Quantum Monte Carlo: the helium atom

This results in $$ \frac{1}{{\cal R}_T(r_1)}\frac{d {\cal R}_T(r_1)}{dr_1}=-Z, $$ and $$ {\cal R}_T(r_1)\propto e^{-Zr_1}. $$ A similar condition applies to electron 2 as well. For orbital momenta \( l > 0 \) we have $$ \frac{1}{{\cal R}_T(r)}\frac{d {\cal R}_T(r)}{dr}=-\frac{Z}{l+1}. $$ Similarly, studying the case \( r_{12}\rightarrow 0 \) we can write a possible trial wave function as $$ \psi_T(\boldsymbol{R})=e^{-\alpha(r_1+r_2)}e^{\beta r_{12}}. \tag{5} $$ The last equation can be generalized to $$ \psi_T(\boldsymbol{R})=\phi(\boldsymbol{r}_1)\phi(\boldsymbol{r}_2)\dots\phi(\boldsymbol{r}_N) \prod_{i < j}f(r_{ij}), $$ for a system with \( N \) electrons or particles.