Question 4.P.8: Using the uncertainty principle, show that the lowest energy...

The motion of the particle is confined to the region -a/2 ≤ x ≤ a/2; that is, \Delta x\simeq a.

Then as a result of the uncertainty principle, the lowest value of this particle’s momentum is \hbar (2\Delta x)\simeq \hbar /(2a). The total energy as a function of a is

E(a)\simeq \frac{1}{2m}\left(\frac{\hbar }{2a} \right) ^{2} +\frac{1}{2}m\omega ^{2}a^{2}.                   (4.282)

The minimization of E with respect to a,

0=\frac{dE}{da}\mid _{a=a_{0} } =-\frac{\hbar ^{2} }{4m a^{3}_{0} } +m\omega ^{2}a_{0},                             (4.283)

gives a_{0}=\sqrt{\hbar /2m\omega } and hence E(a_{0})\simeq \hbar \omega /2; this is equal to the exact value of the oscillator’s zero-point energy.