Integrals in k Space

Why is the Brillouin zone so important in plane-wave DFT calculations The simple answer is that in a practical DFT calculation, a great deal of the work reduces to evaluating integrals of the form 

The key features of this integral are that it is defined in reciprocal space and that it integrates only over the possible values of k in the Brillouin zone. Rather than examining in detail where integrals such as these come from, let us consider instead how we can evaluate them numerically. 

TABLE 3.1 Approximations to the Integral J j sinf-n-x) dx = 1 Using the Trapezoidal and Legendre Quadrature Methods 

As a simple test case, we can evaluate (ttx/2) sin(ttx) dx. This integral 

Two features of the trapezoidal method are that we use a uniform spacing between the positions where we evaluate fix) and that every evaluation of fix) (except the end points) is given equal weight. Neither of these conditions is necessary or even desirable. An elegant class of integration methods called Gaussian quadrature defines methods that have the form 

---

Our first example, the DOS for bulk Ag, is shown in Fig. 8.1. There are several points to notice associated with the numerical details of this calculation. First, a large number of k points were used, at least relative to the number of k points that would be needed to accurately determine the total energy of Ag (see, e.g., Fig. 3.2). Using a large number of k points to calculate the DOS is necessary because, as described above, the details of the DOS come from integrals in k space. 

The third numerical detail that is important in Fig. 8.1 arises from the fact that Ag is a metal and the observation that the DOS is obtained from integrals in k space. In Section 3.1.4 we described why performing integrals in k space for metals holds some special numerical challenges. The results in Fig. 8.1 were obtained by using the Methfessel and Paxton smearing method to improve the numerical precision of integration in k space. 

We commented above that special techniques must be applied to accurately perform integrals in k space for metals. It is important to note that the smearing... 

The Maxwell divergence equations in k space therefore integrate to... 

The presence of a crack or other discontinuity presents a serious difficulty for the standard V z) theory, because the reflectance function is defined for infinite plane waves that are reflected into infinite plane waves, and this requires a reflecting surface that is uniform. If a surface contains a crack then this requirement is violated, and an incident plane wave may be scattered into a whole family of waves (Tew et al. 1988). This scattering can be described in k-space by a scattering function S(kx, k x), where the prime refers to incident waves and the unprime to scattered waves. The x-direction is taken as tangential to the surface, and at this stage the theory is confined to two dimensions in the plane normal to both the surface and the crack. The response of the microscope can then be written in terms of the scattering function by integrating over the incident and reflected waves separately... 

In many applications we encounter such sums of contributions from different modes, and because in the limit Q oo the spectrum of modes is continuous, such sums are converted to integrals where the density of modes enters as a weight function. An important attribute of the radiation field is therefore the density of modes per unit volume in k-space, Pk, per unit frequency range, p, or per unit energy, ps (E = hoP). We find (see Appendix 3A)... 

For this FID ky =k = 0, so that t otk. Consequently, yi (t) in (5.4.12) defines a cross-section pikjc) in k space along the kx-axis. This cross-section is the Fourier transform of P(x), and P(x) is a ID projection of the 3D distribution Mo(x, y, z) of longitudinal magnetization onto the x-axis corresponding to the integral of Mo(x, y, z) over the space dimensions y and z (cf. Fig. 1.1.5). In the mathematical sense, a projection is just the integral over a function of many variables, so that fewer variables remain. Fourier transformation over t leads to... 

The dynamic problem of vibrational spectroscopy must be solved to find the normal coordinates as linear combinations of the basis Bloch functions, together with the amplitudes and frequencies of these normal vibrations. These depend on k, and therefore the problem must be solved for a number of k-points to ensure an adequate sampling of the Brillouin zone. Vibrational frequencies spread in k-space, just as the Bloch treatment of electronic energy gave a dispersion of electronic energies in k-space. The number of vibrational levels whose energy lies between E and fc +d E is called the vibrational density of states. Vibrational contributions to the heat capacity and to the crystal entropy can be calculated by appropriate integrations over the vibrational density of states, just like molecular heat capacities and entropies are obtained by summation over molecular vibration frequencies. 

Here, the integration is over the surface in k-space at which E = fico. By evaluating the second integral explicitly, we obtain... 

When comparing the states in the k-space for 3-D and 2-D materials (Figure 2.3 and 2.5), for a 2-D solid that is extended in the x-y-plane, only discrete values are allowed for k. The thinner the solid in z-direction is, the larger the spacing Akz between those allowed states. On the other hand, the distribution of states in the kx — ky plane remains quasi-continuous. Therefore, it is possible to describe the possible states in the k-space as planes parallel to the kx- and ky-axes, with a separation Akj between the planes in the k -direction the individual planes can then be numbered as Uz. Since within one plane the number of states is quasi-continuous, the number of states is proportional to the area of the plane. This means that the number of states is proportional to k2 = k + k. The number of states in a ring with radius k and thickness Ak is therefore proportional to k-Ak, and integration over all rings yields the total area of the plane in k-space. Here, in contrast to the case of a 3-D soUd, the density of states scales linearly with k ... 

The k appears as a parameter in the equation similarly to the nuclear positions in molecular Hartree-Fock theory. The solutions are continuous as a function of k, and provide a range of energies called a band, with the total energy per unit cell being calculated by integrating over k space. Fortunately, the variation with k is rather slow for non-metaUic systems, and the integration can be done numerically by including relatively few points. Note that the presence of the phase factors in eq. (3.76) means that the matrices in eq. (3.79) are complex quantities. 

The classical bath sees the quantum particle potential as averaged over the characteristic time, which - if we recall that in conventional units it equals h/k T- vanishes in the classical limit ft -+ 0. The quasienergy partition function for the classical bath now simply turns into an ordinary integral in configuration space. 

A plane wave basis set with an orbital cutoff distance of 3.9 A was used. All-electron calculations and a double numerical basis set with polarization functions (DNP) were employed. Wave function integration in reciprocal space was performed via fine grid sampling of k points with a separation of 0.02 A . For the calculation of the density of states (DOS), a 5 X 8 X 3 Monkhorst-Pack grid was used. Charge transfer was calculated based on the Mulliken Population Analysis (MPA) (Mulliken 1955). 

Let us now return to our problem of evaluating the probability of emission. Integrating Equation 6.28 over frequencies kc, all propagation directions in k space, and summation over the polarizations e, we obtain [144]... 

VVc can now see why the normalisation factor of the Slater determinantal wavefunction is I v/N . If each determinant contains N terms then the product of two Slater determinants, ldeU rminant][determinant], contains (N ) terms. However, if the spin orbitals form an oi lhonormal set then oidy products of identical terms from the determinant will be nonzero when integrated over all space. We Ccm illustrate this with the three-electron example, k ljiiiidering just the first two terms in the expansion we obtain the following ... 

---