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Free and Bound Electrons, Dimensionality Effects
                                                ψ
                                                 I    = ψ II                     (3.30a)
                                                  x =  0   x =  0
                                               ψ '    = ψ '                      (3.30b)
                                                 I       II
                                                  x =  0    x =  0
                                               ψ      = ψ                        (3.30c)
                                                II       III
                                                  x =  a    x =  a
                                               ψ '    = ψ '                      (3.30d)
                                                II       III
                                                  x =  a    x =  a
                             Inserting (3.29a)-(3.29c) into (3.30a)-(3.30d), we obtain a linear system
                             of equations to determine A-D:


                                          1  – 1    – 1   0    A
                                          0 e ika  e  i – ka  e –  – κa  B
                                                                  =  0            (3.31)
                                          κ – ik    ik    0    C
                                          0 ike ika  – ike  i – ka  κe – κa  D

                             A solution can only be found if the determinant of the coefficient matrix
                             in (3.31) is zero:

                                                      2
                                                          2
                                        2κkcos ( ka) +  ( κ –  k )sin ( ka) =  0  (3.32)
                                                                     2
                                                             2
                                                                            2
                             With ak =  ξ   we have aκ =  ( 2mU a ) —⁄  2  –  ξ =  C –  ξ 2   and we
                                                           0
                             may write (3.32) as
                                                   2
                                             2ξ C –  ξ  2
                                             ---------------------------- +  tan ξ =  0  (3.33)
                                               C –  2ξ  2
                                                2
                             This equation must be solved either graphically, see Figure 3.4 and
                             Figure 3.5, or by numerical methods. A numerical solution of the dis-
                             cretized model for the discussed eigenvalue scenario may be counter-
                                                             2
                                                                   2
                             checked in the special case with  U «  — ⁄  ( ma )   where there is exactly
                                                         0
                                                          (
                             one bound state found with E ≈  U 1 ma ⁄  ( 2— ))  .
                                                               2
                                                                    2
                                                            –
                                                    0    0
                             Note that only in special limiting cases can an analytical solution be
                             obtained easily. Usually, analytical methods end at a certain point and
                             Semiconductors for Micro and Nanosystem Technology    109
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