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Quantum Physics 2005 Notes-2 The State Function and its Interpretation Notes 2 Quantum Physics F2005 1 In a particle-detection scheme, the wave-like interference pattern builds up particle-by-particle from Morrison Notes 2 Quantum Physics F2005 • Note that the arrival of each particle appears to be random, until enough particles have arrived for a statistically useful sample. • This suggests that we should use a probability distribution approach for quantum physics calculations 2 Discussion of probability • The word event refers to obtaining a making a measurement of an observable. • To define the probability of an event result, we consider the fraction of times that result will occur for measurements on a very large number of identical systems. Such a collection of systems is called an ensemble. • To perform an ensemble measurement of an observable, we perform precisely the same measurement on the members of an ensemble of identical systems. Notes 2 Quantum Physics F2005 3 Normalized Probability The probability of a specific outcome is the number of times that it occurs, divided by the total number of experiments: n P(a) = a N The probability of a set of outcomes is the sum of the probabilities of each outcome: c 1 c P (a or b or ...) = ∑ P ( j ) = ∑ n j j =a N j =a By this definition, all possible outcomes ∑ P( j ) = 1 j =a Notes 2 Quantum Physics F2005 4 Ensemble Averages • Given the probabilities for a set of outcomes, we can find useful quantities, like the average or expectation value of a result q, through a simple sum: M ∑ q j P(q j ) "mean of q"= q = q = j =1 M ∑ P(q j ) j =1 • Another statistic that is useful in describing distributions is the dispersion or variance: variance=dispersion= ( q - q Notes 2 Quantum Physics F2005 ) 2 5 Continuous distributions • In many physical cases, the result of a measurement can be any value within a physical range, for example, the position of a particle. • We can define the probability of finding the particle within a certain infinitesimal range as P(x)dx and we call P(x) the probability density. • Given a probability density, we can compute the expectation value of any parameter q(x). ! x = ∫ xP( x)dx "! ! q( x) = ∫ q ( x) P( x)dx "! Notes 2 Quantum Physics F2005 6 Matter waves • Just adding probability distributions does not give interference. • In order to get an interference pattern, we must be able to superpose two waves. • A working hypothesis: – Observations occur in proportion to the probability that a particle will be in a state (r(t), p(t), E(t)) when it is measured. – The probability is proportional to the square of a wave amplitude. Notes 2 Quantum Physics F2005 7 The State Function • Every physically realizable state of a system is described in quantum mechanics by a state function # that contains all accessible information about the system in that state. – The phrase “wave function” is usually used to mean a state function that is specified as a function of position and time #(x,t). Notes 2 Quantum Physics F2005 8 Born’s Postulate If, at time t, a measurement is made to locate the particle r associated with the wave function # (r , t ), then the probability r P(r , t )dv that the particle will be found in the volume r dv around position r is equal to: r r r P (r , t ) dv = # *(r , t ) # (r , t ). r r Where # *( r , t ) denotes the complex conjugate of # (r , t ). Notes 2 Quantum Physics F2005 9 Principle of Superposition If #1 and # 2 represent two physically realizable states of a system, then the linear combination # = c1#1 + c2 # 2 is also a physically realizable state of the system. Notes 2 Quantum Physics F2005 10 Normalization • Sometimes you will be given wave functions that lead to an integrated probability that is not already 1. (In fact, lazy professors do this a lot.) • Before you try to do any calculation with a wavefunction, you should make sure it is properly ! r r normalized, that is: ∫ # *( r , t )# (r , t )dv = 1 "! ! r r Given a un-normalized wavefunction #' such that ∫ # '*(r , t )# '(r , t )dv = M "! just multiply #' by the constant #(x,t)= 1 to create a new function: M #'(x,t) that will be normalized. M Notes 2 Quantum Physics F2005 11 Physical constraints on wavefunctions • A physical wavefunction must be normalizable. (The particle must be somewhere.) • A physical wavefunction must be single valued. (It can’t have two probabilities