Calculate expectation values for odd powers of \(x\) in state \(n\) using ladder operators


 * 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) Calculate \(x\): \(x=\sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}+\hat{a}^{\dagger}\right)\)
 * 3) Substitute this into \(\langle n|\hat{x}^k|n\rangle\): \(\langle n|\hat{x}^k|n\rangle = \left(\frac{\hbar}{2m\omega}\right)^{n/2}\langle n|\left(\hat{a}+\hat{a}^{\dagger}\right)^k|n\rangle\)
 * 4) Find the general form of the terms of the expansion: \(\left(\frac{\hbar}{2m\omega}\right)^{n/2} (\langle n| \left(\hat{a}\right)^{\alpha_1} \left(\hat{a}^{\dagger}\right)^{\beta_1}|n\rangle\ + \cdots + \langle n|\left(\hat{a}\right)^{\alpha_m} \left(\hat{a}^{\dagger}\right)^{\beta_m}|n\rangle)\) where \(\alpha_n + \beta_n = k\), (\(\alpha_n,\beta_n \in \mathbb{N}\)).
 * 5) Sum the terms.
 * 6) Simplify: \(\left(\frac{\hbar}{2m\omega}\right)^{n/2} (c_1\langle n|n - \alpha_1 + \beta_1\rangle\ + \cdots + c_m\langle n|n - \alpha_m + \beta_m\rangle)\)
 * 7) Since \(\langle a|b \rangle = \delta_{ab}\), eliminate all terms where \(\alpha_n \neq \beta_n\). As \(k\) is odd and \(\alpha_n + \beta_n = k\), this is all terms. \(\langle n|\hat{x}^k|n\rangle=0\)