Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 10 / c

Differentiating the Gaussian variational bound for an attractive Gaussian well givesThus an interior stationary point obeysThe left side of the first equation increases strictly from zero to infinity, so there is exactly one such point. More explicitly,so $E$ is a strictly convex function on $a>0$ and $a_{\ast }$ is its global minimum. At that minimum the stationarity equation givesThere is a continuous limiting zero at $a=0$, although $y_T=1$ at exactly $a=0$ is not a square-integrable function and hence is not an admissible trial. For a positive zero, division by $\sqrt{a}$ and squaring gives $a(1+a)=4V_0^2$, sois the unique positive zero. The other quadratic root is negative and inadmissible. Near zero,so the graph leaves the origin with slope $-\infty$, falls to its unique negative minimum, rises through $a_z$, and grows as $a/2-V_0+O(a^{-1})$ at infinity.

By the Rayleigh-Ritz variational principle,Thus the attractive well supports a negative-energy bound state for every $V_0>0$.