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Rigged Hilbert space, separable space, domain of CSCO, mapping 
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#19
Jan3113, 09:04 PM

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I don't have access to Antoine, sadly, but I'll look on the internet for a more complete definition. 


#20
Jan3113, 09:11 PM

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I'll also dig out some other references and post them shortly. 


#21
Jan3113, 09:42 PM

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"On the Mathematical Basis of the Dirac Formulation of Quantum Mechanics", IJTP, vol 42, no 10, 2003, p2225. (If you can't access that, send me a PM... ;) 2) Gadella, Gomez, "Eigenfunction Expansions and Transformation Theory", Available as: math/0607548 and of course, there's always the reliable, but less rigorous: 3) Rafael de la Madrid, "The role of the rigged Hilbert space in Quantum Mechanics", Available as: quantph/0502053 Ref (3) is more physicistoriented, but it does give a more intuitive picture of how the small space ##\Omega## is constructed in terms of an initial Hilbert space, one or more unbounded operators with continuous spectra, and the sequence of seminorms (and hence nested Hilbert spaces) induced by the latter. [And now I need to go over your post #17 again more closely... ] 


#22
Feb113, 12:36 AM

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#23
Feb113, 01:08 AM

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For a quantum theory, we must have a space with (+vedefinite) inner product which admits a probability measure. This is the Hilbert space ##H_0## (possibly after a completion in the norm induced by the inner product). The meaning of the "0" subscript will become clear below. In a useful physical theory of dynamics, we essentially start from differential equations (of motion), and determine the maximal invariance group of this (system of) differential equation(s). This is the "dynamical group"  it maps solutions of the dynamics among themselves. Depending on the degree of sophistication in the model, we might be content with restriction to a subgroup, rather than the maximal group. So for a quantumdynamical theory, this group must be represented as operators on the Hilbert space [note #1]. One might use group operators (i.e., in exponential form which are bounded), but more usually one finds the generators of the group and attempts to represent these on the Hilbert space. Typically, these generators are unbounded operators with continuous spectrum, hence cannot be defined everywhere on ##H_0##. Take a typical such generator ##P## (ordinary momentum), represented by (e.g.,) ##i\hbar\partial_x## on ##H_0##, being a Hilbert space of squareintegrable wave functions. Now define $$ H_1 ~:=~ \left\{ \psi\in H_0 : P\psi \in H_0 \right\} ~. $$ Clearly ##H_1 \subset H_0##. Further, if the norm on ##H_0## is denoted ##\\psi\_0##, we can define another norm on ##H_1## : $$ \\psi\_1 ~:=~ \\psi\_0 + \P\psi\_0 ~, $$ so that ##\\psi\_1 \le \\psi\_0##. Clearly, both norms ##\\cdot\_0## and ##\\cdot\_1## are welldefined on ##H_1## and one can define a new topology ##\tau_1## on ##H_1## via the 1norm. Continuing in this way we can define a sequence of nested spaces, e.g., $$ H_{n+1} ~:=~ \left\{ \psi\in H_n : P\psi \in H_n \right\} ~, $$ and corresponding ##(n+1)##norms similarly. Thus, one has an infinite sequence of nested spaces with a corresponding sequence of seminorms and topologies. Gel'fand & Vilenkin call this setup a "countablyHilbert" space. Then, one puts $$ \Omega ~:=~ \bigcap_{n=0}^\infty H_n ~, $$ thus arriving at the "small" space in the triple (which hopefully is not trivially empty). Anyway,... the point of posting all this is because I'm not sure whether the extra structure implied by the above might affect what you've been saying about duals... [Edit: I also think this construction guarantees that each canonical embedding map from ##H_{n+1}## into ##H_n## is nuclear (HilbertSchmidt). It appears (eg Wiki) that "nuclear" requires only traceclass embedding operators, but the nuclear spectral theorem seems to require HilbertSchmidt embedding operators. Can you shed any more light on this?] [Note #1: Algebraic quantum theory handles this differently: one constructs abstract functionals over a ##C^*##algebra related to the dynamical group.] 


#24
Feb113, 02:05 AM

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[tex]\{\mathbf{x}_1,\mathbf{x}_2,...\}[/tex] of [itex]\ell^\infty[/itex]. Then we can write this as [tex]\mathbf{x}_1: ~x_1^1 ~ x_1^2~ x_1^3 ~ ...[/tex] [tex]\mathbf{x}_2: ~x_2^1 ~ x_2^2~ x_2^3 ~ ...[/tex] [tex]\mathbf{x}_3: ~x_3^1 ~ x_3^2~ x_3^3 ~ ...[/tex] and so on But consider the following diagonal sequence: [itex](x_n^n +1)_n[/itex]. This is an element of [itex]\ell^\infty[/itex], but it has distance at least 1 from any of the [itex]\mathbf{x}_k[/itex]. So our set isn't dense. 


#25
Feb113, 02:51 AM

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GelfandVilenkin seem to define a rigged Hilbert space as a nuclear countably Hilbert space [itex]\Phi[/itex] together with a continuous, injective, linear map [itex]T:\Phi\rightarrow H[/itex] into a Hilbert space.
However, since [itex]\Phi[/itex] is nuclear countably Hilbert, it is separable. This seems very much nontrivial, but Gelfand and Vilenkin do prove it (in an earlier volume). Furthermore, any nuclear countably Hilbert space is reflexive. This means that [itex]\Phi^\times[/itex] is separable as well. So if we use the definition from GelfandVilenkin, then we can answer (1) and (2) with yes. 


#26
Feb113, 02:56 AM

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BTW, I'm really impressed that you were able to absorb that much from G+V in such a disgustingly short time. 


#27
Feb113, 03:12 AM

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But in any case, your language seems to imply that traceclass is a weaker property that being of HilbertSchmidt type. But it is actually a stronger property, isn't it? It seems to me that traceclass operators are always of HilbertSchmidt type. So if you say that the spectral theorem requires HilbertSchmidt operators, then it holds for nuclear operators as well. 


#28
Feb113, 10:53 AM

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Yes, the traceclass property is stronger than HilbertSchmidt.
From page 61 of R.de la Madrid PhD thesis: Definition A compact operator A = U]A] is called HilbertSchmidt if ynC__l/2 n OD, where the An are the eigenvalues of the operator ]A]. Therefore, an operator is of HilbertSchmidt type if[ admits a decomposition of the form (2.5.18) such that the series ynC__l/2 n converges. One can also see that in order an operator A be of HilbertSchmidt type, it is necessary and sufficient that the series ynC__l ]]Aen]] 2 converge for at least one orthonormal basis ˝1, ˝2,... in 7/. An even more restrictive requirement that the operator A be HilbertSchmidt is that it be a nuclear operator. Definition A compact operator is called nuclear (or trace class) if ynřC=l/n < O(), where the An are the eigenvalues of the operator IAI appearing in the decomposition A = UIAI. Since the convergence of the series Y]i A} follows from the convergence of Y]i An, every nuclear operator is of HilbertSchmidt type. 


#29
Feb113, 08:36 PM

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#30
Feb113, 08:46 PM

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http://physics.lamar.edu/rafa/dissertation.htm (I really must read that through again.) 


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