Calculation of the tail energy

A simulation of a homogeneous system is done by considering a finite box of size $L$. One restricts interaction between the particles to a distance of $L/2$. Larger distances should be avoided in order not to have a double counting of a same particle which would leave to artificial correlation. Thus one introduces a cut-off at $L/2$ and a proper calculation of the energy is necessesary.

The situation is different for a Bijl-Jastrow construction of the wave function and a Slater determinant. We will consider a generalization of the wave function containing a product of both terms. The energy per particle in the thermodynamic limit $N\rightarrow \infty$ is given by the integral of the interaction energy from the cut-off length $L/2$ to infinity.

$$E_{pot}^{tail}=\sum\limits_i\sum\limits_{ j<i,\vert{{\vec r}_i-{\... ...\vec r}_i-{\vec r}_j}\vert)\rightarrow n\int\limits_{L/2}^{\infty }V(r)\,d^{3}r$$ (2.108)

In the thermodynamic limit $N\rightarrow \infty$ the Jastrow force becomes zero as the summation on $j$ is approximated by a symmetric uniform distribution of particles outside a sphere of $L/2$ radius. So, the tail of a kinetic energy for a Jastrow wave function is

$$E_{J}^{tail}=\frac{\hbar ^{2}n}{m}\int\limits_{L/2}^{\infty }{\ca... ...prime }(r)}{f(r)} +\left( \frac{f^{\prime }(r)}{f(r)}\right) ^{2}\right] d^{3}r$$ (2.109)

On the opposite, there is no similar cancellation due to the symmetry in the Slater term, but instead due to linearity the square of the first derivative is exactly cancelled by the force squared term, thus

$$E_{Det}^{tail}=\frac{\hbar ^{2}n}{m}\int\limits_{L/2}^{\infty }\l... ...^{\prime \prime }(r)}{g(r)}-\frac{2}{r}\frac{g^{\prime }(r)}{g(r)}\right]d^{3}r$$ (2.110)