# Phonon waves

1. Feb 21, 2013

### hokhani

Are phonons in crystals standing waves or traveling waves?

2. Feb 21, 2013

### Darwin123

Any vibrational wave can be described as a superposition of either standing waves or traveling waves. One can expand the wave function in terms of one or the other but not both. Both expansions are mathematically equivalent. The physics derived from these expansions is physically equivalent. So the question isn't well-posed.

You need to re-express the question. A little understanding of linear algebra would help.

A vibrational wave can be characterized by a wave function. A function is a type of generalized vector. Any function can be expanded as a sum of basis functions, or as an integral of basis functions. The Fourier series or Fourier transform is a type of expansion. However, one has a choice with the basis. One can use a basis for the expansion comprised of standing waves. One can also use a basis for the expansion with traveling waves.

A basis is the set of linearly independent vectors (i.e., functions) which is complete. By complete, I mean that any function satisfying certain boundary conditions can be expanded in terms of the basis.

The two types of waves satisfy different boundary conditions. With vibrations, standing waves satisfy boundary conditions where the displacement is zero on a specified surface. Traveling waves satisfy a periodic boundary condition.

Therefore, the choice of basis must be determined by the boundary conditions on the crystal. If the crystal is set up in such a way as the vibrational displacement is zero on the surface, then one should use standing waves. If the crystal is set up in such a way as the surface is impedance matched, then one should use traveling waves.

Fortunately, one rarely has to specify the boundary conditions on the surface. So for most problems, the choice is arbitrary.

For most introductory solid state problems, the boundary conditions don't affect the physical results. In statistical thermodynamics, for instance, the distribution of energy satisfies the ergodic theorem independent of the boundary conditions. So one can analyze specific heat and conductivity without specifying boundary conditions.

The solution to problems is often presented in terms of standing waves. I would say that 95% of the time, the standing wave hypothesis is adequate to solve the problem. Such a solution shouldn't be taken as a unique solution to the problem. Usually, one could have solved the same problem with a basis of traveling waves. However, the boundary conditions don't matter once a statistical assumption is made about the waves.

3. Feb 21, 2013

### hokhani

Thanks for the reply. But I wanted to know whether or not phonons are really the waves (free particles) made by crystal's atoms? In other words, do atoms oscillate on the phonon waves? If yes, according to BVK boundary conditions, they must be travelling waves. And also I got confused by what is the meaning of a number of phonons in the same mode? For example if we have 5 phonons in the same mode, does it mean that these 5 phonons strengthen 5 times the effect of one of them?

Last edited: Feb 21, 2013
4. Feb 21, 2013

### Darwin123

It is a wave-duality thing.

Phonons are the quantized manifestation of vibrational waves. As such, they act a lot like particles. However, the associated waves do move atoms.

You may be able to resolve the imaginary paradox by considering the wave nature of a vibrational wave this way. According to quantum mechanics, the amplitude of a vibrational wave has to change in discrete steps. The wave moves atoms, but the amplitude of the wave can't change continuously.

Thus, think of quantization as acting on amplitude of the wave, not the frequency. The discrete change of amplitude is a phonon.

The phonon is not the entire wave. The phonon is a step in amplitude.

The boundary conditions are specified by the experimental configuration or the observational environment. Although boundary conditions can be ignored for many calculations, they are physical conditions. They are not just mathematical conditions.

I don't know what you mean by BVK boundary conditions. I suspect you mean impedance-matching conditions on the boundary.

If the crystal is somehow set up to have impedance matching on both sides of the boundary, then the waves will be traveling waves. If the crystal were set up with zero displacement boundary conditions, they will be standing waves. Either nature or the experimenter sets up the boundary conditions.

How would you set up BVK boundary conditions?

This is important for understanding your question. The boundary conditions can be set up. Therefore, what you are asking is similar to asking how we want the experiment set up. There is no a priori way we can answer that question.

I will answer using the approximation that the zero point energy is zero. In an exact calculation, one would need to include the zero-point energy. However, it is best to wait until you get to problems where the zero point energy is important.

