Position expectation values in quantum harmonic oscillator I have read that $<x^n> $ is zero for all odd values of $n$ in any state of quantum harmonic oscillator. What is the reason behind this?
I can imagine for $n=1$ ( because wavefunction is symmetric about center), but I cannot get any conclusion for other odd values of $n$.
 A: The eigenstates of the harmonic oscillator are
\begin{equation}
\psi_{m}(x) = \frac{1}{\sqrt{\mathcal{N}}} e^{-x^{2}/2l_{B}^{2}}H_{m}(x/l_{B})
\end{equation}
where the magnetic length is defined as $l_{B}^{2} = \frac{m\omega^{2}}{\hbar}$, $H_{m}(z)$ is the $\text{m}^{\text{th}}$ Hermite polynomial and $\mathcal{N}$ is some normalisation constant. The expectation value of $x^{n}$ is therefore
\begin{equation}
\begin{split}
\langle x^{n} \rangle &= \frac{1}{\mathcal{N}}\int^{\infty}_{-\infty}H^{*}_{m}(x/l_{B})e^{-x^{2}/2l_{B}^{2}} x^{n} e^{-x^{2}/2l_{B}^{2}} H_{m}(x/l_{B}) \text{d}x \\ 
&= \frac{1}{\mathcal{N}}\int^{\infty}_{-\infty} x^{n} e^{-x^{2}/l_{B}^{2}} H^{2}_{m}(x/l_{B}) \text{d}x
\end{split}
\end{equation}
since Hermite polynomials are real-valued functions. For odd $n$ the integrand is odd and the integral will evaluate to zero when taken over a symmetric range.
A: This is not true for any state $\psi$. However, since the harmonic oscillator potential $V(x)=V(-x)$ is even, we know that we can pick a basis $\psi_m$ of solutions to the TISE that has definite parity $\psi_m(-x)=\pm\psi_m(x),$ cf. e.g. this and this Phys.SE posts. Then OP's statement follows:$$\int_{\mathbb{R}} \mathrm{d}x~ \underbrace{x^n|\psi_m(x)|^2}_{\text{odd integrand}}~=~0\quad\text{for }n\text{ odd}. $$
