Prove that the raising operator raises the state of the 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 \(\hat{a}^{\dagger}|n\rangle = \sqrt{n + 1}|n+1\rangle\)
 * 3) Express \(|n+1\rangle\) in terms of \(|0\rangle\) and the raising operator: \(|n+1\rangle = \frac{1}{N_{n+1}}\left(\hat{a}^{\dagger}\right)^{n+1}|0\rangle\)
 * 4) Separate \(\left(\hat{a}^{\dagger}\right)^{n+1}\): \(|n+1\rangle = \frac{1}{N_{n+1}}\hat{a}^{\dagger}\left(\hat{a}^{\dagger}\right)^{n}|0\rangle\)
 * 5) Substitute in \(|n+1\rangle = \frac{1}{N_{n}}\left(\hat{a}^{\dagger}\right)^{n}|0\rangle\): \(|n+1\rangle = \frac{1}{N_{n+1}}\hat{a}^{\dagger}N_n|n\rangle\)
 * 6) Rearrange: \(\frac{N_{n+1}}{N_n}|n+1\rangle = \hat{a}^{\dagger}|n\rangle\)
 * 7) Calculate \(\frac{N_{n+1}}{N_{n}}\).
 * 8) Multiply the expression by its conjugate: \(\langle n+1|n+1\rangle = \left(\frac{N_n}{N_{n+1}}\right)^2\langle n|\hat{a}\hat{a}^{\dagger}|n \rangle\)
 * 9) Use the commutator to rearrange \(\hat{a}\hat{a}^{\dagger}\): \(\langle n+1|n+1\rangle = \left(\frac{N_n}{N_{n+1}}\right)^2\langle n|1+\hat{a}^{\dagger}\hat{a}|n \rangle\)
 * 10) Substitute the number operator \(\hat{n}\) for \(\hat{a}^{\dagger}\hat{a}\) and calculate: \(\langle n+1|n+1\rangle = \left(\frac{N_n}{N_{n+1}}\right)^2 (n+1) \langle n|n \rangle\)
 * 11) Simplify: \(\frac{N_{n+1}}{N_n}=\sqrt{1+n}\)
 * 12) Substitute for \(\frac{N_{n+1}}{N_{n}}\): \(\hat{a}^{\dagger}|n\rangle = \sqrt{n + 1}|n+1\rangle\)