Schrodinger equation for Fermi pseudo potential This may very well be a very basic question, but I am somewhat confused by a version of the Schrodinger equation I have encountered studying quantum scattering.
Let us assume we have some potential that is sharply peaked around the origin, and vanishes otherwise. Intuitively this problem was introduced as contact scattering, e.g. particles only feel each other when they are located at the same position. Also assume that the wave function is spherically symmetric and can be written as $\psi(\vec{r}) = u(r)/r$. Then the two body Schrodinger equation reads, $$\left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}) - E\right] \frac{u(r)}{r} = 0.$$ To simulate a contact potential the Fermi-pseudo potential is introduced, $V(\vec{r}) = g \delta^{(3)}(\vec{r})\frac{\partial}{\partial r}(r\psi(\vec{r})) = g \delta^{(3)}(\vec{r}) u '(r)$, which is inserted in the Schrodinger equation. So far so good, however the author then introduces the following permutation on the kinetic energy term, $$ \nabla^2 \left[\frac{u(r)}{r}\right] = -4\pi u(0) \ \delta^{(3)}(\vec{r}) + \frac{1}{r} u''(r).$$
This is where I get confused. My first intuition would be to simply evaluate the laplacian in spherical coordinates, but this only gives a $u''(r)$ term. Where does this $u(0)\delta^{(3)}(\vec{r})$ term come from? I suppose there is some separation in terms of the origin and the remaining space happening here, but how does one justify this mathematically?
Thanks for the help!
 A: This is a problem I had as well. During your expansion of the Laplacian, I assume you had something that looked like this:
$$\nabla^2 \left( \frac{u(r)}{r} \right) = \frac{\nabla^2 u}{r} + u(r) \nabla^2 \left( \frac{1}{r}\right) + 2\,\, \vec{\nabla}\left(\frac{1}{r}\right)\cdot\vec{\nabla}u.$$
The important thing to notice is that you have the Laplacian of $1/r$, which might seem to be zero, but which actually isn't. In fact, it is zero almost everywhere, except at the origin $r=0$ where it isn't specified. There is a famous result that show that:
$$\nabla^2\left(\frac{1}{r}\right) = -\vec{\nabla}\cdot\left(\frac{\hat{r}}{r^2}\right) = - 4 \pi \delta^3(\vec{r}),$$
which is exactly what we described: a function that's zero everywhere except at the origin. (A fun way to prove this is to plug in the potential and the charge density for a point charge into Poisson's Equation $\nabla^2\phi=-\rho(r)/\epsilon_0$, and this identity falls right out.)
Plugging this in, we get:
$$\nabla^2 \left( \frac{u(r)}{r}\right) = \frac{1}{r} u''(r) - 4\pi u(r) \delta^3(\vec{r}).$$
Of course, the delta function is zero for all values of $r\neq 0$, and so we can easily say $$u(r) \delta^3(\vec{r}) \equiv u(0) \delta^3(\vec{r})$$ wherever it matters, and so
$$\nabla^2 \left( \frac{u(r)}{r}\right) = \frac{1}{r} u''(r) - 4\pi u(0) \delta^3(\vec{r}).$$
