We discussed how the energy ground state of the infinite square well, (left|psi_{1} ightangle), was the state

We discussed how the energy ground state of the infinite square well, \(\left|\psi_{1}\rightangle\), was the state that minimized the uncertainty relation. That is, in this state, the product of the variance of the momentum and position, \(\sigma_{x}^{2} \sigma_{p}^{2}\), was minimized. Further, we demonstrated that for all energy eigenstates

\[\begin{equation*}\sigma_{x}^{2} \sigma_{p}^{2} \geq \frac{\hbar^{2}}{4} \tag{5.71}\end{equation*}\]

as required. However, consider the momentum eigenstate

\[\begin{equation*}\psi(x)=\frac{e^{i \frac{p x}{\hbar}}}{\sqrt{a}} \tag{5.72}\end{equation*}\]

for some momentum \(p\), in the infinite square well. Verify that it is indeed \(L^{2}\) normalized on the well and calculate both variances \(\sigma_{x}^{2}\) and \(\sigma_{p}^{2}\). Is the uncertainty principle satisfied? Why or why not?