of being at the same point.) • A physical wavefunction and its spatial derivatives must be continuous. Notes 2 Quantum Physics F2005 12 Examples of allowed and disallowed wavefunctions (from Morrison) Notes 2 Quantum Physics F2005 13 Expectation value • In quantum mechanics, the ensemble average of an observable for a particular state of the system is called the expectation value of that observable. ! * # ∫ Q#dv Q = "! ! * # #dv ∫ "! • The uncertainty in the observable is the square root of the dispersion. $Q = ( Q " Q Notes 2 ) 2 1/ 2 = Q Quantum Physics F2005 2 2 1/ 2 " Q 14 Einstein-de Broglie relations As we continue to analyze behavior of wavefunctions it is useful for us to keep in mind two basic relations for matter waves, relating particle properites to wave properties: E = hf = h% and h p = = hk & Notes 2 Quantum Physics F2005 15 A quick peek at the Schrodinger Equation Later in this course we will deduce and solve the wave equation for matter waves (The Schrodinger Equation). Here I will show it to you, make some simplifying assumptions, and discuss the nature of the solutions. Here it is, in one dimension, h2 ' 2 '# ( x, t ) " 2m 'x 2 + V ( x) # ( x, t ) = ih 't Notes 2 Quantum Physics F2005 16 Free space solution for # • For V(x)=0, we can illustrate some physical points by working through the wave function logic of this physical situation without fully solving the differential equation. A. If # is to be normalizable, it must go to zero as r(infinity. B. # must be complex, because the right side of the SE has an i in it. C. # must not violate the homogeneity of free space. (The probability must be the same after we translate # by an arbitrary distance.) Notes 2 Quantum Physics F2005 17 Three trial functions 1.) #( x, t ) = A cos(kx " %t + ) ) :a real harmonic wave 2.) #( x, t ) = Aei (kx"%t ) :a complex harmonic wave 3.) #( x, t ) = a wavepacket made up of complex harmonic waves A. # normalizable? B. # complex? C. # must not violate the homogeneity of free space. Notes 2 Quantum Physics F2005 18 Test the harmonic function #( x, t ) = A cos(kx " %t + ) ) :a real harmonic wave A. Normalizable? No, unless A goes to zero. B.) Complex? No. Unlikely to satisfy the wave equation. C.) Homogeneous P? No. P( x, t ) = #( x, t )* #( x, t ) = A2 cos2 (kx " %t + ) ) P( x + a, t ) = #( x + a, t )* #( x + a, t ) #( x + a, t ) = A[ cos(kx " %t )cos ka + sin(kx " %t )sin ka] P( x + a, t ) = P( x, t )cos2 ka + + A2 sin2 (kx " %t )sin2 ka + 2cos ka sin ka cos(kx " %t )sin(kx " %t ) Notes 2 Quantum Physics F2005 19 Test the complex harmonic function #( x, t ) = Aei ( kx"%t ) A. Normalizable? Only if A goes to zero. "i ( kx "%t ) ∫ P(x,t)dx = ∫ Ae i ( kx"%t ) Ae ! dx = ∫ Adx = A[ 2!] "! B.) Complex? Yes. It is indeed a solution to the SE. C.) Homogeneous P? Yes. P( x, t ) = #( x, t )* #( x, t ) = A2 (independent of position) We seem to be on the right track. Notes 2 Quantum Physics F2005 20 Notes on the complex harmonic function # ( x, t ) = Aei ( kx "%t ) = A [ cos(kx " %t ) + i sin(kx " %t ) ] = A cos ( ( px " Et ) / h ) + i sin ( ( px " Et ) / h ) Represents a state with a single value of momentum and a single value of energy, therefore it is called a pure momentum state. •It extends over all space. •The phase fronts move in the +x direction % E at a speed v p of (or ). p k E p 2 / 2m p .) = (For a classical particle, v = = 2m p p Notes 2 Quantum Physics F2005 21 The complex wave packet We can construct a localized/normalizable wavepacket out of complex harmonic waves using the Fourier transform: 1 ! i ( kx "%t ) # ( x, t ) = dk ∫ A(k )e 2* "! ⇒ Because this wavefunction can be localized, it is likely to be normalizable. (We will need to test specific cases.) ⇒ Because each term in it is a solution to the wave equation, the wave function must also be a solution. ⇒ Is the shape of the probability of the sum of complex harmonics translation invariant? ( % ik2 x % ik1 x + be P( x, t ) = ae * ) ( ae% ik1 x ) ( % ik2 x = a% *e " ik1x + b%*e " ik2 x + be )( ae% ik1 x % ik2 x + be ) % " i ( k1 " k2 ) x + b%*ae % " ik ' x + b%*ae % i ( k1 " k2 ) x = a 2 + b 2 + a% *be % ik ' x = a 2 + b 2 + a% *be % ik2 x +ik2+ + b%*e " ik2 x "ik2+ ae % ik1x +ik1+ P( x + + , t ) = a 2 + b 2 + a% *e " ik1x "ik1+ be % " ik '( x ++ ) + b%*ae % ik '( x ++ ) = a 2 + b 2 + a% *be This function is just the old one, moved over. Notes 2 Quantum Physics F2005 22 In-class exercise 1 •Morrison 3.1 p. 89 This is a normalized state function at t=0. The magnitude of #(x,0) continues to decay smoothly to zero as x(±!. 