Brillouin zone translation vectors

The quantity x is a dimensionless quantity which is conventionally restricted to a range of —-ir < x < tt, a central Brillouin zone. For the case yj = 0 (i.e., S a pure translation), x corresponds to a normalized quasimomentum for a system with one-dimensional translational periodicity (i.e., x s kh, where k is the traditional wavevector from Bloch s theorem in solid-state band-structure theory). In the previous analysis of helical symmetry, with H the lattice vector in the graphene sheet defining the helical symmetry generator, X in the graphene model corresponds similarly to the product x = k-H where k is the two-dimensional quasimomentum vector of graphene. 

In the case of a perfect crystal the Hamiltonian commutes with the elements of a certain space group and the wave functions therefore transform under the space group operations accorc g to the irreducible representations of the space group. Primarily this means that the wave functions are Bloch functions labeled by a wave vector k in the first Brillouin zone. Under pure translations they transform as follows... 

Because of the translational symmetry of the reciprocal lattice (Section 16.3) and the definition of the Brillouin zone (BZ), the BZ faces occur in pairs separated by a reciprocal lattice vector. For example, the cubic faces of the first BZ of the simple cubic (sc) lattice occur in pairs separated by the reciprocal lattice vectors b (2rc/a)[[l 0 0]] (see eq. (16.3.27)). In general, for every k vector that terminates on a BZ face there exists an equivalent vector k (Figure 17.1) such that... 

We now study the disordered effective hamiltonian (4.4). Since a direct diagonalization of (4.4) is too hard, we shall have to use approximations which are conveniently expressed in the resolvent (or Green s function) formalism. The translation-invariant K sum in HeU is restricted to the optical wave vectors only (for K oj/c, RK / K 0I)- Therefore, it is possible to restrict the problem to this small part of the Brillouin zone using the projector operator... 

What we have learned is that our solutions may be labeled in terms of the vector q, which is known as the wavevector of the mode of interest. Each such mode corresponds to a displacement wave with wavelength X = 27r/ q. This interpretation makes it clear that there is a maximum q above which the oscillations are associated with lengths smaller than the lattice parameter and hence imply no additional information. This insight leads to an alternative interpretation of the significance of the first Brillouin zone relative to that presented in chap. 4. As an aside, we also note that had we elected to ignore the translational periodicity, we could just as well have solved the problem using the normal mode idea developed earlier. If we had followed this route, we would have determined 3 A vibrational frequencies (as we have by invoking the translational periodicity), but our classification of the solutions would have been severely hindered. 

Since the irreducible representations of the group of translations are onedimensional and are determined by the given wavevector, among the set of quantum numbers characterizing a surface exciton there is always a quasicontinuous quantum number - the wavevector. This vector differs from the corresponding one in the case of bulk excitons it may be directed only along the crystal surface and assumes values only within a two-dimensional Brillouin zone. In... 

Thus these points in a small but well-defined region of k space include all possible irreducible representations of the translation group the vectors of the reciprocal lattice transform points in the Brillouin zone into equivalent points. The Brillouin zone therefore contains the whole symmetry of the lattice, each point corresponding to one irreducible representation, and no two points being related by a primitive translation. The smallest value of k ki, k2, kz) belonging to the rep is called the reduced wave-vector. The set oi reduced wavevectors is called the first Brillouin zone. 

Thus, the translational symmetry of the potential leads to the eigenfunctions being characterized by a wave vector k (the Bloch vector). It is only defined modulo 2nfa since k + p (2 Jt/a) results in the same phase vector in (6.12) as k alone (p is an integer). It is, therefore, customary to label the eigenfunction fe(x) by restricting k to lie within the first Brillouin zone that is defined by... 

The calculation of the diffraction pattern for a periodic system revolves around the construction of the reciprocal lattice and subsequent placement of the first Brillouin zone however, in this case the aperiodicity of the pentagonal array requires a different approach due to the lack of translational symmetry. The reciprocal lattice of such an array is densely filled with reciprocal lattice vectors, with the consequence that the wave vector of a transmitted/reflected light beam encounters many diffraction paths. The resultant replay fields can be accurately calculated by taking the FT of the holograms. To perform the 2D fast Fourier transform (FFT) of the quasi-crystalline nanotube array, a normal scanning electron micrograph was taken, as shown in Fig. 1.13. 

The WignerSeiiz unit cell of the inverse lattice is called the First Brillouin Zone (FBZ). The vectors k inside the FBZ label all possible irreducible representations of the translation group. 

In order to construct the reducible representation of the translations or of the librations, at a particular point in the Brillouin zone (for a particular wave vector q), we must investigate the transformations of the basis functions under the pertinent symmetry operations. As basis functions we use the translationally symmetrized functions which already belong to irreducible representations of the group of pure translations by integral lattice vectors. The basis functions were given in Eq. (2.11). They are... 

We restrict the attention to periodic solids, molecular crystals. The excitations are characterized by wave vectors q, that lie in the first Brillouin zone of the lattice considered. These excitations are not necessarily pure translational phonons, librons or vibrons, in general they will be mixed. Much experimental information has been collected about such excitations, by infrared and Raman spectroscopy and, in particular, by inelastic neutron scattering. Due to the optical selection rules infrared and Raman spectra can only probe the = 0 excitations. By neutron scattering one can excite states of any given q and thus measure the complete dispersion (wave vector dependence) of the phonon and vibron frequencies. 

Let m be a subset of m corresponding to the stars of vectors gam 9 F) As is proved in [86], for symmetrical transformation (4.77) L points in the initial Brillouin zone (related to the initial basic translation vectors Cj)... 

To generate the set of points for any of the 14 Bravais lattices it is sufhcient to find the inverse of the corresponding matrix from Appendix A, to pick out according to (4.84) L points in the Brillouin zone related to basic translation vectors a and to distribute them over stars. The distribution of these points over stars depends on the symmetry gronp F of the function f K) and can not be made in general form. 

To eliminate the results dependence on Brillouin-zone samphng the dense Monk-horst-Pack fc-point mesh was used. For primitive unit cells the 12 x 12 x 12 and 12 x 12 special-point sets have been taken for bulk and slab calculations, respectively. In the case of 3D-slabs the number of points in the third fe-direction depends on the chosen value of the c translation vector in direct space. The latter was chosen to provide a similar size in aU directions of the corresponding cycUc model of the crystal (the crystal is composed of equidistant superceUs). The increasing of the 3D unit cell in direct space for producing the 2D supercell was accompanied by the corresponding reduction in the fe-points mesh in the reciprocal space. 

Consider the sum over all points in the first Brillouin Zone (BZ) J2k BZ exp(ik R). Let us assume that all k vectors lie within the first BZ. If we shift k by an arbitrary vector ko 6 BZ, this sum does not change, because for a vector k + ko which lies outside the first BZ there is a vector within the first BZ that differs from it by G and exp(iG R) = 1. Moreover, all points translated outside the first BZ by the addition of ko have exactly one image by translation inside the first BZ through G, since all BZs have the same volume. Therefore we have... 

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