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If we have one particle we first look at an orthonormal basis of the one-particle Hilbert space $|n\rangle$. Here $n$ is the abbreviation for a compete set of quantum numbers, for example $n = (\vec{x}, s, m)$, i.e. the eigenvalues of the operators $\vec{\hat{x}}, \vec{\hat{S}}^2$ and $\hat{S}_3$. As a basis of the $N$-particle Hilbert space we can then use $$|n_1\rangle \otimes |n_2\rangle \otimes \cdots \otimes |n_N\rangle\, .$$

Question: What does "$\otimes$" mean here? Is that the algebraic tensor product?

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    $\begingroup$ Do you know another tensor product that could be defined on Hilbert spaces? $\endgroup$ – ACuriousMind May 31 '15 at 15:14
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    $\begingroup$ @ACuriousMind: I've heard from the topological tensor product. But I don't know what that is. $\endgroup$ – Ystar May 31 '15 at 15:16
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On the level of elements, it doesn't really make sense to ask whether $\otimes$ is the algebraic or the topological tensor product. Here's why:

For Hilbert spaces, the topological tensor product is the completion of the algebraic tensor product with respect to the Hilbert norm. This means the algebraic tensor product is a dense subset of the topological version, so the tensor products of elements of the space is a tensor that lies in both the algebraic and the topological product.

If they claim that the tensor products of elements are a basis for the tensor product space, then one must ask in which sense they are a basis - i.e. are infinite sums of basis vectors allowed. Since the physicist usually allows this (implicitly, by talking about countable bases of infinite-dimensional spaces), we should assume the text means that. Allowing infinite sums, the tensor products of basis elements of the spaces that are tensored are such a (Schauder) basis of the topological tensor product (since that is defined as the completion, i.e. "adding limits" of finite combinations of the basis, which is what an infinite sum/series is).

Therefore, the Fock space that is constructed this way is intended to be the topological tensor product.

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  • $\begingroup$ Doesn't this mean that in case of Hilbert spaces they are both the same because of compactness? Am I missing something? $\endgroup$ – CuriousOne May 31 '15 at 15:26
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    $\begingroup$ @CuriousOne: You mean completeness, and yes, you are missing something because the algebraic tensor product of two complete spaces is not guaranteed to be complete - that is why the topological version has to take the completion. The Hilbert case is far more well-behaved than the general case of tensoring Banach spaces, though, since it is obvious which norm to define on the product. $\endgroup$ – ACuriousMind May 31 '15 at 15:28
  • $\begingroup$ Ah! A light bulb just went on! Thanks! (And yes, I did mean completeness - not fully away yet, I need more coffee!). $\endgroup$ – CuriousOne May 31 '15 at 15:36
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    $\begingroup$ The products of (Schauder) basis elements will form a (Schauder) basis of the topological tensor product. What a physicist calls the basis of the individual Hilbert spaces is already not really an algebraic (= Hamel) basis, the algebraic basis of an inifinite-dimensional space is not countable! It is a Schauder basis, which has just a dense linear hull. The products of the basis elements will by construction have a dense linear hull and therefore be a Schauder basis. $\endgroup$ – Sebastian Riese May 31 '15 at 16:07
  • $\begingroup$ @SebastianRiese: You are correct. Somehow I never realized that the physicist's basis is a Schauder basis. $\endgroup$ – ACuriousMind May 31 '15 at 16:12

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