1. 2. 3. 4. Is this function physically admissible? Why or why not? Where is the particle most likely to be found? Estimate the expectation value of the position. Is it the same as for part 2? Is the uncertainty in momentum of this state equal to zero? Justify. Notes 2 Quantum Physics F2005 23 In-class exercise 2 • Morrison 3.3 p. 90 Notes 2 Quantum Physics F2005 24 (In which I try to clarify what we are doing) • From evidence of interference in “particle” experiments, we introduced the idea that matter must be described by some sort of wave. – • • The wave is not directly observable, but the resulting probability of various outcomes is, hence we introduced a probability interpretation of quantum physics experiments. We then assumed that there existed a wave that could yield the correct observables, with classical relations between time, energy, momentum, and position. The ramifications were then explored. – – – – • De Broglie-Einstein relations: p=h/ &; E=hf The probability of an outcome is proportional to the square of the wavefunction for that outcome. The probability interpretation leads to the ability to compute the expectation (average) value for appropriate observables once the wave function is known. The superposition of waves to form a packet leads to a mathematical statement of the Heisenberg Uncertainty Principle. We could deduce the form of a useful wave equation for matter waves. We will now explore some examples of wave functions and their observable ramifications. – – Notes 2 Wavepackets, momentum-position transforms More general state functions – the momentum state Quantum Physics F2005 25 Important results so far Einstein- deBroglie relations: p = hk = h & ; T = h% = hf The probability interpretation: q = ∫ q ( x) P( x)dx; P( x)dx = # *#dx The general form for a wave packet using waves 1 ! i ( kx "%t ) with well-defined momentum: # ( x, t ) = dk ∫ A( k )e 2* "! Notes 2 Quantum Physics F2005 26 In class exercise 3 – Developing the matter-wave equation (from Morrison Problem 2.7) • Write down an expression in exponential form for a harmonic traveling wave (because our matter wavepacket is composed of simple traveling waves.) • Rewrite your wave function in terms of p(k) and E(%). • Deduce the simplest differential equation that might give rise to such a traveling wave, consistent with E=p2/2m,. ', ' 2, ', ' 2% + a2 2 + ... = b1 + b2 2 + ... a0 + a1 'x 'x 't 't Notes 2 Quantum Physics F2005 27 In-class exercise notes You should now have something like: i 1 2 "i 2 p4 a0 + a1 p, " a2 2 p , = b1 p , + b2 h h 2mh 4m 2 which can only be true if, a "i 2 m a0 = 0; a1 = 0; b2 = 0; and 2 = b1 h so a plausible differential equation for the motion of a free particle is: h 2 ' 2, ', " = ih 2 2m 'x 't Notes 2 Quantum Physics F2005 28 A solution to the matter wave equation – the infinite square well h2 ' 2, ', = h i Starting with " and specifying that the wavefunction must 2 't 2m 'x go to zero at x= ± L/2 leads to some interesting results. Only certain quantized values of wavelength are allowed. * x " i%t is one solution. e L For further work, we want this function to be normalized: , ( x, t ) = N cos *x 2 L dx N 1 = ∫ , *, dx = N ∫ cos 2 = L 2 -L/2 -L/2 L/2 , ( x, t ) = 2 L/2 ⇒ N= 2 L 2 * x " i%t cos e L L This function looks and behaves like this Notes 2 Quantum Physics F2005 29 The square well continued Now, find the expectation value of position. 2 L/2 2 *x x = ∫ xP( x, t )dx = ∫ x cos dx = 0 L "L/ 2 "L/ 2 L You can do this integral directly, or argue its value based on symmetry. L/2 Notes 2 Quantum Physics F2005 30 The square well continued Now, find the variance of position so we can assess the range of positions over which we might expect to find the particle. x x 2 2 2 L/2 2 2 *x x cos = ∫ x P ( x, t )dx = ∫ dx L "L/ 2 "L/2 L L/2 2 2L = L* 3 * /2 2 2 ∫ u cos ( u ) du "* / 2 * /2 2 u cos 2u 2L u3 u 2 1 = 3 + " sin 2u + * 6 4 8 4 "* / 2 x Notes 2 2 * 21 2 L2 * 3 1 = 3 +0" = L " 2 * 24 4 12 2* Quantum Physics F2005 31 Particles and waves: the Gaussian wavepacket We will simplify calculations by assuming that the spatial part of the wavefunction can be described at t = 0 by a Normal Gaussian function centered at x0 and moving with average wavenumber k0 ; 1/ 2 1 ik0 x " ( x " x0 )2 / 4- x2 . # ( x, 0) = e e - x 2* (We have already solved a similar problem in Notes 1. ) 1 So A(k ) = 2* Notes 2 1/ 2 1 - x 2* ! ∫e i ( k0 " k ) x " ( x " x0 )2 / 4- x2 e dx "! Quantum Physics F2005 32 Gaussian wavepacket continued to simplify further let's take x0 = 0, 1/ 2 1 1 ! i ( k0 " k ) x " x2 / 2- x2 so A(k ) = e dx ∫e 2* - x * "! Then letting k ' = k " k0 , we solve by completing the square: 2 1 1 "ik ' x " 2 x 2 + - x2 k '2 " - x2 k '2 = x " i- x k ' + - x2 k '2 2- x 2- x A(k ) = 1 2* 1/ 2 1 * x 2 2 ! e "- x k ' ∫ e " 2 1 2- x x "i- x k ' dx "! 