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Free and Bound Electrons, Dimensionality Effects
                             3.2.3 Periodic Boundary Conditions
                             A practical approach to set boundary conditions would be to have an infi-
                             nite box potential with a large extension from  x =  0   to  x =  L   forcing
                             the wavefunction to vanish on the boundaries. This seems to be artificial:
                             because of the vanishing wavefunction it is hard to imagine how elec-
                             trons get into or out of the sample. There is another way of introducing
                             boundary conditions for free electrons in one dimension that requires the
                             wavefunction to be periodic, i.e., ψ 0() =  ψ L()  . This allows us to have
                             plane wave solutions for the Schrödinger equation, that are normalized to
                                     L
                             the length  :
                                                         /
                                             ψ x() =  L () – 12 exp ( ikx)        (3.43)
                             Applying periodic boundary conditions the wavevector is restricted to

                                                        2πn
                                                    k =  ----------               (3.44)
                                                         L
                                         ⋅
                             Taking L =  N a  , i.e., in multiples of the atomic spacings, there are N
                                            k
                             discrete values for   to fulfill the periodic boundary condition.

                             3.2.4 Potential Barriers and Tunneling


                             Let us discuss the situation inverse to that given in Figure 3.3, i.e., a
                             potential of the form

                                                      U ,  x ∈  [ 0 a]
                                                               ,
                                            Ux() =    0                          (3.45)
                                                      0,  x ∉  [ 0 a]
                                                               ,
                             (see also Figure 3.6). We insert the potential (3.45) into the Hamiltonian.
                                                      ⋅
                             Again we choose  ψ =  const.  exp ( λx)   as the trial solution. Thus we
                             obtain

                                                    – 2mE
                                                                   ,
                                        λ =  i ± k =  ±  --------------- , for x ∉  [ 0 a]  (3.46a)
                                               I       2
                                                      —


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