5.73 F2018 Lecture 3: Motion, Position, Spreading, Gaussian Wavepacket

5.73 Lecture #3 3-1 Lecture #3: |ψ(x,t)|2: Motion, Position, Spreading, Gaussian Wavepacket Reading Chapter 1, CTDL, pages 9-39, 50-56, 60-85 Last lecture: Infinite well and Delta-function well What are the key points? En, �n for 0 L ⎡ h2 ⎤ E n = n2 ⎢ 2 ⎥ ⎣ 8 mL ⎦ ψ n = (2 / L )1/2 sin(nπx ) E=− E, � for ma 2 2! 2 ⎛ ma ⎞ ψ =⎜ 2 ⎟ ⎝! ⎠ 1/2 e− ma|x| ! Do E and � for delta function well behave as you expect? TODAY: Can we construct a �(x,t) for which |�|2 acts like a CM particle, but with correct QM characteristics? * stationary phase point and its motion * stationary phase approximation for evaluating an integral with wiggly integrand Motion requires Ψ(#, %) from TDSE! Motion is encoded in '(#), but we will need to actually observe motion (pages 3-4 thru 3-12). 1 5.73 Lecture #3 3-2 Our goal is to use a well understood function that appears frequently in quantum mechanics, a normalized Gaussian, as a particle-like quantum mechanical state function, a "Gaussian Wavepacket." What we want is to know how the time-evolving center position, center amplitude, center velocity, and the width of this wavepacket are encoded in the mathematical expression. This will guide us in constructing particle-like states with chosen properties and in knowing how to recognize these properties in an unfamiliar state function. From a stationary Gaussian G(x; x0, ∆x) to a moving Gaussian wavepacket |�(x,t)|2 1. 2 − x−x G ( x; x0 ,∆ x ) = (2π)−1/2 (∆ x)−1 e ( 0 ) Normalized: x2 ≡ ∫ ∞ −∞ ∫ You can show by evaluating the integral that G(x; x0, ∆x) is normalized to 1. 2(∆ x )2 G ( x; x0 ,∆ x ) dx = 1 ∞ −∞ G(x;x0 ,∆ x)x 2dx and a similar equation for ⟨x⟩. The width, Δx, the standard deviation of G(x), is the square root of the variance 1/2 2 ⎡ ⎤ ∆ x = ⎢ x2 − x ⎥ ⎣ ⎦ Finally, we want a function that is normalized to 1 at t = 0 ∫ 1= ∞ −∞ G ( x; x0 ,∆ x ) = Ψ ( x,0 ) Ψ(x,0) = Ψ(x,0) * ∞ normalized as ∫ Ψ(x,0) = (2π) −1/4 −∞ 2 Ψ(x,0) dx 2 2 Ψ(x,0) dx = 1 (∆ x)−1/2 e−( x−x0 ) 2 4(∆ x )2 This is also a Gaussian. Ψ(x,0) is broader and not as tall as G(x; x0 ,∆ x) at x = x0 . 2 5.73 Lecture #3 3-3 2. How do we get to the following complicated-looking textbook function? spreading and moving with minimum width at t = 0 1/2 ⎛ 2 ⎞ ⎛ 4! 2t 2 ⎞ Ψ(x,t) = ⎜ 2 ⎟ ⎜ 1+ 2 4 ⎟ e ⎝ πa ⎠ ⎝ ma ⎠ "$$$$$$$$$$$$# 2 2 ⎡ ⎛ !k0 ⎞ ⎤ 2 ⎢ 2a ⎜ x− t⎟ ⎥ ⎠ ⎥ ⎝ ⎢ m −⎢ ⎥ 2 2 ⎥ ⎢ a 4 + 4! t ⎥⎦ ⎢⎣ m2 "$$$$$$$$$$$# time-dependent normalization and magnitude. Since probability is conserved, the normalization factor must be t-dependent because the denominator of the exponential factor is t-dependent. ⎛ 2 ⎞ Ψ(x,0) = ⎜ 2 ⎟ ⎝ πa ⎠ 2 1/2 e−2 x 2 /a 2 at t = 0, by comparison to the normalized Gaussian ⎛ 2 ⎞ ⎜⎝ 2 ⎟⎠ πa 1/2 ⎛ ⎞ 1 =⎜ 2⎟ ⎝ 2π(∆ x) ⎠ 1/2 we have ∆x(t = 0) = a/2. Now, for motion of the center of the wavepacket, x0(t), we expect that p0 t p is momentum at t = 0 m p (0) !k0 (t) v is velocity at t = 0, k is the v0 (0) = 0 = m m wavenumber, k = p / ! at t = 0 !k (t) x0 (t) = x0 + 0 t m ∆ x(0) = a / 2. Width increases as ∣t∣ increases from t = 0. x0 (t) = x0 (0) + 0 0 0 Wavepacket is moving and changing its width. Minimum width is at t = 0. Could shift the t at which minimum width occurs by replacing t by t′ = t + δ in the formula for �(x,t). 