I know that two particles are entangled when their Hamiltonian contains an interaction part, but it seems to me that for the ground state of the hydrogen atom $\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{3/2}e^{-\frac{r}{a_0}}$, when under different reference frames, can be entangled in one frame and separable in another.

Say for the relative coordinate $\mathbf{r}=\mathbf{r_1}-\mathbf{r_2}=\mathbf{r_1'}-\mathbf{r_2'}$, where the former represents coordinates of a fixed reference frame and the latter represents that of the center-of-mass frame, 1 and 2 stand for the electron and the proton respectively. Then the ground state would have two different forms: $$\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{3/2}e^{-\frac{1}{a_0}(r_1^2+r_2^2-2r_1r_2\cos\alpha)^{1/2}}$$ (which is entangled), and $$\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{3/2}e^{-\frac{1}{a_0}(r_1'+r_2')}$$ (which is separable).

How so? Plus, I also know that separability is invariant under unitary operation, which makes me even more confused. As an initiate I really don't understand.

P.S. $\alpha$ is the angle between $\mathbf{r_1}$ and $\mathbf{r_2}$, $r=r_1^2+r_2^2-2r_1r_2\cos\alpha$ is from law cosines.

  • $\begingroup$ What is $\alpha$? $\endgroup$
    – garyp
    Commented Jul 11, 2017 at 11:22
  • $\begingroup$ @garyp: $\alpha$ is the angle between $\mathbf{r_1}$ and $\mathbf{r_2}$, sorry for not denoting properly. $\endgroup$
    – 2ub
    Commented Jul 11, 2017 at 11:46
  • $\begingroup$ Why do you seem to think this switch from $r_1,r_2$ to $r'_1,r'_2$ is a unitary operation? $\endgroup$
    – ACuriousMind
    Commented Jul 11, 2017 at 11:51
  • 1
    $\begingroup$ Actually one can show that not all change of coordinates allow for separation exactly for that reason: some of them do (when you start in an inertial reference frame and the operator that does the change is unitary and preserves the commutation relations), some other don't. $\endgroup$
    – gented
    Commented Jul 11, 2017 at 11:57
  • $\begingroup$ @ACuriousMind: Isn't a change of reference frame akin to unitary transform? I don't really know. $\endgroup$
    – 2ub
    Commented Jul 11, 2017 at 12:10

1 Answer 1


The problem is that you're understanding the meaning of the coordinates in the hydrogen atom wave function incorrectly. The hydrogen atom begins with two coordinates, the positions of the electron and proton, $\mathbf{r}_e$ and $\mathbf{r}_p$, respectively. The very first step in solving the hydrogen atom is to produce the center of mass position, $\mathbf{R} \equiv \frac{m_p \mathbf{r}_p + m_e \mathbf{r}_e}{m_p + m_e}$. This coordinate is completely free, and so it, and its wave functions, are those of a free particle.

The next step is to produce a translation invariant coordinate, so that all changes from spatial translation are isolated in the center of mass coordinate. Since we have three vectors, there are a lot of different choices we could make, but the standard one is to use the difference of the electron and proton position: $$\mathbf{r} \equiv \mathbf{r}_e - \mathbf{r}_p.$$

Under these two pairs of coordinates, the Hamiltonian takes the following forms: \begin{align} H & = \frac{\mathbf{p}_e^2}{2 m_e} + \frac{\mathbf{p}_p^2}{2 m_p} + V(|\mathbf{r}_e - \mathbf{r}_p|) \\ & = \frac{\mathbf{P}^2}{2 (m_e + m_p)} + \frac{\mathbf{p}^2}{2 \mu} + V(|\mathbf{r}|), \end{align} where $\mu$ is the reduced mass of the electron/proton system, $$\mu \equiv \frac{1}{\frac{1}{m_e} + \frac{1}{m_p}},$$ $\mathbf{P}$ is the momentum conjugate to $\mathbf{R}$, and $\mathbf{p}$ is conjugate to $\mathbf{r}$.

You can see why we chose to define $\mathbf{r}$ the way we did - it's the coordinate of the potential, and this choice is the only one that makes the time independent Schrödinger equation (the eigenvalue equation for the Hamiltonian) a separable partial differential equation.

