Class 6: The free particle A free particle is one on which no forces act, i.e. the potential energy can be taken to be zero everywhere. The time independent Schrödinger equation for the free particle is ℏ2d 2 ψ − = Eψ . (6.1) 2m dx 2 Since the potential is zero everywhere, physically meaningful solutions exist only for E ≥ 0. For each value of E > 0, there are two linearly independent solutions ψ ( x) = e ±ikx , (6.2) where 2mE k = . (6.3) ℏ Putting in the time dependence, the solution for energy E is Ψ( x, t) = Aei( kx−ω t) + Be i( − kx − ω t ) , (6.4) where A and B are complex constants, and Eℏ k 2 ω = = . (6.5) ℏ 2m The first term on the right hand side of equation (6.4) describes a wave moving in the direction of x increasing, and the second term describes a wave moving in the opposite direction. By applying the momentum operator to each of the travelling wave functions, we see that they have momenta p= ℏ k and p= − ℏ k , which is consistent with the de Broglie relation. So far we have taken k to be positive. Both waves can be encompassed by the same expression if we allow k to take negative values, Ψ( x, t) = Ae i( kx−ω t ) . (6.6) From the dispersion relation in equation (6.5), we see that the phase velocity of the waves is ω ℏk v = = , (6.7) p k2 m which differs from the classical particle velocity, 1 pℏ k v = = , (6.8) c m m by a factor of 2. Hence, the wave function (6.6) is not suitable for describing a particle. Another problem with this wave function is that it is not normalizable. In other words, the particle is not localized. To remedy these deficiencies, we need to consider the complete solution of the Schrödinger equation, which is a linear superposition of the stationary state solutions. Since E is a continuous quantity, the complete solution involves an integral rather than the sum that is appropriate for discrete energy values. We therefore consider 1 ∞ Ψ()xt, = ∫ φ () keikx()−ω t dk , (6.9) 2π −∞ where the factor 1 2 π is included for convenience. Also it is important to remember that ω depends on k. Equation (6.9) describes a wave packet . The function φ (k ) either can be specified or be determined from initial conditions. In the latter case 1 ∞ Ψ()x,0 = ∫ φ () k eikx dk . (6.10) 2π −∞ Then 1 ∞ φ ()k=∫ e−ikx Ψ () x,0 dx . (6.11) 2π −∞ The transformation in equation (6.11) is a Fourier transform; φ (k ) is the Fourier transform of Ψ ( x,0) . The transformation in equation (6.10) is an inverse Fourier transform; Ψ ( x,0 ) is the inverse Fourier transform of φ (k ). Suppose a particle is initially localized between x = -a and x = a, where a > 0. The initial wave function is 1 , if −a < x < a , Ψ()x,0 = 2a (6.12) 0, otherwise. Using equation (6.11), we have 2 a −ikx a 111ikx e 1 i ika ika φ ()k=∫ edx− = i =−() ee− 2π 22a π ak 2 π a k −a −a (6.13) 1 sinka a sin ka = = . π a kπ ka This is plotted below: 1.0 0.8 0.6 ka )/ 0.4 ka sin( 0.2 0.0 -0.2 -0.4 -10 -5 0 5 10 ka The zeroes occur at ka= n π, where n is a non-zero integer. The width of the central peak corresponds to a range in k of width 2π ∆k = . (6.14) a We see that the more localized the particle, the broader is φ (k ). This is a general result and not limited to a uniform probability distribution. Problem 2.22. The Gaussian wave packet A free particle has initial wave function 2 Ψ()x,0 = Ae −ax , (6.15) 3 where A and a are constants ( a is real and positive). (a) Normalize Ψ ( x,0) . (b) Find Ψ ( x, t ) . (c) Find Ψ 2 ( x, t ) . Express your answer in terms of the quantity a w = . (6.16) 1+ () 2 ℏat m 2 2 2 Sketch Ψ as a function of x at t = 0 and again for some very large t. Qualitatively, what