11 The average distance is =血=（得=州 Problem 2 Griffiths,page 12,Problem 1.2,1.3 E.Normalization We return now to the statistical interpretation of the wave function.which says that lv(r.t)is the probability density for finding the particle at point at timet.It follows that the integral of)mst be1 (the particle's got to be somewhere w.p (2) t)Here is tan 品二器 n is pick this undetermined multiplicative factor s same goes Ic aoTly the vay tunctio ,0)= 0, where A,a and bare constants. (a)Normalize(that is,find A.in terms of a and b). (b)Sketch)as a function ofr. (c)Where is the particle most likely to be found.at t=0? (d)Wh the probability of ding the prticle to the ef ofCheck yor r in theand (e)What is the expectation value of r? Solution: (a) ffo =r{(+。(} =B+g-号
11 The average distance is ⟨x⟩ = ∫ h 0 x 1 2 √ hx dx = 1 2 √ h ( 2 3 x 3/2 )h 0 = h/3 which is somewhat less than h/2, as anticipated. Figure shows the graph of ρ(x). Notice that a probability density can be infinite, though probability itself (the integral of ρ) must of course be finite (indeed, less than or equal to 1). Problem 2 Griffiths, page 12, Problem 1.2, 1.3 E. Normalization We return now to the statistical interpretation of the wave function, which says that |ψ(x, t)| 2 is the probability density for finding the particle at point x, at time t. It follows that the integral of |ψ(x, t)| 2 must be 1 (the particle’s got to be somewhere) ∫ +∞ −∞ |ψ(x, t)| 2 dx = 1 (2) If ψ(x, t) is a solution of Schr¨odinger equation, so too are Aψ(x, t) and e iαψ(x, t). Here A is any complex constant, α is a phase factor of the wave function. What we must do, then, is pick this undetermined multiplicative factor so as to ensure that Equation (2) is satisfied. This process is called normalizing the wave function. For some solutions to the Schr¨odinger equation, the integral is infinite; in that case no multiplicative factor is going to make it 1. The same goes for the trivial solution ψ(x, t) = 0. Such non-normalizable solutions cannot represent particles, and must be rejected. Physically realizable states correspond to the ”square-integrable” solutions to Schr¨odinger’s equation. Example: Problem 1.4 on page 14 of Griffiths. At time t = 0, a particle is represented by the wave function ψ(x, 0) = A x a , A b−x b−a , 0, if 0 ≤ x ≤ a if a ≤ x ≤ b otherwise where A, a and b are constants. (a) Normalize ψ (that is, find A, in terms of a and b). (b) Sketch ψ(x, 0) as a function of x. (c) Where is the particle most likely to be found, at t = 0? (d) What is the probability of finding the particle to the left of a? Check your result in the limiting cases b = a and b = 2a. (e) What is the expectation value of x? Solution: (a) 1 = |A| 2 a 2 ∫ a 0 x 2 dx + |A| 2 (b − a) 2 ∫ b a (b − x) 2 dx = |A| 2 1 a 2 ( x 3 3 )a 0 + 1 (b − a) 2 ( − (b − x) 3 3 )b a = |A| 2 [ a 3 + b − a 3 ] = |A| 2 b 3 ⇒ A = √ 3 b
12 b FIG.9:Sketch of (r,0)as a funetion of r. (c)At r=a. P-trk=g厂a=wr号-{2 倒-jr=w{侣厂+o可o- (+(号-号+》 Problem 3 Griffiths,page 14.Problem 1.5 III.MOMENTUM AND UNCERTAINTY RELATION =高6 0-点e A.Expectation value of dynamical quantities For a particle in state(r).the expectation/mean/average value ofr is =w红P=a心血 五品心 minate)will collapse the wave function to a spike at the vale actually obtained,and e subsequent measurements (if they're performed quickly)will simply repeat that same result.Rather,()is the
12 FIG. 9: Sketch of ψ(x, 0) as a function of x. (c) At x = a. (d) P = ∫ a 0 |ψ(x, 0)| 2 dx = |A| 2 a 2 ∫ a 0 x 2 dx = |A| 2 a 3 = a b { P = 1 if b = a √ P = 1/2 if b = 2a √ (e) ⟨x⟩ = ∫ x |ψ| 2 dx = |A| 2 { 1 a 2 ∫ a 0 x 3 dx + 1 (b − a) 2 ∫ b a x (b − x) 2 dx} = 3 b { 1 a 2 ( x 4 4 )a 0 + 1 (b − a) 2 ( b 2 x 2 2 − 2b x 3 3 + x 4 4 )b a } = · · · · · · = 2a + b 4 Problem 3 Griffiths, page 14, Problem 1.5 III. MOMENTUM AND UNCERTAINTY RELATION In classical mechanics, we learn that position and momentum are canonical variables to each other. In 1D case, this can be seen from the Fourier transformation and its reverse ψ(x, t) = 1 √ 2π~ ∫ +∞ −∞ φ(p, t)e i ~ pxdp φ(p, t) = 1 √ 2π~ ∫ +∞ −∞ ψ(x, t)e − i ~ pxdx A. Expectation value of dynamical quantities For a particle in state ψ(x), the expectation/mean/average value of x is ⟨x⟩ = ∫ +∞ −∞ x |ψ(x, t)| 2 dx = ∫ +∞ −∞ ψ ∗ (x, t) xψ (x, t) dx What exactly does this mean? It emphatically does not mean that if you measure the position of one particle over and over again, ∫ x |ψ| 2 dx is the average of the results you’ll get. On the contrary, the first measurement (whose outcome is indeterminate) will collapse the wave function to a spike at the value actually obtained, and the subsequent measurements (if they’re performed quickly) will simply repeat that same result. Rather, ⟨x⟩ is the
13 average of measurements performed on particles all in the statewhich means that either you must find some way of d the tem ergy V(r) we)=voe,r=ee,vewe,也 But for the momentump,this is not true 创≠广。e,9gt Instead,by means of the Fourier transformation of the wave function,p,),we have 分=度aap脚-P(.prl.e -(aee）p -a∫广veo0 -∫“e()ae Collecting the underlined terms into,t),we have 创-ra(←品)a Definition:For ID case,the momentum operator is defined as =-培 and It is straightforward to generalize the Fourier transformation resut to 3D case The corresponding momentum operators are defined as =-h with the components 在=-协品A=-冰品a=-h品
13 average of measurements performed on particles all in the state ψ, which means that either you must find some way of returning the particle to its original state after each measurement, or else you prepare a whole ensemble of particles, each in the same state ψ, and measure the positions of all of them: ⟨x⟩ is the average of these results. In short, the expectation value is the average of repeated measurements on an ensemble of identically prepared systems, not the average of repeated measurements on one and the same system. Similarly one can get the expectation value of the potential energy V (x) ⟨V (x)⟩ = ∫ +∞ −∞ V (x)|ψ(x, t)| 2 dx = ∫ +∞ −∞ ψ ∗ (x, t) V (x)ψ (x, t) dx But for the momentum p, this is not true ⟨p⟩ ̸= ∫ +∞ −∞ ψ ∗ (x, t) ψ (x, t) pdx Instead, by means of the Fourier transformation of the wave function, φ(p, t), we have ⟨p⟩ = ∫ +∞ −∞ φ ∗ (p, t)φ(p, t)pdp = ∫ +∞ −∞ φ ∗ (p, t)pφ(p, t)dp = ∫ +∞ −∞ dp (∫ +∞ −∞ dx 1 √ 2π~ ψ ∗ (x, t) e i ~ px) pφ(p, t) = 1 √ 2π~ ∫ ∫ +∞ −∞ dpdxψ∗ (x, t) e i ~ pxpφ(p, t) = ∫ ∫ +∞ −∞ dxdpψ ∗ (x, t) ( −i~ d dx) 1 √ 2π~ e i ~ pxφ(p, t) Collecting the underlined terms into ψ (x, t), we have ⟨p⟩ = ∫ +∞ −∞ dxψ∗ (x, t) ( −i~ d dx) ψ (x, t) Definition: For 1D case, the momentum operator is defined as pˆ = −i~ d dx and ⟨p⟩ = ∫ +∞ −∞ ψ ∗ (x, t) ˆpψ (x, t) dx ̸= ∫ +∞ −∞ ψ ∗ (x, t) ψ (x, t) ˆpdx It is straightforward to generalize the Fourier transformation result to 3D case ψ (r, t) = 1 (2π~) 3/2 ∫∫∫ +∞ −∞ φ(p, t)e i ~ p·r d 3p φ(p, t) = 1 (2π~) 3/2 ∫∫∫ +∞ −∞ ψ(r, t)e − i ~ p·r d 3 r The corresponding momentum operators are defined as pb = −i~∇ with the components pˆx = −i~ ∂ ∂x, pˆy = −i~ ∂ ∂y , pˆz = −i~ ∂ ∂z
14 For simplicity we introduce (地，)=drb=dr (,0)=1 The mean value of a dynamical quantity A is (=c,)Aw(c,利r=(中，Ap） If the wave function is not normalized we should compute it as Problem 4 Consider a particle in one dimensional system.Both the wave function(,t)and its Fourier transfor mationp,t)can represent the state of the particle. (包，t)is the wave function(or state function,. probability amplitude)in position representation (p,t)represents the same state in momentum representation,just like a vector can be erpressed in different coordinate systems If we usep,t)to represent the state,what are the operators and p?p is just p itself,but what aboutr? B.Examples of uncertainty relation Let us examine again the example of Gaussian wave packets It is localized in momentum about p=hko.We can check the normalization 以r=臣ao=臣层=i because em=V层 The Fourier transformation =(品)- The probability density in position space
