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?


1 Answer 1


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.

When you do this with the electromagnetic field (see it as a sum of harmonic oscillators) you can interpret these quanta as photons. Particles are now understood as excitations of the respective field. The second state I obtained had one more photon that the one before. The same thing can be done with mechanical oscillations in solids. In that case you would call them phonons.

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.

  • $\begingroup$ Thanks for the precisions. But is what I say correct or does it contain some mistakes ? $\endgroup$
    – StarBucK
    Sep 17, 2016 at 15:46

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