Exploring Quantum Physics - Week 7 Question 2

Question:

What is the first order energy for the Quantum Harmonic Oscillator under gravity?

Solution:

We will use perturbation theory.

First, as usual, we let $E = \sum\limits_{k = 0}^{\infty}{E_k\lambda^k}$, $\psi = \sum\limits_{k = 0}^{\infty}{\psi_k \lambda^k}$.

Substitute these back to the Schrodinger equation. We get this by picking the terms with $\lambda$ to the first order.

$\lambda(H_0\psi_1 + V\psi_0) = \lambda(E_0\psi_1 + E_1\psi_0)$

That allow us to show, $E_1 = \langle \psi_0 | V | \psi_0 \rangle$.

We know from the Quantum Harmonic Oscillator that $\psi_0 = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar})$.

So the rest is just an integration.

$\begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{dx (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar})) mgz (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar}))} \\ &=& \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} mg \int\limits_{-\infty}^{\infty}{dx z (\exp(-\frac{m\omega z^2}{\hbar}))} \\ \end{eqnarray*}$

Lot of typesetting, but it is now obvious the first order energy is 0 because of the symmetry.