Page 56 - Science at the nanoscale
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9:2
                                                     RPS: PSP0007 - Science-at-Nanoscale
                   June 9, 2009
                              Brief Review of Quantum Mechanics
                          46
                                   Since the particle can travel in the forward as well as backward
                                   directions, we can express the wavefunction as
                                                      ψ I (x) = Fe
                                                                  + Ge
                                   where the first term represents the incident wave while the second
                                   term represents the reflected wave. F and G are coefficients that
                                   can be determined using the boundary conditions.
                                     For region II, the potential has a finite height of V o such that
                                   V o > E, where E is the energy of the particle. The Schr¨odinger
                                   equation can be expressed as
                                                       2
                                                            2
                                                      h
                                                           d ψ II
                                                                                        (3.37)
                                                   −
                                                                = (E − V o )ψ II
                                                         2
                                                            dx
                                                     8mπ
                                     We can re-write the above equation as
                                                           2
                                                          d ψ II
                                                                  2
                                                                                        (3.38)
                                                               = κ ψ II
                                                           dx
                                                   2
                                          2
                                             = 8mπ (Vo − E)/h , and hence we can express the
                                   where κ
                                   wavefunction in region II as
                                                                    −κx
                                                                                        (3.39)
                                                         ψ II (x) = He
                                     To determine the coefficients F, G and H in Eqs. (3.36) and
                                   (3.39), ψ(x) and dψ(x)/dx must be continuous at the boundary
                                   points x = 0. We have
                                                                                        (3.40)
                                                         ψ I (0) = ψ II (0)
                                                           dψ I
                                                                                        (3.41)
                                                           dx 2 2 2 = ikx dψ II  −ikx   (3.36)     ch03
                                                                  dx
                                   and thus
                                                  F + G = H   ik(F − G) = −κH
                                     For region I
                                                                  ik + κ
                                                  ψ I (x) = F e ikx  +  e −ikx          (3.42)
                                                                  ik − κ
                                     For region II
                                                                 2ik  −κx
                                                       ψ II (x) = F  e                  (3.43)
                                                                ik − κ
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