= Optimizing the Gaussian trial for an attractive Gaussian well
{title2=$a_*(1+a_*)^3=V_0^2,\quad E(a_*)=-a_*/2-a_*^2$}
For $E(a)=a/2-V_0\sqrt{a/(1+a)}$ with $a,V_0>0$, differentiation gives $E'(a)=1/2-V_0/[2\sqrt a(1+a)^{3/2}]$. The positive stationarity equation has a unique solution because its left side increases strictly. Moreover $E''(a)=V_0(1+4a)/[4a^{3/2}(1+a)^{5/2}]>0$, 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 $E_0\leq-V_0^2/2+\varepsilon$ with
$$
\varepsilon=\frac{V_0^4}{\sqrt{1+V_0^2}(\sqrt{1+V_0^2}+1)}=\frac12V_0^4+O(V_0^6).
$$
This is a variational upper bound, not an equality for the exact weak-depth coefficient.
Back to article page
