Phonons and Crystal Momentum

[WWCY]

Hi all, I have a few questions. Thanks in advance!

Phonons:

Can phonons be thought of as a quantised wave of sound, in the same manner photons being a quantised light wave? Also, can they be taken as particles of sound in the same way that photons can be particles of light?

Crystal momentum:

My text states the following (assuming that this is a system with real-space lattice of $a$):
"If 3 phonons each with crystal momentum $\hbar (2/3) \pi /a$ scatter off each other, they can produce 3 phonons each with a crystal momentum of $\hbar (-2/3) \pi /a$"

Here, is crystal momentum conserved because the collision produces 3 extra phonons in addition to the original 3 and therefore gives a total final crystal momentum of $0$ (which is the same as having a crystal momentum of $3 \hbar \frac{2\pi}{3 a}$ in this system)? Or is crystal momentum because the 3 initial phonons "turn into" 3 phonons of crystal momentum $3 \hbar \frac{-2\pi}{3 a}$, which is also the same as having the initial crystal momentum of $3 \hbar \frac{2\pi}{3 a}$?

 

[WWCY]

Additional Question: Is the idea of Phonons being particles and waves actually literal, or is this a "convenient" way of thinking about things?

Thanks!

 

[Cutter Ketch]

Regarding your first question, yes and no. To some degree phonons can be thought of as quantized sound, and that is the implication of the name. However I think it is more correct to call them quantized vibrations rather than sound despite the name. The distinction is important because sound implies a particular kind of vibration and there are many other vibrations which aren’t related to sound. Are they analogous to photons? Well electric fields are not very similar to vibrations of an elastic medium, but many of the ideas of quantized packets of energy are applicable.

Regarding your last question about phonons being real, have you ever flicked a rope or a long slinky and sent a pulse, a wave, traveling down the rope? There is a discernible lump. It has definite extent. It travels, and it carries momentum. If somebody is holding the other end of the rope their hand moves. You have to accept that the traveling pulse is real, and it is easy to accept it as it’s own identifiable entity, but on the other hand it doesn’t exist independent of the rope. In reality it is a mode of the movement of the rope. So phonons are real. Their wave function description is not just score keeping. However they aren’t really particles. They are modes of the vibration of the material that act like particles in the same way the lump in the rope acts like a particle.

Regarding crystal momentum, you can tell the 3 phonons didn’t become 6 phonons because that wouldn’t conserve energy. Other than that I don't know enough about crystal momentum to help much. I do know that crystal momentum is not quite the same as regular momentum. From what I’ve just read (that gives you some idea of what this “help” is worth) the lattice periodicity makes periodic values of momentum indistinguishable, so crystal momentum isn’t the whole momentum. It’s sort of the Brillouin zone of momentum. Even if I’ve managed to get that right, I have no idea how it helps answer your question except to say crystal momentum is weird.

 

[WWCY]

That cleared things up a lot, cheers!

Edit: When you mentioned that phonons are quantised vibrations, what sort of "vibrations" do you mean? Also, are they only applicable to quantum mechanical systems?

 

[Cutter Ketch]

The vibrations are what you would expect: motions of the atom cores of the material. Because they are all tied together by electric bonds you can’t displace one atom without pushing on others, and the disturbances propagate like ripples through the material. But that probably isn’t telling you anything you don’t already know.

The vibrations aren’t special exotic things that only exist in quantum mechanics. You could illustrate many of the interesting behaviors with springs and balls. Certainly waves and propagation. Even some interference effects would show a kind of quantization. Think standing waves on a string. Some frequencies work others don’t. However at small scales and small energies quantum mechanics requires an energy quantization that produces lots of phenomena that you wouldn’t see with springs and balls. Nevertheless, that’s what we are talking about: ripples of displacement propagating through the material.

A ripple can propagate in a particular direction while the displacement of the cores is perpendicular to the direction of propagation like waves on a string. These are called transverse acoustic phonons. The displacement could be in the same direction as the propagation like pushing on a slinky. These are called longitudinal acoustic phonons. The displacement could include opposed displacement of cores in the unit cell (like a bit of material stretching) transverse to the direction of propagation. These are called optical phonons because in ionic crystals a stretching between the positive and negative cores constitutes a dipole moment and so readily couples with photons. What we think of as sound is carried by low frequency longitudinal acoustic phonons.

 

[WWCY]

Thanks for clearing it up, cheers!

 

[SpinFlop]

Can phonons be thought of as a quantised wave of sound,

This is analogous to saying something along the lines of 'can electromagnetic radiation be thought of as quantised waves of visible light.' Yes electromagnetic radiation travels as quantized waves (photons), however visible light only takes up a small bandwidth of this spectrum. Similarly, lattice vibrations are quantized waves (phonons), however sound only makes up a small bandwidth of the full spectrum of vibrations.

Can phonons be thought of as a quantised wave of sound, in the same manner photons being a quantised light wave?

Perhaps the closest relationship between phonons and photons is that they are both Goldstone bosons produced by spontaneously breaking an underlying symmetry. Photons are the Goldstone mode generated by the breakdown of guauge symmetry in quantum electrodynamics. Whereas phonons in a solid are the result of breaking translational and rotational symmetry when you freeze the atoms into a crystal. A word of caution here, although this is the closest relationship, some may not find it the cleanest. Namely, there is quite a bit of argument as to whether photons should truly be considered Goldstone bosons. Unpacking this is bit too off topic and also too far removed from my field of study for me to give it a proper treatment anyways. However, everyone agrees that lattice vibrations are quantized as phonons, that these phonons are a Goldstone boson, and that they travel with a crystal momentun through the lattice and transport energy along the way. From this point of view, they can clearly be classified as particles. However, there is no escaping the fact that at there very core they exist as a disturbance of the lattice with finite spatial extent. Thus, from this point of view they are also a wave, and so at the end of the day we have a similar wave particle duality picture at play.