If there are 5 phonons in a particular mode, that means that the square of the amplitude of that mode is 5 times as large as the square of the amplitude of a hypothetical mode that has the same frequency but only one phonon.

5. Feb 22, 2013

### hokhani

Born-Von-Karman boundary conditions or cyclic boundary conditions are applied to crystals. If you consider for instance a one-dimensional crystal with length L, According to BVK boundary conditions $\psi(x+L)=\psi(x)$ so the wave functions would be travelling.

Last edited: Feb 22, 2013
6. Feb 22, 2013

### Darwin123

You wrote that BVK conditions mean:
$\psi(x+L)=\psi(x)$

This expression would apply just as well to a standing wave as to a traveling wave. In fact, you are not writing in the time dependence at all.

For all anyone can tell, this may not be harmonic at all. There is no indication that the wave is cyclic in time as well as space.

If I saw that equation, cold, then I would assume that it stood for a standing wave. I would assume that you decided not to write down the time varying factor for convenience.

So I still don't know what BVK conditions are.

7. Feb 22, 2013

### Cthugha

Born-von Karman conditions just consist of assuming periodic boundary conditions for a crystal. In a nutshell you assume that your crystal is infinite, so there are no surface effects.

8. Feb 23, 2013

### Darwin123

In that case, the waves can be either standing waves or traveling waves. If there are no surface effects at all, then the problem is under determined. One could use either a basis of standing waves or a basis of traveling waves.

Therefore, BVK conditions are consistent with both standing waves and traveling waves. The initial conditions at two separate points are necessary in order to uniquely determine the coefficients of the Fourier expansion.

9. Feb 23, 2013

### Darwin123

If you wrote the BVKI conditions correctly, then the BVK conditions are consistent with both standing waves and traveling waves. The waves do not have to be traveling.

I replied to someone else on this issue. "Infinite" crystals are under determined with regards to the phase of the waves.

I think that your original question presupposes a finite crystal. Infinite crystals are very rare !-)

10. Feb 24, 2013

### hokhani

BVK boundary conditions are used because one be able to consider a finite crystal as a part of infinite crystal and hence work with traveling waves. Standing waves would not entail conduction.
For standing waves we have some fixed points that can not move along with the wave while in traveling waves each point is oscillating. Also if you even consider time dependency, nothing would change and only wavefunctions would be multiplied by $exp(i\omega t)$.

11. Feb 24, 2013

### Darwin123

That is not true. That can't be true because each traveling wave can be decomposed into two standing waves. Obviously, if conduction can occur with standing waves then they can occur with the two traveling waves they are made of.

Thermal conduction on a molecular level isn't characterized by vibrational waves "traveling". Thermal conduction is caused by anharmonic lattice interactions. The part of the interatomic potential that is not equivalent to a harmonic oscillator causes thermal conduction.

In terms of quantum mechanics, two phonons collide to form a single phonon with some momentum transferred to the center of mass of the crystal. This is called an umklapp process. The total angular momentum is conserved, but the single phonon has some additional momentum.

This description is found in,
"Introduction to Solid State Physics" 7th ed. by Charles Kittel (Wiley, 1996) page 133-136.

Moderators, hold up. I am describing the process in classical terms using my physical understanding and my own words. I am expanding on Kittel's description which emphasized quantum mechanics. I am not making up my own theory.

In terms of standing waves and classical theory, you can think of the umklapp processes this way. The internal energy of a crystal is stored in a set of standing waves. Each standing wave has peaks which are fixed in position and can't move. The peaks denote places where the energy density is maximum. I am going to describe the "thermal conduction" of energy from the peak of the wave to the nodes.

Look closely at two standing waves that are superimposed. Each of the waves has nodes where no displacement of atoms takes place. Each of them has peaks where the displacement is maximum.

Even without an anharmonic force, the two standing waves interfere with each other. There is a diffraction pattern. This means that the sum of the two waves can be described as one "carrier wave" modulated by an "envelope".

