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In treating the hydrogen-like atoms, we normally neglect the gravitational potential for the system. I am able to justify the above statement by comparing the gravitational potential to the electrostatic potential, I need someone to help me on;

  1. how to obtain the exact eigenvalues of the complete system. Does this mean obtaining the eigenvalues whilst neglecting the gravitational potential?

  2. how to obtain the correction to the eigenvalue as a result of adding gravitational potential. Here, in the expression for energy, I have added $Gmm$ to $\frac{Ze^2}{4 \pi \epsilon_{o}}$ where $G$ is the gravitational constant and $m$ is mass of hydrogen-like particle, I am not sure whether I have done the correct thing.

  3. how can i obtain the exact eigenfunctions of the whole system?

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    $\begingroup$ Why do you want to do this? The electrostatic force between the proton & electron is of the order of $10^{40}$ times stronger than the gravitational force. So unless your atom is extremely isolated from all other electromagnetic influences, those influences will perturb the system more than its self-gravity does. $\endgroup$ – PM 2Ring Aug 18 at 12:33
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The potential energy is still just $1/r$,

$$V=-\frac{Ze^2}{4\pi\epsilon_0r}-\frac{Gm_pm_e}{r}=-\frac{Ze^2}{4\pi\epsilon_0}\left(1+\frac{4\pi\epsilon_0Gm_pm_e}{Ze^2}\right)\frac{1}{r},$$

so instead of the energy levels being

$$E_n=-\left(\frac{Ze^2}{4\pi\epsilon_0}\right)^2\frac{1}{2\hbar^2}\frac{m_pm_e}{m_p+m_e}\frac{1}{n^2}$$

they are

$$E_n=-\left(\frac{Ze^2}{4\pi\epsilon_0}\right)^2\left(1+\frac{4\pi\epsilon_0Gm_pm_e}{Ze^2}\right)^2\frac{1}{2\hbar^2}\frac{m_pm_e}{m_p+m_e}\frac{1}{n^2}.$$

For hydrogen, the dimensionless correction $\frac{4\pi\epsilon_0Gm_pm_e}{Ze^2}$ is only $4.4\times 10^{-40}$ and the energy shift of the ground state energy ($-13.6\,\text{eV}$) is only $-1.2\times 10^{-38}\,\text{eV}$. Since gravity is attractive, just like the electrostatic force between the proton and the electron, it makes the energy very slightly more negative.

As @PM2Ring commented, the gravitational correction is absurdly small and is swamped by many other physical effects being neglected here, so this is a fairly pointless (but amusing) calculation. For example, relativistic corrections to the kinetic energy and corrections due to electron and proton spin are much more important, which is why textbooks cover those effects but ignore this one.

The eigenfunctions follow from a similar correction to the Bohr radius which I will let you work out using

$$\frac{Ze^2}{4\pi\epsilon_0}\to\frac{Ze^2}{4\pi\epsilon_0}\left(1+\frac{4\pi\epsilon_0Gm_pm_e}{Ze^2}\right).$$

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  • $\begingroup$ Is just inserting a Newtonian potential in the Schroedinger equation and solving it valid in a quantum system? $\endgroup$ – user2723984 Aug 18 at 17:42
  • $\begingroup$ We use the electrostatic potential energy from classical electromagnetism in the Schrodinger equation. The analogous thing for gravity is to use the gravitational potential energy from classical Newtonian gravity. I have seen a much more sophisticated treatment using General Relativity and the Dirac equation! $\endgroup$ – G. Smith Aug 18 at 17:45
  • $\begingroup$ we do have QED to back that approximation up, but I see your point. The correction is so ridiculously small that even if there was a problem with doing that it wouldn't be measurable anyway.. $\endgroup$ – user2723984 Aug 18 at 17:49
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    $\begingroup$ That’s right. We can’t test this prediction. It’s just the kind of thing theorists do to amuse themselves. $\endgroup$ – G. Smith Aug 18 at 17:50
  • $\begingroup$ @PM 2Ring I saw this question in an examination paper otherwise you are right the gravitational force is negligible in that case. This is purely theoretical I guess. $\endgroup$ – pierse Aug 18 at 19:01

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