Hydrogen atom in 2 spatial dimensions with logarithmic potential

Question

I was trying to solve for the wave function of an electron in a hydrogen atom confined to only two dimensions (in flatland). First of all I figured out the electrostatic potential energy in flatland. This came out to be $$U=-\frac{Ze^2}{2\pi\epsilon_0}\log{r}.$$ Here I have chosen $r=1$ as the surface of zero potential.

The Schrödinger equation in two dimensions is $$-(\frac{1}{r}\partial_r(r\partial_{r}\Psi)+\frac{1}{r^2}\partial^{2}_{\theta}\Psi) + U\Psi=E\Psi.$$ To avoid excessive typing I have decided to set $\hbar/2\mu$ to one. Using $\Psi=R(r)Y(\theta)$ the equation can be separated into radial and angular equations.

The angular equation is $$\partial^{2}_{\theta}Y=-m^2Y$$ and the radial equation is $$\frac{1}{r}\partial_r(r\partial_{r}R)+(E+\frac{Ze^2}{2\pi\epsilon_0}\log{r}-\frac{m^2}{r^2})R=0.$$

I am able to easily solve and quantise the angular equation but I have been stuck on the radial equation for days now. I do not know how to solve the radial differential equation (I am not very good at solving second order DE). I am also not able to figure out how to quantise the total energy of the electron.

Questions:

1. How to solve the radial wave function equation and how does the energy quantise in such an atom?

2. When I looked for online sources for help I found that everyone considered the potential energy to be of the form $1/r$ instead of $\log(r)$. According to Gauss law the electrostatic potential will be logarithmic in two dimensions. Why do they consider the potential to be of the form $1/r$?

Answer 1

I agree that this equation probably does not admit a solution in terms of elementary functions. However, with a little digging I found this paper: Atabek et. al. Phys. Rev. A 1974 covering your questions (they use a logarithmic 2D potential and they discuss the spectrum of energy eigenvalues).

One useful thing they do is substitute $R_m = r^{-½}f_m$ to eliminate the term proportional to $R^\prime$ in the equation and to generate an equation of the form: $f^{\prime \prime}_m = g(r,m) f_m$ which is nicer to work with.

Regarding a numerical solution:

What you have here is a "2-point boundary value problem (BVP)" (since you specify the value of $R_m$, or equivalently $f_m$, at $r = 0$ and you want it to vanish as $r \rightarrow \infty$) which are often treated with so-called "Shooting Codes".

If you are familiar with simpler (1-step, explicit) numerical methods such as RK4, one simple thing to do is iterate over values of $E$ with a guess for $R^\prime(0)$ and check the value of solution ($R$) at some large, final value of $r$. Depending on your units, even something like $r_f = 10$ seems to suffice. Then you can manually adjust your search-grid of $E$'s until this final value, $R(r_f)$, is approximately zero.

I coded-up this RK4-iterative scheme just to check the qualitative shape of the radial wave functions (with $m = 0$) reported in the linked paper from 1974 and they seem to agree (c.f. Fig. 4 of the paper).

One caveat: don't pay attention to the numerical values of $E$ or $R_{m=0}$ in the following figure, the units are probably a bit confused. Also, I normalized the peaks of $R_m$ to 1 for comparison across $m$'s.

Nevertheless, for qualitative inspection, here are three eigenfunctions of the equation, one for $m = 0$, and two for $m = 1$. These were found using the iterative RK4 method.

Answer 2

I doubt that this equation is solvable, although it would be wise to check in a book on special functions, such as Abramovitz&Stegun or Gradshtein &Ryzhik.

Your equation seems to be correct for a flatland. Many problems however deal with real 3D world where the motion is confined to two or even one dimensions. In this case the potential remains 3D: $1/r$. One-dimensional case is notable, since the binding energy diverges, which long posed a problem for analyzing excitons in carbon nanotubes.