Exercise 4.6: Ground state energy

Although the use of periodic boundary conditions minimizes surface effects, it is also important to choose the symmetry of the central ``unit cell'' to be consistent with the symmetry of the solid phase of the system. This choice of cell is essential if we wish to do simulations of the high density, low-temperature behavior. We know that the equilibrium structure of a crystalline solid at $T=0$ is the configuration of lowest energy.

1. Consider a triangular lattice of with $L_x$ and height $L_y=\frac{\sqrt{% 3}}{2}L_x$. Each column of the triangular lattice is separated by a distance $% a=L_x/n_{c}$ , where $n_{c}$ is the number of particles per row and $% N=n_{c}^{2}$. For a square lattice to have the same density, we have to choose the linear dimension to be $L=\sqrt{L_xL_y}$. Choose $n_{c}=6$ and determine the energy for each lattice using the Lennard-Jones potential. Use $L_x=5$ and $7$. What is the density of the system for each case? Are your results for $E/N$ independent of the since of the lattice? Which lattice has lower energy? Why?

2. Consider $n_{c}=6$ and $L_x=10,20,$ and $2^{1/6}n_{c}$. Which lattice structure has lowest energy in each case? In all these cases the net force on a particle is zero. Why? The latter density correspond to the case where the force between nearest neighbors is zero. Convince yourself that at this density the lattice is unstable, since the force on any particle which deviates from its equilibrium position in the lattice is negative. Consequently, the system will collapse to a higher density, where the triangular lattice is always favored.

3. Repeat the previous calculations for $V(r)=1/r^{6}$ and $1/r^{4}$. Is the triangular lattice still favored?