Isomorphism of rigged Hilbert spaces

Question

In connection with the statement that QM can be formulated in terms of separable complex (rigged) Hilbert spaces, the fact that all infinite dimensional separable complex Hilbert spaces are isomorphic to one another is sometimes emphasized.

I understand why some authors like to add "(rigged)" in order hint how to make this slightly simplified statement more rigorous, but is this modification also compatible the "additional hint" that all "physically relevant" Hilbert spaces are isomorphic? For a "physically relevant" rigged Hilbert space, the ket space has to be nuclear:

A nuclear space is a topological vector space with a topology defined by a family of Hilbert seminorms, such that for any Hilbert seminorm $p$ we can find a larger Hilbert seminorm $q$ so that the natural map from $V_q$ to $V_p$ is trace class.

So I have some questions:

1. I have the impression that rigged Hilbert spaces don't add anything for finite dimensional Hilbert spaces, especially all finite dimensional rigged Hilbert spaces of the same dimension should be isomorphic. Is this correct?
2. Are there simple examples of non-isomorphic infinite dimensional separable rigged Hilbert spaces where the ket space is nuclear?
3. Are the most common examples of the $\mathbb R$- and $\mathbb R^3$-Schwartz-space (i.e. $\mathscr{S}(\mathbb{R}) \subset L^2(\mathbb{R}) \subset \mathscr{S'}(\mathbb{R})$ and $\mathscr{S}(\mathbb{R}^3) \subset L^2(\mathbb{R}^3) \subset \mathscr{S'}(\mathbb{R}^3)$) examples of non-isomorphic rigged Hilbert spaces?
4. Is the $\mathbb R^3$-Schwartz-space at least isomorphic to a subspace of the $\mathbb R$-Schwartz-space?

Answer 1

To specify a Gel'fand triple $(\Phi^*,\mathscr{H},\Phi)$ it is sufficient to specify the Hilbert space $\mathscr{H}$ and the topological vector space $\Phi\subset \mathscr{H}$. The necessary requirement is that the imbedding of $\Phi$ into $\mathscr{H}$ is continuous with respect to the topology of $\Phi$, so this gives the imbedding of $\mathscr{H}^*=\mathscr{H}\subset \Phi^*$.

Therefore, we consider the triple $(\mathscr{H},\Phi,\mathscr{T})$, where $\mathscr{T}$ is the topology of $\Phi$, as defining the Gel'fand triple $(\Phi^*,\mathscr{H},\Phi)$.

Consider the class of Gel'fand triples; we may define a morphism $f$ between objects of the class $(\mathscr{H},\Phi,\mathscr{T})$ and $(\mathscr{K},\Psi,\mathscr{R})$ as following:

• $f(\mathscr{H},\Phi,\mathscr{T})$ is a triple of sets $(A,B,C)$;
• $f\rvert_{Hilb}:\mathscr{H}\to \mathscr{K}$ is a morphism of Hilbert spaces, $f\lvert_{Hilb}(\mathscr{H})=A$;
• $f\rvert_{Top}:(\Phi,\mathscr{T})\to (\Psi,\mathscr{R})$ is a morphism of topological vector spaces, $f\lvert_{Top}(\Phi,\mathscr{T})=(B,C)$;
• $(\Psi,\mathscr{R})$ is continuously imbedded in $\mathscr{K}$ (i.e. $(\mathscr{K},\Psi,\mathscr{R})$ is a triple).

The map is an isomorphism if each of the reduced maps is an isomorphism (i.e. if it is one-to-one and $A=\mathscr{K}$, $B=\Phi$, $C=\mathscr{T}$). [The inverse map $f^{-1}=(\,f\rvert_{Hilb}^{-1}\,,\,f\rvert_{Top}^{-1}\,)$]

Obviously there are non-isomorphic triples, simply choose a triple $(\mathscr{H},\Phi,\mathscr{T}_1)$ and the triple $(\mathscr{H},\Phi,\mathscr{T}_2)$ with $\mathscr{T}_1\subset \mathscr{T}_2$ (i.e. $\mathscr{T}_2$ a topology strictly finer than $\mathscr{T}_1$) and $(\Phi,\mathscr{T}_1)$, $(\Phi,\mathscr{T}_2)$ not homeomorphic (I suppose that e.g. different cardinality of $\mathscr{T}_1$ and $\mathscr{T}_2$ would do the trick).

Given $L^2(\mathbb{R}^d)$ and the spaces $\mathscr{S}(\mathbb{R}^d)$ (rapid decrease smooth functions) and $\mathscr{D}(\mathbb{R}^d)$ (compactly supported smooth functions) we have that $(\mathscr{S}',L^2,\mathscr{S})$ and $(\mathscr{D}',L^2,\mathscr{D})$ are not isomorphic w.r.t. the definition above ($\mathscr{D}$ and $\mathscr{S}$ are not homeomorphic, because $\mathscr{S}$ is metrizable while $\mathscr{D}$ is not).

Answer 2

Item 1. has answered affirmative by Phoenix87 in the comments. Item 2. has been answered affirmative by yuggib in the other answer.

The answers to item 3. and 4. have been given without proof by Phoenix87 in the comments, namely that the $\mathbb R$- and $\mathbb R^3$-Schwartz-space are isomorphic as rigged Hilbert spaces. A nice proof of this fact is given in this answer to a simplified versions of item 3. and 4. from this question. It is the "obvious" candidate isomorphism, which maps the Hermite functions basis of $\mathscr{S}(\mathbb{R}^3)$ to the Hermite functions basis of $\mathscr{S}(\mathbb{R})$.

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It seems that the question has been answered completely. But what about the motivation behind this question: "..., but is this modification also compatible the "additional hint" that all "physically relevant" Hilbert spaces are isomorphic?" As rigged Hilbert space, $\mathscr{D}(\mathbb{R}^d)$ is not really "physically relevant". My impression is that indeed many physically relevant rigged Hilbert spaces are isomorphic, i.e. the "additional hint" is not necessarily wrong.