1/ 2 1 1 x let x ' = -i- x k ', then A(k ) = 2- x 2* - x * 1/ 2 1 1 A(k ) = 2* - x * Notes 2 2- x e "- x2 k '2 1 2- x e "- x2 k '2 ! 2 " x' ∫ e dx ' "! 1 4 * = - x2 e * Quantum Physics F2005 " - x2 ( k " k0 )2 2 33 Gaussian wavepacket continued 1 4 - x2 ( k " k0 )2 1 2 A(k ) = - x2 e * This is a Normal Gaussian k " distribution, " centered at k0 with width - k = 1 -x . • Although we specified a wave momentum k0 we find that the wavefunction actually contains a spread of wave momenta, 1 -x , which increases as the $p . h • Waves of differing k move at different speeds, so the original wavefunction spatial Gaussian narrows. Remember that $k = will spread with time. Notes 2 Quantum Physics F2005 34 Inversely correlated widths Note that if # is properly normalized ( ∫ # * #dx = 1) then the wavevector amplitude function is also normalized to 1: 2 ∫ A(k ) dk = 1 Notes 2 Quantum Physics F2005 35 An aside on the definition of the width of a Gaussian The width of a Gaussian can be defined in various ways. Two conventional ways are by the Full Width at Half Maximum (FWHM) and twice the standard deviation. Morrison uses $L=standard deviation of the position. Notes 2 Quantum Physics F2005 36 Time evolution of the wavefunction The general form of the wavefunction is: 1 ! " i ( kx "% t ) # ( x, t ) = A ( k ) e dk . ∫ 2* "! This means that if you are given # ( x, 0) you can find # ( x, t ). 1 ! " i ( kx ) A(k ) = # ( x , 0) e dx. ∫ 2* "! Note that to do the top integral, you must know % ( k ). Time evolution of the wavefunction Notes 2 Quantum Physics F2005 37 In class exercise 4 – Constructing a wave packet from momentum amplitudes Consider the amplitude function: 0 1 A(k ) = $k 0 1 k < k0 " $k 2 1 1 k0 " $k < k < k0 + $k 2 2 1 k > k0 + $k 2 a ) Find # ( x, 0). b) Find P( x, 0). c) Sketch P( x, 0) and find the positions of the first zeros. d) Estimate $x$k . Notes 2 Quantum Physics F2005 38 Formal derivation of group velocity for a wave-packet , ( x, t ) = ∫ A(k )e " i ( kx "% ( k ) t ) dk Assuming that A(k ) has a fairly narrow range of important k 's. we will allow ourselves to expand % (k ) : d% % ( k ) . % ( k0 ) + ( k " k 0 ) + ... dk k0 Letting k ' = k - k0 , ( x, t ) . ei ( k x "% ( k ) t ∫ dk 'e 0 Notes 2 0 d% " ik ' x " t dk 0 Quantum Physics F2005 39 Momentum Probability Amplitude Note that A(k ) clearly contains information about momentum because k = p / h. It is useful (later) to be able to describe the state function directly in terms of momentum states. 1 p A (the momentum probability amplitude) h h Then we can rewrite the wavefunction as; 1 ! ipx / # ( x, 0) = ∫ / ( p )e dp and 2* h "! 1 ! " ipx / /( p) = dx. ∫ # ( x, 0)e 2* h "! /( p) 0 Note that the factor of 1/ h is required for proper normailzation of # and /. Notes 2 Quantum Physics F2005 40 Interpretation of the space and momentum state functions # ( x, t ) = the wavefunction and directly gives information about the probability of finding a particle at a specific position, but it is difficult to directly calculate properties like the expected momentum because momentum is not written as a function of position. / ( p ) = the momentum probability amplitude and directly gives information about the probability of finding a particle with a specific momentum. They both describe the same quantum state. If the members of an ensemble are in the quantum state #(x,t) with Fourier transform / ( p) = F [ # ( x, 0)] then P ( p) dp = / *( p )/ ( p )dp is the probability that in a momentum measurement at time t a particle's momentum will be found to be between p and p + dp. Notes 2 Quantum Physics F2005 41 Expectation values Given the momentum amplitude function and the associated probability distribution, we can compute expectation values: ! p = ∫ / ( p ) * p/ ( p )dp = expectation value (average) of p. "! $p = momentum uncertainty= ( p " p p 2 ) 2 1/ 2 = p 2 2 1/ 2 " p ! = ∫ / ( p ) * p 2 / ( p ) dp "! Notes 2 Quantum Physics F2005 42