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Lecture 16 Tunneling (c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has been developed and made available online by work supported jointly by University of Illinois, the National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the sponsoring agencies. Tunneling We will consider a problem in which a particle of mass m and energy E hits a potential barrier of height V and width L. V is greater than E. Classically, the particle cannot overcome the barrier, but quantum mechanically it can “tunnel” through it. Tunneling The Schrödinger equation to solve: d 2Y +V Y = EY 2 2m dx 2 Energy E is given (continuous) and assumed smaller than V. Tunneling The situation we describe: A particle flies in from the left. Sometimes it is bounced back by the left barrier; sometimes it passes through it. The particle can be bounced back by the right barrier. What is the ratio of transmission versus reflection |A|2/|A’|2? Tunneling The shape of the potential naturally divides the space into three regions. In each region, the potential is flat. For a flat potential, eikx (or equivalently sin and cos) and ekx are the most promising functional forms. 2 2 ikx d ikx ikx e : - 2 e = - ( ik ) e = k 2eikx dx 2 d ekx : - 2 ekx = -k 2ekx dx Tunneling Our intuitive picture of the solution: eikx represents an incoming particle, with momentum +ħk e–ikx is a reflected particle, with momentum –ħk eikx is a transmitted particle, with momentum +ħk e–κx, e+κx is a decaying function which connects the left and right wave functions. Tunneling The wave function within the barrier: Here, V > E and e–κx and e+κx are more suitable. - 2 2m 2 d e dx 2 d e dx d Y Middle 2 dx 2 = (E -V )Y Middle x e 2 2 x Negative ikx k e 2 2 ikx Positive Tunneling Left A e - 2 d 2 Y Left 2m dx 2 ikx Be ikx R ig h t A e k2 2 = Y Left 2m - M iddle C e E - 2 2m d 2 Y Middle dx 2 x +V Y Middle De 2 2m d 2 Y Right dx 2 k2 2 = Y 2m Right x æ k2 2ö = çV Y Middle ÷ 2m è ø =E<V ikx E Boundary conditions The wave functions must be continuous and smooth (no infinity in potentials). Continuous AB C D Continuous Ce L k2 2 = E We know k and κ. 2m L A e ikL Smooth Smooth ikA ikB C D De Ce V- L De k2 2 2m L =E ikA e ikL Tunneling Energy, thus, k and κ are known. We have five unknowns: A, B, C, D, A΄. We have four boundary conditions. The fifth condition comes from the normalization, determining all five unknowns. We want to determine the transmission probability |A΄|2 / |A|2. In fact, with the four boundary conditions alone (without normalization), we can find four ratios: B/A, C/A, D/A, A΄/A. Tunneling A¢ 2 A 2 µe -2k L V- k2 2 2m = E ®k = 2m(V - E) For a high, wide barrier, the transmission probability decreases exponentially with Thickness of barrier, Square root of particle mass, Square root of energy deficit V–E. Tunneling Tunneling occurs when the particle does not have enough energy to overcome the barrier. Classically, this is impossible. Within the barrier, the kinetic energy would be negative. Quantum mechanically, this is possible according to the Schrödinger equation. (Some argue that the particle can acquire a higher energy necessary to cross the barrier momentarily thanks to the time-energy uncertainty principle). Permeation We have already seen (in harmonic oscillator problem) that a wave function can permeate in the classically forbidden region. This is by the same mechanism that tunneling occurs. Ammonia inversion Scanning tunneling microscope Gold (100) surface Public image from Wikipedia Summary A particle of energy E can tunnel through a barrier of potential height V > E. A particle of energy E can also permeate into the potential wall of V > E. The thinner the barrier or the lower the barrier height or the lighter the particle, the more likely the particle can tunnel.