Question

USING the uncertainty principle,
estimate the minimum energy of a particle in a simple harmonic potential U= ½kx² .

Answer 1

The energy value can be calculated on the basis of the Hamiltonian of the system as follows:

$E = <\hat{H}>=<\hat{T}> + <\hat{U}> = \frac{<\hat{p}^2>}{2m} + \frac{k <\hat{x}^2>}{2}$

By definition, the uncertainties of the momentum and position operators are equal to:

$(\Delta{p})^2 = <\hat{p}^2> - <\hat{p}>^2, \quad (\Delta{x})^2 = <\hat{x}^2> - <\hat{x}>^2$

Due to the symmetry of potential $<\hat{p}> = 0 = <\hat{x}>$ .

Hence, we obtain:

$E = \frac{(\Delta p)^2}{2m} + \frac{k (\Delta x)^2}{2}$ .

The uncertainty principle states the following: $\Delta p \Delta x \geq \frac{\hbar}{2}$ . Hence, $\Delta p \geq \frac{\hbar}{2 \Delta x}$ . As a result:

$E \geq \frac{(\hbar)^2}{8 m (\Delta x)^2} + \frac{k (\Delta x)^2}{2}$

The expression on the right side is minimal when $\frac{d}{dx} \left[ \frac{(\hbar)^2}{8 m (\Delta x)^2} + \frac{k (\Delta x)^2}{2} \right] = 0$. Doing this explicitly leads to the following result: $(\Delta x)^2 = \frac{\hbar}{2 \sqrt{km}}=\frac{\hbar}{2 m \omega},$ where we utilized the relation $\omega^2 = \frac{k}{m}.$ Substituting this into the expression for energy results in the following:

$E_{min}= \frac{\hbar^2 2 m\omega}{8m \hbar}+ \frac{m \omega^2 \hbar}{4 m \omega} = \frac{\hbar \omega}{2}$