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Lecture%2010 - EEE 352 Lecture 12 Tunneling Through a...

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1 EEE 352: Lecture 12 Tunneling Through a Potential Barrier * Tunneling through a potential barrier Finite barrier length Scanning tunneling microscope There is total internal reflection of this light in this media, yet … we see some light coming out the back side. The evanescent wave decays exponentially. In fact, we can use the evanescent wave to couple into another propagating media, such as a surface wave shown here.

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2 This makes for wave couplers in optical fibers. How does this appear for quantum waves? Electron Motion in the Presence of a Barrier Let us first recall the solutions to the barrier problem * Which was solved by breaking the problem into TWO different regions [ ] ) ( 2 exp ) ( 2 2 2 E V m x C x = = h γ ψ [ ] [ ] 2 2 1 1 1 1 2 exp exp ) ( h mE k x ik B x ik A x = + = FREE ELECTRON PROPAGATION PROPAGATION IN THE BARRIER REGION (V>E) 2 1 2 1 k k k k A B + = = 1 1 2 ik ik A C Tunneling Through a Potential Barrier If the barrier EXCEEDS the total energy, motion in the barrier is FORBIDDEN classically, but quantum mechanics allows the electron to LEAK into the barrier Even though its energy remains CONSTANT This effect is referred to as TUNNELING 2 ) ( 2 : h E V m E V = > WHILE CLASSICALLY FORBIDDEN FOR A PARTICLE, THE WAVEFUNCTION MAY PENETRATE INTO A BARRIER WHOSE HEIGHT E XCEEDS ITS TOTAL ENERGY Ψ ( x ) x E < V If the barrier stops at this point, there remains a non-zero amplitude of the wave function. This can excite a wave in the region to the right of the barrier. This is a wave property of the electron . Infinite Barrier Case
3 Finite, Small Barrier Case X Tunneling Through a Potential Barrier X =0 X = a x V 0 2 2 1 1 2 ) ( 1 1 h mE k Be Ae x x ik x ik = + = ψ REGION 1 2 2 1 3 2 ) ( 1 1 h mE k Fe Ee x x ik x ik = + = REGION 3 – the propagation is identical to that of region 1, since the energy and the potential are the same. 2 2 2 ) ( 2 ) ( h E V m De Ce x x x = + = γ REGION 2 – need to consider both exponentials due to reflection from boundary with region 3. Tunneling Through a Potential Barrier We have 5 constants (A, B, C, D, E). One is related to normalization , so we need 4 boundary conditions. (1) ) 0 ( ) 0 ( 2 1 = D C B A + = + (2) 0 2 0 1 = = = x x x x ( ) ( ) D C B A ik = 1 (3) ) ( ) ( 3 2 a a = a ik a a Ee De Ce 1 = + (4) a x a x x x = = = 3 2 ( ) a ik a a Ee ik De Ce 1 1 =

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4 Working with eqns. (3) and (4), the sum and difference give: a ik e ik E C ) ( 1 1 1 2 γ +
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Lecture%2010 - EEE 352 Lecture 12 Tunneling Through a...

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