3 5.73 Lecture #3 3-4 How do we know that the width of the wavepacket is t-dependent? If the value of Ψ at the t-dependent center is changing and Ψ(#, %) is normalized, then the wavepacket must be spreading or contracting. We will have to look at the tdependent Schrödinger equation to see how the momentum depends on x. NON-LECTURE The Fourier transform of a Gaussian is another Gaussian. This means that if you have a wavepacket, |Ψ($, 0) |!, with a Gaussian shape, the momentum distribution of this wavepacket, |Φ(), 0)|! , will also be a Gaussian. This Gaussian distribution of the momentum will cause the time-dependent spatial shape of the wavepacket to be either stretching or compressing. If the wavepacket shape, |Ψ($, *)|!, expands as t advances, it compresses as t decreases until it reaches the minimum possible width and then re-expands. The widths, Δ$ and Δ), are reciprocally related and the minimum uncertainty wavepacket, at the t when Δ$Δ) reaches its minimum value, is of particular interest. It is at this instant that the quantum mechanical wavepacket maximally resembles a classical particle. 2 How do we get from G(x; x0,∆x) to |Ψ(x,t)| ? Time Dependent Schrödinger Equation (TDSE) HΨ = i! ∂Ψ ∂t Time Independent Schrödinger Equation (TISE) HΨ = Eψ Hψ n = En ψ n or Special very useful case: if H is independent of time and if we know the solutions to the TISE, hen it is trivial to go from {"!, $!} to Ψ((, )). Suppose we create an arbitrary state at ) = 0. It is always possible to express this arbitrary state as a linear combination of eigenstates of H, ψ(x) = ∑ an ψ n n because the set of {"! } is “complete”. We can convert this "(%) to Ψ(%, )) very simply: Ψ(x,t) = ∑ cn ψ n e− iEnt n 4 ! 5.73 Lecture #3 3-5 Show that this satisfies the TDSE: i! ∂Ψ − iE t/! = i!∑ (−iE / !)cn ψ n e n ∂t n = ∑ En cn ψ n e − iEnt/! HΨ = H∑ cn ψ n e− iEnt/! n = ∑ En cn ψ n e same, so the TDSE is satisfied − iEnt/! n It is clear that Ψ(x,t) “moves”, but we still need help in understanding that motion. (i) ∫ ∞ −∞ Ψ * (x,t)Ψ(x,t)dx = 1 no motion because all ∆n ≠ 0 integrals involving ∞ * ∫−∞ ψ nψ n′ dx = 0 by orthogonality. (ii) Ψ * (x,t)Ψ(x,t) evolves in time if eigenstates that belong to at least two different En are included. For example, Ψ1,2 = c1ψ 1e− iE1t/! + c2 ψ 2 e− iE2t/! 2 2 2 2 2 Ψ1,2 = c1 ψ 1 + c2 ψ 2 + c1*ψ 1*c2 ψ 2 eiE1t/! e− iE2t/! +c1ψ 1c2*ψ *2 e− iE1t/! eiE2t/! ω12 ≡ ( E1 − E2 ) ! 2 2 2 2 2 Ψ1,2 = c1 ψ 1 + c2 ψ 2 + c1*c2 ψ 1*ψ 2 eiω12t +c1c2*ψ 1ψ *2 e − iω12t 5 5.73 Lecture #3 3-6 The first two terms are t-independent and the second two terms are tdependent and their sum is definitely a real number: ( 2 Re c1*c2 ψ 1*ψ 2 e iω 2t ). Now let us consider the particle in a constant potential. {ψ eigenfunctions k = eikx , ψ − k = e− ikx } p2 "2k 2 E|k| = + E0 = + E0 ! 2m 2m arbitrary zero of energy E|k| − E0 ! ≡ ωk ⎡ Ae ikx + Be −ikx ⎤ ⎣ ⎦ = Ae i (kx−ωt ) + Be −i(kx+ωt ) Ψ|k| (x,t) = e −iω k t 6 5.73 Lecture #3 3-7 Stationary phase kxφ − ωt = 0 ωt xφ = k ωt xφ = − k (could choose any constant instead of 0) xf is the constant phase point. A-term B-term phase velocity dxφ dt = vφ = ±ω / k ωt xφ (t) = xφ (0) ± k 7 5.73 Lecture #3 3-8 Some arbitrarily chosen constant point on Ψ(x,t) movesonatΨ(#, a velocity ω/k. Some arbitrarily phase point %) moves at a velocity '/). What about ∫ ∞ 2 −∞ Ψ(x,t) dx ? The t-dependent term integrates to zero due to ∫ ∞ −∞ e ±2ikx dx = 0 . So there is no motion in |Ψ(x,t)|2, only a constant term and standing waves. But Ψ(x,t) encodes motion through p̂ and x̂ . For example: p̂x = ∫ ∞ −∞ Ψ* ! ∂ Ψdx i ∂x ! ∂ ! ! Ψ(x,t) = e− iω k t [ Aikeikx − Bike− ikx ] = ike− iω k t [ Aeikx − Be− ikx ] i ∂x i i Now the whole thing: ! * ∂ Ψ (x,t) Ψ = !k [ A*e− ikx + B*eikx ][ Aeikx − Be− ikx ] ∂x i Now integrate ∫ ∞ −∞ dx ∫ e ±2ikx dx = 0 2 2 p = !k ⎡⎣ A − B ⎤⎦ as expected! Motion, just like Classical Mechanics! To get motion, it is necessary that A ≠ B 8 5.73 Lecture #3 3-9 Now for the payoff. ikx Consider a superposition of e for many values of k: Ψ(x,0) = ∫ g(k)eikx dk We can experimentally produce any g(k) we want. Let g(k) be a Gaussian in k g(k) = e But ∫ ∞ −∞ −( a 2 /4 )( k−k0 ) 2 g(k)eikx dk is the Fourier Transform of a Gaussian in k. Fourier ⎡ f (x) = (2π)−1/2 ∞ g(k)eikx dk Transform ⎢ −∞ and Inverse ∞ ⎢ Fourier −1/2 g(k) = (2π) f (x)eikx dx Transform ⎢ ⎣ −∞ ∫ ∫ get rid of k get rid of x So let us build Ψ(x,0) as a superposition of eikx. We can write g(k) in amplitude, argument form: eiα(iαk(k)) g(k) g(k)== g(k) |g(k)|e complex function of real variable We want |g(k)| to be sharply peaked near k = k0, so use a Gaussian g(k) = e−( a 2 /4)( k−k0 )2 center k = k0 1/2 width ∆k = 2 a 9 5.73 Lecture #3 3 - 10 dα α(k) = α(k0 ) + (k − k0 ) !##" dk k=k0 power series expansion α0 g(k) = e dα −( a 2 /4)( k−k0 )2 iα 0 i( k−k0 ) dk e e 2 2 ⎡ ⎤ dα i ⎢( k−k0 ) +kx ⎥ ⎣ ⎦ dk −( a /4)( k−k0 ) iα 0 g(k)eikx = e!########" e e rapidly oscillating in x except independent of x at a special region of x To find findthe thevalue value at which the phase is stationary, To of xofatx which the phase is stationary, we want we want ⎤ d ⎡ dα + kx k − k ( ) ⎥⎦ = 0 0 dk ⎢⎣ dk dα +x=0 dk dα so ififwe wechoose choose dα = −x have stationary in kk0 and near so wewe have stationary phasephase in k near 0 = −x 0 dk k = k dk 0 k=k0 ∞ dx integral k0 and x =means x0. This that thatmeans the ∫−∞ g(k) eikxthe dx integral near near x = near x0. This ∫−∞ g(k)eaccumulates to its exact value near x = x0. xaccumulates =x. ∞ 0 10 idx 5.73 Lecture #3 3 - 11 How does an integral over a rapidly oscillating integrand accumulate? It accumulates near the stationary phase point, x0. I(k) g(k 0 ) I(k ) 0 • x = x 0 + δx x = x0 0• 0• x = x 0 − δx g(k ) k0 k Integral accumulates near k = k0 but only when x ≈ x0. I(k) = ∫ k −∞ f (x,k) dk. If you examine the integrand and can identify the stationary phase region, you can determine the value of the integral without actually evaluating the integral. Amaze your friends! 11 5.73 Lecture #3 3 - 12 NON-LECTURE Joel Tellinghuisen, “Reflection and Interference Structure in Diatomic FranckCondon Factors,” J. Mol. Spectrosc. 103, 455-465 (1984). The figures in this paper show how an integral accumulates at a stationary phase point of the integrand. The stationary phase point, xsp, is the coordinate at which the vibrational wavefunctions for states 1 and 2 have the same classical momentum, #/% pclassical=!2#(% − '((!")* . The stationary phase point is located at the crossing of the V1 and V2 potential curves, V1(xsp)=V2(xsp). The semiclassical approximation for calculating vibrational overlap integrals is discussed on pages 278-285 of H. Lefebvre-Brion and R. W. Field, The Spectra and Dynamics of Diatomic Molecules. a1/2 Ψ(x,0) = (2π )3/4 ∫ ∞ −∞ e −( a 2 /4)( k−k0 )2 − i( k−k0 ) x0 ikx e!#### e dk " − k ( x−x ) ik x e!#####0" e 0 0 This is a �-function. It causes �(x,0) to be localized near x0. So we get |Ψ|2 localized at x0(t), k0, ∆x(t), ∆k if g(k) is Gaussian. –1/2 ∆x = 2 a 1/2 ∆k = 2 /a ∆x∆k = 1 at t = 0 We have constructed a Gaussian wavepacket, Ψ(x,t), from Ψ(x,0) with localization of x0(t), ∆x(t) minimum at t = 0, Gaussian in x, Gaussian in k. We askhow howthis this Ψ(#,can %) can be modified features of any Wecan can now now ask Ψ(x,t) be modified by anyby timetime-independent independent V(x). '(#). 12