Apropos of nothing in particular, I thought I would play around with the Schrödinger equation in 2D with a delta function potential. To keep things simple I thought I would concentrate on the bound state(s) with zero angular momentum. The equation should then be $$ -\frac{\hbar^2}{2m} \frac{1}{r}\partial_r(r\partial_r \Psi) - U \frac{\delta(r)}{2\pi r}\Psi= -\epsilon \Psi $$ where $\epsilon$ and $U$ are both greater than zero and $U$ has dimensions of length squared times energy. I try to simplify and non-dimensionalize (Is that a word?) and get $$\frac{1}{\rho}\partial_\rho(\rho \partial_\rho \Psi) + u \frac{\delta(\rho)}{\rho}\Psi -\Psi = 0$$ where $$\rho = \frac{\sqrt{2m\epsilon}}{\hbar}; u = \frac{mU}{\pi\hbar}.$$

Away from the origin, this immediately reduces to a modified Bessel equation, solvable by some combination of $I_0(\rho)$ and $K_0(\rho)$. But $\Psi$ has to be normalizable, so $K_0(\rho)$ (or some multiple thereof) is the solution. $K_0(\rho)$ diverges logarithmically at the origin, but the divergence is slow enough to allow normalization.

So now I want to find the energy eigenvalue(s) of the allowed bound state(s). So I integrate in a small disk around the origin $$\int_0^{\rho^\prime} 2\pi \rho \left(\frac{1}{\rho}\partial_\rho(\rho \partial_\rho \Psi) + u \frac{\delta(\rho)}{\rho}\Psi -\Psi\right)\mathrm{d}\rho$$ and try to take the limit as $\rho^{\prime}$ goes to zero.

And here's where my problem comes. The solution $K_0(\rho) \sim \ln(\rho) $ near the origin. So the first term integrates to a constant value no matter how the radius of the disk shrinks. The third term integrates to zero since a log diverges slower than $\rho$ goes to zero, and the second term produces an infinity. I know that $K_0$ has other terms, of course. But as I understand it none of them are going to eliminate the infinity in the second term or supply a canceling infinity in the other terms. So how is this equation correct? And, to get back to what I was trying to do in the first place, what is the energy eigenvalue? How to calculate it?

UPDATE AND RESPONSE TO CLOSING: So sorry, but the question tagged as a duplicate to this one involves a 3D delta function, while I am dealing with 2D, a different problem. Also, I asked for some help on some mathematical peculiarities which might be briefly touched on in the other question with its reference to regularization, but there is no way to really tell. Finally, I asked if there is a way to calculate the energy eigenvalue(s). The referenced question has much of interest to say about energies, but does not give a way to calculate them. I am puzzled as to why this question would be flagged as having a duplicate.

  • $\begingroup$ Usually the energy eigenvalues are found for the far-from-origin limit, not the near-origin limit. $\endgroup$ Commented Apr 12 at 2:14
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    $\begingroup$ It's possible/likely that the 2D delta function potential requires regularization, in which case the "simple-minded" treatment that you do for the 1D system doesn't work anymore. You might be running into this issue. In fact, a quick search for research papers indicates that 2D and 3D delta-function potential systems have an infinite number of bound states with the lowest having infinite binding energy. $\endgroup$
    – march
    Commented Apr 12 at 4:27
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    $\begingroup$ The 2D delta potential is discussed in my Phys.SE answer here: physics.stackexchange.com/q/24009/2451 $\endgroup$
    – Qmechanic
    Commented Apr 12 at 7:36

1 Answer 1


Congratulations! You (probably) just stumbled for the first time in your life with the fine art of regularization and renormalization, the craft of making sense and getting rid of infinities in quantum mechanics and, most importantly, in quantum field theories. These are advanced and non-intuitive ideas that are fundamental for QFT, and are usually introduced then, so do not worry too much if you do not understand this on the first try (nor on the fifth!).

As you have seen, using a 2D delta function, your physical predictions diverge (for example, the energy). So you need to regularize it, i.e. make it finite. There are several approaches to this:

  • One option is to modify your potential to $$ V(r,R) = \frac{\lambda}{2\pi R}\delta^{(1)}(r-R),$$ where $\delta^{(1)}$ is a one dimensional delta function (you can see that as $R\to0$ this diverges and you "recover" a 2D delta). You then shall study how you must change $\lambda$ as you change $R$ so that the bound state energy remains fixed. The main idea of this process is that your theory is now accurate up to an energy $\propto 1/R$. If you want your theory to work at higher energy scales, you need to modify your parameters. You have thus modified your theory to make it work at "low" energies.

  • Another option is to modify your Hamiltonian to add something called counterterms, which "get rid" of the divergent part of your physical predictions. Your new Hamiltonian would be $$ H(\vec{p},\vec{r}) = \frac{\vec{p}^2}{2m} + U\delta^{(2)} (\vec{r}) + \delta_U \delta^{(2)} (\vec{r}), $$ where $\delta_U$ is the counterterm and the deltas are again 2D. This is a renormalized Hamiltonian. Then you shall perform all your physical predictions in $D=2-\epsilon$ dimensions, a process called dimensional regularization. You must then identify the divergent part of these predictions as $\epsilon\to0$. Your counterterm shall then be fixed as minus the divergent part as $\epsilon\to0$ (that's why it's called a counterterm, it counters the divergence).

Now, the involved calculations might get a bit messy. I recommend taking a look at this paper, where they cover this in more detail. I hope this answer helps you!

This is quite a difficult topic to introduce, so please ask me anything in the comments, and I hope the SE community can provide with links to other novice-friendly posts on renormalization.

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    $\begingroup$ Wow! I thought I was just going to get a little SE practice in! Well, this is really interesting. Will take a look at your reference and give it some thought. Accepted your answer. Thanks! $\endgroup$ Commented Apr 20 at 18:56

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