Quantum Monte Carlo: the helium atom

The hamiltonian becomes then $$ \hat{H}=-\frac{\hbar^2\nabla_1^2}{2m}-\frac{\hbar^2\nabla_2^2}{2m} -\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2}+ \frac{ke^2}{r_{12}}, $$ and Schroedingers equation reads $$ \hat{H}\psi=E\psi. $$ All observables are evaluated with respect to the probability distribution $$ P(\boldsymbol{R})= \frac{\left|\psi_T(\boldsymbol{R})\right|^2}{\int \left|\psi_T(\boldsymbol{R})\right|^2d\boldsymbol{R}}. $$ generated by the trial wave function. The trial wave function must approximate an exact eigenstate in order that accurate results are to be obtained.