Picturing Phonons

Question

Consider vibrations of a One-dimensional monatomic chain of atoms. What I'm trying to do is to picture phonons?

First where I am? So I have a one-dimensional monoatomic chain. With little calculation, I can find out dispersion relation for the system $$\omega =2\sqrt{\frac{\kappa}{m}}\left|\sin\left(\frac{ka}{2}\right)\right|$$ Further $k=2\pi p/Na$ where $p$ is an integer.

So far I have the following picture in my mind. That is atom jiggling with the same frequency.

Now If a classical harmonic oscillator has a normal oscillation mode at frequency $\omega$ then the quantum system will have eigenstates with energy $$E_n=\hbar \omega \left(n+\frac{1}{2}\right)$$ If I have a many-particle system, I can always decouple them to write the above expression. For a particle to jump from one state to its adjacent state, $\hbar \omega $ amount of energy must be supplied.

Each excitation of normal mode by a step up the harmonic oscillator excitation ladder is known as a "phonon".

From above statement, I don't know how to think of phonon. Is it like particle in some state, a phonon particle collide sort of this particle and goes to next level? And from where this phonon come from? If system is in some definite energy state with any excitation, Does it mean no phonon? If the particle get excitated by a phonon, is it mean the phonon destroyed?

It is said that phonons are bosons because you can put many phonons in the same state. What do we mean by putting phonons in state because these were excitation and should appear during the transition?

Answer 1

Thinking of atoms oscillating independently near their equilibrium positions is incorrect. A better picture is a chain of balls connected by springs - this is easily solvable and produces a phonon-like dispersion relation.

One can also think of continuous systems: longitudinal elastic oscillations of a beam are longitudinal phonons (acoustic waves), while the transverse oscillations of a beam ir a string are transverse phonons.

In other words, there is nothing special about phonons and elastic oscillations. Moreover, when the dispersion relation is expanded near the extremum, these are the oscillations of a continuum, as those of a beam or string. What really makes these oscillations different is quantizing them.

Update

From above statement, I don't know how to think of phonon. Is it like particle in some state, a phonon particle collide sort of this particle and goes to next level? And from where this phonon come from? If system is in some definite energy state with any excitation, Does it mean no phonon? If the particle get excitated by a phonon, is it mean the phonon destroyed?

Perhaps what is worth mentioning is that when we call phonons "particles" we mean particles in quantum sense, not in classical one. Indeed, in classical physics particles are point-like objects, distinct from waves - and phonons, being elastic waves are certainly not particles.

Basic quantum mechanics then teaches us about an outdated concept of particle-wave duality, where a particle behaves sometimes as a particle and sometimes as a wave (like in a two-slit experiment), but this still conjures an image of a classical particle/fermion - which is the main object of basic QM books.

QFT then adopts a completely abstract view of a particle being an elementary excitation of a quantum field, or an irreducible representation, or a pole of a Green's function. This seems too abstract/mathematical, and leaves one thinking of particles as point-like objects which sometimes behave like waves.

However, in reality QM completely redefines a particle, or more precisely the properties that define what particles or wave are: notably waves interfere, while particles can be counted... and in this sense phonons (quantized elastic waves) are particles like any other quantum particles.

Answer 2

Perhaps a different point of view might help. I am a chemist and I was initially quite confused by the phonon point of view.

I would say that phonons excite the vibrational states of the lattice. By "absorbing" a phonon with energy $\hbar \omega_k$ we can excited the vibrational normal mode $k$ of the lattice. The lattice can "absorb/accept" $n$ phonons to reach the $n$-th excited state of mode $k$.

I guess the physicists point of view focuses more on the phonons and he might say that $n$ phonons populate mode $k$. Or that the phonons are in state $k$. I find that point of view a bit strange since it makes it sound as if the excitation state of mode $k$ is a property of the phonons, disregarding the underlying lattice. But that point of view might be more convenient when you talk about occupation numbers.

Answer 3

In short: Don't forget that phonons are sound, something you can hear - the lattice is emitting a sound

It's confusing because you have atoms (particles) and "phonons" (also particles) - and getting the picture ~ language ~ precision is confusing; and I'm sure people use the word "phonon" in a variety of ways. Here, I will use the word "particle" meaning "point-like" in the same way that we think of electrons as particles or points. Electrons, protons, "phonons", "photons" - are all 'dots on your page'. If you want to add more to this you can, but this is a starting point.

Briefly and for comparison, an oscillating charge produces light which comes to you in the form of a wave. But given wave-particle duality, for every wave, you can suppose a particle, and vice versa. So instead of a wave, I will say, "an atom" emits a "particle of light" - a photon. We learn this occurs in a quantized/discrete way.

A phonon is something you can hear (if you could hear a single phonon, in the same way if you could see a single photon - you're probably receiving many phonons, and many photons in any time interval). Your lattice of atoms is vibrating - that is, it is producing a sound. This sound comes to your ears. Tap on your table (you shook the table). You can either view this as a wave, or a bunch of discrete particles "phonons" being emitted. As atoms can only receive discrete and specific particles of light, so too, if you want to vibrate the lattice, it can only receive discrete and specific particles of sound, which it can then reemit (again, as sound, that is, you'd have to hear it, not see it). What makes this confusing is that, you also see the oscillations to the real/material object (which would be photons) - but phonons deal with sound - that are related to the actual oscillations of the physical object (this aside, you can really confuse yourself)

The air for the (sound) wave, is equivalent to the electric field for the (light) wave. Or you can disregard the mediums and think of particles being emitted - "whatever that means".

(He would know more than me) But I think I would agree with Hans Wurst. I wouldn't think too much about this response (it's confusing me too, because now I'm questioning "what is a wave"? "what is a medium"? - given I'm trying to make a distinction between the classical and quantum) However, I believe the basics of this response, that your lattice is emitting sound is correct; In the same way an atom can emit light, and receive light.

(actually, 'that your lattice is emitting sound' really is no more than a basic, common sense thing - but in the jargon and complexity of the math/language we can sometimes forget this) So if you're OK with the quantized atom - you should be OK with the quantized lattice. The atom isn't called a photon - nor should the lattice oscillations be called phonons. But they can receive and emit phonons. But again, there may be examples of "population this" and whatnot that I'm not familiar with (as in a "photon can be stored")

Also for clarity: I disagree with what your wikipedia link calls a phonon. (They give themselves a very broad liberty with that word - but it's not right). It's wrong in the sense of teaching/honesty to how they were taught/accumulation of knowledge. Would you call an atom in an excited state a "photon"? Show a photo of an atom and call it a "photon"? No - therefore, you would not call a lattice oscillation a "phonon". What you're looking at is not a phonon - but it can emit and absorb phonons

Also - I disagree with Roger V (though he's probably much smarter than me). I think he's playing around with words, thinking QFT says one thing, but if you were to pin him down on how he was taught/his training/accumulation of knowledge/what he actually thinks when he solves problems/how he would teach (if he could teach)/how he would explain truth (and not just brush it under the rug as "advanced stuff") - you could either pin him down as not really understanding what he's talking about, or that what he's talking about contradicts himself, and thus is just playing with words. (But I think towards the end of his answer he comes full circle)