Optimizing the Gaussian trial for an attractive Gaussian well ($a_{\ast }(1+a_{\ast }{)}^3=V_0^2,E(a_{\ast })=-a_{\ast }/2-a_{\ast }^2$)

For $E(a)=a/2-V_0\sqrt{a/(1+a)}$ with $a,V_0>0$, differentiation gives $E^{\prime }(a)=1/2-V_0/[2\sqrt{a}(1+a{)}^{3/2}]$. The positive stationarity equation has a unique solution because its left side increases strictly. Moreover $E^{\prime \prime }(a)=V_0(1+4a)/[4a^{3/2}(1+a{)}^{5/2}]>0$, so it is the global minimum. Substituting the stationarity equation gives the displayed negative energy. The only positive zero is $(\sqrt{1+16V_0^2}-1)/2$; $a=0$ is only a limiting zero, since the limiting constant trial is not a square-integrable function. The Rayleigh-Ritz variational principle therefore proves a negative-energy bound state for every positive well depth. For weak depth, the simpler choice $a=V_0^2$ yields an explicit upper bound $E_0\le -V_0^2/2+\epsilon$ withThis is a variational upper bound, not an equality for the exact weak-depth coefficient.