined rigorously in 1 and 2 space dimensions for example). However, most of the strategies even "forget" about the Hamiltonian, since in relativistic theories the generator of time translations does not play a preferred role. Quadratic scalar theories and the Yukawa theory in 2 spatial dimensions have been however defined rigorously using the Hamiltonian approach (works by Ginibre and Velo, and Glimm, respectively).

yuggib
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  • And that's the problem: the regularizing doesn't let you even solve the hydrogen atom - becuase it's nonperturbative - without making further assumptions beyond the fields and the interactor equation, as I seem to have heard. But it shouldn't require extra assumptions. It should be a complete theory by itself that extends NRQM even if not "infinitely far" i.e. absolutely valid forever any more than any other theory. Since NRQM does hydrogen atoms, molecules etc. just fine, RQFT should also handle them with equal grace. – The_Sympathizer Jul 07 '22 at 05:10