A Wave properties of a phonon

[eneacasucci]

I am currently reading Kittel's Introduction to Solid State Physics and am confused by the terminology regarding phonons. On page 99 (8th ed.), regarding Eq. 27, Kittel writes:

"The energy of an elastic mode of angular frequency $\omega$ is $\epsilon = (n + 1/2)\hbar\omega$ when the mode is excited to quantum number $n$; that is, when the mode is occupied by $n$ phonons.

This definition implies that:

The mode (the harmonic oscillator) is the entity that possesses the wave properties (frequency $\omega$, wavevector $\vec{k}$, and polarization). The phonon is simply the unit of excitation (the quantum number $n$). However, in many other resources and later in the same text, I see phrases like "a phonon of frequency $\omega$" or "a phonon of wavevector $\vec{k}$."

My Question:

Is attributing $\omega$ and $\vec{k}$ to the phonon itself just linguistic shorthand for "an excitation of the mode characterized by $\omega$ and $\vec{k}$? Or is there a physical justification for treating the phonon as a distinct particle that carries these wave properties, rather than just being a counter of energy within a pre-existing mode?

 

[pines-demon]

Is attributing $\omega$ and $\vec{k}$ to the phonon itself just linguistic shorthand for "an excitation of the mode characterized by $\omega$ and $\vec{k}$? Or is there a physical justification for treating the phonon as a distinct particle that carries these wave properties, rather than just being a counter of energy within a pre-existing mode?

You are right it is a short hand to avoid "excitation of mode $\omega,\mathbf k$". But in second quantization treatment, this is exactly what we do with particles, the particles are excitations of a mode with a given energy-momentum, and we say particles with momentum $\mathbf p$. The only difference is that phonons are quasi-particles.

 

[eneacasucci]

You are right it is a short hand to avoid "excitation of mode $\omega,\mathbf k$". But in second quantization treatment, this is exactly what we do with particles, the particles are excitations of a mode with a given energy-momentum, and we say particles with momentum $\mathbf p$. The only difference is that phonons are quasi-particles.

Indeed, in Kittel's discussion of phonons, it states:

"A phonon of wavevector $\mathbf{k}$ will interact with particles such as photons, neutrons, and electrons as if it had a momentum $\hbar\mathbf{k}$. However, a phonon does not carry physical momentum."

Is it correct what I'm writing in the following:

The normal mode of vibration is the wave itself.
We are attributing the wavevector (of the corresponding mode!) $\mathbf{k}$ to the phonon to emphasize its wave-like nature (since $\mathbf{k}$ is a wave property). We then define its quasi-momentum using the de Broglie-like relation, $\mathbf{p} = \hbar\mathbf{k}$.
Finally, we treat the phonon as a particle when considering it as a discrete packet and in the scattering processes (such as with photons, neutrons, or electrons) by using this $\hbar\mathbf{k}$ value in the conservation of momentum equation for the interaction.


In the end we remember that phonons are not elementary particles since they are emergent excitations that arise from the collective motion of many atoms in a periodic structure.

 

[pines-demon]

The normal mode of vibration is the wave itself.

How? What wave?

Finally, we treat the phonon as a particle when considering it as a discrete packet and in the scattering processes (such as with photons, neutrons, or electrons) by using this $\hbar\mathbf{k}$ value in the conservation of momentum equation for the interaction.

Have you ever seen how we quantize the electromagnetic field to get photons? It is the same idea, photons might be real but the description is the same. We say a "photon with momentum $\hbar \mathbf k$" and we do not worry about semantics if its a wave/mode or whatever (techically photons are excitations of a given mode of the electromagnetic field).

 

[eneacasucci]

How? What wave?

I was referring to the collective lattice displacement field (the deviation of atoms from equilibrium).The reason I phrased it as 'the normal mode is the wave' is that in a periodic crystal structure, the normal modes of vibration take the form of traveling plane waves (characterized by wavevector $\mathbf{k}$ and polarization).

we do not worry about semantics if its a wave/mode or whatever (techically photons are excitations of a given mode of the electromagnetic field).

My point was simply that the phonon is the quantization of the normal mode, and to treat it as a particle in scattering events, we attribute to it a 'momentum' $\mathbf{p} = \hbar \mathbf{k}$ derived directly from the wavevector of that underlying mode.

I was for sure in a semantics loop because I'm preparing for an exam and I'm afraid to use the wrong words to describe the various physical entities :( e.g., if my professor asked me "is the phonon a wave? does it have a wavevector? can we attribute a wave lenght to it?" or some tricky questions like that. Maybe I'm just overthinking