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lecture_17 - 6.7: Barriers and Tunneling I Much of what we...

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Unformatted text preview: 6.7: Barriers and Tunneling I Much of what we know about the structure of atoms and nuclei, and about the interaction between particles, has been derived from the results of scattering experiments. A full analysis of scattering events must, in principle, embody the description of a localized free particle that travels toward a center of force and is scattered (deflected) by it. It is possible, however, to obtain many of the results of such scattering problems with the help of a time- independent treatment, using pure momentum states. In wave mechanics this means that one can fix attention on incident particles of sharply defined momentum and energy. The wave function can then be written without any mention of time dependence; for a one-dimensional problem one can put Then obtain a steady-state solution with the aid of boundary conditions, just as is done for the truly time-independent bound-state energy eigenfunctions (potential well, harmonic oscillator, hydrogen atom, ) ( x ) ~ e ikx h k = 2 mE 6.7: Barriers and Tunneling I Consider the problem of scattering by a one-dimensional potential step (for example, an electron inside a metal and incident normally on its surface from within). We can assume that, just as classical propagating waves scatter only from abrupt changes in the medium, the de Broglie waves scatter only from abrupt changes in the potential. Region I : x Region II : x > 0 Region I : x Region II : x > 0 E We make the physical assumption that in region II we have only a traveling wave in the +x direction, whereas in region I we have a superposition of the incident wave and a reflected wave. d 2 dx 2 + 2 m h 2 ( E- V ( x )) = A o e ik 1 x Ae- ik 1 x Be ik 2 x I ( x ) = A o e ik 1 x + Ae- ik 1 x k 1 = 2 mE h II ( x ) = Be ik 2 x k 2 = 2 m ( E- V ) h Boundary Conditions The values of the constants A and B relative to A are found by applying the usual boundary conditions. I ( x ) = A o e ik 1 x + Ae- ik 1 x d I ( x ) dx = ik 1 A o e ik 1 x- ik 1 Ae- ik 1 x II ( x ) = Be ik 2 x d II ( x ) dx = ik 2 Be ik 2 x A + A = B ik 1 A- ik 1 A = ik 2 B A = k 1- k 2 k 1 + k 2 A and B = 2 k 1 k 1 + k 2 A Reflection Coefficient R = A 2 A 2 = k 1- k...
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lecture_17 - 6.7: Barriers and Tunneling I Much of what we...

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