# Entanglement and continuous basis sets

I've usually seen entanglement discussed when dealing with discrete basis sets. For example, if we consider the Hilbert space $$\mathcal{H}_0\otimes\mathcal{H}_1$$, where both $$\mathcal{H}_j$$ are two dimensional, the state $$|\psi\rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle\otimes|\downarrow\rangle - |\downarrow\rangle\otimes|\uparrow\rangle)$$ is entangled, since it can't be written as $$|\alpha\rangle\otimes|\beta\rangle$$, with $$|\alpha\rangle \in \mathcal{H}_0$$ and $$|\beta\rangle \in \mathcal{H}_1$$.

I think that a similar thing can happen in position basis; it may not be possible to express the associated wave function $$\psi(x,y)$$ as $$f(x)g(y)$$. I believe an example of this to be $$\psi(x,y) = A(x + y) \mathrm{e}^{-x^2 - y^2}$$ where $$A$$ is a normalization constant. In Dirac notation I would probably write something like: $$|\psi\rangle = \int \mathrm{d}x\mathrm{d}y |x y\rangle A(x + y) \mathrm{e}^{-x^2 - y^2}$$

where $$|xy\rangle = |\mathbf{r}\rangle = |x\rangle\otimes|y\rangle$$. My question is: Is it meaningful to call such a state "entangled" in the same way? Is "entanglement" between coordinates different than entanglement when the underlying Hilbert spaces are discrete?

$$\psi(x,y)=\int dp\ e^{-i(x-y)p}$$
is an entangled state. Basically, any function that is non-separable would represent an entangled wavefunction. The example you provide is also non-separable and thus, would represent an entangled state. It would be rather violently non-normalizable tho due to the linear dependence on $$x,y$$. But, you get the idea. A relatively more normalizable wavefunction $$\psi(x,y)\sim\sin(xy)$$ would be another example of an entangled wavefunction.
• I really like the answer, but I believe you may have misread my example, it has a factor (x + y) before the gaussian to prevent the factorization, $e^{(-x^2 -y^2)}$ is factorizable, but I think $(x+y)e^{(-x^2-y^2)}$ is not. I will accept it if you correct the issue or show how $(x+y)e^{(-x^2-y^2)}$ is factorizable in case I'm missing something. I will certainly look at the paper, I had always thought the proposed wavefunction was a discrete one. Sep 5, 2021 at 3:25
• @Ignacio Sorry, I read it as $A(x,y)$ -- as in arguments of the normalization function $A$. It didn't make much sense because the normalization factor would be a constant anyway but I thought it was your idiosyncratic way of writing things ;-) Anyway, yes, your example is non-separable. I have edited the answer.