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my question can be quickly stated as follows: is the tensor product of two bosonic Fock spaces again a Fock space?

More precisely, given a separable Hilbert space $\mathfrak{h}$, I have the following Hilbert space \begin{equation} \mathcal{F}_1:=\Big(\bigoplus_{j=0}^\infty \mathfrak{h}^{\otimes_{\operatorname{sym}}j} \Big)\otimes \Big(\bigoplus_{k=0}^\infty \mathfrak{h}^{\otimes_{\operatorname{sym}}k} \Big), \end{equation} which I interpret as the space of two distinguishable species of identical bosons. As it is written, it has no clear Fock space structure, but my claim is that it is somehow naturally isomorphic to the space \begin{equation} \mathcal{F}_2:=\bigoplus_{n=0}^\infty(\mathfrak{h}\oplus\mathfrak{h})^{\otimes_{\operatorname{sym}}n}. \end{equation} For this, I have a good ``dimension counting'' argument that compares the first levels of both Fock spaces, as well as a map between creation and annihilation operators in the two cases.

I cannot find any references on that, but I cannot even believe it's a new result, somebody need have discovered that in the past. Could you maybe help me? Thanks a lot

popoolmica
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  • Well, they really are not that different as far as I can understand. Comparing $\mathfrak{h}^{\otimes_{\operatorname{sym}}2}\oplus\mathfrak{h}^{\otimes_{\operatorname{sym}}2}\oplus (\mathfrak{h}\otimes\mathfrak{h})$, which is the two-particle content of $\mathcal{F_1}$, with $(\mathfrak{h}\oplus\mathfrak{h})^{\otimes_{\operatorname{sym}}2}$, which is the two-particle content of $\mathcal{F_2}$ it seems that their dimensions are the same. This occurs for 3 particles as well, suggesting that it might be general; however, the proof could be very combinatoric in absence of a more abstract argument – popoolmica Jul 10 '17 at 17:50
  • Your right, this should be true, at least algebraically. Let $V$ and $W$ be any two (not necessarily finitely generated) vector spaces over a field. Then the map $${\rm sym}(V)\otimes {\rm sym}(W)\to {\rm sym}(V\oplus W);,$$ sending a $v\otimes w\mapsto (v,0)\otimes (0,w)$, defines an isomorphism. –  Jul 10 '17 at 19:51
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    Relevant?: Section 5.6. – Keith McClary Jul 11 '17 at 03:51
  • Yes! It is exactly what I was looking for, thank you very much – popoolmica Jul 11 '17 at 07:57

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