Physics 2D Lecture Slides Lecture 24: Feb 28th 2005 Vivek Sharma - PDF document

Physics 2D Lecture Slides Lecture 24: Feb 28th 2005 Vivek Sharma UCSD Physics Simple Harmonic Oscillator: Quantum and Classical Pictures Compared Motion of a Classical Oscillator (ideal) Spring with Ball originally displaced from its equilibirium Force Const position, mot ion confined between x=0 & x=A k 1 1 k = ω ω = = 2 2 2 U(x)= ; . k x m x Ang Freq 2 2 m 1 = ⇒ 2 Changing amplitude A ch ang es E E k A 2 m → → X=0 E can take any value & if A 0, E 0 ± Max. KE at x = 0, KE= 0 at = x A x 1

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 - ( ) 1 x + ω ψ = ψ 2 2 Time Dependen t Schrodinger Eqn: ( ) ( ) m x x E x ∂ 2 2 m x 2 ψ 2 ( ) 2 1 d x m ⇒ = − ω x ψ = ψ 2 2 ( ) ( ) 0 What (x) solves this? E 2 m x � 2 2 dx Two guesses about the simplest Wavefunction: ψ ψ → → ∞ 1. (x) should be symmetric about x 2. (x) 0 as x ψ (x) d ψ + (x) should be continuous & = continu o s u dx ψ e α − 2 α x My guess: (x) = C ; Need to find C & : 0 0 What does this wavefu nct ion & PDF l ook like? Quantum Picture: Harmonic Oscillator − α 2 2 2 x P(x) = C e 0 C 2 ψ e α − 2 x (x) = C 0 C 0 x How to Get C 0 & α ?? …Try plugging in the wave-function into the time-independent Schr. Eqn. 2

Time Independent Sch. Eqn & The Harmonic Oscillator ∂ ψ 2 ( ) x 2 m 1 = ω − ψ 2 2 Master Equation is : [ ] ( ) m x E x ∂ 2 � 2 x 2 ψ ( ) d x ψ = − α 2 = − α − α 2 x x Since ( ) , ( 2 ) , x C e C x e 0 0 dx ψ − α 2 ( ) ( 2 ) d x d x = − α 2 + − α − α 2 = α − α − α 2 x 2 x 2 2 x ( 2 ) [4 2 ] C e C x e C x e 0 0 0 2 dx dx 2 1 m ⇒ α − α − α 2 = ω − − α 2 2 2 x 2 2 x [ 4 2 ] [ ] C x e m x E C e 0 0 � 2 2 2 Match t he coeff of x and the Constant terms on LHS & RHS ω 2 1 m m ⇒ α = ω α 2 2 4 or = m � 2 � 2 2 2 m E α α ⇒ & the other match gives 2 = , substituing � 2 1 ω � E= =hf !!!!...... Planck's Oscil ( la tor s ) 2 What about C ? We learn about that from the Normalization cond. 0 SHO: Normalization Condition +∞ +∞ − ω 2 m x ∫ ∫ ψ = = 2 2 � | ( ) | 1 x dx C e dx 0 0 −∞ −∞ +∞ π ∫ − 2 = ax Since (dont memorize this) e d x a −∞ ω m Identi fying a= and using the identity above � 1 ω ⎡ ⎤ m 4 ⇒ = ⎢ C ⎥ 0 π ⎣ � ⎦ k Hence the Comple te NORMA L IZED wave function is : groundstate 1 − ω 2 ω ⎡ ⎤ m x m 4 ψ � (x) = e 2 Ground State Wavefunction ⎢ ⎥ 0 π ⎣ � ⎦ m e X=0 has ene rgy E = hf x Planck's Oscillators were electrons tied by the "spr ing" of the mutually attractive Coulomb Force 3

Quantum Oscillator In Pictures Quantum Mechanical Prob for particle = + > To live outside classical turning points ( ) 0 for n=0 E KE U x Is finite ! U(x) C 0 U -A +A +A -A Classically particle most likely to be at the turning point (velocity=0) Quantum Mechanically , particle most likely to be at x=x 0 for n=0 Classical & Quantum Pictures of Harmonic Oscillator compared • Limits of classical vibration ⇒ Turning Points Classical oscillator : at x = ± A, changes all KE into potential energy of spring 1 ω 2 2 Total energy E (x = ±A) = KE (x = ±A) + U (x = ±A) = 0 + 2 m A 1 ω � For Quantum Oscillator : T otal Energ y E = ; 2 1 1 ⇒ ω = ω � 2 2 c omp a ring classical and quantum energies m A 2 2 � ⇒ A = ; ω m � ≤ ≤ Classical oscillator bound within -A x A= m ω Cannot venture outside x = ±A because it has no K E l eft • But due to Uncertainty principle, the Quantum Probability for particle outside classical turning points P(|x|>A) >0 !! 4

