For a particle in a box, the mean value and mean square value of the linear
momentum are given by \(\int_{0}^{L} \psi^{*} p \psi \mathrm{d} x\) and
\(\int_{0}^{L} \psi^{*} p^{2} \psi \mathrm{d} x,\) respectively. Evaluate these
quantities. Form the root mean square deviation \(\Delta p=\left\\{\left\langle
p^{2}\right\rangle-\langle p\rangle^{2}\right\\}^{1 / 2}\) and investigate the
consistency of the outcome with the uncertainty principle. Hint. Use \(p=(\hbar
/ \mathrm{i}) \mathrm{d} / \mathrm{d} x .\) For \(\left\langle
p^{2}\right\rangle\) notice that \(E=p^{2} / 2 m\) and we already know \(E\) for
each \(n\). For the last part, form \(\Delta x \Delta p\) and show that \(\Delta x
\Delta p \geq \frac{1}{2} \hbar,\) the precise form of the principle, for all
\(n\) evaluate \(\Delta x \Delta p\) for \(n=1.\)