Show that number states are eigenstates of the Hamiltonian operator for a simple harmonic oscillator

Method

 * 1) Find the Hamiltonian that describes a simple harmonic oscillator in ladder operator form as described here: \(\mathbf{\hat{H}}=\omega\hbar \left (\hat{a}^{\dagger}\hat{a} + \frac{1}{2} \right )\) \(\hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} + i\frac{\hat{p}}{m\omega} \right ) \) \(\hat{a}^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} - i\frac{\hat{p}}{m\omega} \right ) \)
 * 2) Show that \(\mathbf{\hat{H}}|n\rangle = \lambda|n\rangle\), where \(|n\rangle = \frac{1}{N_n}\left ( \hat{a}^{\dagger} \right )^{n}|0\rangle\)
 * 3) Multiply \(|n\rangle\) by \(\mathbf{\hat{H}}\): \(\mathbf{\hat{H}}|n\rangle = \frac{\omega\hbar}{N_n} \left (\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^n|0\rangle + \frac{1}{2}\left(\hat{a}^{\dagger}\right)^n|0\rangle\right)\)
 * 4) Show by induction that \(\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^{n}|0\rangle=n\left(\hat{a}^{\dagger}\right)^n|0\rangle\).
 * 5) Consider \(\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle\): \(\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle=\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}\left(\hat{a}^{\dagger}\right)^n|0\rangle\)
 * 6) Substitute out \(\hat{a}\hat{a}^{\dagger}\) using the commutator \([\hat{a},\hat{a}^{\dagger}]=1\): \(\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle=\hat{a}^{\dagger}(\hat{a}^{\dagger}\hat{a} + 1)\left(\hat{a}^{\dagger}\right)^n|0\rangle\)
 * 7) Expand: \(\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle=\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^n|0\rangle\ + \hat{a}^{\dagger}\left(\hat{a}^{\dagger}\right)^n|0\rangle\)
 * 8) Simplify: \(\hat{a}^{\dagger}\hat{a}\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle=(n+ 1)\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle\)
 * 9) Substitute \(\alpha_n\) into the expression for \(\mathbf{\hat{H}}|n\rangle\): \(\mathbf{\hat{H}}|n\rangle = \frac{\omega\hbar}{N_n} \left (\alpha_n\left(\hat{a}^{\dagger}\right)^n|0\rangle + \frac{1}{2}\left(\hat{a}^{\dagger}\right)^n|0\rangle\right)\)
 * 10) Simplify: \(\mathbf{\hat{H}}|n\rangle = \omega\hbar \left (n + \frac{1}{2}\right)^n|n\rangle\)