We want to perform a Variational Monte Carlo calculation of the ground state of two electrons in a quantum dot well with different oscillator energies, assuming total spin \( S=0 \). Our trial wave function has the following form
$$ \begin{equation} \psi_{T}(\boldsymbol{r}_1,\boldsymbol{r}_2) = C\exp{\left(-\alpha_1\omega(r_1^2+r_2^2)/2\right)} \exp{\left(\frac{r_{12}}{(1+\alpha_2 r_{12})}\right)}, \tag{13} \end{equation} $$where the $\alpha$s represent our variational parameters, two in this case.
Why does the trial function look like this? How did we get there? This will be one of our main motivations for switching to Machine Learning later.
To find an ansatz for the correlated part of the wave function, it is useful to rewrite the two-particle local energy in terms of the relative and center-of-mass motion. Let us denote the relative distance between the two electrons as \( r_{12} \). We omit the center-of-mass motion since we are only interested in the case when \( r_{12} \rightarrow 0 \). The contribution from the center-of-mass (CoM) variable \( \boldsymbol{R}_{\mathrm{CoM}} \) gives only a finite contribution. We focus only on the terms that are relevant for \( r_{12} \) and for three dimensions.
The relevant local energy becomes then
$$ \lim_{r_{12} \rightarrow 0}E_L(R)= \frac{1}{{\cal R}_T(r_{12})}\left(2\frac{d^2}{dr_{ij}^2}+\frac{4}{r_{ij}}\frac{d}{dr_{ij}}+\frac{2}{r_{ij}}-\frac{l(l+1)}{r_{ij}^2}+2E \right){\cal R}_T(r_{12}) = 0. $$Set \( l=0 \) and we have the so-called cusp condition
$$ \frac{d {\cal R}_T(r_{12})}{dr_{12}} = -\frac{1}{2(l+1)} {\cal R}_T(r_{12})\qquad r_{12}\to 0 $$The above results in
$$ {\cal R}_T \propto \exp{(r_{ij}/2)}, $$for anti-parallel spins and
$$ {\cal R}_T \propto \exp{(r_{ij}/4)}, $$for anti-parallel spins. This is the so-called cusp condition for the relative motion, resulting in a minimal requirement for the correlation part of the wave fuction. For general systems containing more than say two electrons, we have this condition for each electron pair \( ij \).