Let $H$ be a Hilbert space. I am interested in isometries $f\colon H\to L^2(X,\mu)$ where $\mu$ is a probability measure on some measure space $X=(X,\mathcal F)$ where $\mathcal F$ is a $\sigma$-algebra on $X$. My question is whether there exists a canonical choice of $(X,\mu,f)$ that depends only on $H$ and not "arbitrary" choices like choosing an orthonormal basis for $H$.


For an example of what I mean, here is a construction that only works when $H$ is finite dimensional. Let $X=H$, and for $h\in H$ consider the function $f\_h\colon X\to\mathbb C$ given by $f\_h(x)=\langle h,x\rangle$. Let $\mu$ be the unique Borel measure on $H$ satisfying $$\int\_X e^{i f\_h(x)}\ d\mu(x) =e^{-\|h\|^2/2}$$
for all $h\in H$, from which it follows that $h\mapsto f\_h$ is an isometry. Note that the $\mu$-expectation of the function $\|x\|^2\colon X\to\mathbb C$ equals $\dim H$.


In fact when $\dim H=\infty$, the analogue of this construction yields a measure supported on "elements of $H$ with infinite magnitude", which can be made sense of rigorously as distributions - elements of the space $\Phi^\*$ in a [Gelfand triple](https://en.wikipedia.org/wiki/Rigged_Hilbert_space#Formal_definition_(Gelfand_triple)) $(\Phi,H,\Phi^\*)$.


Thus, a close variant of my question is to ask whether there exists a canonical construction of a Gelfand triple using only the space $H$ as input.



Canonical embedding of Hilbert space in $L^2$ space
This is impossible, at least in the case when $X=H$, as in your "finite-dimensional" example. 


Indeed, suppose that $H$ is infinite dimensional. If you want the measure $\mu$ not to depend on the choice of an orthonormal basis, you have to make $\mu$ spherically invariant. But such a probability measure can only be $\delta\_0$, the Dirac measure at $0$. (See details below.) And if $\mu=\delta\_0$, then $L^2(H,\mu)$ is one dimensional and hence not isometric to the infinite-dimensional Hilbert space $H$. 


**Details:** Suppose that $\mu\ne\delta\_0$ is a spherically invariant probability measure on $H$. Let $\nu$ be the conditional distribution of the vector $x/|x|$ given $x\ne0$, assuming that the distribution of $x$ is $\mu$ (here, $|x|$ denotes the norm of $x$). 
Then $\nu$ is a spherically invariant probability measure on the unit sphere in $H$, and such a probability measure does not exist -- see e.g. [Sudakov, page 24](https://books.google.com/books?id=I5ehm9W58LoC&pg=PA24&lpg=PA24&dq=%22spherically%20invariant%20measure%22%20%22hilbert%22&source=bl&ots=L1LPR5rBW0&sig=ACfU3U2tfXddEdiPkUJkYKmk3tbdxgIkhg&hl=en&sa=X&ved=2ahUKEwj0ssTH8sDmAhVTK80KHY0LBUAQ6AEwBnoECAsQAQ#v=onepage&q=%22spherically%20invariant%20measure%22%20%22hilbert%22&f=false). So, indeed, the only spherically invariant probability measure on $H$ is $\delta\_0$. 


