Download Reflection and Transmission of Waves in Quantum Mechanics: Step Potential - Prof. Ana Jofr and more Study notes Physics in PDF only on Docsity! Lecture 19 Last class: Solved Time-Independent Schrödinger equation for 1-D infinte square well. L xn L x πψ sin2)( = 2 222nEn πh = 2mL TODAY: Reflection and Transmission of Waves Here we consider the type of problems where particles are incident on an energetic barrier. Examples: • A beam of particles hitting some material A l i l i l ( f l ) hi i i f hi h i • n e ectr ca s gna stream o e ectrons tt ng a reg on o g er res stance (impedance mismatch) Recall – A single particle is described by a wavepacket. Here we consider a stream of particles, we describe them with a single harmonic d f d wave over a e ine region. General Solutions fl d b Region I, x < 0: xikxikI BeAe 11 −+=ψ Incident beam moving from left to right. Re ecte eam moving from right to left Region II x > 0: xikxik II DeCe 22 −+=ψ, Transmitted beam moving from left Incident beam moving from right to left to right. Does not exist! Therefore, D=0 Now apply continuity condition on ψ and on dψ/dx xikxik I BeAe 11 −+=ψ xik II Ce 2=ψ At x=0: III ψψ = dd ψψ CBA =+ CkBkAk 211 =−dxdx III = Solve CkBkAk 211 =−CBA =+ and * See problem 6-47 * AkkB 21 − 2mE VEE −− kk 21 + = 21 h k = A VEE B o o −+ = AkC 12= )(2 oVEmk − A EC = 2 kk 21 + 2 2 h = VEE o−+ Important Notes 1) Even though E>Vo, R≠0 (There is always some reflection) D ff f l l l ! i ers rom c assica resu t (Analogy: light reflected between two media) ) d d |k k | d ’ h h l 2 R epen s on 1- 2 ; oesn t matter w ic is arger. Step down in potential (-Vo) produces same reflection as step up. 3) For Vo>0, wavelength in region II is longer than the wavelength in region I (less momentum). Tilustration
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Now consider situation where E < Vo Region II (x>0, V(x) = Vo) : )( )( 2 22 2 xk d xd ψψ −= 2 2 2 )(2 where h oVEmk −= x Since Vo > E, k2 is imaginary! αi EVm ik o = − = )(2 2 h )()( 2 2 xxd ψαψ =Rewrite equation: 2dx l xC α− )(2 EVm −So ution in region II: II eψ = h o=α Now consider situation where E < Vo xikxik I BeAe 11 −+=ψ xII Ce αψ −= Particles are not all reflected at x=0! Boundary conditions require CBikAik α−=− 11CBA =+ and A ik ikB α α − + = 1 1 21 2 h mEk = AikC 12 )(2 EVm ik α− = 1 h o −=α Barrier Potential: Tunneling V(x) = 0 for x<0 and x>a V(x) = Vo for 0<x<a xikxik BeAe 11 −+ψ xik III Fe 1=ψ General Solutions: I = 21 2 h mEk = 21 2 h mEk = xx II DeCe ααψ −+= )(2 EVm − Lower amplitude means fewer h o=α particles. Transmission Coefficient 1− ⎤⎡ 2 2 2 )1(4 sinh1 ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ − +== EE a A F T αDerived using continuity conditions. ⎦⎣ oo VV ae V E V ET α2)1(16 −−≈For αa >> 1 : Tunneling: oo Particle has a finite probability of crossing an energetic barrier! Application: Scanning Tunneling Electron Microscope - Sample is coated in gold - Air gap between sample and tip serves as barrier - Tunneling current is very sensitive to size of gap (width of barrier) - resolution ~0.5 nm (size of an atom)
