Wave Packets

PHY3101 Spring 2008 Wave Packets

A. Trigonometry

Euler’s identity, eiθ = cos θ + i sin θ, may be new to you, but it provides convenient easy ways to derive some of the trig identities. For example

ei(α+β) = eiαeiβ cos(α + β) + i sin(α + β) = (cos α + i sin α) × (cos β + i sin β) cos(α + β) + i sin(α + β) = cos α cos β − sin α sin β + i(cos α sin β + sin α cos β)

The real and the imaginary parts of this give the well known trig identities:

cos(α ± β) = cos α cos β ∓ sin α sin β

and sin(α ± β) = sin α cos β ± cos α sin β. Use the Euler identity to show that

sin2 θ + cos2 θ = 1.

Hint: start with eiαe−iα = 1 and use the Euler Identity.

B. Wave packets

One application of the previous trig identities involves wave interference. We want to use trig identities to rewrite the algebraic sum of two waves, A cos(k1x − ω1t) + A cos(k2x − ω2t) Start with cos(α + β) + cos(α − β) = 2 cos α cos β. Let α =α ¯, and β = ∆α/2, for reasons that will be clear below, so that

cos(¯α + ∆α/2) + cos(¯α − ∆α/2) = 2 cos(∆α/2) cos(¯α).

Also, let k + k ω + ω ∆α k − k ω − ω ∆k ∆ω α¯ = 1 2 x − 1 2 t and = 1 2 x − 1 2 t ≡ x − t. 2 2 2 2 2 2 2 The bar onα ¯ refers to the average. We now have

α¯ + ∆α/2 = k1x − ω1t andα ¯ − ∆α/2 = k2x − ω2t. 2

Finally, we have

∆k ∆ω A cos(k x − ω t) + A cos(k x − ω t) = 2A cos x − t cos(kx¯ − ωt¯ ). 1 1 2 2 2 2

In dealing with wave packets, often the pairs k1 and k2 and also ω1 and ω2 are nearly equal. So, ∆k k¯ and ∆ω ω¯. In this case the terms on the right hand side of this last big equation have a specific interpretation. The higher frequency waving all comes ¯ ¯ from the cos(kx − ωt¯ ) part, and k ≈ k1 ≈ k2. And the amplitude of the wave is given by 2A cos(∆kx/2 − ∆ωt/2). Now this “amplitude” changes in space and in time, but it changes only slowly because ∆k k¯ and ∆ω ω¯. We often say that the cos(∆kx/2 − ∆ωt/2) modulates the cos(kx¯ − ωt¯ ) wave. Figure 5-15 on page 215 of the textbook does a good job of demonstrating this. The relationships of k and ω with λ and f, are k = 2π/λ and ω = 2πf. These are easy to remember because the units of k are radians/distance while λ has units of cycles/distance, and 1 cycle = 2π radian. Similarly, the units of ω are radians/time while f has units of cycles/time. The previous big equation demonstrates that the sum of two waves (with nearly the same frequency and wavelength) can be written as the product of two waves. But, in the product form the first part has a small frequency ∆f = ∆ω/2π and a long wavelength ∆λ = 2π/∆k. (Here ∆k is small so that 1/∆k is big.) This first part changes slowly in space and in time. But, the second part has a larger frequency f¯ =ω/ ¯ 2π, a shorter wavelength λ¯ = 2π/k¯ and changes relatively rapidly in space and in time. The speeds of the high frequency and low frequency parts of the wave need not be the same. The speed of any sinusoidal wave is

v = λf = ω/k.

For the combined wave above, the speed of the high frequency, second part of the wave is called the phase velocity, ¯ vphase =ω/ ¯ k. While, the speed of the ﬁrst part, which creates the modulation (or “packet”), is called the group velocity, dω v = ∆ω/∆k = , group dk in the limit that the wavelengths and frequencies of the initial waves are very close to each other. For many kinds of waves, the speed of the wave is the same for all frequencies and wavelengths. This is the case for light in a vacuum and for a wave traveling down a string under tension. We say that these waves are non-dispersive. For other waves, such as water waves, light waves through glass and sound waves, when we look at very high and very low frequencies, the speed of the wave depends upon the frequency and wavelength. We say that these waves are dispersive. For a non-dispersive wave, the phase velocity and the group velocity are always the same. But, for a dispersive wave the two velocities might be very diﬀerent. Here is a website http://physics.usask.ca/~hirose/ep225/animation/dispersion/anim-dispersion.html that has some nice animations as well as some of the mathematics of creating wave packets. 3

C. The dispersive nature of de Broglie’s electron waves.

De Broglie hypothesized that the position of an electron could be described by a wave 1 2 with a frequency f = E/h = 2 mv /h and wavelength λ = h/p. Equivalently, we can use 1 2 2 ω = 2π( 2 mv )/h = πp /mh and k = 2πp/h. We can create a wave packet which localizes an electron’s wave by adding up a combination of diﬀerent frequencies and wavelengths. In that case the phase velocity vphase = λf = ωk = v/2. The phase velocities of de Broglie’s electron wave is half the velocity of the actual electron! But, use the chain rule for derivatives to evaluate the group velocity, and see that

dω/dp 2πp/mh p v = dω/dk = = = = v group dk/dp 2π/h m accurately tracks the position of the electron. You might be wondering why we used the Newtonian kinetic energy in the equation 1 2 f = E/h = 2 mv /h. Einstein might have preferred that we use the total, relativistic energy E = pp2c2 + m2c4, in f = E/h. These two possibilities clearly give two diﬀerent frequencies, but they give the same group velocity! First note that

dk d 2πp 2π = = , dp dp h h and also dω d 2πE = dp dp h d 2πpp2c2 + m2c4 = dp h 2πpc2 = hpp2c2 + m2c4 2πpc2 = hE 2πc2mv/p1 − v2/c2 = hmc2/p1 − v2/c2 2πv = . h Now, use these results in the expression for the group velocity

dω/dp v = dω/dk = group dk/dp 2πv 2π = h h = v.

Our conclusion is that the group velocity of the electron’s wave packet is, in fact, equal to the actual velocity of the electron when we use the correct relativistic expressions for the energy 4 and the momentum of the electron. When de Broglie ﬁrst proposed his expressions for the frequency and wavelength of the electron probability wave he used these correct relativistic expressions. Here are some facts about wave packets: You can algebraically add together sinusoidal waves of diﬀerent amplitudes, frequencies and wavelengths to obtain a wave packet. If you try to create a wave packet that is very tightly restricted in position x, then this requires a broad range of wave numbers k. And, if you desire a very narrow range of wave numbers k, then the wave packet will necessarily be spread out in position x. Similarly, if you try to create a wave packet that is very tightly restricted in time t, then this requires a broad range of wave frequencies ω. And, if you desire a very narrow range of wave frequencies ω, then the wave packet will necessarily be spread out in time t. These are elementary consequences of Fourier analysis. One way to express these ideas is through the expressions

∆k ∆x ≈ 1 and ∆ω ∆t ≈ 1.