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?


1 Answer 1


Yes, entangled states certainly do exist for (products of) Hilbert spaces spanned by a continuous basis. In fact, the first example of entanglement that was presented by Einstein, Podolsky, and Rosen in their famous 1935 EPR paper was of this kind, which was only later popularized in the form the simpler example of the entanglement of spin-states by Bohm, Bell, Mermin, and others in the mid-century. As you rightly recognize, the defining property of an entangled state is that it cannot be written as a product of two (or more) states, each belonging to the two (or more) factors of the given factorization of the Hilbert space. There is no reason as to why such states won't exist when the Hilbert space (and its factors in a given factorization) are spanned by continuous bases.

For example, to cite from the original EPR paper referenced above (i.e., the first example of an entangled state recognized in human history ;-)),

$$\psi(x,y)=\int dp\ e^{-i(x-y)p}$$

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.

  • $\begingroup$ 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. $\endgroup$
    – Ignacio
    Sep 5, 2021 at 3:25
  • $\begingroup$ @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. $\endgroup$
    – user87745
    Sep 5, 2021 at 3:32

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