1
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ What do you mean by "continuous basis"? What do you mean by "next one space"? How do you define $\bigotimes$? Usually we define tensor products if Hilbert spaces in a way that makes the result a Hilbert space. $\endgroup$
    – Nik Weaver
    Commented Mar 8, 2020 at 0:04
  • $\begingroup$ @NikWeaver: Sorry for my mistakes in English. I mean "counting basis" and "the following space". As for the definition of this tensor product, I'm not sure about its correctness. However, I suggested that construction like this one may exist. $\endgroup$ Commented Mar 8, 2020 at 8:10
  • $\begingroup$ It still sounds like you're not using the right word. Do you perhaps mean "countable basis" instead of "counting basis"? In any case, there is a construction for a possibly infinite tensor product of any collection of Hilbert spaces, indexed by any set, which could be the real numbers or any other set of the power of the continuum in your case. $\endgroup$ Commented Mar 8, 2020 at 10:39
  • 2
    $\begingroup$ The place to look here is the oeuvre of A. Guichardet. You can find online his notes ”Tensor products of $C^\ast$-algebras II. Infinite tensor products” which contains a chapter on such products of Hilbert space. $\endgroup$
    – user131781
    Commented Mar 8, 2020 at 11:12

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Thank you for the answer. This construction looks very interesting! $\endgroup$ Commented Mar 9, 2020 at 10:59
  • $\begingroup$ You are welcome! This is a basic idea in quantum field theory --- the idea is that you have a Hilbert space at each point in space (state of the field at that point) and you want to take their tensor products to get the "composite system". $\endgroup$
    – Nik Weaver
    Commented Mar 9, 2020 at 12:02
  • 1
    $\begingroup$ Why is von Neumann original definition not used anymore? $\endgroup$
    – ceillac
    Commented Mar 3, 2022 at 3:44
  • $\begingroup$ Yes, let me second @ceillac's question: why has von Neumann's construction fallen out of fashion? The currently fashionable construction looks a lot like an inductive limit of tensor products over finite subsets of the index set. Why is this approach "the good one"? Is it only because von Neumann's construction provided a non-separable space? How do we settle on what "good" means? (I suppose that if the Hilbert spaces involved are also algebras, then the natural choice is to take $u_t = 1 \in H_t$?) $\endgroup$
    – Alex M.
    Commented Apr 28, 2022 at 17:57
  • $\begingroup$ I wouldn't say it's disliked because it's nonseparable, but the fact that it's nonseparable is a strong hint that it's not the "right" definition. Ultimately people use the constructions that help them prove theorems. In the case of infinite tensor products, as far as I know the main application is physics, where the part I describe is the physically relevant part and the rest is irrelevant. $\endgroup$
    – Nik Weaver
    Commented Apr 28, 2022 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.