Show that the uncertainty in the momentum of a ground state harmonic oscillator is $\left(\frac{\hslash }{\sqrt{\mathbf{2}}}\right){\left(\mathbf{mk}\right)}^{\mathbf{1}\mathbf{/}\mathbf{4}}$.

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

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

$\left(\Delta x\right)\left(\Delta p\right)=\frac{\hslash }{2}......\left(1\right)$

The angular frequency of the harmonic oscillator is given as:

$\omega =\sqrt{\frac{k}{m}}$

The energy of a quantum harmonic oscillator is given as:

$$\begin{gathered} E=\frac{{\left(\Delta p\right)}^2}{2m}+\frac{1}{2}m{\omega }^2{\left(\Delta x\right)}^2 \\ E=\frac{1}{2m}{\left(\frac{\hslash }{2\Delta x}\right)}^2+\frac{1}{2}m{\omega }^2{\left(\Delta x\right)}^2 \\ E=\frac{{\hslash }^2}{8m{\left(\Delta x\right)}^2}+\frac{1}{2}m{\omega }^2{\left(\Delta x\right)}^2 \end{gathered}$$

Differentiate the above expression with respect to position.

role="math" localid="1660045601661" $$\begin{gathered} \frac{dE}{dx}=\frac{d}{dx}\left[\frac{{\hslash }^2}{8m{\left(\Delta x\right)}^2}+\frac{1}{2}m{\omega }^2{\left(\Delta x\right)}^2\right] \\ \frac{dE}{dx}=-\frac{2{\hslash }^2}{8m{\left(\Delta x\right)}^3}+\frac{1}{2}m{\left(\sqrt{\frac{k}{m}}\right)}^2\left(2\Delta x\right) \\ \frac{dE}{dx}=-\frac{{\hslash }^2}{4m{\left(\Delta x\right)}^3}+k\left(\Delta x\right) \end{gathered}$$

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

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

Substitute the value of uncertainty in position of oscillator in equation (1) to find the uncertainty in momentum of oscillator.

$\begin{array}{rcl}\left(\left(\frac{1}{\sqrt{2}}\right){\left(\frac{{\hslash }^2}{mk}\right)}^{1/4}\right)\left(\Delta p\right)& =& \frac{\hslash }{2}\\ \Delta p& =& \left(\sqrt{\frac{\hslash }{2}}\right){\left(mk\right)}^{1/4}\\ & & \end{array}$

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