Exploring Quantum Physics - Week 7 Question 4

Question:

Compute the second order energy.

Solution:

Now we pick the terms for second order.

$H_0\psi_2 + V\psi_1 = E_0\psi_2 + E_1\psi_1 + E_2\psi_0$.

To find $E_2$, we take inner product with $\phi_0$. That gives

$\langle \phi_0 | H_0 |\psi_2 \rangle + \langle \phi_0 | V | \psi_1 \rangle = \langle \phi_0 | E_0 | \psi_2 \rangle + \langle \phi_0 | E_1 | \psi_1 \rangle + \langle \phi_0 | E_2 | \psi_0 \rangle$.

As in lecture, the terms cancels, and we already establish $E_1 = 0$, so we have

$\langle \phi_0 | V | \psi_1 \rangle =  E_2$.

But we still don't know what is $\psi_1$, what can we do now?

Note that $V = mgz$, so we can relate $\langle \phi_0 | V$ with $\phi_1$ as follow.

$\begin{eqnarray*} \langle \phi_1 | &=& \frac{1}{\sqrt{2}}\phi_0(2\sqrt{\frac{m\omega}{\hbar}}z) \\ &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\phi_0 mgz \\ &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\langle \phi_0 | V \\ \langle \phi_1 | \psi_1 \rangle &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\langle \phi_0 | V | \psi_1 \rangle \\ -\frac{mg}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}} &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\langle \phi_0 | V | \psi_1 \rangle \\ \frac{mg^2}{2\omega^2} &=& \langle \phi_0 | V | \psi_1 \rangle \\ \end{eqnarray*}$

The last equal sign involve messy cancelation, but at the end that's the perfect result I needed!