# What is exactly a phonon? [duplicate]

What is exactly a phonon? I have my explanation of what it is but it's probably wrong and I would like that you correct me.

I see phonons like this:

We take a crystal. It allows some waves to propagate on.

In first approximation we can say that our crystal can be modelled as an ensemble of harmonic oscillators having the same frequency. The properties of the wave propagating on it will be defined by :

The frequency of the wave $w$

The wavevector $k$

The polarisation of the wave (is it longitudinal, transversal).

But all we said is purely classical and phonons are linked to Quantum Mechanics. So in fact to define a phonon we have to quantize our model of harmonic oscillators.

We assume that all oscillators have the same oscillation frequency $w(k)$. Then the hamiltonian can be written as a sum of harmonic oscillators (a first sum on the possible $k$ and a second for each node of the crystal). A phonon will be an eigenvector of this Hamiltonian. So it represents an excitation of the system (and if we want a link to the classical world, as it is an excitation of an ensemble of harmonic oscillators, it represents a wave travelling on the crystal).

And because we have a wave function we can say that the phonon is a particle : if we do a measure at a give $\overrightarrow{r}$ of our wave function we would have a probability $| \Psi(\overrightarrow{r}) |^2$ to find a phonon here.

Could you correct me if I am wrong, or give me more details?

• – UKH
Sep 17, 2016 at 15:42
• Why do we need phonons? physics.iisc.ernet.in/~aveek_bid/PH208/…
– UKH
Sep 17, 2016 at 15:43
• Thank you for the links. But I would like to know at least if what I say is true or incorrect (even if it is not complete) ? Sep 17, 2016 at 15:45

You can get any solution to the Quantum Harmonic Oscillator using creation and destruction operators. If you have a solution $|n>$ of energy $E_n=\hbar \omega (n+1/2)$, and you apply the creation operator you'll get a solution with energy $E_{n+1} = \hbar \omega (n+3/2)$. You could say you are adding little packets (quanta) of energy.
The number of phonons, as in the case of photons, is not fixed. You can also have a degeneracy of phonons for mode, i.e. number of phonons with energy $E$. The difference with phonons is that there's a maximum frequency at which you can excite the system. This is because there's a minimum wavelength, the distance between particles (you can't excite where there's nothing there). This is called the Debye frequency.