Physics 2D Lecture Slides Nov 19 Vivek Sharma UCSD Physics Where - PDF document

Physics 2D Lecture Slides Nov 19 Vivek Sharma UCSD Physics

Where Do Wave Functions Come From ? ∂ Ψ ∂Ψ 2 2 � ( , ) x t ( , ) x t − + Ψ = U x ( ) ( , ) x t i � ∂ ∂ 2 2 m x t U(x) = characteristic Potential of the system

Factorization Condition For Wave Function Leads to: ∂ ψ 2 2 - � ( ) x + ψ = ψ U x ( ) ( ) x E ( ) x TI SE ∂ 2 2m x ∂ φ ( ) t = φ i � E ( ) t ∂ t What is the Constant E ? How to Interpret it ? Consider the free particle situation : ω Ψ ψ ikx -i t ikx (x,t)= Ae e , (x)= Ae U(x,t) = 0 ⇒ Plug it into the Time Independent Schrodinger Equ ation (T ISE ) − 2 2 ( ikx ) 2 2 2 � d ( A e ) � k p + = ⇒ = = = ( ikx ) 0 E Ae E (NR Energy) 2 2 m dx 2 m 2 m Ψ ψ ω -i t Stationary states of the free particle: (x,t)= (x)e 2 2 ⇒ Ψ = ψ ( , ) x t ( ) x ψ Probability is static in time t, character of wave functio n depends on ( ) x

A More Interesting Potential : Particle In a Box Write the Fo rm of Potential: Inf inite Wal l U(x) ∞ ≤ ≥ U(x,t) = ; x 0, x L U(x,t) = 0 ; 0 > X > L • Classical Picture: •Particle dances back and forth •Constant speed, const KE •Average <P> = 0 •No restriction on energy value • E=K+U = K+0 •Particle can not exist outside box •Can’t get out because needs to borrow infinite energy to overcome potential of wall What happens when the joker is subatomic in size ??

Ψ (x) for Particle Inside 1D Box with Infinite Potential Walls ⇒ Inside the box, no force U=0 or constant (same thing) Why can’t the 2 2 ψ � - d ( ) x ⇒ + ψ = ψ 0 ( ) x E ( ) x particle exist 2 2m dx 2 ψ d ( ) x 2 mE Outside the box ? ⇒ = − ψ = 2 2 k ( ) ; x k 2 2 dx � � E Conservation ψ 2 d ( ) x + ψ = ⇐ ψ 2 or k ( ) x 0 fig ure out what (x) solves this diff e q. ∞ ∞ 2 dx ψ = + In General the solu t io n is ( ) x A sinkx B coskx (A,B are constants) Need to figure out values of A, B : How to do that ? A p pl y BO UNDA R Y Conditions on the Physical Wav efunction ψ We said ( ) must be continuous everywhe x re So match the wavefunction just outside box to the wavefunction value just inside the box ⇒ ⇒ ψ = = ⇒ ψ = = At x = 0 ( x 0) 0 & At x = L ( x L ) 0 ∴ ψ = = = ( x 0) B 0 (Continuity condition at x =0) X=L ψ = = ⇒ X=0 & ( x L ) 0 A Sin kL = 0 (Continuity condition at x =L) π n ⇒ π ⇒ = ∞ kL = n k = , n 1,2,3,... L π 2 2 2 n � So what does this say about Energy E ? : E = Quantized (not Continuous)! n 2 2 mL

Quantized Energy levels of Particle in a Box

What About the Wave Function Normalization ? The particle's Energy and Wavefu nct ion a re determi ned by a nu mb er n → We will call n Quantum Number , just like in Bohr's Hydrogen atom W hat about the wave functions cor res pondi ng to each of these e ner g y states? π n x ψ = = A sin( kx ) A sin( ) for 0<x < L n L ≥ ≥ = 0 for x 0, x L Normalized Condition : L L π n x ∫ ∫ ψ ψ = θ = − θ * 2 2 2 1 = dx A S in ( ) Use 2Sin 1 2 Cos 2 n n L 0 0 L π 2 ⎛ ⎞ A 2 n x ∫ ∫ = − θ θ 1 1 c os( ) and since cos = sin ⎜ ⎟ 2 ⎝ L ⎠ 0 2 A L 2 = ⇒ = 1 A 2 L π 2 2 n x ψ = = So sin( kx ) sin ( ) ...What does this look l ike? n L L L

Wave Functions : Shapes Depend on Quantum # n Probability P(x): Where the Wave Function particle likely to be Zero Prob

Where in The World is Carmen San Diego? • We can only guess the probability of finding the particle somewhere in x – For n=1 (ground state) particle most likely at x = L/2 – For n=2 (first excited state) particle most likely at L/4, 3L/4 • Prob. Vanishes at x = L/2 & L – How does the particle get from just before x=L/2 to just after? » QUIT thinking this way, particles don’t have trajectories Classically, where is the particle most » Just probabilities of Likely to be : Equal prob of being being somewhere anywhere inside the Box NOT SO says Quantum Mechanics!

