91影视

The uncertainty in the position of harmonic oscillator is the deflection from the actual position of the oscillator.

The relation between uncertainty in position and momentum is given as:

$\left(螖x\right)\left(螖p\right)=\frac{\mathrm{鈩�}}{2}$

The angular frequency of the harmonic oscillator is given as:

$蝇=\sqrt{\frac{k}{m}}$

The energy of a quantum harmonic oscillator is given as:

$$\begin{gathered} E=\frac{{\left(螖p\right)}^2}{2m}+\frac{1}{2}m{蝇}^2{\left(螖x\right)}^2 \\ E=\frac{1}{2m}{\left(\frac{\mathrm{鈩�}}{2螖x}\right)}^2+\frac{1}{2}m{蝇}^2{\left(螖x\right)}^2 \\ E=\frac{{\mathrm{鈩�}}^2}{8m{\left(螖x\right)}^2}+\frac{1}{2}m{蝇}^2{\left(螖x\right)}^2 \end{gathered}$$

Differentiate the above expression with respect to position.

$$\begin{gathered} \frac{dE}{dx}=\frac{d}{dx}\left[\frac{{\mathrm{鈩�}}^2}{8m{\left(螖x\right)}^2}+\frac{1}{2}m{蝇}^2{\left(螖x\right)}^2\right] \\ \frac{dE}{dx}=-\frac{2{\mathrm{鈩�}}^2}{8m{\left(螖x\right)}^3}+\frac{1}{2}m{\left(\sqrt{\frac{k}{m}}\right)}^2\left(2螖x\right) \\ \frac{dE}{dx}=-\frac{{\mathrm{鈩�}}^2}{4m{\left(螖x\right)}^3}+k\left(螖x\right) \end{gathered}$$

The uncertainty in position in the ground state will be zero so substitute $\frac{dE}{dx}=0$in the above expression.

$\begin{array}{rcl}0& =& -\frac{{\mathrm{鈩�}}^2}{4m{\left(螖x\right)}^3}+k\left(螖x\right)\\ \frac{{\mathrm{鈩�}}^2}{4m{\left(螖x\right)}^3}& =& k\left(螖x\right)\\ {\left(螖x\right)}^4& =& \frac{{\mathrm{鈩�}}^2}{4mk}\\ 螖x& =& \left(\frac{1}{\sqrt{2}}\right){\left(\frac{{\mathrm{鈩�}}^2}{mk}\right)}^{1/4}\\ & & \end{array}$

Therefore, the uncertainty in position in the ground state of harmonic oscillator is $\left(\frac{1}{\sqrt{2}}\right){\left(\frac{{\mathrm{鈩�}}^2}{mk}\right)}^{1/4}$.