Physics 2D Lecture Slides Lecture 25: March 1st 2005 Vivek Sharma UCSD Physics 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 < >= = = = �� x 1 1 2 2 3 3 i i i 1 � + + + n n n ... n N � 1 2 3 i P x dx ( ) � � For a general Fn f(x) Sharpness of a distribution : � n � � = scatter around the average � � * ( ) ( ) ( ) x f x x dx n f x ( ) i i < >= = � f x ( ) = �� i 1 � 2 (x x ) � N � i = � P x dx ( ) N �� � � 2 2 = ( x ) ( ) x � � = small Sharp distr. � � Uncertainty X = 1
Particle in the Box, n=1, find <x> & Δ x ? � 2 � � � (x)= sin x � � L � L � � � � 2 sin � � 2 � � � <x>= x x sin x dx � � � � L � L � L � L � - � L � � 2 � � � � � � = x sin 2 x dx , change variable = x � � � � L � L � � L � 0 � 2 1 � � � 2 � 2 � = � � <x>= sin , use sin (1 cos2 ) � L 2 2 0 � � � � 2L � � � � � � � � � � <x>= d - cos2 d use ud v=uv- vdu � � � 2 2 � � 0 0 � � = � 2 L L � � 2 <x>= (same result as from graphing ( )) x � � � 2 2 2 � � L � 2 2 L L � 2 2 2 = � Similarly <x >= x s in ( x dx ) � 2 L 3 2 0 2 2 2 L L L � 2 > � < > = 2 � � = and X= <x x 0.18 L � 2 3 2 4 � 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 + � � ˆ < >= � * � * Q ( , ) [ ] x t Q ( , ) x t d x �� ˆ Q is the observable, [ ] is the operator Q < > & Q is the Expectation va lue � d Exam p les : [X] = x , [P] = i dx 2 2 � 2 � [P] - � = [K] = 2 2 [E] = i � � � m 2m x t 2
Operators Information Extractors � d [p] or p = ˆ Momentum Operator i dx gives the value of average mometum in the following way: � � + + � � � � d � � � * � � * <p> = (x) [ ] ( ) p x dx = (x) 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 � � + + � � � � 2 d 2 ( ) x � � � * � = � * � <K> = (x)[ K ] ( ) x dx (x) dx � � 2 2m dx � � � � - - Similerly + � � � * � <U> = (x )[ U x ( )] ( ) x dx : plug in the U(x) fn for that case � - � � + + � � � � 2 d 2 ( ) x � � � * + � = � * � + an d <E> = (x)[ K U x ( )] ( ) x dx (x) U x ( ) dx � � 2 2m dx � � � � - - Hamiltonian Operator [H] = [K] +[U] � The Energy Operator [E] = i � informs you of the averag e energy � t 3
[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 � � � � � + � = � � 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 Example :[ ] The momentum Extractor (operator) p : Consider as an example: Free Particle Wavefu nction � 2 h p � i(kx-wt) � = � = (x,t) = Ae ; k = , k � p � p �� p (x,t) p p i( x-wt) i( x-wt) � = = � rewrit e (x,t) = Ae ; i A e i (x,t) � � � x � � � � � � � � � (x,t) = p (x,t) � � � � i x � � � � � So it is not unreasonable to associate [p]= with observable p � � � � i x � 4
Example: Average Momentum of Particle in Rigid 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! � � 2 n 2 n � = � * = ( ) x sin( x ) & ( ) x si n( x ) n n L L L L + � � � � d � [ ] � � < >= � � = � � p * p dx * dx � � � i dx � �� �� � � � � � 2 n n n � < >= p sin( x )cos( x dx ) i L L L L �� � 1 n � Since sinax cosax dx = sin 2 ax ...here a = 2a L = x L � � � � n � < >= = = n � = p sin 2 ( x 0 since Sin (0) 2 Sin ( 2 ) 0 � � iL � L � = x 0 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 < >= p 0 so what about expectation value of K= ? 2m < >= < >= � K 0 because p 0; clearly not, since we showed E=KE 0 Why ? What gives ? � n � = ± = ± ± Because p 2 mE ; " " is the key! n n L The AVERAGE p =0 , since particle i s moving back & forth > p 2 <p 2 � <KE> = < > 0 ; not ! 2m 2 m Be careful when being "lazy" Quiz: what about <KE> of a quantum Oscillator? Does similar logic apply?? 5
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 � � 2 2 - � ( ) x + � = � U x ( ) ( ) x E ( ) x � 2 2m x � = � � ( , ) x t ( ) ( ) x t � � ( ) t = � 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 Stationary State: Putting Humpty Dumpty Back Togather d 1 d ( ) f t [ ] = Since ln f t ( ) dt f t ( ) dt � � � � ( ) t 1 ( ) t E iE = � = = � In i � E ( ) , rewrite as t � � � t ( t ) t i � � and integrate both sides w.r.t. time = t t � � t t � 1 ( ) t iE 1 d ( ) t iE � � � = � � = � dt dt dt � � � ( ) t t � ( ) t dt � t=0 0 0 i E � � � � = � ln ( ) t ln (0) t , no w exponentiate both sides � iE t � � � = � � = ( ) t (0) e � ; (0) constant= initial condition = 1 (e.g) iE t i E t � � � � = � � ( ) t e & T h us (x,t)= (x) e where E = ener gy of syst m e � � 6
Schrodinger Eqn: Stationary State Form iE iE iE iE + � � t t t t = � � = * � * � = � * � = � 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 Os cill ator (why?) Total energy of the system depends on the spatial o rientation of the system : charteristic of the pote ntia l U(x,t) ! The Case of a Rusty “Twisted Pair” of Naked Wires & How Quantum Mechanics Saved ECE Majors ! Oxide layer Wire #1 Wire #2 • 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 ! 7
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