= Solution
For a normalized <top-hat wavefunction> of width $L$ centred at zero, $\psi(x)=L^{-1/2}$ inside the interval and zero outside. Its <position> <distribution> has $\langle x\rangle=0$ and $\Delta x=L/\sqrt{12}$. With Fourier convention $(2\pi\hbar)^{-1/2}$, the <momentum> <amplitude> is
$$
\widetilde\psi(p)=\sqrt{\frac L{2\pi\hbar}}\,
\operatorname{sinc}\left(\frac{pL}{2\hbar}\right),\qquad
\operatorname{sinc}z=\frac{\sin z}{z}.
$$
The <momentum> density has long oscillatory $1/p^2$ tails. Consequently $\int p^2|\widetilde\psi|^2dp$ diverges: \b[the ideal discontinuous top hat has infinite r.m.s. <momentum> spread]. Its symmetric <momentum> density has formal principal-value <mean> zero, but no finite absolute first moment either. The finite-variance form $\Delta x\Delta p\ge\hbar/2$ is therefore not saturated and cannot be evaluated using the width of the central diffraction lobe as $\Delta p$.
The first <momentum> zeros occur at $p=\pm2\pi\hbar/L$, showing the inverse-width scale. Smoothing the edges sufficiently gives a finite <momentum> <variance> and restores the ordinary finite-variance uncertainty comparison; sharper edges require larger high-momentum components. It is the ideal jump, not a violation of the uncertainty principle, that causes the divergence.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2014/paper-1-top-hat.png]
{title=Original top-hat position density and sinc-squared momentum density; the latter has tails giving divergent momentum variance}
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