Exploring Quantum Physics - Final Exam Part 2 Question 5

Question:

What is the exact energy for the Hamiltonian of the previous problem.

Solution:

So we get started with the original equation again

$\hat{H}\Psi\left(\vec{r}\right) = -\frac {\hbar^2} {2 \mu} \nabla^2 \Psi\left(\vec{r}\right) + kr \Psi\left(\vec{r}\right) = E \Psi\left(\vec{r}\right)$

We were given a wonderful substitution$\Psi(r)= \frac{\psi(r)}{r}$ so that $\nabla^2 \Psi(r) = \frac{1}{r} \frac{\partial^2 \psi(r)}{\partial r^2}$. Using it, we get

$-\frac {\hbar^2} {2 \mu}\frac{1}{r} \frac{\partial^2 \psi(r)}{\partial r^2}+ kr\frac{\psi(r)}{r}= E\frac{\psi(r)}{r}$

Multiply by $r$ on both side, we get the standard form

$-\frac {\hbar^2} {2 \mu} \frac{\partial^2 \psi(r)}{\partial r^2}+ kr\psi(r)= E\psi(r)$

This is simply the bouncing ball potential if we set $\mu = M$ and $k = Mg$, therefore the ground state energy is simply

$\begin{eqnarray*} E_0 &=& - \alpha \rho_0 \\ &=& \left(\frac{\hbar^2 M g^2}{2}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 M \left(\frac{k}{M}\right)^2}{2}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 \frac{k^2}{M}}{2}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 k^2}{2M}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 k^2}{2\mu}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 k^2}{\mu}\right)^{1/3}\left(\frac{1}{2}\right)^{1/3} \rho_0 \\ \end{eqnarray*}$

So that's the answer - I have got this wrong just because I am not using the numerically accurate enough $\rho_0$ - Ooops - too bad.