Exploring Quantum Physics - Final Exam Part 2 Question 2

Question:

What is the value of $\langle j | x | k \rangle$ where $| j \rangle$ and $| k \rangle$ are the j and k eigen function of the Quantum Harmonic Oscillator?

Solution:

As a disclaimer, I didn't quite solve the problem completely in the exam. But I get the correct answer.

The key annoying piece is the $x$ inside the sandwich. We just break it down into ladder operators.

$\hat{x} = \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a})$.

Substitute this back into $\langle j | x | k \rangle$, we have got


$\begin{eqnarray*} & & \langle j | x | k \rangle \\ &=& \langle j | \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} \langle j | (\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\langle j | \hat{a}^{\dagger} | k \rangle + \langle j | \hat{a} | k \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\langle j | k + 1 \rangle + \sqrt{k} \langle j | k - 1 \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\delta_{j, k + 1} + \sqrt{k} \delta_{j, k - 1})\\ \end{eqnarray*}$