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The Electronic System
                                                           (
                                                 (
                                   ux() =     ( Ae – ik –  κ)x  +  Be – ik + κ)x ), for  0<x<a  (3.96)
                                            ( Ce  ( –  ik – λ)x  +  De  ( –  ik +  λ)x ), for  -b<x<0
                                            ⁄
                                                                   ⁄
                                                           (
                             where κ =  2mE —   and λ =  2mV –  E) —  . To solve for the coef-
                                                             0
                             ficients A, B, C and D in (3.96) we apply the boundary. The wavefunc-
                             tions and their derivatives must be equal at  x =  0   (u 0() =  u 0()   and
                                                                        1
                                                                                2
                             u′ 0() =  u′ 0()  ), and the wavefunctions and their derivatives at
                               1
                                       2
                                                                              (
                             x =  – b    must equal those at  x =  a   (u a() =  u – b)   and
                                                                      1
                                                                             2
                                        (
                             u′ a() =  u′ – b)  ). This gives us a homogeneous system of four linear
                                       2
                               1
                             equations, which has only non-trivial solutions if its determinant is zero,
                             which yields
                                 2
                                λ –  κ 2
                                -----------------sinh ( λb)sin ( κa) +  cosh ( λb)cos ( κa) =  cos ( kL)  (3.97)
                                 2κλ
                             We see that already for the simple periodic potential barrier model (3.97)
                             cannot be solved analytically.  A graphical solution is given in
                             Figure 3.11.






                                                  1
                                                                                  κa

                                                  -1
                                                     i  = 1  2   3    4   5   6  ...
                Figure 3.11. Graphical solution
                for equation (3.97) that shows the
                emergence of allowable energy
                bands.








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