What does the wavenumber in a group velocity represent?

In summary: You will get a certain group velocity. In summary, when constructing a wavepacket from multiple traveling waves with different wavelengths, the phase and group velocities can be calculated using the equations provided. The dispersion relation for a free quantum mechanical particle can also be used to understand the spreading out of a wavepacket. When dealing with a crystal in thermal equilibrium, the individual modes can be described using standing waves or traveling waves depending on the situation. The dispersion relation for phonons can be obtained for a 1D monoatomic chain of atoms connected by springs and this can cause confusion when trying to understand the concept of group velocity. However, the group velocity can be calculated by taking into account the bandwidth of various wave numbers in the dispersion equation.
  • #1
confused_man
16
1
I'm trying to wrap my head around the dispersion relation ##\omega(k)##. I understand how you can construct a wavepacket by combining multiple traveling waves of different wavelengths. I can then calculate the phase and group velocities of this wavepacket:
\begin{align*}
v_p &= \frac{\omega}{k} \\
v_g &= \frac{d\omega}{dk} .
\end{align*}

Suppose that I'm looking at a free quantum mechanical particle, the dispersion relation is
\begin{align*}
E &= \hbar\omega =\frac{\hbar^2 k^2}{2m}\\
\omega &= \frac{\hbar k^2}{2m}
\end{align*}
Since ##\omega## depends the wavenumber in a non-linear way, I understand that a wavepacket made of a combination of these plane waves will experience dispersion and will spread out.

If I'm talking about a pure plane wave with frequency ##k##, then I wouldn't use the group velocity to describe how fast it's moving, I'd just use the phase velocity.

My question is if I'm looking at a wave packet built up from a number of different plane waves and I want to calculate its group velocity, what value of ##k## should I use? The group velocity equation only has a single ##k## in it.
 
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  • #2
I initially started getting confused about this when thinking about phonons. A single phonon is an excitation of a single mode with frequency ##\omega##. In a crystal, there are a vast number of phonons corresponding to the excitations of multiple modes of different frequencies. The number of phonons in each mode follows the Bose Einstein distribution function.

These "modes" can be different types of functions, depending on what's most convenient to describe the situation. If we are dealing with a crystal that is in thermal equilibrium, the simplest set of functions that I can use as the individual modes are the various standing waves that I can fit into the crystal of length ##L## (the longest mode having a wavelenth ##\lambda = 2L## and the shortest with ##\lambda = 2a##, where ##a## is the distance between atoms in the lattice). If we are dealing with a situation where one end of the crystal is being excited, say by heating it, it makes more sense to use a set of traveling waves as a basis to describe the crystal vibration.

If you read a solid state textbook, at this point it's usually the case that the book discusses the dispersion relation of phonons around this point. The easiest thing to do is to start with a 1D monoatomic chain of atoms of mass ##m## connected by springs with a spring constant ##K##. You then get the following dispersion relation: $$\omega(k) = \sqrt{\frac{4K}{m}}\left|\sin\left(\frac{ka}{2}\right)\right|. $$
This is where I start to go bonkers. Like I said above, I think of a phonon as an excitation of a single vibration mode of frequency ##\omega## and energy ##\hbar \omega##, and corresponding to this, a single value of ##k##. The textbooks I have then process to discuss the group velocity of the phonons based on the dispersion relation and talk about how at the boundary of the first Brillouin zone (i.e., when ##k = \pm \pi/a##) the phonons have a group velocity of 0 since ##d\omega/dk=0## at the boundary. BUT HOW CAN PHONONS HAVE A GROUP VELOCITY?? Don't they by definition, represent the vibration of a single pure mode with a single frequency ##\omega## and ##k##?

OR, when they say that the group velocity is zero at ##k=\pm \pi/a##, are they referring to a wavepacket made of a bunch of traveling waves with various wavenumbers. If so, then it comes back to my initial question, how can this wavepacket get defined by a single wavenumber? What wavenumber do I put in there? Is it the average? Does this dispersion relation mean that I can construct a wavepacket with any number of traveling waves with various ##k## values, but if the average ##k## is equal to ##\pm \pi/a## then it doesn't move? How do you even calculate the average ##k##? Would you need to use the Bose Einstein distribution function or is it just the simple arithmetic mean?

Please help. The more I think about this the more I am getting messed up!
 
  • #3
confused_man said:
My question is if I'm looking at a wave packet built up from a number of different plane waves and I want to calculate its group velocity, what value of kk should I use? The group velocity equation only has a single kk in it.
As you have already said you need more than one wave to have a wave packet. So you have a certan bandwidth of ##\Delta k## around a center wave number ## k ## and you have to put this in the dispersion equation.
 

Related to What does the wavenumber in a group velocity represent?

1. What is the definition of wavenumber in relation to group velocity?

The wavenumber, represented by the symbol k, is defined as the number of complete wave cycles per unit distance. In the context of group velocity, it represents the spatial frequency of a wave packet, which is a localized disturbance made up of a group of waves with different frequencies and wavelengths.

2. How is the wavenumber related to the speed of a wave packet?

The wavenumber is directly proportional to the speed of a wave packet. This means that as the wavenumber increases, the speed of the wave packet also increases. Conversely, as the wavenumber decreases, the speed of the wave packet decreases.

3. Can the wavenumber be negative?

Yes, the wavenumber can be negative. In fact, negative wavenumbers are commonly used in physics and mathematics to represent waves that are traveling in the opposite direction compared to waves with positive wavenumbers.

4. How is the wavenumber related to the wavelength of a wave packet?

The wavenumber is inversely proportional to the wavelength of a wave packet. This means that as the wavenumber increases, the wavelength decreases, and vice versa. This relationship is described by the equation k = 2π/λ, where λ is the wavelength.

5. What is the significance of the wavenumber in understanding the behavior of waves?

The wavenumber is a crucial parameter in understanding the behavior of waves. It is used to calculate important quantities such as the speed, frequency, and wavelength of a wave. It also plays a role in determining the dispersion of waves, which is the dependence of wave velocity on wavenumber. In addition, the wavenumber is used in various mathematical and physical models to describe the properties and interactions of waves.

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