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Phys 3750
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 ψ ( x , 0 )that corresponds to a particle initially
localized near the origin. We saw that ψ ( x , t ) 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 ω( ) k. 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 ω ( ) k 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
( ) ( ) [^ (^ )]
∞
−∞
xt = dkCk eikx −^ ω kt π
ψ 2
where the coefficients C ( ) k are the Fourier transform of the initial condition ψ ( x , 0 ),
∞
−∞
C k = dx x , 0 e − ikx 2
ψ π
and the dispersion relation (for the SE) is given by
m
k k 2
h^2
The example that we previously considered was for the initial condition
ψ (^ x , 0 )^ = ψ 0 e − x^2 σ^2.^ (4)
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We saw that for increasing positive time ψ ( x , t ) 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:
( ) 0 2 2 , (^00)
ψ x = ψ eik^ xe − x^ σ. (5)
As will be demonstrated below, you may think of k 0 as some average wave vector (or
momentum h k 0 through deBroglie's relation p = h k ) associated with the state ψ ( x , t ).
As we did in the last lecture, let's find an expression for ψ ( x , t ). We start by using Eq.
(2) to calculate C ( ) k , so we have
( ) (^) ∫ (^ )
∞
−∞
C k = dxe − x^^22 e − ik − k^0 x 2
0 σ
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
k ′^ = k − k 0 , so that the rhs of Eq. (6) becomes
∫
∞
−∞
dx e −^ x^^22 e − ik ′ x 2
0 σ
This equals the Gaussian (in the variable k ′ )
0 224 2
ψ σ e − k ′ σ , (8)
and now reusing the relation k ′^ = k − k 0 we can write
( ) 0 (^ )^4 02 2 2
C k = ψ^ σ e − k − k^ σ. (9)
(^1) As we stated in the last lecture, the Fourier transform of the Gaussian e −^ x^2 σ^2 is another Gaussian
4 2
σ k^2 σ^2 e −^.
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of the uncertainty principle. We shall discuss this in great detail in Lecture 29.
Let’s now look at the time dependence of ψ (^ x , t ). With Eq. (9) we can now use Eq. (1)
to write
( ) (^) ∫ (^ )^ [^ (^ )]
∞
−∞
xt = dke − k^ − k σ eikx −^ ω kt
02 2 2
keeping in mind that the dispersion relation ω( k )is given by Eq. (3). This looks
complicated, so let's look at some graphs of ψ ( x , t )to see what is going on with this
solution. The preceding figure, which contains snapshots of the video SE Wavepacket
3.avi , illustrates ψ ( x , t )as a function of time (for a positive value of k 0 ). Notice that
the wave packet moves in the + x direction with a constant velocity. Notice also that
ψ ( x , t )is not simply a translation in time of the function ψ ( x , 0 ). That is, the solution is
not of the form g ( x − vt ), 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 ψ ( x , t )can be thought of as having an average wave
vector k 0 , you might guess that the velocity is simply the phase velocity v (^) ph = ω( k ) k
evaluated at the average wave vector k 0. That is, you might think that the packet's
velocity is simply the velocity of the normal-mode traveling-wave solution
ψ (^) k (^) 0 ( x , t ) = ψ 0 ei [^ k^0 x −ω^ (^ k^0 ) t ]^ =ψ 0 eik^0 [^ x −(^ ω^ k^0 ) t ], (11)
which propagates in the + x direction at the phase velocity v (^) ph =ω ( k 0 ) k 0 =h k 0 2 m.
However, this is not correct!
