Let's suppose that with number $\mu\_1 \in \mathbb{R}$ we associate a Hilbert space $\mathcal{H}\_{\mu\_1}$ with countable basis $|1\rangle \_{\mu\_1}$, $|2\rangle \_{\mu\_1}$, $|3\rangle \_{\mu\_1}$, $\ldots$ Analogically


$\mathcal{H}\_{\mu\_2}$: $\mu\_2 \in \mathbb{R}$ with basis $|1\rangle \_{\mu\_2}$, $|2\rangle \_{\mu\_2}$, $|3\rangle \_{\mu\_2}$, $\ldots$


$\mathcal{H}\_{\mu\_3}$: $\mu\_3 \in \mathbb{R}$ with basis $|1\rangle \_{\mu\_3}$, $|2\rangle \_{\mu\_3}$, $|3\rangle \_{\mu\_3}$, $\ldots$


$\vdots$


And so on.


Could you tell me please, if we formally create the following space (continuum tensor product of separable Hilbert spaces)
$$
\mathcal{H} = \bigotimes\limits\_{k\in \mathbb{R}} \mathcal{H}\_{\mu\_k},
$$
would it be a Hilbert space (obviously, not separable)?


P.S. Yes, I know that we can't enumerate all real numbers and chosen notation for indexes of $\mu$ isn't really good, but I did not come up with something better.



Result of continuum tensor product of Hilbert spaces
There are several ways you can define a Hilbert space tensor product $\bigotimes\_{t \in X} H\_t$, if each $H\_t$ is a Hilbert space. The "full" tensor product is generated by all functions $h: X \to \bigcup H\_t$ with $h(t) \in H\_t$ for each $t$ and such that $\prod \|h(t)\|$ converges. We write $h$ as $\bigotimes h(t)$ and define the inner product by $\langle \bigotimes h(t), \bigotimes k(t)\rangle = \prod \langle h(t),k(t)\rangle$. (The convergence condition on $h$ is chosen to ensure this product converges, modulo some technicality.) Then take the linear span, factor out null vectors, and complete to get a Hilbert space. If infinitely many $H\_t$ have dimension at least $2$ then this won't be separable.


That was the definition originally given by von Neumann, but it's not much used any more. The standard way to do things now is to fix a unit vector $u\_t \in H\_t$ for each $t$ and consider only those functions $h$ with $h(t) = u\_t$ for all but finitely many $t$. If you do this then a countable tensor product of separable Hilbert spaces will be separable.



> 
> If $H\_t = l^2$ for all $t$ as in your case, the most appropriate definition of a tensor product over $\mathbb{R}$ may be to take $\bigotimes\_{t \in \mathbb{R}} H\_t$ to be the symmetric Fock space $$\mathcal{F}\_s(L^2(\mathbb{R})) = \bigoplus\_{n \in \mathbb{N}} L^2(\mathbb{R})^{\otimes\_s n}.$$
>  Here $\otimes\_s$ denotes the symmetric part of the tensor product. This is a separable Hilbert space.
> 
> 
> 


How is this a tensor product, you ask? Let $(e\_n)$ be the standard basis of $l^2$. Then taking $u\_t = e\_0$ for all $t$, and using the second definition of tensor product given above, we get a Hilbert space with an orthonormal basis indexed by all functions $m: \mathbb{R} \to \mathbb{N}$ which are zero except at finitely many points. The basis itself will consist of the vectors $\bigotimes e\_{m(t)}$. Using this basis, the tensor product naturally decomposes into infinitely many summands, one for each value of $\sum m(t)$. Only $\bigotimes e\_0$ gives the value zero, so that is a summand of $\mathbb{C}$. To get the sum $1$ we need one $m(t) = 1$ and the rest zero, so those basis vectors span a copy of $l^2(\mathbb{R})$. To get the sum $2$ we need either one $m(t) = 2$ and the rest zero, or $m(t) = 1$ for exactly two values of $t$ and the rest zero; those basis vectors span a copy of $l^2(\mathbb{R})\otimes\_s l^2(\mathbb{R})$. And so on. You end up with the Fock space over $l^2(\mathbb{R})$. Thinking about tensor products in that way, it is natural to incorporate the measurable structure of $\mathbb{R}$ by putting $L^2(\mathbb{R})$ in place of $l^2(\mathbb{R})$.


Full details are given in Section 2.5 of my book *Mathematical Quantization*.


