# Sphereically Symmetric potential well

## Homework Statement

A particle that movies in three dimensions is trapped in a deep spherically symmetric potential V(r):

V(r) = 0 at r < r$_{}0$
--> ∞ at r ≥ r$_{}0$

where r$_{}0$ is a positive constant. The ground state wave function is spherically symmetric, so the radial wave function u(r) satisfies the one-dimensional Schroedinger energy eigenvalue equation (6.17) with the boundary condition u(0) = 0 (eq. 6.18).

Using the known boundary conditions on the radial wave function u(r) at r = 0 and r = r$_{}0$, find the ground state energy of the particle in this potential well.

## Homework Equations

6.17
-$\hbar^{2}/2m$ $d^{2}/dr^{2}$u(r) + V(r)u(r) = Eu(r)

6.18
u(r) = rψ(r)

## The Attempt at a Solution

I know that there is no potential V(r) to deal with, since it is zero inside the well and infinite outside.

I also know that ψ is zero at r$_{}0$ and finite at zero.

So, that leaves me with finding a solution to

-$\hbar^{2}/2m$ $d^{2}/dr^{2}$u(r) = Eu(r)

which I am a bit confounded with. Even then, how do I go from a solution to that to finding the ground state energy?

Thanks.

$$\frac{\partial^2}{\partial r^2} u(r) = -A^2 u(r)$$
Where $A = \sqrt{2mE/\hbar^2}$. See if you can work out the basic type of function that will satisfy this equation (hint: what function yields minus itself when differentiated twice?)