Physics 2D Lecture Slides Lecture 21: Feb 23 rd Vivek Sharma UCSD - PDF document

Confirmed: 2D Final Exam:Thursday 18 th March 11:30-2:30 PM WLH 2005 Physics 2D Lecture Slides Lecture 21: Feb 23 rd Vivek Sharma UCSD Physics

Quiz 6 35 Number of Students 30 25 20 15 10 5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Score Ψ : The Wave Function Of A Particle • The particle must be some where The Wave Function is a mathematical +∞ ∫ ψ = function that describes a physical 2 | ( , ) | 1 x t dx object � Wave function must have some −∞ Any Ψ satisfying this condition is • rigorous properties : NORMALIZED • Prob of finding particle in finite interval • Ψ must be finite = ∫ b ≤ ≤ ψ ψ * ( ) ( , ) ( , ) P a x b x t x t dx • Ψ must be continuous fn of x,t a • Ψ must be single-valued • Fundamental aim of Quantum Mechanics • Ψ must be smooth fn � – Given the wavefunction at some instant (say t=0) find Ψ at some ψ d subsequent time t must be continuous – Ψ (x,t=0) � Ψ (x,t) …evolution dx – Think of a probabilistic view of particle’s “newtonian trajectory” WHY ? • We are replacing Newton’s 2 nd law for subatomic systems

A Simple Wave Function : Free Particle • Imagine a free particle of mass m , momentum p and K=p 2 /2m • Under no force , no attractive or repulsive potential to influence it • Particle is where it wants : can be any where [- ∞ ≤ x ≤ + ∞ ] – Has No relationship, no mortgage , no quiz, no final exam….its essentially a bum ! – how to describe a quantum mechanical bum ? • Ψ (x,t)= Ae i(kx- ω t) =A(Cos(kx- ω t)+ i sin (kx- ω t)) E p = ω k ; = Has definite momentum � � For non-relativistic particles and energy but location 2 � 2 p k unknown ! ⇒ ω E= (k)= 2m 2m X � Wave Function of Different Kind of Free Particle : Wave Packet Sum of Plane Waves: +∞ ∫ Ψ = ikx ( ,0) ( ) x a k e dk Combine many free waves to create a −∞ Localized wave packet (group) +∞ ∫ Ψ = − ω i kx ( t ) ( , ) ( ) x t a k e dk −∞ Wave Packet initially localized ∆ ∆ in X, t undergoes dispersion The more you know now, The less you will know later Why ? Spreading is due to DISPERSION resulting from the fact that phase velocity of individual waves making up the packet depends on λ (k)

Normalization Condition: Particle Must be Somewhere x − ψ = x : ( , 0) , C & x are constants Example x Ce 0 0 This is a symmetric wavefunction with diminishing amplitude ⇒ The Amplitude is maximum at x =0 Prob ability is max too Norma lization Condition: How to figure ou t C ? A real particle must be somewhere: Probability of finding ∞ ∞ x + + − 2 ∫ ∫ 2 ∞ ≤ ≤ ∞ ψ = = x 2 particle is finite P(- x + ) = ( ,0 ) 1 x dx C e 0 d x ∞ ∞ - - ∞ x ⎡ ⎤ − 2 x ∫ ⇒ = = = 2 x 2 2 0 1 2 2 C e dx C C x 0 ⎢ ⎥ 0 ⎣ ⎦ 2 0 x − 1 ⇒ ψ = x ( ,0) x e 0 x 0 Where is the particle within a certain location x ± ∆ x Lets Freeze time (t=0) Prob | Ψ (x,0)| 2 ? x x +x +x − 0 0 2 ∫ ∫ ≤ ≤ ψ 2 = 2 x P(-x x +x ) = ( ,0) x dx C e 0 dx 0 0 -x -x 0 0 ⎡ ⎤ ⎡ x ⎤ ⎡ ⎤ = − − = − − = ⇒ 2 2 2 0 2 1 1 0.865 87% C e e ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2

Where Do Wave Functions Come From ? • Are solutions of the time Schrodinger had an interesting life dependent Schrödinger Differential Equation (inspired by Wave Equation seen in 2C) ∂ Ψ ∂Ψ � 2 2 ( , ) ( , ) x t x t − + Ψ = � ( ) ( , ) U x x t i ∂ ∂ 2 2 m x t • Given a potential U(x) � particle under certain force ∂ ( ) U x − – F(x) = ∂ x Introducing the Schrodinger Equation ∂ 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 • Different potential for different forces • Hence different solutions for the Diff. eqn. • � characteristic wavefunctions for a particular U(x)