The carrier and the envelope are not like the two waves we considered. Both the envelope function and the carrier wave may be nonzero at the nodes of the original waves. Both the envelope function and the carrier sometimes are zero where the original waves reach maxima.

No thermal conduction takes place without anharmonic terms because the decomposition just described is basically a geometric illusion. The two waves aren't interacting without anharmonic terms. No energy can really be transferred.

Due to anharmonic forces, the envelope itself becomes a new standing wave. The carrier wave itself becomes a new standing wave. Energy is transferred from the original modes of vibration to new modes of vibration corresponding to carrier and envelope.

The new waves are still standing waves. However, they are shifted with respect to the original modes. The atoms vibrate in the regions where there was no vibration before the anharmonic interaction. The atoms in in the original peaks vibrate less because energy was transferred out of the original modes.

The shift in the new waves relative to the old waves corresponds to thermal conduction from the peaks of the old waves to the nodes of the new waves. The thermal conduction is a manifestation of nonlinearity in the potential, not the "travel" of vibration.

Therefore, one can mathematically describe thermal conduction as easily using standing waves as one can using traveling waves. For certain situations, maybe it may be convenient to describe thermal conduction in terms of traveling waves. However, my conjecture is that it doesn't matter even in terms of convenience. What is important in thermal conductivity is the anharmonic component of force between atoms.

12. Mar 2, 2013

### zhanghe

in my opinion, travelling wave, because hear (as energy) could be transported from hot part to cold part by phonon.

13. Mar 2, 2013

### zhanghe

I agree with you that these 5 phonons strengthen 5 times the effect of one of them.

14. Mar 2, 2013

### DrDu

In a real (=finite) crystal, only standing waves are exact eigenstates. However, the energetic splitting between energetically neighbouring cosine and a sine wave with the same direction k/|k| is minute and a coherent superposition of the two will behave like a travelling wave for a very long time. Given the size of the crystal being L, two neighbouring k values differ by ~1/L and their energies by $\Delta E\sim \hbar^2 k/(mL)$. Using time energy uncertainty, the superposition will behave like a travelling wave for times up to $t\sim m L/(\hbar k)$. That's basically the time a phonon of wavevector k needs to travel through the whole crystal.

Edit: Just realised that, at least for long wave phonons, a linear dispersion is more appropriate, i.e. $E=v\hbar k$.
Then $\Delta E\sim v\hbar/L$ and $t \sim L/v$ directly.

Last edited: Mar 2, 2013
15. Mar 2, 2013

### zhanghe

dear DrDu, so you mean, standing or travelling depends on the condition?

16. Mar 4, 2013

### hokhani

In fact my main goal of performing this question was to understand this:
Consider one 1D-harmonic oscillator. Instead of saying for example "it is in it's second exited state" we can say that "it has two phonons" . Also the oscillation amplitude is limited between two end points at A and -A. Now if the phonons are travelling waves, how can they travel beyond the end points? On the other hand if they are standing waves limited between end points, how can we regard them as free particles?

17. Mar 4, 2013

### DrDu

It is not the atoms which are travelling but the excitation.

18. Mar 5, 2013

### zhanghe

Dear hokhani,
Remember we used the periodic boundary condition, just because we have decided to ignorate the effect of the boundary, so you of course can not get any results about "travelling beyond end points". Anyway, in most semiconductor textbook when we consider interaction btw. phonon with electron, photon etc. we never consider the end point case.

And you said "it is in it's second exited state", "it has two phonons", could you tell me who is the "it"??

19. Mar 5, 2013

### Darwin123

We were talking about the vibrational wave. Of course, the same logic applies to a harmonic oscillator consisting of a single atom attached to one very heavy-atom.

The single-atom harmonic oscillator is what is called a "limiting case" for the extended wave. The extended wave comes about because the atoms are coupled to each other by elastic forces. If the two atoms are connected by an elastic force, one atom being much more massive than the other, then the light atom can be considered a harmonic oscillator by its.

20. Mar 5, 2013

### hokhani

By "it" I mean 1D-harmonic oscillator which has only one oscillation mode and has for example two phonons in that mode.