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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 $n$.

Qmechanic
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sawan kumawat
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    It's the same argument as for $n=1$. $x^n$ is odd if $n$ is odd, and the square of the ground-state wave function (in fact, any of the squares of the energy eigenfunctions) is even, and so the integral of the product is zero. – march Jun 01 '22 at 15:39

2 Answers2

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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.

Niall
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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:$$\int_{\mathbb{R}} \mathrm{d}x~ \underbrace{x^n|\psi_m(x)|^2}_{\text{odd integrand}}~=~0\quad\text{for }n\text{ odd}. $$

Qmechanic
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