Question

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

Short answer

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

Step-by-step solution

Uncertainty

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

Determination of uncertainty in the momentum of harmonic 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 $\left(\sqrt{\frac{\hslash }{2}}\right){\left(mk\right)}^{1/4}$.