40 From the finite to the infinite matrix clements. We have ac, + Bcz = Ec, (for p = 1) Bc, + ac2 + Bcz = Ec2 (for p= 2) Bc2 + ac, + Bc, = Ec, (for p = 3) Bcj-1 + ac, + Bc,+1 = Ec, (for p = j) (3.5)...

FROM BOOK SUTTON, ELECTRONIC STRUCTURE OF MATERIALS,

HOW TO GET 3.10

Extracted text: 40 From the finite to the infinite matrix clements. We have ac, + Bcz = Ec, (for p = 1) Bc, + ac2 + Bcz = Ec2 (for p= 2) Bc2 + ac, + Bc, = Ec, (for p = 3) Bcj-1 + ac, + Bc,+1 = Ec, (for p = j) (3.5) Bex-2 + aCN-1 + BCN = EcN-1 (for p = N - 1) BCN-1+ acN = EcN (for p = N). There are N coupled linear equations here, and we have to solve them! But, it is not as bad as it looks. First of all we note that the equations have all the same form except for p = 1 and p = N, the end atoms. All the equations can be simplified by first dividing by B. For example, for p =j we get Cj-1 - xc, + c,+1 = 0 (for p = j) (3.6) where x = (E – a)/B. (3.7) Let us try the solution c, = e where i = (-1)2 and 0 is to be determined (do not confuse i in the exponential with the label of atom i). At this stage 0 may be real, imaginary, or complex. Inserting this guess into eqn (3.6) we get elU- 1)0 - x ev + eU+1]® = 0 or x = (e" + e-10) = 2 cos 0. (3.8) If we replace e by -0 we get the same solution for x. Therefore we could have c, = A e + Be-, where A and B are arbitrary constants and still eqn (3.8) would be satisfied. How do we determine A and B? Well, do not forget the end atoms. For p =1 we have c, = xc. Therefore, A e210 + Be-2io = 2 cos 8(A eº + Be-) (3.9) and this equation is satisfied by setting A = - B. Thus the equation c2 = xc, pins down A and B to the extent that c, = A(e – e-iv) where A is still an arbitrary constant and 0 satisfies eqn (3.8). We may rewrite c, as D sin je where D = 2iA, i.e. yet another arbitrary constant. The equation for the other end atom, cN -1 = xCN, is used now to pin down the allowed values of 0 D sin(N – 1)0 = xD sin(N6) = 2D cos 8 sin(NO) or sin(N – 1)0 = 2 cos 0 sin(N8) = sin(N8) cos 0 - cos(NO) sin 0 (3.10)