happen to Ψ as time goes on? 2 2 (d) Find x,, p x , p ,,σ x and σ p . (e) Does the uncertainty principle hold? At what time t does the system come closest to the uncertainty limit? (a) To normalize the wave function, we need ∞ ∞ 2 2 ∫ΨΨ∗ ()()x,0 x ,0 dxA = ∫ e− 2ax dx = 1. (6.17) −∞ −∞ A useful result is ∞ b2 2 π +c ∫ e−ax + bx + c dx= e 4a . (6.18) −∞ a For the integral in equation (6.17), b = c = 0, and we find 2 π A =1. (6.19) 2a Hence 2a 1 4 A = . (6.20) π (b) To find Ψ ( x, t ) , we first need to find φ (k ). From equation (6.11), we have ∞ 1 2 φ ()k= A∫ e−ikx − ax dx . (6.21) 2π −∞ Using the useful result in equation (6.18), 4 k2 k 2 1π − 1 − φ ()k= Ae4a = Ae 4 a . (6.22) 2π a2 a Next, we use equation (6.9) to get 2 2 ℏk 1 ℏ ∞ k i kx− t ∞ − +i t k2 + ixk 1− 2m 1 Ψ()xt, = Aee∫4a dk = Ae ∫ 4a 2 m dk . (6.23) 4πa−∞ 4 π a −∞ Using the useful result, we find ax 2 − ∞ 1 ℏ 2 2ℏat 1− +i t k + ixk 1 1+i Ψ()xt, = Ae4a 2 m dkA = e m 4π a ∫ 2ℏat −∞ 1+ i m ax 2 (6.24) − 2ℏat 1+i 2a 1 4 e m = . π 2ℏat 1+ i m For convenience, let 1 u = . (6.25) 2ℏat 1+ i m Then 1 4 2a au2 x 2 Ψ()xt, = ue − . (6.26) π Note that 1 1 1 w u∗ u = = = , (6.27) 2ℏat 2 ℏ at2ℏat 2 a 1−i 1 + i 1+ m m m and 1 1 22 w2 u2+= u ∗ 2 + = = . (6.28) 2ℏat 2 ℏ at 2ℏat 2 a 1+i 1 − i 1+ m m m 5 (c) The probability distribution is 14 14 12 2∗ 2 2 ∗ 2a∗−aux∗22 2 a − aux 22 2 a ∗ −au() + u x ΨΨ=()()xtxt, , ue ue = uue (6.29) π π π In terms of w, this is 1 2 2 2 2 Ψ∗ ()()xt, Ψ xt , = we − 2w x . (6.30) π We see that the probability density retains its Gaussian shape. At t = 0, w= a , so that 1 2 2a 2 Ψ∗ ()()x,0 Ψ x ,0 = e − 2ax . (6.31) π and for t≫ m(2 ℏ a ) , w a= m(2ℏ at ) , 1 2 mx 2 2 m −2a Ψ∗ ()()xtxt, Ψ , = e 2ℏat . (6.32) π a 2ℏ t The figure below shows how a Gaussian wave packet spreads with time. 1.0 Equal time intervals 0.8 0.6 ψ ∗ ψ 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 x 6 We see that the width of the wave packet increases with time. (d) Because the probability density is an even function of x, we see that x = 0 and p = 0. Now ∞ 1 2 ∞ 2 2 2 x2=Ψ∫∗ ()() xtx, 2 Ψ xtdx , = wxe ∫ 2− 2 w x dx . (6.33) −∞ π −∞ Consider the integral ∞ 2 π Ib() =∫ e−bx dx = . (6.34) −∞ b Taking the derivative with respect to b, we get ∞ 2 1 π ′ 2 −bx (6.35) Ib() =−∫ xe dx =− 3 . −∞ 2 b Hence ∞ 2− 2 w2 x 2 1π 1 π 1 (6.36) ∫ x e dx =6 = 3 . −∞ 28w 42 w Finally, we have 1 x2 = . (6.37) 4w2 Also ∞ 2 ∞ ∗ 2ℏ 2 ∗ ∂ ℏ 2 ∂Ψ ∂Ψ p=−Ψ∫() xt,2 Ψ() xtdx , = ∫ dx −∞ ∂x−∞ ∂ x ∂ x 1 2 ∞ 2a ∂∗22 ∂ 22 = ℏ2 ∫ ()u∗ e −au x() ue − au x dx π −∞ ∂x ∂ x 1 2 ∞ 2a ∗22 22 =ℏ2 ∫ e−aux()() −2 auxe ∗3 − aux − 2 auxdx 3 (6.38) π −∞ 1 2 ∞ ∗2 2 2 22a 2∗ 3 2 −au() + u x = ℏ 4auu() ∫ xe dx . π −∞ 1 2 ∞ 2 2 2 = ℏ2 4aw 322∫ xe − w x dx π −∞ = ℏ2a. 7 The uncertainties in x and p are 1 σ =x2 = , (6.39) x 2w and ℏ σ p = a. (6.40) (e) Now ℏa ℏ2 ℏ at 2 σx σ p = =1 + . (6.41) 2w 2 m We see that the uncertainty principle does hold and the uncertainty limit is met at t = 0. Phase and group velocity of a wave packet A general expression to describe a wave packet is ∞ i kx−ω() k t Ψ()()xt, = ∫ Ake dk . (6.42) −∞ The relationship between ω and k is called the dispersion relation. For example for waves on a string, the dispersion relation is ω = kc , where c is the speed of propagation of the wave. Suppose that the amplitude function A(k) is uniform over a narrow range of wave numbers (k0, k 0 + ∆ k ) and zero outside that range. Then k0 +∆ k Ψ = A∫ ei kx−ω() k t dk . (6.43) k0 Let k= k 0 + κ, so that ∆k Ψ = A∫ eikx0+κ x − ω() k 0 + κ t d κ. (6.44) 0 Because ∆k is small, we can expand ω(k0 + κ ) in a Taylor series and discard second and higher order terms.