Physics 2D Lecture Slides UCSD Physics Vivek Sharma Nov 26 - PowerPoint PPT Presentation

Physics 2D Lecture Slides UCSD Physics Vivek Sharma Nov 26

Expectation Values & Operators: More Formally • Observable: Any particle property that can be measured – X,P, KE, E or some combination of them,e,g: x 2 – How to calculate the probable value of these quantities for a QM state ? • Operator: Associates an operator with each observable – Using these Operators, one calculates the average value of that Observable – The Operator acts on the Wavefunction (Operand) & extracts info about the Observable in a straightforward way � gets Expectation value for that observable +∞ ∫ < >= Ψ ˆ Ψ * * ( , ) [ ] ( , ) Q x t Q x t d x −∞ ˆ is the observable, [ ] is the operator Q Q < > & is the Expectation va lue Q � d Exam p les : [X] = x , [P] = i dx ∂ ∂ 2 � 2 2 [P] - = � [K] = 2 2 [E] = i ∂ ∂ m 2m x t

Operators � Information Extractors � d ˆ [p] or p = Momentum Operator i dx gives the value of average mometum in the following way: ∞ ∞ ψ + + ⎛ ⎞ � d ∫ ∫ ψ ψ ψ * * <p> = (x) [ ] ( ) = (x) ⎜ ⎟ p x dx dx ⎝ ⎠ i dx ∞ ∞ - - Similerly : Plug & play form � 2 2 d ˆ [K] or K = - gi ves the value of average K E 2 2m dx ∞ ∞ ⎛ ⎞ ψ + + � 2 2 ( ) d x ∫ ∫ ψ ψ = ψ − * * <K> = (x)[ ] ( ) (x) ⎜ ⎟ K x dx dx 2 ⎝ 2m ⎠ dx ∞ ∞ - - Similerly ∞ + ∫ ψ ψ * <U> = (x )[ ( )] ( ) : plug in the U(x) fn for that case U x x dx ∞ - ∞ ∞ ⎛ ⎞ ψ + + � 2 2 ( ) d x ∫ ∫ ψ + ψ = ψ − + * * ⎜ ⎟ an d <E> = (x)[ ( )] ( ) (x) ( ) K U x x dx U x dx 2 ⎝ 2m ⎠ dx ∞ ∞ - - Hamiltonian Operator [H] = [K] +[U] ∂ � The Energy Operator [E] = i informs you of the averag e energy ∂ t

[H] & [E] Operators • [H] is a function of x • [E] is a function of t …….they are really different operators • But they produce identical results when applied to any solution of the time-dependent Schrodinger Eq. [H] Ψ (x,t) = [E] Ψ (x,t) • ⎡ ⎤ ∂ ∂ ⎡ ⎤ � 2 2 − + Ψ = Ψ � ( , ) ( , ) ( , ) ⎢ ⎥ U x t x t i x t ⎢ ⎥ ∂ ∂ 2 ⎣ ⎦ ⎣ 2 ⎦ m x t • Think of S. Eq as an expression for Energy conservation for a Quantum system

Where do Operators come from ? A touchy-feely answer :[ ] The momentum Extractor (operator) : Example p Consider as an example: Free Particle Wavefu nction π 2 h p Ψ λ = ⇒ = i(kx-wt) (x,t) = Ae ; k = , k λ � p ∂Ψ p p (x,t) p p i( x-wt) i( x-wt) Ψ = = Ψ � � (x,t) = Ae ; A e (x,t) rewrit e i i ∂ � � x ∂ ⎡ ⎤ � ⇒ Ψ Ψ (x,t) = p (x,t) ⎢ ⎥ ∂ ⎣ ⎦ i x ∂ ⎡ � ⎤ So it is not unreasonable to associate [p]= with observable p ⎢ ⎥ ∂ ⎣ ⎦ i x

Example : Average Momentum of particle in box • Given the symmetry of the 1D box, we argued last time that <p> = 0 : now some inglorious math to prove it ! – Be lazy, when you can get away with a symmetry argument to solve a problem..do it & avoid the evil integration & algebra…..but be sure! π π 2 2 n n ψ = ψ = * ( ) sin( ) & ( ) si n( ) x x x x n n L L L L +∞ ∞ ⎡ ⎤ � d [ ] ∫ ∫ < >= ψ ψ = ψ ψ * * p p dx dx ⎢ ⎥ ⎣ ⎦ i dx −∞ −∞ ∞ π π π � 2 n n n ∫ < >= sin( )cos( ) p x x dx i L L L L −∞ π 1 n ∫ 2 Since sinax cosax dx = sin ...here a = ax 2a L = π x L � ⎡ ⎤ n ⇒< >= = = n π = 2 2 2 sin ( 0 since Sin (0) Sin ( ) 0 p x ⎢ ⎥ ⎣ ⎦ iL L = 0 x We knew THAT befor e doing any ma t h ! Quiz 1: What is the <p> for the Quantum Oscillator in its symmetric ground st ate Quiz 2: What is the <p> for the Qua ntum Osc i l la t or in it s asymmetric first excited state

