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Feb 1, 2023 at 16:39 comment added Valter Moretti Actually the notion of rigged Hilbert space can be extended to the case of an abstract Hilbert space equipped with a certain couple of topological subspace and sup space analogous to $\cal S$ and ${\cal S}'$ (nuclear spaces). However, I find that construction quite cumbersome in the general case.
Feb 1, 2023 at 15:25 comment added Plop In my opinion, the "von Neumann formalism" aims at treating quantum mechanics in a general way, and the rigged Hilbert space formalism only aims at providing a rigourous formalization of quantum mechanics of particles evolving on $\mathbb{R}^n$. I have seen attempts (I cannot judge if those were successful or not...) of using rigged Hilbert spaces to rigourously define path integrals, see for example the book of Kuo, White Noise Distribution Theory.
Feb 1, 2023 at 4:35 history became hot network question
Feb 1, 2023 at 3:01 history tweeted twitter.com/StackPhysics/status/1620618259606212608
Feb 1, 2023 at 1:09 vote accept EE18
Jan 31, 2023 at 22:20 answer added Valter Moretti timeline score: 32
Jan 31, 2023 at 22:04 comment added EE18 Thanks for your response Prof. Moretti. Is it correct to say that the rigged H space formulation is useful in the manner you describe because it effectively lets us make precise sense of all the crazy "momentum eigenstates" etc. that we take for granted in a "physics textbook"? @ValterMoretti
Jan 31, 2023 at 21:56 comment added Valter Moretti Personally, I sometimes pretend to use the rigged H space approch and finally I rigorously prove the found result with the vN formulation which is quite easy to use once guessed the formal result. I think this is the only safe way to exploit the rigged H. space formulation...
Jan 31, 2023 at 21:51 comment added Valter Moretti The approach based on the rigged Hilbert space structure requires more mathematical hypotheses, and it is much more delicate to rigorously handle, than the von Neumann formulation. For that reason I strongly prefer the latter. Physically speaking, in all concrete situations, they are equivalent. I think that nobody uses the rigged H space for rigorous computations: an impossible heavy work to obtain formally evident results.
Jan 31, 2023 at 20:56 comment added EE18 @CosmasZachos No, my understanding is that what I am calling the "von Neumann formulation" goes through a generalized spectral theorem which lets us make sense of an eigendecomposition of unbounded operators (in terms of some integral over a projection-valued measure?). This is strikingly different than simply admitting the "eigenfunctions" of such operators which exist in a space conjugate to the nuclear space of our Hilbert space as in the rigged Hilbert space formalism.
Jan 31, 2023 at 20:40 history edited Tobias Fünke
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Jan 31, 2023 at 20:37 comment added Cosmas Zachos von Neumann? Density matrices?
Jan 31, 2023 at 20:29 history asked EE18 CC BY-SA 4.0