Bom—von Karman boundary condition

Let us imagine that the inhnite periodic 3D solid discussed in Section 1.7 is separated into two halves, leading to two semi-inhnite 3D solids, preserving their 3D bulk periodicity but becoming aperiodic in the direction perpendicular to the generated surfaces. Because the translation symmetry is lost in this direction, the Bom-von Karman boundary conditions can no longer be applied, hence the apparent paradox that a semi-infinite problem becomes more complex than the infinite case. This fact inspired W. Pauli to formulate his famous sentence God made solids, but surfaces were the work of the Devil. 

Let us consider the ID lattice (1) with repeat distance a, where each unit cell contains just one atomic orbital (AO) /. The simplest way to describe the energy levels of this system is to impose Bom von Karman boundary conditions. Essentially, this means that we are bending our system as shown in (2), that is, we are transforming the chain into a loop. However, since the number N) of sites is very... 

The boundary conditions are a problem because there is a finite number of atoms, N, in any real crystal. The most adequate choice is the Bom-von Karman boundary condition, that is [2,3],... 

Bom von Karman Boundary Condition in 3-D Crystal Orbitals from Bloch Functions (LCAO CO Method)... 

The same may be done in 2-D and 3-D cases. We introduce usually the Bom-von Karman boundary conditions for a finite N and then go with N to oo. After such a procedure is carried out, we are pretty sure that the solution we are going to obtain will not only be hue for an infinite cycle, but also for the mass (bulk) of the infinite crystal. This stands to reason, provided... 

Here V is the volume of the Bom-von Karman region, i.e. that part of position space which is repeated as a result of the fundamental periodic boundary conditions. The integration in (11.1) is carried out over that region, which we denote by BK. 

Equation (18) is a statement of the Bom-von Karman periodic boundary conditions N = NiN2N3 is the number of unit cells in the crystal lattice. 

We again assume Bom-von Karman periodic boundary conditions for the motion The Nth atom has displacement equal to that of the zeroth atom (closed loop). Then the two equations of motion are... 

Consider a one-dimensional (usually almost infinite) set of N atoms, molecules, or point masses, all equally spaced at inter-particle distances d along the real-space coordinate x, with Bom-von Karman periodic boundary conditions for the potential energy ... 

In consequence of the three-dimensional translational symmetry of the polymer and of the Bom-von Karman periodic boundary conditions, matrices H and S are cyclic hypermatrices. For the sake of simplicity we show this for the one-dimensional case the generalization to two- and three-dimensional cases is straightforward. In the one-dimensional case, if we take into account the translational symmetry, the hypermatrices H and S have the form... 

Here, submatrix denotes interactions within the elementary cell, submatrices m and E3-QE] correspond to first-neighbor interactions, and so on (all the submatrices have dimension mXm only). Introduction of the Bom-von Karman periodic boundary conditions... 

It was seen above that if we take into account the translational symmetry of the system and introduce the Bom-von Karman periodic boundary conditions, our matrix equations (1.3) reduce to relationship (1.17). In the Hartree-Fock-Roothaan case the elements of the Fock matrices F(q) occurring in the expression... 

We now consider a three-dimensional periodic system with (2N -f 1) unit cells and m orbitals within the cell. If we again apply the Bom-von Karman periodic boundary conditions the matrices F , F, and S become cyclic hypermatrices of order m(2N -h 1). Therefore we can again apply to equations (1.97) the unitary transformation described in Section 1.1. Hence... 

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