Physics 2D Lecture Slides Mar 5 Vivek Sharma UCSD Physics - PowerPoint PPT Presentation

Physics 2D Lecture Slides Mar 5 Vivek Sharma UCSD Physics

Potential Barrier U E<U Transmitted ? x Description of Potential U = 0 x < 0 (Region I ) U = U 0 < x < L (Region II) U = 0 x > L (Region III) Consider George as a “free Particle/Wave” with Energy E incident from Left Free particle are under no Force; have wavefunctions like Ψ = A e i(kx-wt) or B e i(-kx-wt)

Tunneling Through A Potential Barrier Region I Region III U Prob ? II E<U •Classical & Quantum Pictures compared: When E>U & when E<U •Classically , an particle or a beam of particles incident from left encounters barrier: •when E > U � Particle just goes over the barrier (gets transmitted ) •When E<U � particle is stuck in region I, gets entirely reflected, no transmission (T) •What happens in a Quantum Mechanical barrier ? No region is inaccessible for particle since the potential is (sometimes small) but finite

Beam Of Particles With E < U Incident On Barrier From Left Region I II Region III U A Incident Beam F B Reflected Beam Transmitted Beam x 0 L Description Of WaveFunctions in Various regions: Simple Ones first − ω − − ω Ψ = + = i kx ( t ) i ( kx t ) In Region I : ( x t , ) Ae Be incident + reflected Waves I 2 2 = � k ω with E = � 2 m 2 |B| = def ine Reflection Coefficient : R = frac of incident wave intensity reflected back 2 |A| − ω − − ω Ψ = i kx ( t ) + i ( kx t ) = In Region III: ( , ) x t F e G e transmitted III − − ω i ( kx t ) Note : Ge corresponds to wave incident from righ t ! This piece does not exist in the scattering picture we are thinking of now (G=0) 2 |F| Ψ = − ω i kx ( t ) So ( , ) x t Fe represents transmitted beam. Define T = |A| III 2 ⇒ Unitarity Condition R + T= 1 (particle i s either reflected or transmitted)

Wave Function Across The Potential Barrier In Region II of Potential U ψ 2 2 � d ( ) x + ψ = ψ TISE: - U ( ) x E ( ) x 2 2m dx ψ 2 d ( ) x 2 m U ⇒ = − ψ ( E ) ( ) x 2 2 dx � α ψ 2 = ( ) x 2m(U-E) α ⇒ α > 2 with = ; U> E 0 � ± α ψ ∝ x Solutions are of for m ( x ) e + α − ω − α − ω Ψ = + x i t x i t ( , x t ) C e De 0<x< L II ⇒ To determ ine C & D apply mat chi ng cond. Ψ = ( , ) x t continuous acros s bar rier (x=0, L) II Ψ d ( , ) x t II = continuous across barrier (x=0,L) dx

Continuity Conditions Across Barrier ψ ⇒ At x = 0 , continuity of (x) A+B=C+D (1) ψ d (x) ⇒ At x = 0 , continuity of dx − = α − α ikA i kB C D (2 ) ψ ⇒ Similarly at x=L continuity of (x) − α + α + = L L ikL Ce De F e (3) ψ d (x) ⇒ at x=L, continuity of dx − α α α α = L L ikL -( C) e + ( D) e ikFe (4) Four equations & four unknow ns Cant determine A,B,C,D but if you Divide thruout by A in all 4 eq uations : ⇒ → ratio of amplitudes rel ations f or R & T That' s wh at we ne ed a ny way

Potential Barrier when E < U Expression for Transmissi on Coeff T=T ( ) E : Depends on barrier Height U, barrier Width L and particle Energy E − 1     − 2 2 m U ( E ) 1 U 2 α α = T(E) = 1+ sinh ( L ) ;     − � 4 E U ( E )     and R(E)=1- T(E).. .......what's not transmitted is reflected Above equation holds only for E < U For E>U, α =imaginary# Sinh( α L) becomes oscillatory This leads to an Oscillatory T(E) and Transmission resonances occur where General Solutions for R & T: For some specific energy ONLY, T(E) =1 At other values of E, some particles are reflected back ..even though E>U !! That’s the Wave nature of the Quantum particle

