We note that at \( \alpha=1 \) we obtain the exact result, and the variance is zero, as it should. The reason is that we then have the exact wave function, and the action of the hamiltionan on the wave function $$ H\psi = \mathrm{constant}\times \psi, $$ yields just a constant. The integral which defines various expectation values involving moments of the hamiltonian becomes then $$ \langle H^n \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})H^n(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})}= \mathrm{constant}\times\frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})}=\mathrm{constant}. $$ This gives an important information: the exact wave function leads to zero variance! Variation is then performed by minimizing both the energy and the variance.