Why were Fock Space and Hilbert Space used in quantum field theory? What was the motivation for choosing them over other mathematical techniques?
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You should read something of the history of quantum mechanics. In particlular Heiseberg's discovery of matrix mechanics, and the realization by Born and others that Heisenberg's matrices were representing operators acting on a Hilbert space. – mike stone Mar 19 '21 at 12:18
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1Does this answer your question? Why does non-commutativity in quantum mechanics require us to use Hilbert spaces? – ohneVal Mar 19 '21 at 12:30
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In more general structures such as Banach spaces, you do not need to have an inner product. So, if you want to define a possibly infinite-dimensional vector space and assign probability amplitudes to the coefficients on some basis, then a Hilbert space is the simplest structure. – Lucas Freitas Mar 19 '21 at 13:18
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I voted to reopen because there are two questions here 1) why Hilbert spaces? and 2) why Fock spaces? I agree that for 1) this duplicates the linked question. But the latter does not at all address 2). – Abdelmalek Abdesselam Mar 19 '21 at 17:51
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Hilbert spaces are used throughout all Quantum Mechanical theories. In fact all Hilbert spaces are isomorphic, so we might as well just call this structure "the abstract Hilbert space". That observables are represented by operators acting on the Hilbert space is one of the axioms of Quantum Mechanics.
The Fock space is a Hilbert space, so it is mathematically isomorphic to the abstract Hilbert space. What makes it special is that on the Fock space, the Canonical (Anti-)Commutation Relations of the field theory are represented by field operator-valued distributions, defined by picking a very special basis that consists of $n$-particle subspaces; then defining the creation and annihilation operators and making quantum field operator-valued distributions from them.
For interacting QFT, the Fock space is not the correct representation of the CCR/CAR, due to Haag's theorem. So contrary to the widespread belief, interacting QFT does not use the Fock space at all – instead it uses a different (read, unitarily inequivalent) representation of the CCR/CAR on the abstract Hilbert space.
However, it is still possible to use the Fock space to label approximately the states of the interacting QFT for which the particles are far enough apart for the interaction to be neglected. We can then make these particles come close to each other, interact, and fly far apart again. This process can be approximately described by the scattering operator that acts on the Fock space. Usually the scattering operator is evaluated perturbatively by summing up expressions that correspond to Feynman diagrams.
Prof. Legolasov
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Perhaps you should elaborate a more on why we want operators acting on Hilbert spaces instead of say, general Banach spaces. – Jakob Elias Mar 19 '21 at 12:49
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@JakobElias I don't think there is a single specific reason for this. You are probably thinking about either the probabilistic interpretation and convergence, or about the representation theory of the CCR, Weyl quantization and SvN theorem. I can't honestly say that any of these reasons is "the one reason" to choose Hilbert spaces. Ultimately, like for all similar questions, the answer is "because it works so well". – Prof. Legolasov Mar 19 '21 at 12:51
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Yes I was thinking along those lines. It was just a suggestion because OP was inquiring on alternatives to HS to which there are none due to the reasons you mentioned. – Jakob Elias Mar 19 '21 at 12:54
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Is it the case that in the interaction picture renormalized Hilbert space cannot have Heisenberg's Uncertainty principles assigned to it satisfactorily? – Mar 19 '21 at 14:37
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@AlistairBain it will be a representation of the CCR (and hence will satisfy the heuristic uncertainty principle to some extent), but a different (read unitarily inequivalent) one from the one used in the nonperturbative QFT. – Prof. Legolasov Mar 19 '21 at 14:55