Exploring Quantum Physics - Week 4 Question 2

Question:

Recalling our definition of the creation, $\hat a^\dagger$, and annihilation, $\hat a$, operators, how can we express the momentum operator in terms of these?

Solution:

Let's actually recall the operators as follow:

$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \frac{i\hat{p}}{\sqrt{2m\hbar \omega}}$
$\hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} - \frac{i\hat{p}}{\sqrt{2m\hbar \omega}}$

The key idea is to take the difference of the two operators above, we have

$\begin{eqnarray*} \hat{a} - \hat{a}^\dagger &=& \frac{2i\hat{p}}{\sqrt{2m\hbar \omega}} \\ \frac{-i}{2}\frac{2i\hat{p}}{\sqrt{2m\hbar \omega}} &=& \frac{-i}{2}(\hat{a} - \hat{a}^\dagger) \\ \frac{\hat{p}}{\sqrt{2m\hbar \omega}} &=& \frac{i}{2}(\hat{a}^\dagger - \hat{a}) \\ \sqrt{2}\frac{\hat{p}}{\sqrt{2m\hbar \omega}} &=& \sqrt{2}\frac{i}{2}(\hat{a}^\dagger - \hat{a}) \\ \frac{\hat{p}}{\sqrt{m\hbar \omega}} &=& \frac{i}{\sqrt{2}}(\hat{a}^\dagger - \hat{a}) \\ \hat{p} &=& i\frac{\sqrt{m\hbar \omega}}{\sqrt{2}}(\hat{a}^\dagger - \hat{a}) \\ &=& i\sqrt{\frac{m\hbar \omega}{2}}(\hat{a}^\dagger - \hat{a}) \\ \end{eqnarray*}$