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.


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 $1/r$?

  • $\begingroup$ If a differential equation is not analytically solvable, you can always solve it numerically. $\endgroup$ – G. Smith Apr 15 '20 at 16:40
  • $\begingroup$ @G.Smith How exactly? $\endgroup$ – Manvendra Somvanshi Apr 15 '20 at 18:03
  • $\begingroup$ Wikipedia’s Numerical methods for ODEs mentions several algorithms. $\endgroup$ – G. Smith Apr 15 '20 at 18:17
  • $\begingroup$ Never think you're the first one to have thought about a problem: newton.ex.ac.uk/research/qsystems/portnoi/JMP_43_4681.pdf and web.math.ku.dk/~schlicht/DL/2013/HydrogenAtom.pdf $\endgroup$ – Gert Jun 4 '20 at 15:28
  • $\begingroup$ @Gert I have seen both of the papers you have sited here already. Both of them are different from the problem that I am (or was) solving. The potential used in both papers is the coulomb potential, which not applicable in flat land, it is only applicable in confined two dimensions in 3D space ( as explained by @Vadium). Also I never said that I am the first one to think of this in my post, all I said that I was having a problem here and needed help. $\endgroup$ – Manvendra Somvanshi Jun 4 '20 at 16:03

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.

enter image description here


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.