Ceparated in Coppertino Oxide layer Wire #1 Wire #2 Solved Example 6.1 (...that I made such a big deal about yesterday) Q: 2 Cu wires are seperated by insulating Oxide layer. Modeling the Oxide layer as a square barrier of height U=10.0eV, estimate the transmission coeff for an incident beam of electrons of E=7.0 eV when the layer thickness is (a) 5.0 nm (b) 1.0nm Q: If a 1.0 mA current in one of the intwined wires is incident on Oxide layer, how much of this current passes thru the Oxide layer on to the adjacent wire if the layer thickness is 1.0nm? What becomes of the remaining current? − 1     2 1 U 2m(U-E) 2mE α α = 2 = , k T(E) = 1+ sinh ( L )     � � − 4 E U ( E )    

− 1     2 1 U α 2 T(E) = 1+ sinh ( L )     − 4 E U ( E )     � = 2 Use =1.973 keV.A/c , m � 511 keV/c e Oxide layer − × × − 2 m U ( E ) 2 3 2 511 kev c / (3.0 10 keV ) � ⇒ α = = = e -1 0.8875A � � 1.973 keV.A/c Substitute in expression for T=T(E) Wire #1 Wire #2 − 1     2 1 10 � � = × − 2 -1 38 T = 1+ si nh (0.8875 A )(50A ) 0.963 10 ( small )!!     − 4 7 (10 7)     � × -7 However, for L=10A; T=0.657 1 0 × Reducing barrier width by 5 leads to Trans. Coeff enhancement by 31 orders of ma gnitude !! ! Q=Nq ⇒ × 15 1 mA current =I= e N =6.25 10 electrons t N =# of electrons that escape to the adjacent wire (past oxide layer) T = = × × 15 Oxide thickness makes all the difference ! N N . T (6.25 10 electrons ) T ; T That’s why from time-to-time one needs to � × -7 ⇒ = × ⇒ = For L=10 A, T=0.657 1 0 N 4.11 10 I 65.7 pA !! T T Scrape off the green stuff off the naked wires Cur en r t Measured on the first wire is sum of incident+reflected currents and current measured on "adjacent" wire is the I T

A Special Case That is Instructive & Useful: U>>E A Given the 4 equations from Continuity Conditions: Solve for F  α   α      A 1 i k 1 i k + α − α = + − ( ik ) L + − − ( ik ) L e e         α α F 2 4  k  2 4  k      2m(U-E) 2mE α = Remember = , k , � � α α α k k α >> − ≈ α when U>>E, >>k & So ; F or large Barrier L, L>>1 α α k k k *  α   α  A 1 i    A  1 i   = + + α = − − + α ( ik ) L ( ik ) L e ; e           F 2 4  k   F  2 4  k        *   α 2   A A 1 1 1 α -1 = = + (2 ) L = T e ; now invert & consolidate       2  F  F 4 16 k T ( ) E         * F F 16   − α = 2 L T = A A e ; now watch the variables emplo yed   2 * α   +   4    k   

A Special Case That is Instructive & Useful: U>>E π 2m(U-E) 2mE p 2 α = = , k = = λ � � � 2 α − 2   = 2 m U ( E ) / � U U = − 1 �   2  k  2 mE / � E E         16 16     α ⇒ = → -2 L varies slowly compared with e term   2   U α     + − 4 1 +   4        E     k    keeing in mind only the Order of magnitude, I sug g est         16 16     ≈ − α =  2 L 1; back to T e substituting   U  2   α   + − 4 1 +    4       E     k    α ≈ -2 L So approxima tely T e Transmi ssion Prob is fn of U,E,L → Why subject you to this TORTURE? Estimate T for complicated Potentials See next (example of Cu oxide layer, radioactivity, blackhole blowup etc)

A Complicated Potential Barrier Can Be Broken Down Multiplicative Transmission prob, ever decreasing but not 0 U(x) Calculus : Integration …… x Can be broken down into a sum of successive Rectangular potential barriers for which we learnt to find the Transmission probability T i The Transmitted beam intensity thru one small barrier becomes incident beam intensity for the following one So on & so forth …till the particle escapes with final Transmission prob T   2 m ∫ − − 2 U x ( ) Edx   ∫ �   = =   T T dx e i

The Great Escape ! My Favorite Movie Story involves an Allied plan for a massive breakout from a Nazi P.O.W. camp, during World War Two. The Nazis had created a high-security, escape -proof prisoner of war camp for those annoying detainees who have attempted escape from their other prison of war camps. These prisoners are not discouraged at all, as they plan a huge escape of 100 men.