USU PHYS 3750 - A Propagating Wave Packet

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Lecture 27 Phys 3750 D M Riffe -1- 4/3/2013 A Propagating Wave Packet – The Group Velocity Overview and Motivation: Last time we looked at a solution to the Schrödinger equation (SE) with an initial condition ()0,xψ that corresponds to a particle initially localized near the origin. We saw that ()tx,ψ broadens as a function of time, indicating that the particle becomes more delocalized with time, but with an average position that remains at the origin. To extend that discussion of a localized wave (packet) here we look at a propagating wave packet. The two key things that we will discuss are the velocity of the wave packet (this lecture) and its spreading as a function of time (next lecture). As we shall see, both of these quantities are intimately related to the dispersion relation ()kω. This discussion has applications whenever we have localized, propagating waves, including solutions to the SE and the wave equation (WE). Key Mathematics: Taylor series expansion of the dispersion relation ()kω will be central in understanding how the dispersion relation is related to the properties of a propagating wave packet. The Fourier transform is again key because the localized wave packet will be described as a linear combination of harmonic waves. I. A Propagating Schrödinger-Equation Wave Packet In the last lecture we found the formal solution to the initial value problem for the free particle SE, which can be written as () ()()[]∫∞∞−−=tkxkiekCdktxωπψ21,, (1) where the coefficients ()kC are the Fourier transform of the initial condition ()0,xψ, () ()∫∞∞−−=xikexdxkC 0,21ψπ, (2) and the dispersion relation (for the SE) is given by ()mkk22h=ω. (3) The example that we previously considered was for the initial condition ()2200,σψψxex−=. (4)Lecture 27 Phys 3750 D M Riffe -2- 4/3/2013 We saw that for increasing positive time ()tx,ψ becomes broader (vs x), but its average position remains at the origin. So, on average the particle is motionless, but there is increasing probability that it will be found further away from the origin as t increases. So you might ask, what initial condition would describe a particle initially localized at the origin, but propagating with some average velocity? Well, here is one answer: ()22000,σψψxxikeex−= . (5) As will be demonstrated below, you may think of 0k as some average wave vector (or momentum 0kh through deBroglie's relation kp h=) associated with the state ()tx,ψ. As we did in the last lecture, let's find an expression for ()tx,ψ. We start by using Eq. (2) to calculate ()kC, so we have ()()∫∞∞−−−−=xkkixeedxkC02220σπψ. (6) This almost looks like the Fourier transform of a Gaussian, which we can calculate.1 Indeed, we can make it be the Fourier transform of a Gaussian if define the variable 0kkk −=′, so that the rhs of Eq. (6) becomes ∫∞∞−′−− xkixeedx2220σπψ. (7) This equals the Gaussian (in the variable k′) 40222σσψke′−, (8) and now reusing the relation 0kkk−=′ we can write ()()402202σσψkkekC−−= . (9) 1 As we stated in the last lecture, the Fourier transform of the Gaussian 22σxe− is another Gaussian 4222σσke−.Lecture 27 Phys 3750 D M Riffe -3- 4/3/2013 Note that if 00=k , then we obtain ()()40222σσψkekC−= , the result from the last lecture. Equation (9) tells us several important things. Recall that we are describing the state ()tx,ψ as a linear combination of normal-mode traveling-wave states ()[]tkxkieω−, each of which is characterized by the wavevector kλπ2= and phase velocity mkkkvph2)( h==ω. As Eq. (1) indicates, the function ()kC is the amplitude (or coefficient) associated with the state with wavevector k. As Eq. (9) indicates, the coefficients ()kC are described by a Gaussian centered at the wave vector 0k . Thus, you may think of the state ()tx,ψ as being characterized by an average wave vector 0k . The width of the function ()kC , with width parameter σ2 , is also key to describing the state ()tx,ψ. Because this width parameter is inversely proportional to the localization (characterized by σ) of the initial wave function ()0,xψ, we see that a more localized wave function ()0,xψ requires a broader distribution (characterized by σ2 ) of (normal-mode) states in order to describe it. Insofar as momentum is equal to kh , this inverse relationship between the widths of ()0,xψ and ()kC is the essence 20 0 20 4010.500.5t = 0Re ψ xt,()()Im ψ xt,()()2 ψ xt,()ψ xt,()⋅()⋅ 1−x20 0 20 4010.500.5t = 5x20 0 20 4010.500.5t = 10Re ψ xt,()()Imψ xt,()()2ψ xt,()ψ xt,()⋅()⋅ 1−x20 0 20 4010.500.5t = 15xLecture 27 Phys 3750 D M Riffe -4- 4/3/2013 of the uncertainty principle. We shall discuss this in great detail in Lecture 29. Let’s now look at the time dependence of ()tx,ψ. With Eq. (9) we can now use Eq. (1) to write ()()()[]∫∞∞−−−−=tkxkikkeedktxωσπσψψ402202,, (10) keeping in mind that the dispersion relation ()kω is given by Eq. (3). This looks complicated, so let's look at some graphs of ()tx,ψ to see what is going on with this solution. The preceding figure, which contains snapshots of the video SE Wavepacket 3.avi, illustrates ()tx,ψ as a function of time (for a positive value of 0k ). Notice that the wave packet moves in the x+ direction with a constant velocity. Notice also that ()tx,ψ is not simply a translation in time of the function ()0,xψ. That is, the solution is not of the form ()vtxg−, where v is some velocity. This can be seen in the video by noticing that the center of the wave packet travels faster than any of the individual oscillation peaks. II. The Group Velocity We now want to determine the velocity of the propagating wave packet described by Eq. (10). Because this solution ()tx,ψ can be thought of as having an average wave vector 0k , you might guess that the velocity is simply the phase velocity ()kkvphω= evaluated at the average wave vector 0k . That is, you might think that the packet's velocity is simply the velocity of the normal-mode traveling-wave solution ()()[]()[]tkxiktkxkikeetx0000000,ωωψψψ−−== , (11) which propagates in the x+ direction at the phase velocity ()mkkkvph2000h==ω. However, this is not correct! To figure out the packet's velocity we must carefully analyze the propagating-pulse solution described by Eq. (10).


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