To figure out the packet's velocity we must carefully analyze the propagating-pulse
solution described by Eq. (10). This solution lends itself to some approximation
because part of the integrand, (^ )^4
k k 02 σ^2
e −^ − , is peaked at k = k 0 , and for values of
k − k 0 >> 2 σ this part of the integrand is nearly zero. The importance of this is that
we only need to know what ω ( k ) is for k − k 0 less than a few times the width
parameter 2 σ. That is, we only need to know what ω ( k )is for values of k close to
k (^) 0. If ω( ) k is does not vary too much for these values of k , then it makes sense to
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approximate ω( ) k by the first few terms of a Taylor series expansion about the point
k 0. So we write
( ) ( ) ( )( ) ( )( ) ... 2
ω k = ω k 0 +ω′ k 0 k − k 0 + ω′′ k 0 k − k 0 + (12)
If we now approximate ω ( ) k in Eq. (10) by the first two terms of the series,
ω ( ) k ≈ω( k 0 ) +ω′( k 0 )( k − k 0 ), then we obtain
( ) [^ (^ )^ (^ ) ]^ ∫ (^ )^ [^ (^ )]
∞
−∞
x t ≈ 0 e − i^ k^0 − ′ k^0 k^0 t dke − k − k^0224 eikx − ′ k^0 t 2
, ω ω σ^ ω
This looks rather messy, but the integral can be calculated exactly to yield
( ) {^ [^ (^ ) ]}^ [^ (^ )]^ 0 0 0 0 2 2 , 0
ψ x t ≈ ψ eik^ x −ω k k t e − x −ω′ k t^ σ. (14)
We can now easily see what is going on. This (approximate) solution is the product of
the normal-mode traveling wave solution ψ (^) k (^) 0 ( x , t ) = ψ 0 eik^0 { x^ −[^ ω(^ k^0 ) k^0 ] t }(at the wave vector
k 0 ), which travels at a speed equal to the phase velocity
( )
( ) k
k v (^) phk
(evaluated at k 0 ) and a Gaussian "envelope" function [^ (^ )]^
x ω k 0 t^2 σ^2
e −^ − ′ , which travels at a
speed equal to ω′(^ k 0 ). The derivative ω′(^ k )is known as the group velocity
( )
( ) dk
d k v (^) gr k
and so the envelope function, which describes the position of the packet, travels at
the group velocity (evaluated at k = k 0 ). Note that the group and phase velocities are
not necessarily equal. The group velocity is typically more important than the phase
velocity because the average position of the particle is given by the peak of the
envelope function.
With these definitions of phase and group velocities we can now write Eq. (14) as
( )
0 (^ )^ (^ )^22
ψ x t ≈ψ eik^ x − vph^ t e − x − vgrt ax , (17)
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figure on the preceding page shows snapshots of the video WE Wavepacket 1.avi. In
contrast to the SE solution both the center of the wave packet and the oscillation
peaks travel at the same velocity, consistent with the solution being a function of
x − ct.
Exercises
* 27.1 Show that Eq. (13) follows from Eq. (10) with the linear Taylor's-series
approximation described in the notes.
* 27.2 Equation (12) is the Taylor's-series expansion of the dispersion relation about
the point k = k 0. For the dispersion relation appropriate to the WE, find all terms in
this expansion. Then argue why Eq. (20), is an exact solution to the WE.
20 0 20 40 1
0
t = 15
x
20 0 20 40 1
0
t = 5
x
20 0 20 40 1
0
t = 0
Re (ψ ( x t,)) Im (ψ ( x t,))
2 ψ (x t ,) ψ ( x t,)
⋅( ⋅ )− 1
x
20 0 20 40
1
0
t = 10
Re (ψ ( x t,)) Im (ψ ( x t,))
2 ψ (x t ,) ψ ( x t,)
⋅( ⋅ )− 1
x
Phys 3750
** 27.3 SE Approximate Solution
( a ) Calculate the integral on the rhs of Eq. (13) and show that Eq. (13) simplifies to
Eq. (14). (Hint: Transform the integral to be the Fourier transform of a Gaussian,
and then use the fact that the Fourier transform of
x^2 σ^2 e −^ is 2 4 σ k^2 σ^2
e −^ .)
( b ) Show that Eq. (14) consistent with Eq. (5), the initial condition.
* 27.4 EM Waves For electromagnetic waves traveling in a dielectric material such as
glass the dispersion relation is ω ( k ) =( cn ) k , where n is the index of refraction, which
is often assumed to be a constant.
( a ) If n is indeed a constant, calculate the phase and group velocities for these waves.
( b ) Often, however, the index of refraction depends upon the wave vector k.
Assuming that n ( ) k = n 0 + n 1 k , find the phase and group velocities.
( c ) For n ( ) k given in (b) show that v (^) gr = vph [ 1 − n 1 k ( n 0 + n 1 k )].
* 27.5 Calculate ψ * ψ^ for the approximate wave function given by Eq. (17) and show
that^ ψ^ *^ ψ^ travels at the group velocity^ v^ gr.
