Page 76 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers
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The Vibrating Uniform Lattice
                                                            qk a) ωt]}
                                                  {
                                                    [
                                                     (
                                                   j pk a +
                                                                                  (2.66)
                                        u =
                                                                 –
                                             cexp
                                          p
                                                              y
                                                        x
                             We can write the ansatz for any neighboring atom in terms of the central
                             atom. For a parallel and diagonal atom we obtain
                                                                           – jk x a
                                             {
                                 u  =  cexp {  j [ ( i 1)k a +  jk a] ωt}} =  u e
                                                –
                                                              –
                                  ,
                                                                        ,
                                  ij                 x     y            ij
                                                  {
                                 u       =  cexp {  j [ ( i 1)k a +  ( j 1)k a] ωt}}  (2.67)
                                                                       –
                                                                –
                                                      –
                                     ,
                                       –
                                   i 1 j 1                x         y
                                   –
                                                        (
                                                      – jk x a +  k y a)
                                                =  u e
                                                    ,
                                                    ij
                             which, when inserted into (2.65) becomes
                                      2       E      –  jk x a  jk x a  –  jk y a  jk y a
                                                ( [
                                                  –
                                    – ω mc =  c--- 4 e   –  e  –  e  –  e  )
                                              a                                   (2.68)
                                        (
                                                              (
                                                                         (
                                                   (
                                      –  jk x a +  k y a)  – j –  k x a + k y a)  jk x a + k y a)  – jk x ak y a)
                                                                            –
                                 (
                               +  f 4 e        –  e        –  e      –  e      )]
                                    –
                             This single parameter eigenvalue equation has two solutions
                                        2   E      – jk x a  jk x a  – jk y a  jk y a
                                      ω =   ------- 4 e–[  –  e  –  e  –  e
                                            am
                                                                                  (2.69)
                                                                         (
                                                              (
                                                   (
                                        (
                                      –  jk x a +  k y a)  – j –  k x a + k y a)  jk x a + k y a)  – jk x ak y a)
                                                                            –
                                 (
                               +  f 4 e        –  e        –  e      –  e      )]
                                    –
                                                      jφ
                                                              ⁄
                             Using the identity  cos φ =  ( e +  e – jφ ) 2   we transform the equation
                             above to
                                             E
                                           +
                                      ω =  −  ------- [{  4 –  cos ( 2k a) –  cos ( 2k a)]
                                                          x
                                                                     y
                                             am
                                                                                  (2.70)
                                                                    1
                                                                    ---
                                           f 4 –[  cos ( 2k a)cos ( 2k a)]}] 2
                                                     x
                                                              y
                             The dispersion relation, equation (2.70), represents two  ω   surfaces in
                             ( k k )   space, and are plotted in Figure 2.22. Because not only nearest,
                               ,
                                 y
                               x
                             but also next nearest neighbor interactions are included, the upper disper-
                             sion surface shows a new feature, in that the maximum is not at the
                             boundary of the primitive cell, i.e., its gradient is zero in the interior of
                             the cell.
                             Semiconductors for Micro and Nanosystem Technology      73
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