This question was motivated by a question asked yesterday (How to find the electrostatic potential of a hydrogen-like charge density?). This got me wondering about what would result from solving the Schrodinger equation for a point proton in the electrostatic potential resulting from the charge distribution of an electron in the 1s orbital of hydrogen. Does anyone know of such an attempt, and if so what is the result?

My guess is that the eigenvalue obtained from this solution (which may need to be obtained numerically) will be very close to the eigenvalue of the 1s electron orbital of hydrogen, but I have never heard of this problem being attempted.

  • $\begingroup$ I'd be interested to hear a qualified answer too. My small quip would be to say that to me inverse problem means something like we: "measure the scattered light from hydrogen and deduce the structure of electron shells from it". Different terminology, I guess. $\endgroup$ – Cryo Sep 25 '19 at 16:21
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    $\begingroup$ This is the same problem as the ordinary hydrogen atom. Note that in the H atom derivation you use the center of mass and relative coordinates. This means the 1s orbital represents not the position of the electron, but the distribution of relative positions of the electron and proton. $\endgroup$ – KF Gauss Sep 25 '19 at 16:55
  • $\begingroup$ @Cryo I was also bothered by using "inverse" but I couldn't think of another name for it. If anyone has a better (more descriptive) suggestion I will edit the title. $\endgroup$ – Lewis Miller Sep 25 '19 at 18:09
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    $\begingroup$ @KFGauss This is not quite the same problem, but I agree it is similar. For the standard problem we assume point particles for both electron and proton. For this problem we continue to treat the proton as a point particle but assume the electron is a smeared out charge distribution. What interests me is how much difference this makes for the gs eigenvalue. The wave function will be quite different because the radial coordinate will be the displacement of the proton from the cm whereas the radial coordinate for the 1s orbital represents the displacement of the electron from the cm.. $\endgroup$ – Lewis Miller Sep 25 '19 at 18:22

Normally we think of the hydrogen atom as an electron moving in the coulomb potential of the proton

For the folks that weren't paying attention, yes, this is a common outlook. For the people that did read their textbooks with care, no, we don't.

The Schrödinger equation for the hydrogen atom, within the electrostatic non-relativistic approximation, reads $$ \left[ -\frac{\hbar^2}{2m_p} \nabla^2_p -\frac{\hbar^2}{2m_e} \nabla^2_e - \frac{e^2}{|\mathbf r_e - \mathbf r_p|} \right] \Psi(\mathbf r_e, \mathbf r_p) = E \Psi(\mathbf r_e, \mathbf r_p), $$ where $\mathbf r_e$ and $\mathbf r_p$ are the positions of the proton and the electron, respectively, and $\nabla^2_e$ and $\nabla^2_p$ are the corresponding laplacians. To simplify this equation, we change coordinates to a center-of-mass + relative coordinate set, i.e. using \begin{align} \mathbf R & = \frac1{m_p+m_e}\left(m_p\mathbf r_p + m_e\mathbf r_e\right) \\ \mathbf r & = \mathbf r_e-\mathbf r_p, \end{align} whose inverse transformation is \begin{align} \mathbf r_e & = \mathbf R + \frac{m_p}{m_p+m_e} \mathbf r \\ \mathbf r_p & = \mathbf R + \frac{m_e}{m_p+m_e} \mathbf r , \end{align} and in terms of which the Schrödinger equation separates into $$ \left[ -\frac{\hbar^2}{2(m_p+m_e)} \nabla^2_\mathbf{R} -\frac{\hbar^2}{2\mu} \nabla^2 - \frac{e^2}{|\mathbf r|} \right] \Psi(\mathbf r, \mathbf R) = E \Psi(\mathbf r, \mathbf R), $$ i.e. with a free-particle Schrödinger equation for the center-of-mass coordinate, and a 'standard-form' hydrogenic Schrödinger equation for the relative coordinate $\mathbf r$ with its corresponding laplacian $\nabla^2$, with the mass set to the reduced mass $$ \mu = \frac{1}{\frac1{m_e}+\frac1{m_p}} \approx m_e(1-m_e/m_p). $$

In other words, the standard Schrödinger equation already accounts for the motion of the proton.

As such, regarding your statement that

Normally we think of the hydrogen atom as an electron moving in the coulomb potential of the proton, but an alternative is to solve for the eigenvalue and wave function of the proton in the potential associated with the smeared-out charge distribution of the 1s electron orbital

no, this is not an "alternative", unless by that term you mean something like "this is an alternative calculation that one might embark on if one is simply looking for things to calculate instead of trying to describe reality". The calculation you propose is indeed a valid calculation one might attempt, but it has nothing to do with the hydrogen atom, and its results will be basically meaningless.

Because of this, while it is plausible that people have attempted such a calculation, I find it extremely hard to envision a scenario in which such a calculation would have found its way to the published record (though you do find solutions for the Yukawa potential, which is similar, so maybe it's not hopeless).

That said, though:

My guess is that the eigenvalue obtained from this solution (which may need to be obtained numerically) will be very close to the eigenvalue of the 1s electron orbital of hydrogen.

That's a terrible guess. If you mean that you're solving the Schrödinger equation $$ \left[ -\frac{\hbar^2}{2m_p} \nabla^2_p - \frac{e^2}{4\pi\epsilon_0} \frac{e^{-\alpha r}}{r} \left(1+\frac{\alpha r}{2}\right) \right] \Psi(\mathbf r_p) = E \Psi(\mathbf r_p) $$ (with the potential pulled straight from the question you linked to, which, I should note, probably corresponds to an excited state of the hydrogenic equation, and not the ground state), then for small $r$ the behaviour will be dominated by the Coulombic singularity and the ground state will correspond to that.

However, it will do so with the proton mass, rather than the electron, which means that the wavefunction will find it much easier to localize to a smaller region, and it will therefore shrink much further down in the potential well. In other words, the energy eigenvalue here will have nothing to do with the 1s energy of hydrogen.

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    $\begingroup$ This is a good, correct answer and I upvoted it. But I think it's unnecessary to belittle the question ("that's a terrible guess" etc). OP has a common misconception and came here for help. $\endgroup$ – doublefelix Sep 25 '19 at 18:32
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    $\begingroup$ I am quite aware that the use of the reduced mass means that the proton's motion is already included in the regular solution. Indeed, I not only read that part of my textbook with care but I also taught it. As for the use of the word "alternative" I did mean an alternative calculation and the reality I have in mind is something other than the hydrogen atom. I await a solution to decide whether the result is meaningless. $\endgroup$ – Lewis Miller Sep 25 '19 at 18:48
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    $\begingroup$ "(former) proton radius puzzle?": If you know of a resolution to the proton radius puzzle, please provide me with a link. $\endgroup$ – Lewis Miller Sep 25 '19 at 20:16
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    $\begingroup$ @Emilio Pisanty, I simply don't understand why you have such an aggressive tone. $\endgroup$ – KF Gauss Sep 26 '19 at 16:06
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    $\begingroup$ I highly doubt you speak to people like that in person, but maybe I'm just old-fashioned. Let's leave it at that. $\endgroup$ – KF Gauss Sep 27 '19 at 0:47

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