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RPS: PSP0007 - Science-at-Nanoscale
                             9:2
                   June 9, 2009
                                                            3.2. Basic Postulates of Quantum Mechanics
                                                                    ’
                                           ’
                                                       V=0
                                       Ȍ
                                                                       Ȍ
                                                        Ȍ
                                                               L
                                                                                  X
                                               0
                                        Figure 3.2. One-dimensional potential box.
                             which can be expressed as
                                                     2
                                                    d ψ
                                                             2
                                                                                  (3.17)
                                                        = −k ψ
                                                       2
                                                     dx
                             where
                                                             2
                                                         8mπ E
                                                     2
                                                                                  (3.18)
                                                    k =
                                                            2
                                                           h
                             Equation 3.17 is a typical standing wave equation, and the solu-
                             tions to the differential equation take the form
                                                                   −ikx
                                                     ikx
                                             ψ(x) = e
                                                                                  (3.19)
                                                       and ψ(x) = e
                             One can easily verify these solutions to the differential equation by  39  ch03
                             direct substitution. Since the particle moves back and forth inside
                             this region 0 < x < L, we can use a linear combination of the two
                             functions in Eq. (3.19) as the general solution in this case, i.e.
                                                ψ(x) = Ae ikx  + Be −ikx          (3.20)
                               Note that the wavefunction should satisfy the boundary condi-
                             tion that ψ(x = 0) = 0, this leads to the requirement that B = −A.
                             Hence we have
                                  ψ(x) = A(e ikx  − e −ikx ) = 2iA sin(kx) = C sin(kx)  (3.21)
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