Exploring Quantum Physics - Week 6 Question 4

Question:

Given an isotropic Quantum Harmonic Oscillator, what is the ground state energy.

$\hat{H} = -\frac{\hbar^2}{2M} \nabla^2 + V\left(x_1,x_2,x_3\right)= -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right).$

Solution:

The Hamilitonian can be written as a sum
$\begin{eqnarray*} \hat{H} &=& -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& -\frac{\hbar^2}{2M}(\frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} + \frac{\partial^2}{\partial x_3^2}) + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& \sum\limits_{k=1}^{3}(-\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x_k^2} + \frac{M\omega^2x_k^2}{2}) \end{eqnarray*}$

In this form, we can easily see the energy is just the sum of three harmonic oscillator, so the answer is $\hbar\omega(n_1+n_2+n_3+\frac{3}{2})$.
Phew - finally an easier question - blocked on question 3 for so long.