How can i obtain the correction to the eigenvalues as a result of adding the gravitational potential to the hydrogenlike atom?

Question

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?

Answer 1

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

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