Ideas of Quantum Chemistry P48 shows how quantum mechanics is applied to chemistry to give it a theoretical foundation. The structure of the book (a TREE-form) emphasizes the logical relationships between various topics, facts and methods. It shows the reader which parts of the text are needed for understanding specific aspects of the subject matter. Interspersed throughout the text are short biographies of key scientists and their contributions to the development of the field. | 436 9. Electronic Motion in the Mean Field Periodic Systems tonian cf. Appendix C . In order to obtain such a function we may use the corresponding projection operator see eq. . There is also another way to construct a function k r of a given k from an auxiliary function u r satisfying an equation similar to eq. for the potential V T Ri u r u r - R u r . Then k r exp ikr u r . Indeed let us check T Rj k r T Rj exp ikr u r exp ik r - Rj u r - Rj exp -ikRj k r INVERSE LATTICE Let us now construct the so called biorthogonal basis b _ b2 b3 with respect to the basis vectors a1 a2 a3 of the primitive lattice . the vectors that satisfy the biorthogonality biorthogonality relations bi aj 2n8ij. The vectors bi can be expressed by the vectors ai in the following way bi 2n E aj S-1 ji j Sij ai aj. The vectors b1 b2 and b3 form the basis of a lattice in a 3D space. This lattice will be called the inverse lattice. The inverse lattice vectors are therefore i 3 Kj Y gjti i 1 where gij represent arbitrary integers. We have Kj Ri InMjj where Mij are integer numbers. Indeed 3 3 3 3 Kj Ri E j bi E nikak EE nikgjlbl ak 33 3 nikgjl 2n lk 2n E nilgjl 2nMij l 1 k 1 l 1 Inverse lattice 437 with nik gji and therefore also Mj as integers. The inverse lattice is composed therefore from the isolated points indicated from the origin by the vectors Kj. All the vectors that begin at the origin form the inverse space. Examples Let us see how we obtain the inverse lattice 1D 2D 3D in practice. 1D We have only a single biorthogonality relation b1 a1 2n . after skipping the index ba 2n. Because of the single dimension we have to have b 2n f where a a. Therefore the vector b has length 2n and the same direction as a. 2D This time we have to satisfy biai 2n b2a2 2n bia2 0 b2ai 0. This means that the game takes place within the plane determined by the lattice vectors ai and a2. The vector bi has to be perpendicular to a2 while b2 has to be
