# Root-mean-square displacement

The goal of this tutorial is to calculate the root-mean square zero-point displacement of a quantum-mechanical harmonic oscillator.

The definition is

$r.m.s\, displacement=\sqrt{\langle\,x^2\rangle}$

where the expectation value of x2 is given by

$\langle\,x^2\rangle=\sqrt\frac{\alpha}{\pi}\int_{-\infty}^{\infty}\,e^{-\alpha x^2/2}x^2e^{-\alpha x^2/2}dx=\sqrt\frac{\alpha}{\pi}\int_{-\infty}^{\infty}\,e^{-\alpha x^2}x^2dx$

Notice that this can be solved using the identity

$\frac{d^vI_0(\alpha )}{d\alpha^v}=(-1)^vI_v(\alpha )$

where I0(a) is

$I_0(\alpha) =\int_{-\infty}^{\infty}\,e^{-\alpha x^2}\,dx=\sqrt\frac{\pi }{\alpha }$

Therefore

$\frac{\partial\,I_0(\alpha ) }{\partial \alpha }=\frac{\partial }{\partial \alpha }\frac{\pi ^{1/2}}{\alpha ^{1/2}}=-\frac{1}{2}\frac{\pi ^{1/2}}{\alpha ^{3/2}}$

and

$\langle\,x^2\rangle=\sqrt\frac{\alpha}{\pi}\int_{-\infty}^{\infty}\,e^{-\alpha x^2}x^2dx=\sqrt\frac{\alpha}{\pi}I_1(\alpha )=\sqrt\frac{\alpha}{\pi}\biggr(-1\biggr)\frac{\partial\,I_0(\alpha ) }{\partial \alpha }$

$\langle\,x^2\rangle=\frac{1}{2\alpha },\sqrt{\langle\,x^2\rangle}=\frac{1}{\sqrt{2\alpha }}$

Given that the classical turning point is 1/Öa the r.m.s. displacement is 1/Ö2 of the classical turning point. The classical turning point is the point at which all of the energy is potential energy (i.e. when the mass stops and turns around its velocity is zero and so all of its energy is potential).

$\frac{1}{2}\hbar\omega=\frac{1}{2}kx^2$

Therefore

$x=\sqrt\frac{\hbar\omega}{k}=\sqrt\frac{\hbar}{(\mu k)^{1/2}}=\frac{1}{\sqrt{\alpha }}$