The energy of a linear harmonic oscillator is $E=p_{x}^{2} / 2 m+C x^{2} / 2$. (a) Show, using the uncertainty relation, that this can be written as
$$
E=\frac{h^{2}}{32 \pi^{2} m x^{2}}+\frac{C x^{2}}{2}
$$
(b) Then show that the minimum energy of the oscillator is $h v / 2$ where
$$
v=\frac{1}{2 \pi} \sqrt{\frac{C}{m}}
$$
is the oscillatory frequency. (Hint: This result depends on the $\Delta x \Delta p_{x}$ product achieving its limiting value $\hbar / 2$. Find $E$ in terms of $\Delta x$ or $\Delta p_{x}$ as in part (a), then minimize $E$ with
respect to $\Delta x$ or $\Delta p_{x}$ in part (b). Note that classically the minimum energy would be zero.)