But what about the <KE> of the Particle in Box ? 2 p < >= 0 so what about expectation value of K= ? p 2m < >= < >= ≠ 0 because 0; clearly not, since we showed E=KE 0 K p Why ? What gives ? π � n = ± = ± ± Because p 2 ; " " is the key! mE n n L AVERAGE p =0 , particle i s moving b ack & forth > 2 2 p <p ≠ <KE> = < > 0 not ! 2m 2 m Be careful when being "lazy" Quiz: what about <KE> of a quantum Oscillator? Does similar logic apply??

Schrodinger Eqn: Stationary State Form • Recall � when potential does not depend on time explicitly U(x,t) =U(x) only…we used separation of x,t variables to simplify Ψ (x,t) = ψ (x) φ (t) & broke S. Eq. into two: one with x only and another with t only ∂ ψ � 2 2 - ( ) x + ψ = ψ ( ) ( ) ( ) U x x E x ∂ 2 2m x Ψ = ψ φ ( , ) ( ) ( ) x t x t ∂ φ ( ) t = φ � ( ) i E t ∂ t How to put Humpty-Dumpty back together ? e.g to say how to go from an expression of ψ (x) →Ψ (x,t) which describes time-evolution of the overall wave function

Schrodinger Eqn: Stationary State Form d 1 d ( ) f t [ ] = Since ln ( ) f t dt ( ) dt f t ∂ φ ∂ φ ( ) 1 ( ) t t E iE = φ = = − � In i ( ) , rew rite as E t ∂ φ ∂ � � t ( ) t t i integrate both sides w.r.t. time and = ∂ φ φ t t t t 1 ( ) 1 d ( ) t iE t iE ∫ ∫ ∫ = − ⇒ = − dt dt dt φ ∂ φ � � ( ) t ( ) dt t t t=0 0 0 iE ∴ φ − φ = − ln ( ) ln (0) , n ow exponentiate both sides t t � iEt − ⇒ φ = φ φ = � ( ) (0) ; (0) constant= initial condition = 1 (e.g) t e iE i E t − − t ⇒ φ = Ψ ψ � � ( ) & T hus (x,t)= (x) where E = energy of system t e e

Schrodinger Eqn: Stationary State Form iE iE iE iE + − − t t t t = Ψ Ψ = ψ ψ = ψ ψ = ψ * * * 2 � � � � ( , ) ( ) ( ) ( ) ( ) | ( ) | P x t x e x e x x e x In such cases, P(x,t) is INDEPENDENT of time. These are called "stationary" states because Prob is independent of tim e Examples : Pa rtic le in a box (why?) : Quantum Oscil lator ( why?) Total energy of the system depends on the spatial orie ntation of the system : charteristic of the potential situat i on !

The Case of a Rusty “Twisted Pair” of Naked Wires & How Quantum Mechanics Saved ECE Majors ! • Twisted pair of Cu Wire (metal) in virgin form • Does not stay that way for long in the atmosphere •Gets oxidized in dry air quickly Cu � Cu 2 O •In wet air Cu � Cu(OH) 2 (the green stuff on wires) • Oxides or Hydride are non-conducting ..so no current can flow across the junction between two metal wires No current means no circuits � no EE, no ECE !! • • All ECE majors must now switch to Chemistry instead & play with benzene !!! Bad news !

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 : ( , ) incident + reflected Waves x t Ae Be I = � 2 2 k ω � with E = 2 m 2 |B| = Reflection Coefficient : R = of incident wave intensity reflected back def ine frac 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 : corresponds to wave incident from righ t ! Note Ge This piece does not exist in the scattering picture we are thinking of now (G=0) 2 |F| Ψ = − ω ( ) i kx t So ( , ) represents transmitted beam. Define T = |A| x t Fe III 2 ⇒ Condition R + T= 1 (particle i s either reflected or transmitted) Unitarity