= Solution
Squaring the states gives $\boxed{p_0=e^{-x^2/\sigma^2}/(\sqrt\pi\sigma),\quad p_1=2x^2e^{-x^2/\sigma^2}/(\sqrt\pi\sigma^3)}$. The first peaks at zero; the second has a node there and peaks at $x=\pm\sigma$. For a centered interval of width $\delta x\ll\sigma$, $P_0\simeq\delta x/(\sqrt\pi\sigma)$, whereas $P_1\simeq(\delta x)^3/(6\sqrt\pi\sigma^3)$. Its leading linear term vanishes, but the <probability> in a finite interval is not exactly zero.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016-oscillator-and-franck-hertz.png]
{title=Ground and first-excited oscillator distributions, and an original Franck-Hertz apparatus and signal schematic}
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