Page 110 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers
P. 110

Free and Bound Electrons, Dimensionality Effects
                             the spectrum of the Hamiltonian will have both discrete and continuous
                             eigenvalues. The one-dimensional time independent Schrödinger equa-
                             tions read
                                         d 2   2m
                                                                   ,
                                        --------Ψ +  -------EΨ =  0 , for x ∈  [ 0 a]  (3.26a)
                                        dx 2   — 2
                                      d 2   2m
                                                                      ,
                                     --------Ψ +  ------- E –(  U )Ψ =  0 , for x ∉  [ 0 a]  (3.26b)
                                                    0
                                     dx 2   — 2
                             The Schrödinger equation, as given in (3.26a) and (3.26b), are differen-
                             tial equations of second order. To find their solutions we must provide
                             additional information, i.e., conditions to hold in the physical domain Ω
                             or on its boundary.
                             • The probability density must be smooth at x =  0   and x =  a  , and so
                               must be the wavefunction. This is because there are no sources or
                               sinks for the probability density, i.e., there is neither particle creation
                               nor destruction going on, i.e., (3.24) holds:
                                                                 (
                                              (
                                                                         (
                                      (
                                 lim  ( Ψ +ε) Ψ – ε)) =  0   and  lim  ( Ψ a+ε) Ψ a ε)) =  0
                                                                           –
                                           –
                                                                      –
                                ε →  0                     ε →  0
                             • For the same reason the flux of probability density must also be con-
                               tinuous at the interface and so
                                             d       d      
                                        lim  ------Ψ +ε(  ) –  ------Ψ –(  ε) =  0   and  (3.27a)
                                        ε →  0  dx   dx     
                                              d        d       
                                        lim   ------Ψ a+ε(  ) –  ------Ψ a ε–(  ) =  . 0  (3.27b)
                                                                
                                        ε → 0  dx      dx
                             The trial function for the solution of (3.26a) and (3.26b) is
                                       ⋅
                             ψ =  const. exp ( λx)  . Inserting this, we obtain for λ
                                                    – 2mE
                                                                   ,
                                        λ =  ± ik =  ±  --------------- , for x ∈  [ 0 a]  (3.28a)
                                                      —  2
                                                    (
                                                 2mU –   E)
                                                      0
                                                                     ,
                                      λ =  κ =  ±  ----------------------------- , for x ∉  [ 0 a]  (3.28b)
                                                     — 2

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