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Questions about Rigged Hilbert Space 
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#1
Jan912, 10:31 PM

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Dirac's braket formalism implicitly assumed that there was a Hilbert space of ket vectors representing quantum states, that there were selfadjoint linear operators defined everywhere on that space representing observables, and that the eigenvectors of any such operator formed an orthogonal basis of the ket space. But this runs into a few problems: first of all, some operators like position and momentum are unbounded, so they cannot be defined everywhere on the ket space due to the HellingerToeplitz theorem. Second, the criterion that you have an orthogonal basis seems to require the ket space to be seperable, but seperable Hilbert spaces can only have countably many dimensions, which is incompatible with the idea that position eigenstates form an uncountable basis.
The RHS construction tries to alleviate these concerns as follows (this is mostly gleamed from reading old PF threads on the topic). We start with the standard Hilbert space [itex]H=L^{2}(ℝ^{3})[/itex], and we pick out a socalled nuclear subspace [itex]\Phi[/itex] which is dense in [itex]H[/itex] and on which any continuous function [itex]f(P,Q,H)[/itex] (defined via Taylor series) of the position, momentum, and Hamiltonian operators is defined. We then basically use [itex]\Phi[/itex] as the space of test functions for a theory of Schwartz distributions  the space [itex]\Phi^{\times}[/itex] of continuous antilinear functionals will be our ket space, and the space [itex]\Phi'[/itex] of continuous linear functionals will be our bra space. My question is, does RHS sucessfully implement all of Dirac's formalism unscathed, so that we can completely forget about the original Hilbert space once we've set up the rigged Hilbert space? I've heard that the GelfandMaurin nuclear spectral theorem proves that every selfadjoint linear operator acting on [itex]H[/itex] has an eigenbasis in [itex]\Phi^{\times}[/itex], but can the same be said of all selfadjoint linear operators acting on [itex]\Phi^{\times}[/itex]? Is it possible to define a distributionvalued inner product on [itex]\Phi^{\times}[/itex] (or do we run into the ageold problem of multiplication of distributions?), and does this make [itex]\Phi^{\times}[/itex] an uncountabledimensional nonseperable Hilbert space? Can [itex]\Phi'[/itex] also be considered as the dual space of [itex]\Phi^{\times}[/itex], and is there a onetoone correspondence between these two spaces? Finally, can things like deltafunction potentials be rigorously justified in RHS framework? Sorry for asking so many questions, but I only recently discovered Rigged Hilbert Spaces, and I'm absolutely fascinated by them. Can anyone recommend a good book on the subject? (I've already read the couple pages Ballentine has to say on the subject, and I wish there was a whole quantum mechanics text using this approach.) Any help would be greatly appreciated. Thank You in Advance. 


#2
Jan912, 11:42 PM

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PF Gold
P: 9,258

Every Hilbert space has an orthormal basis. It has a countable orthonormal basis if and only if the space is separable.
I don't know this stuff well, but I think the references in this thread will be a good place to start. 


#3
Jan1012, 01:56 AM

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Rafael de la Madrid, "The role of the rigged Hilbert space in Quantum", Available as: arXiv:quantph/0502053 Hilbert spaces have a scalarvalued inner product. But Dirac's formalism has a distributionvalued inner product in general. Rigged Hilbert space can be regarded as a version of Dirac's formalism made rigorous and generalized. If you really want to learn the maths, try Gelfand & Vilenkin vol 4. :) Another (more mathmatical) paper is: M. Gadella, F. Gomez, "On the Mathematical Basis of the Dirac Formulation of Quantum Mechanics", IJTP, vl. 42, No. 10, October 2003, p2225. Also do a search on Google Scholar for papers/books by JP Antoine. 


#4
Jan1012, 01:44 PM

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Questions about Rigged Hilbert Space



#5
Jan1012, 02:20 PM

P: 1,583

I've found a nice article from the Stanford Encyclopedia of Philosophy which among other things discusses Rigged Hilbert Space. But it has a few surprising assertions:



#6
Jan1012, 05:13 PM

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PF Gold
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The material of this fourth volume represents a complete unit in itself, and, as we have said, the exposition is practically independent of the preceding volumes. In spite of the relation of one chapter to another, one can begin a reading of this book with the first chapter, which contains the general theory of nuclear and rigged Hilbert spaces, or with the second chapter, which discusses the more elementary theory of positive definite generalized functions.So it looks like it's just chapter 1 of volume 4. *) He supported his claim by linking to an article that apparently explains the details, but I didn't read it. 


#7
Jan1012, 10:53 PM

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#8
Jan1012, 11:18 PM

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Cf. http://en.wikipedia.org/wiki/Weak_derivative [tex] \def\<{\langle} \def\>{\rangle} E\>\<E \, dE [/tex] where E ranges over the (continuous) range of energies in the spectrum of the selfadjoint operator. (That's why a Hilbert space  the completion of [itex]\Phi[/itex]  sits in the middle of a Gel'fand triple.) But we're dealing with a case where [itex]\Phi^\times[/itex] is typically much larger than [itex]\Phi[/itex]. [tex] H ~\in~ Lin(\Phi, \Phi') [/tex] but this gets tricky since [itex]H^2[/itex] might not be defined everywhere, and one must retreat to [itex]\exp(itH)[/itex] and take a t derivative to get the generator. I vaguely remember that Thirring treats things this way, but I don't have a copy of his book here. JP Antoine's stuff on PIPspaces (partial inner product spaces) is also relevant here. PIPspaces allow a finer grained treatment than the (rather coarse) RHS  but I haven't yet studied it deeply. 


#9
Jan1012, 11:40 PM

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especially the details for handling nuclear spaces and the various theorems. 


#10
Jan1012, 11:47 PM

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#11
Jan1112, 12:58 AM

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PF Gold
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#12
Jan1112, 10:47 AM

P: 1,583

EDIT: On second thought I think I may be good with abstract integration: if I understand correctly you just have to replace the Lebesgue measure with another measure, and the rest is pretty straightforward. What I'm not familiar with might be more of a notational thing: I'm not used to treating what looks like a differential form, like [tex] \def\<{\langle} \def\>{\rangle} E\>\<E \, dE [/tex], as a measure. I've seen things like [itex]∫fd\mu[/itex], where [itex]\mu[/itex] is the Lebesgue measure, but I thought that was just fancy notation. 


#13
Jan1112, 10:51 AM

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#14
Jan1112, 11:02 AM

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#15
Jan1112, 11:59 AM

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PF Gold
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#16
Jan1212, 01:55 AM

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There's been some history here over this subject. I remember sending people who have questions on this to first make a solid read of the most informative source on RHS I know of: Rafael de la Madrid's PhD thesis, which should still be freely available online. So our OP is kindly invited to perform some further reading.



#17
Jan1212, 02:33 AM

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PF Gold
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I never looked at his thesis before. The table of contents certainly looks good. Here's a link. http://physics.lamar.edu/rafa/webdis.pdf



#18
Jan1212, 04:11 AM

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[tex] 1 ~=~ \int E\>\<E \, dE [/tex] 


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