I stress again, $\mathbf{r}$ is completely invariant under translations, so it doesn't transform when you shift the origin of your coordinate system. It's also invariant under Galilean boosts, $\mathbf{r}_{e/p} \rightarrow \mathbf{r}_{e/p} + \mathbf{v} t$, which are just time parameterized translations. It is not invariant under Lorentz boosts, but we didn't expect it to be.

In the hydrogen ground and excited states, $r$ is the length of $\mathbf{r}$, so it is always $r^2 = r_e^2 + r_p^2 - 2r_e r_p \cos \alpha$, but in changing back to that basis you have to include the wave function for the position of the center of mass. Otherwise you have a mismatch between what is being described by the wave function (in this case, a probability amplitude for the relative positions) and the variables in it.

To be concrete, suppose a hydrogen atom is in its ground state and its center of mass is spread over a Gaussian wave packet, so: \begin{align} \Psi(\mathbf{R},\, \mathbf{r}) & = \psi_{\mathrm{com}}(\mathbf{R}) \, \psi_{\mathrm{internal}}(\mathbf{r}) \\ & = \frac{1}{\sqrt{\sigma_c^3 (2\pi)^{3/2}}}\mathrm{e}^{-R^2 / (4 \sigma_c^2)} \frac{2}{\sqrt{a_0^{3}}} \mathrm{e}^{-r / (2 a_0)}. \end{align} Changing variables to $\mathbf{r}_p$ and $\mathbf{r}_e$ requires multiplication by the square root of the determinant of the Jacobian matrix ($\Psi^2$ is part of a differential form): \begin{align} \Psi(\mathbf{R},\, \mathbf{r}) & = \Psi(\mathbf{r}_p,\, \mathbf{r}_e) \sqrt{\left|\frac{\partial (\mathbf{r}_p,\,\mathbf{r}_e)}{\partial (\mathbf{R},\, \mathbf{r}) }\right|} \\ & = \frac{2}{\sqrt{\sigma_c^3 (2\pi)^{3/2} a_0^3}} \exp\left(\frac{m_e^2 r_e^2 + m_p^2 r_p^2 + 2 m_e m_p r_e r_p \cos \alpha}{4 \sigma_c^2 (m_e+m_p)^2} - \frac{\sqrt{r_e^2 + r_p^2 - 2 r_e r_p \cos\alpha}}{2 a_0}\right). \end{align} Notice that since $\mathbf{r}_p \cdot \mathbf{r}_e = r_e r_p \cos\alpha$ the angle $\alpha$ depends on where in space both the proton and electron are. There is no special coordinate system where $\alpha = 0$, always.

  • $\begingroup$ Thank you for your extensive insight, I have received much help, though a few questions linger. 1, is there no way to fix an origin on the center of mass, is my change of coordinates to $\bf{r'}$ invalid? 2, I know that the hydrogen wavefunction is derived through isolating the relative $\bf{r}$ and expelling the center of mass $\bf{R}$ to its own equation, but why does dealing with the internal part have to concern the external motion, since they are already separated from one another? 3, is there a way to rewrite a hydrogen eigenstate in single particle form, like a bell state? $\endgroup$
    – 2ub
    Commented Jul 13, 2017 at 8:38
  • $\begingroup$ @L.Quen The problem arises because there is no fixed center of mass - it has a wave function of it's own. You could change from $(\mathbf{R},\, \mathbf{r})$ to $(\mathbf{R},\, \mathbf{r}_e)$ or $(\mathbf{R},\, \mathbf{r}_p)$, but the formulae are very different from what you used: $r_e = m_p r / (m_e + m_p)$ and $r_e = m_e r / (m_e + m_p)$. This because when $\mathbf{R}=0$ $m_e\mathbf{r}_e = - m_p \mathbf{r}_p$. $\endgroup$ Commented Jul 13, 2017 at 13:28
  • $\begingroup$ Believe it or not, @L.Quen, what you're trying to do here is construct something analogous to a conditional probability (amplitude) in quantum mechanics. Thus, this question is relevant. $\endgroup$ Commented Jul 13, 2017 at 13:55

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