Consider two identical particles $A$ and $B$. The combined Hilbert space $\mathcal{H}_A\otimes \mathcal{H}_B \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} $ is a valid Hilbert space. But what about the physical Hilbert spaces: $$\mathcal{H}_A \odot \mathcal{H}_B=\textrm{Span}\{ \ket{i}\otimes \ket{j}+\ket{j}\otimes\ket{i}\}\tag{Bosons}$$ $$\mathcal{H}_A \wedge \mathcal{H}_B=\textrm{Span}\{ \ket{i}\otimes \ket{j}-\ket{j}\otimes\ket{i}\}\tag{Fermions}$$ How do we go about showing that they are valid Hilbert spaces (if indeed they are)?

It is trivial to show they are inner product spaces but I wouldn't know where to start in show completeness.


The (anti-)symmetrized spaces are the quotient of the full tensor product space $H\otimes H$ (since the particles are identical I'm dropping the subscript on the spaces) by a closed subspace, and the quotient of a Hilbert space by a closed subspace is again a Hilbert space, cf. e.g. this math.SE question.

What's left is to justify the claim that we are indeed quotienting by a closed (or "complete") subspace. In the case of the anti-symmetrized space, we're quotienting by the ideal $I$ generated by elements of the form $v\otimes v$ for all $v\in H$. Elements of this ideal have the form $\sum_i (v_i\otimes v_i)$, and if a limit of this exists at all, it will be of the form $v\otimes v$ for $v = \lim_{i\to\infty} v_i$, which is clearly again an element of the ideal, so the ideal is closed. A very similar reasoning works for the ideal we have to divide out in the symmetric case.

| cite | improve this answer | |
  • $\begingroup$ Hmm, what about the initial justification that the tensor product of Hilbert spaces is actually a Hilbert space? $\endgroup$ – Mozibur Ullah Nov 11 '18 at 12:10
  • $\begingroup$ @MoziburUllah The tensor product of Hilbert spaces is defined to be the completion of the inner product space you get from the ordinary tensor product of vector spaces, cf. en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces, i.e. it is a Hilbert space by definition. $\endgroup$ – ACuriousMind Nov 11 '18 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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