Also, can they be taken as particles of sound in the same way that photons can be particles of light?

The two above answers also answer this question.

My text states the following (assuming that this is a system with real-space lattice of $a$):
"If 3 phonons each with crystal momentum $\hbar (2/3) \pi /a$ scatter off each other, they can produce 3 phonons each with a crystal momentum of $\hbar (-2/3) \pi /a$"

Here, is crystal momentum conserved because the collision produces 3 extra phonons in addition to the original 3 and therefore gives a total final crystal momentum of $0$ (which is the same as having a crystal momentum of $3 \hbar \frac{2\pi}{3 a}$ in this system)? Or is crystal momentum because the 3 initial phonons "turn into" 3 phonons of crystal momentum $3 \hbar \frac{-2\pi}{3 a}$, which is also the same as having the initial crystal momentum of $3 \hbar \frac{2\pi}{3 a}$?

Conservation of crystal momentum requires

$\sum_{i=1}^3 \mathbf k_{i} = \sum_{i=1} ^3 \mathbf k'_{i} + \mathbf K$

Where $\mathbf k$ is the phonons wavevectors which is related to the crystal momentum by $\mathbf p = \hbar \mathbf k$ and $\mathbf K$ is some reciprocal lattice vector which can be written as $\mathbf K = \frac{2\pi n}{a}$

Our goal is to find an integer n that satisfies this equation

We can immediately identify that $\mathbf k_{i} = \frac{2\pi}{3a}$ for each initial phonon and $\mathbf k'_{i} = -\frac{2\pi}{3a}$

Thus, we may now write

$\sum_{i=1}^3 \left( \frac{2\pi}{3a} \right) = \sum_{i=1}^3 \left( -\frac{2\pi}{3a} \right) + \frac{2\pi n}{a}$

$\frac{2\pi}{a} = -\frac{2\pi}{a} + \frac{2\pi n}{a}$

$n = 2$

The key point here is that crystal momentum, although similar to real momentum, follows a slightly different conservation rule. Namely, conservation is always satisfied when the initial and final sums are equal up to a reciprocal lattice vector. I think your confusion stemmed from applying real momentum conservation. You may ask, "why is crystal momentum conservation different?" The answer is that real and crystal momentum conservation both stem from Hamiltonian symmetries. The additional flexibility in crystal momentum conservation is due to the periodicity of the lattice, thus a plane deformation $\mathbf u (\mathbf R )$ at location $\mathbf R$ repeats identically if you translate by a Bravais lattice vector $\mathbf R_{0}$. ie: $\mathbf u (\mathbf R ) \to \mathbf u (\mathbf R -\mathbf R_{0} )$

 

[WWCY]

Thanks for the response!

Thus, we may now write

$$
\sum_{i=1}^3 \left( \frac{2\pi}{3a} \right) = \sum_{i=1}^3 \left( -\frac{2\pi}{3a} \right) + \frac{2\pi n}{a}
$$
$$
\frac{2\pi}{a} = -\frac{2\pi}{a} + \frac{2\pi n}{a}
$$

Does it mean then, that the 3 phonons, each with initial crystal momentum of $\frac{2\pi}{3a}$ collide, and then "turn into" 3 phonons with crystal momentum of $-\frac{2\pi}{3a}$ each?

 

[SpinFlop]

Does it mean then, that the 3 phonons, each with initial crystal momentum of $\frac{2\pi}{3a}$ collide, and then "turn into" 3 phonons with crystal momentum of $-\frac{2\pi}{3a}$ each?

That is correct. However, to provide a little more insight into this. The different phonon processes are generated via an expansion of the lattice potential whose equilibrium value places each ion at a specific location in the crystal. The expansion is of Taylor form. Keeping only the first term results in the static approximation of the crystal (ie: no phonons, all atoms rigidly fixed in place). Adding in the second term is equivelant to approximating the vibrations with a Hooke's law term and leads to the harmonic cross-section where single phonons are stationary states that propogate with infinite lifetime through the lattice (unless they hit a crystal defect or the crystal surface). Additional terms are referred to as anharmonic terms and are responsible for multi-phonon creation and anhialation processes. These higher order terms account for increasing number of multi-phonon processes. For this reason, the terms are often classified as the elastic (zero phonon) cross-section for the first term, the one phonon for the quadratic term, two-phonon for the cubic (1 in -> 2 out, or 2 in -> 1 out), and so on. In your case you have 3 in -> 3 out, and thus should correspond to a sixth order process. I expect this transition rate is extremely low and would probably not lead to any noticeable effects in the crystal properties. Indeed, it is rare to ever go beyond the quartic term when modeling properties of solids.

 

[fluidistic]

However, everyone agrees that lattice vibrations are quantized as phonons, that these phonons are a Goldstone boson, and that they travel with a crystal momentun through the lattice and transport energy along the way.

Reference https://www.physicsforums.com/threads/phonons-and-crystal-momentum.954022/

Bold emphasis mine. I did not know this. I actually thought electrons (or quasielectrons) were the ones that traveled with such a momentum.