Show that the ladder operator commutator is consistent with the momentum and position commutator

Method

 * 1) Find the Hamiltonian that describes a simple harmonic oscillator: \(\mathbf{\hat{H}}=\frac{1}{2m}\hat{p}^2 + \frac{1}{2}m\omega^2\hat{x}^2\)
 * 2) Identify the ladder operators for this Hamiltonian: \(\hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} + i\frac{\hat{p}}{m\omega} \right ) \) \(\hat{a}^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} - i\frac{\hat{p}}{m\omega} \right ) \)
 * 3) Substitute these into the ladder operator commutator, \([\hat{a},\hat{a}^{\dagger}]=1\): \([\sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} + i\frac{\hat{p}}{m\omega} \right ),\sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} - i\frac{\hat{p}}{m\omega} \right )]=1\)
 * 4) Expand the commutator: \(\frac{m\omega}{2\hbar} \left ( \left ( \hat{x} + i\frac{\hat{p}}{m\omega} \right ) \left ( \hat{x} - i\frac{\hat{p}}{m\omega} \right ) - \left ( \hat{x} - i\frac{\hat{p}}{m\omega} \right ) \left ( \hat{x} + i\frac{\hat{p}}{m\omega} \right ) \right )=1\)
 * 5) Expand the brackets: \(\frac{m\omega}{2\hbar} \left ( \hat{x}^2 + i\frac{\hat{p}\hat{x} - \hat{x}\hat{p}}{m\omega} + \frac{\hat{p}^2}{m^2\omega^2} - \hat{x}^2 + i\frac{\hat{p}\hat{x} - \hat{x}\hat{p}}{m\omega} - \frac{\hat{p}^2}{m^2\omega^2} \right ) =1\)
 * 6) Simplify: \(\frac{i}{\hbar} (\hat{p}\hat{x} - \hat{x}\hat{p}) =1\)
 * 7) Write the expression in terms of a commutator of \(p\) and \(x\) to produce the standard commutator: \([x,p]=i\hbar\)