Can it be shown in experiment that the momentum (or position) states of the electron and proton in hydrogen are entangled? The states of the electron and proton in hydrogen are entangled. Which means that the momentum and position of both are entangled. Can this be shown in an experiment, so if you measure the momentum (or a small range of them) of the proton you will know in advance what the momentum of the electron will be?
If you first have to separate them, to do a measurement on them, will the information of the entanglement show itself in the measurement? As the information of entanglement of two spin states can show itself in measurement?
 A: In a two body bound state, (quantum mechanical or not) from conservation of energy and momentum, once the masses are known, measuring  the four vector of one particle , the four vector of the other is known.
A: Indeed, the states of the electron and the proton in a hydrogen atom are entangled. The wave function of a hydrigen atom has form
$$
\psi(\mathbf{x}_e, \mathbf{x}_p) = e^{i\mathbf{k}\mathbf{R}}\phi(\mathbf{x}_e- \mathbf{x}_p),
$$
where
$$
\mathbf{R}=\frac{m_e\mathbf{x}_e+ m_p\mathbf{x}_p}{m_e+m_p}
$$
is the position of the center-of-mass of the atom, whereas $\phi(\mathbf{x}_e- \mathbf{x}_p)$ is one of the hydrogen eigenfunctions discussed in any quantum mechanics text. Function $\psi(\mathbf{x}_e, \mathbf{x}_p)$ can be expanded in terms of momentum eigenstates of electron and proton
$$
\psi(\mathbf{x}_e, \mathbf{x}_p) = \int d^3\mathbf{k}_e \int d^3\mathbf{k}_pe^{i\mathbf{k}_e\mathbf{x}_e}e^{i\mathbf{k}_p\mathbf{x}_p} \varphi(\mathbf{k}_e, \mathbf{k}_p),
$$
which is an entangled state. We could further calculation the coefficients $\varphi(\mathbf{k}_e, \mathbf{k}_p)$ using the form of the wave function given above as:
$$
\varphi(\mathbf{k}_e, \mathbf{k}_p) = \int d^3\mathbf{x}_e \int d^3\mathbf{x}_pe^{-i\mathbf{k}_e\mathbf{x}_e}e^{-i\mathbf{k}_p\mathbf{x}_p}\psi(\mathbf{x}_e, \mathbf{x}_p) =\\
\int d^3\mathbf{q}\tilde{\phi}(\mathbf{q}) \delta^3\left(\mathbf{q}+\mathbf{k}\frac{m_e}{m_e+m_p}-\mathbf{k}_e\right)
\delta^3\left(\mathbf{q}-\mathbf{k}\frac{m_p}{m_e+m_p}-\mathbf{k}_p\right)=\\
\delta^3\left(\mathbf{k}-\mathbf{k}_e-\mathbf{k}_p\right)
\tilde{\phi}\left(\frac{m_p\mathbf{k}_e + m_e\mathbf{k}_p}{m_e+m_p}\right),
$$
where $\tilde(\mathbf{q})$ is the Fourier transform of $\phi(\mathbf{x})$.
Separating the proton and the electron would mean changing the state by ionizing the atom - whether the two will remain entangled depends on the process of separation. Measuring a momentum of one of the two particles is hard due to their strong binding, but this probably could be done in accelerator experiments. We then see from the equation above that the momentum state of one particle is related to the momentum state of the other - if we do manage to measure the momentum of the electron, we woudl obtain the momentum of proton as $\mathbf{k}_p=\mathbf{k}-\mathbf{k}_e$. It could be much easier to emasure their angular momenta.