What Was Sesame Street Trying to Teach you ! This particle in the box is brought to you by the letter Its the Big Boss Quantum Number

How to Calculate the QM prob of Finding Particle in Some region in Space Consider n =1 state of the particle L 3 L ≤ ≤ Ask : What is P ( x )? 4 4 3 L 3 L 3 L π π 4 4 ⎛ ⎞ 4 2 x 2 1 2 x ∫ 2 ∫ ∫ ψ = = − 2 P = dx sin dx . (1 cos ) dx ⎜ ⎟ 1 L L ⎝ L ⎠ 2 L L L L 4 4 4 3 /4 L π π π ⎡ ⎤ ⎡ ⎤ ⎛ ⎞ 1 L L 2 x 1 1 2 3 L 2 L = − = − − P sin sin . sin . ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ π π L ⎣ 2 ⎦ ⎣ 2 L ⎦ 2 2 ⎝ L 4 L 4 ⎠ L /4 1 1 ( 1 1) = − − − = ⇒ P 0.818 8 1. 8 % 2 π 2 ⇒ Classically 50% (equal prob over half the box size) ⇒ Substantial difference between Class ical & Quantu m predictio n s

When The Classical & Quantum Pictures Merge: n →∞ But one issue is irreconcilable: Quantum Mechanically the particle can not have E = 0 This is a consequence of the Uncertainty Principle The particle moves around with KE inversely proportional to the Length Of the 1D Box

Finite Potential Barrier • There are no Infinite Potentials in the real world – Imagine the cost of as battery with infinite potential diff • Will cost infinite $ sum + not available at Radio Shack • Imagine a realistic potential : Large U compared to KE but not infinite Region III Region II Region I U E=KE X=L X=0 Classical Picture : A bound particle (no escape) in 0 < x < L Quantum Mechanical Picture : Use ∆ E. ∆ t ≤ h/2 π Particle can leak out of the Box of finite potential P(|x|>L) ≠ 0

Finite Potential Well ψ 2 2 - � d ( ) x + ψ = ψ U ( ) x E ( ) x 2 2m dx ψ 2 d ( ) x 2 m U ⇒ = − ψ ( E ) ( ) x 2 2 dx � 2m(U-E) α ψ α 2 = ( ); = x 2 � ⇒ ψ = + α + − α x x General Solutions : ( ) x Ae Be ψ Require finiteness of ( ) x ⇒ ψ = + α x ( x ) A e .....x<0 (region I) ψ = − α x ( x ) Ae .....x>L (regi on III ) Again, coefficients A & B come from matching conditions at the edge of the walls (x =0, L) ψ ≠ But note th at wave fn at ( ) at (x =0, L) x 0 ! ! (why?) ψ d ( ) x ψ Further require Continuity of ( ) and x dx These lead to rather different wave funct ions

Finite Potential Well: Particle can Burrow Outside Box

Finite Potential Well: Particle can Burrow Outside Box Particle can be outside the box but only for a time ∆ t ≈ h/ ∆ E ∆ E = Energy particle needs to borrow to Get outside ∆ E = U-E + KE The Cinderella act (of violating E Conservation cant last very long Particle must hurry back (cant be caught with its hand inside the cookie-jar) 1 � δ Penetration Length = = α 2m(U-E) ⇒ If U>>E Tiny penetration → ∞ ⇒ δ → If U 0

Finite Potential Well: Particle can Burrow Outside Box 1 � δ Penetration Length = = α 2m(U-E) ⇒ If U>>E Tiny penetration → ∞ ⇒ δ → If U 0 π 2 2 2 n � = E = , n 1,2,3,4... n + δ 2 2 ( m L 2 ) When E=U then solutions blow up ⇒ < Limits to number of bound states(E U ) n When E>U, particle is not bound and can get either reflected or transmitted across the potential "b arrier"

Simple Harmonic Oscillator: Quantum and Classical Spring with Force Const k m X=0 x

U(x) Stable Equilibrium: General Form : Unstable 1 − 2 U(x) =U(a)+ k x ( a ) 2 1 ⇒ = − 2 R escale U x ( ) k x ( a ) 2 c Motion of a Classical Os cillator (ideal) x a b Ball originally displaced from its equilib irium Stable Stable position, mo tion co nfined betw e en x =0 & x=A 1 1 k 2 = ω 2 2 ω = = U(x)= kx m x ; Ang F . req Particle of mass m within a potential U(x) 2 2 m � dU x ( ) 1 F(x)= - = 2 ⇒ E kA Changing A changes E dx 2 � → → dU x ( ) E can take any value & if A 0, E 0 = F(x=a) = - 0, ± Max. KE at x = 0, KE= 0 at x= A dx � � F(x=b) = 0 , F(x=c)=0 ...But... look at the Cur vature: ∂ ∂ 2 2 U U > 0 (stable), < 0 (uns tabl ) e ∂ 2 ∂ 2 x x

Quantum Picture: Harmonic Oscillator ψ Find the Ground state Wave Function (x) 1 ω 2 2 Find the Ground state Energy E when U(x)= 2 m x ∂ ψ 2 2 - � ( ) x 1 + ω ψ = ψ 2 2 Time Dependen t Schrodinger Eqn: m x ( ) x E ( ) x ∂ 2 2 m x 2 ψ 2 d ( x ) 2 m 1 ⇒ = − ω x ψ = ψ 2 2 ( E 2 m ) ( ) x 0 What (x) solves this? 2 2 dx � Two guesses about the simplest Wavefunction: ψ ψ → → ∞ 1. (x) should be symmetric about x 2. (x) 0 as x ψ d (x) ψ + (x) should be continuous & = continu o s u dx 2 ψ e α − α x My guess: (x) = C ; Need to find C & : 0 0 What does this wavefu nct ion & PDF l ook like?