Wavefunction symmetrization for bosons, or antisymmetrization for fermions, renders the wavefunction no longer a simple tensor product, i.e. it is no longer separable. This is the same thing that entanglement does more generally. Is there some deeper relationship between these two aspects?
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$\begingroup$ Search for entanglement of identical particles or so. It is a subtle topic, but there is literature. $\endgroup$– Tobias FünkeCommented May 21 at 14:17
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2$\begingroup$ Does this answer your question? Are identical particles always entangled even when not interacting? See also the links therein. $\endgroup$– Tobias FünkeCommented May 21 at 14:18
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One subtlety here is that entanglement between subsystems can occur when the Hilbert space has a tensor product decomposition $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$. But for a boson (and similarly for a fermion, adding an anti- in appropriate places), the two-particle Hilbert space does not have this kind of decomposition. The only physical states are the symmetric ones, so there's no way to actually split up $\mathcal{H}$ into "particle 1's Hilbert space" and "particle 2's Hilbert space" in such a way that they could be entangled in the first place.
That being said, there's an obvious isometry $V: \mathcal{H} \hookrightarrow \mathcal{H}_1 \otimes \mathcal{H}_2$ so you can view the state as entangled after applying that isometry. But then the entanglement will depend on the specific map you choose, and unlike "actual" forms of entanglement, it doesn't have a clear operational meaning, because the isometry $V$ is not physical.
Edit: Apparently I was wrong! The post linked in the comment above includes this paper which says you CAN give an operational meaning to this kind of entanglement.