14 For simplicity we introduce (ψ, ψ) ≡ ∫ dτψ∗ψ = ∫ dτ |ψ| 2 where dτ means dx for 1D, dxdydz for 3D, and dx1dy1dz1 · · · dxN dyN dzN for N particles. The normalization takes a very simple form (ψ, ψ) = 1 The mean value of a dynamical quantity A is ⟨A⟩ = ∫∫∫ +∞ −∞ ψ ∗ (r, t) Aψˆ (r, t) d 3 r = ( ψ, Aψˆ ) If the wave function is not normalized we should compute it as ⟨A⟩ = ( ψ, Aψˆ ) (ψ, ψ) Problem 4 Consider a particle in one dimensional system. Both the wave function ψ(x, t) and its Fourier transformation φ(p, t) can represent the state of the particle. ψ(x, t) is the wave function (or state function, probability amplitude) in position representation φ(p, t) represents the same state in momentum representation, just like a vector can be expressed in different coordinate systems If we use φ(p, t) to represent the state, what are the operators x and p? p is just p itself, but what about x? B. Examples of uncertainty relation Let us examine again the example of Gaussian wave packets φ(k) = ( 2α π )1/4 e −α(k−k0) 2 . It is localized in momentum about p = ~k0. We can check the normalization ∫ +∞ −∞ |φ(k)| 2 dk = √ 2α π ∫ +∞ −∞ e −2α(k−k0) 2 dk = √ 2α π √ π 2α = 1 because ∫ +∞ −∞ e −αx2 dx = √ π α The Fourier transformation ψ(x) = ( 1 2πα)1/4 e ik0x e − x 2 4α The probability density in position space |ψ(x)| 2 = 1 √ 2πα e − x 2 2α
15 while in wave vector space ker=Va-w From the standard definition of Gaussian distribution we easily identify the standard deviation n andspaces,respectively =Va=△r We thus have △x·△k=1/2 △x·△p=h/2 More strict calculations show that the e 1eV=1.602×10-12crg=1.602×10-1 Joule he 1973eV.A,ao =0.53A,me =0.51MeV/c2 and finally the fine structure constant o=e2/hc=1/137. IV.PRINCIPLE OF SUPERPOSITION OF STATES Superposition is a common property of all kinds of linear waves.The most striking characters,the interference and 时 A.Superposition of 2-states Letand represent two different states of one micro-particle.The principle of superposition asserts that =a1+a2 he experiments 1.The probability of an event in an ideal exp riment is given by the square of the absolute value of a complex number which is called the probability amplitude P-proability
15 while in wave vector space |φ(k)| 2 = √ 2α π e −2α(k−k0) 2 From the standard definition of Gaussian distribution P(x) = 1 √ 2πσ2 e − (x−X) 2 2σ2 we easily identify the standard deviation in x and k spaces, respectively σx = √ α = ∆x σk = 1 √ 4α = ∆k We thus have ∆x · ∆k = 1/2 ∆x · ∆p = ~/2 More strict calculations show that ∆x · ∆p ≥ ~/2 which is Heisenberg’s Uncertainty Principle. So the Gaussian wave packets seem to saturate the bound! We need in near future some constants to evaluate the experiments related quantities 1eV = 1.602 × 10−12erg = 1.602 × 10−19Joule ~c = 1973eV · A, a ˚ 0 = 0.53A, m ˚ e = 0.51MeV /c2 and finally the fine structure constant α = e 2/~c = 1/137. IV. PRINCIPLE OF SUPERPOSITION OF STATES Superposition is a common property of all kinds of linear waves. The most striking characters, the interference and the diffraction, are results of superposition of waves. The distinction between wave motion and particle dynamics lies on whether there exists superposition. The wave nature of micro-particle is recognized and confirmed just through experimental observation of interference of matter waves. It is not surprised that the wave funtions (the states) obey superposition principle. A. Superposition of 2-states Let ψ1 and ψ2 represent two different states of one micro-particle. The principle of superposition asserts that ψ = a1ψ1 + a2ψ2 is also a possible state of the micro-particle - the superposition state. Here we show an interesting animation of the double slit experiment for bullets, water waves and electrons which is from http://www.upscale.utoronto.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSlit.html We summarize the main conclusions of the experiments. 1. The probability of an event in an ideal experiment is given by the square of the absolute value of a complex number ψ which is called the probability amplitude P = probability ψ = probability amplitude P = |ψ| 2