Write an arbitrary number state as an excitation of the ground state for the simple harmonic oscillator


 * 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) From the excited state, lower the state once: \(\hat{a}|n\rangle = \sqrt{n}|n-1\rangle\)
 * 3) Multiply on the left by \(\hat{a}^{\dagger}\): \(\hat{a}^{\dagger}\hat{a}|n\rangle = \sqrt{n}\hat{a}^{\dagger}|n-1\rangle\)
 * 4) Substitute the number operator \(\hat{n}\) for \(\hat{a}^{\dagger}\hat{a}\) and calculate: \(n|n\rangle = \sqrt{n}\hat{a}^{\dagger}|n-1\rangle\)
 * 5) Divide through by the coefficient of \(|n\rangle\): \(|n\rangle = \frac{1}{\sqrt{n}}\hat{a}^{\dagger}|n-1\rangle\)
 * 6) If the ket on the right hand side is not \(|0\rangle\), go to step 2. Otherwise, continue to step 7. \(|n\rangle = \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n-1}}\frac{1}{\sqrt{n-2}}\cdots\left(\hat{a}^{\dagger}\right)^n|0\rangle\)
 * 7) Simplify: \(|n\rangle = \frac{1}{\sqrt{n!}}\left(\hat{a}^{\dagger}\right)^n|0\rangle\)