3
$\begingroup$

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?

$\endgroup$
1

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ There is no picturing in your solution. $\endgroup$ Commented Apr 16, 2021 at 7:40
  • $\begingroup$ You mean you want literally a picture? I am however answering to the questions posed in the text of the OP. $\endgroup$
    – Roger V.
    Commented Apr 16, 2021 at 7:51
  • $\begingroup$ I don't find where you have answered my questions. In the first part, you said about the chain of ball that I already knew, and I have mentioned in the question and also the dispersion relation. and then you said it's phonon like dispersion relation. This doesn't suppose to help me. $\endgroup$ Commented Apr 16, 2021 at 9:03
  • $\begingroup$ What I was trying to explain that phonons are quantized elastic waves. You need to understand elastic waves, which does not necessarily have to do with quantum physics, whether phonons are bosons etc. $\endgroup$
    – Roger V.
    Commented Apr 16, 2021 at 9:08
  • $\begingroup$ The text I'm following regard phonon as an excitation of vibration and equally the same definition is in Wikipedia so can you make me understand the phonon in this text. $\endgroup$ Commented Apr 16, 2021 at 9:21
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.