Quantum Oscillator In The Classically Forbidden Territory Calculate probability of Quantum oscillator where a Classical oscillator can't dare be ! 2 ∞ 2 � -A ∫ ∫ ⇒ ψ ψ Calculate P(|X|>A = ) ; P(|X|>A)= ( ) x d x + ( ) x d x ω 0 0 m ∞ - A 1 ω ⎛ ⎞ m − 2 ⎡ ω ⎤ x ⎜ ⎟ m 4 � = ⎢ 2 ψ ⎝ ⎠ e is symmetric about x=0 Since ( ) x ⎥ 0 π ⎣ � ⎦ 1 ω C 2 ⎛ ⎞ m ∞ 2 ∞ − ω 2 ⎡ ⎤ 2 2 ⎜ ⎟ x m ⇒ 2 ∫ ∫ � ψ ⎝ ⎠ P(|X|>A)=2 ( ) =2 e x dx dx ⎢ ⎥ π 0 ⎣ � ⎦ A A ω � m Change variable: z = x and write A = ω � m x -A A ∞ 2 ∫ − 2 ⇒ z = P(|X|>A)= e Error Fn=erfc(1)=0.157 dz π 1 P(|X|>A) =16% ! !! Large prob ability to go on to the "other side" ! Excited States of The Quantum Oscillator ω 2 m x − ψ = � ( ) x C H ( ) x e 2 ; n n n = ( ) Hermite Polynomials H x n with H (x)=1 0 H (x)=2x 1 − 2 H (x)=4x 2 2 − 3 H ( x)=8x 1 2 x 3 − 2 n x d e 2 n x H (x )=(-1) e n n d x and 1 1 = + ω = + � ( ) ( ) E n n hf n 2 2 ∞ Again n=0,1,2,3... Qu antum # 5

Excited States of The Quantum Oscillator As n � ∞ classical and quantum Ground State Energy >0 always probabilities become similar Measurement Expectation: Statistics Lesson • Ensemble & probable outcome of a single measurement or the average outcome of a large # of measurements ∞ n ∑ ∫ ( ) xP x dx n x + + + i i .... n x n x n x n x < >= = = = −∞ 1 1 2 2 3 3 1 x i i i + + + ∞ ... n n n n N ∫ 1 2 3 i P x dx ( ) − ∞ For a general Fn f(x) Sharpness of a distribution : ∞ n ∑ ∫ ψ ψ = scatter around the average * ( ) ( ) ( ) x f x x dx ( ) n f x i i ∑ < >= = = −∞ ( ) i 1 − f x 2 (x ) x ∞ σ N i ∫ = ( ) P x dx N −∞ σ − 2 2 = ( ) ( ) x x σ → = small Sharp distr. Δ σ Uncertainty X = 6

Particle in the Box, n=1, find <x> & Δ x ? ⎛ π ⎞ 2 ψ (x)= sin ⎜ ⎟ x ⎝ ⎠ L L ∞ ⎛ π ⎞ ⎛ π ⎞ 2 sin 2 ∫ <x>= ⎜ ⎟ sin ⎜ ⎟ x x x dx ⎝ ⎠ ⎝ ⎠ L L L L ∞ - π π L ⎛ ⎞ ⎛ ⎞ 2 ∫ θ 2 = sin , change variable = x ⎜ x dx ⎟ ⎜ x ⎟ ⎝ ⎠ ⎝ ⎠ L L L 0 π 2 1 ∫ ⇒ θ θ θ = − θ 2 2 <x>= sin , use sin (1 cos2 ) π 2 L 2 0 ⎡ π π ⎤ 2L ∫ ∫ ∫ ∫ ⇒ θ θ θ θ θ ⎢ ⎥ <x>= d - cos2 d use ud v=uv- vdu π 2 2 ⎣ ⎦ 0 0 ⎛ ⎞ = π 2 L L ⇒ ψ 2 <x>= ⎜ ⎟ (same result as from graphing ( )) x π 2 ⎝ 2 ⎠ 2 L π 2 2 L L ∫ = − 2 2 2 Similarly <x >= x s in ( x dx ) π 2 3 2 L 0 2 2 2 L L L Δ > − < > = − − = 2 2 and X= <x 0.18 x L π 2 3 2 4 Δ X= 20% of L, Particle not sharply confi ned in Box 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 7

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 )[ U x ( )] ( ) x dx : plug in the U(x) fn for that case ∞ - ∞ ∞ ⎛ ⎞ + + 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 8