Schrodinger Wave Equation ψ Wavefunction which is a sol. of the Sch. Equation embodies all modern physics experienced/learnt so fa r: h ∆ ∆ ∆ ∆ ∼ � ∼ � E=hf, p= , . , . , quantiza tion etc x p E t λ Schrodinge r Equation is a D ynamical Equation � � much like Newton's Equation F = a m → ψ → → ψ (x,0) Force(potentia l ) (x,t) Evolves the System as a function of space-time The Schrodinger Eq. propogates the system forwar d & backward in time: ψ ⎡ ⎤ d ψ δ ψ ± ⎢ δ (x, t) = (x,0) t ⎥ ⎣ ⎦ dt = 0 t Where does it come from ?? ..."First Principles"..no real derivation exists Time Independent Sch. Equation ∂ Ψ ∂Ψ � 2 2 ( , ) x t ( , ) x t − + Ψ = � ( ) ( , ) U x x t i ∂ ∂ 2 2 m x t Sometimes (depending on the character of the Potential U(x,t)) The Wave function is factorizable: can be broken up ( ) Ψ = ψ φ x,t ( ) ( ) x t Ψ ω = ω i(kx - t) i(kx) -i( t) : Plane Wave (x,t )=e e e Exa mple In suc h cases, use seperation of variables to get : ∂ ψ ∂ φ � 2 2 - ( ) ( ) x t φ + ψ φ = ψ � ( ). ( ) ( ) ( ) ( ) t U x x t i x ∂ ∂ 2 2m x t Ψ ψ φ Divide Throughout by (x,t )= ( x) ( t) ∂ ψ ∂ φ � 2 2 - 1 ( ) 1 ( ) x t ⇒ + = � . ( ) U x i ψ ∂ φ ∂ 2 2m ( ) ( ) x x t t L HS i s a function of x; RHS is fn of t x and t are independent variables, hence : ⇒ RHS = LHS = Constant = E

Factorization Condition For Wave Function Leads to: ∂ ψ � 2 2 - ( ) x + ψ = ψ ( ) ( ) ( ) U x x E x ∂ 2 2m x ∂ φ ( ) t = φ � ( ) i E t ∂ t What is the Constant E ? How to Interpret it ? Back to a Free particle : Ψ ω ψ ikx -i t ikx (x,t)= Ae e , (x)= Ae U(x,t) = 0 ⇒ Plug it into the Time Independent Schrodinger Equation (TISE) − � � 2 2 ( ikx ) 2 2 2 ( ) d Ae k p + = ⇒ = = = ( ikx ) 0 E A e E (NR Energy) 2 2 2 2 m dx m m Ψ ψ ω -i t Stationary states of the free particle: (x,t)= (x)e ⇒ Ψ 2 = ψ 2 ( , ) x t ( ) x ψ Probability is static in time t, character of wave function depends on ( ) x A More Interesting Potential : Particle In a Box Write the Form of Potential: Infinite Wall ∞ ≤ ≥ 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 ??

Example of a Particle Inside a Box With Infinite Potential (a) Electron placed between 2 set of electrodes C & grids G experiences no force in the region between grids, which are held at Ground Potential However in the regions between each C & G is a repelling electric field whose strength depends on the magnitude of V (b) If V is small, then electron’s potential energy vs x has low sloping “walls” (c) If V is large, the “walls”become very high & steep becoming infinitely high for V →∞ (d) The straight infinite walls are an approximation of such a situation U= ∞ U(x) U= ∞ 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 ψ � 2 2 - ( ) d x ⇒ + ψ = ψ 0 ( ) ( ) x E x particle exist 2 2m dx ψ 2 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 ( ) 0 fig ure out what (x) solves this diff e q. or k x ∞ ∞ 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 ( 0) 0 & At x = L ( ) 0 x x L ∴ ψ = = = ( 0) 0 (Continuity condition at x =0) x B ψ = = ⇒ X=L & ( x L ) 0 A Sin kL = 0 (Continuity condition at x =L) X=0 π n ⇒ π ⇒ = ∞ kL = n k = , 1,2,3,... n L π 2 2 � 2 n So what does this say about Energy E ? : E = Quantized (not Continuous)! n 2 2 mL