# Does spin entanglement imply position entanglement?

My question is whether two electrons can be entangled only with respect to their spins but not with respect to some other observable, such as position.

I initially believed that spin-entanglement doesn't entail position or momentum entanglement. But if particles $$a$$ and $$b$$ have a composite state given by $$|\Psi\rangle_{ab} = \frac{1}{\sqrt2}(|1\rangle|0\rangle + |0\rangle|1\rangle)$$, where $$|0\rangle$$ and $$|1\rangle$$ are distinct z-spin eigenstates, then it would mean that a single particle, say, $$a$$, does not have a quantum state in the single-particle Hilbert space for $$a$$, $$\mathscr{H}_a$$.

Since the state of the particle $$a$$ can only be represented by a reduced density matrix and does not correspond to any vector in $$\mathscr{H}_a$$, its position or momentum would not be expressible in terms of position or momentum eigenstates in $$\mathscr{H}_a$$. It seems that this can only mean that $$a$$ and $$b$$ are entangled with respect to any (single-particle) observable.

Is this correct? If so, however, what about their mass or charge? It seems weird to think that the particle doesn't have an eigenstate concerning mass and charge in $$\mathscr{H}_a$$.

• I'm not going to write a full answer because I expect someone will do a much better job than I am capable of. But when I represent a single electron wave function, don't I usually need to include both a spatial part and a spin part, if I'm working in position space? And for two electrons I can for instance imagine both electrons sharing the same spatial part of the wave function (symmetric in space) but antisymmetric in their spin parts. Commented Apr 21, 2023 at 18:17
• @MariusLadegårdMeyer this goes even further: electrons, being fermions, must have an overall antisymmetric state, so if the spatial (spin) part is symmetric (antisymmetric) then the spin (spatial) part must be antisymmetric (symmetric). We know that the antisymmetric part must be entangled (see the singlet state for spins), while a symmetric part may or may not be entangled (contrast the states $|x\rangle|x\rangle$ vs $(|x\rangle|y\rangle+|y\rangle|x\rangle)/\sqrt{2}$). Commented Apr 21, 2023 at 20:25
• may be my answer here will help physics.stackexchange.com/questions/439450/… Commented Apr 22, 2023 at 5:57