What is the mass of collective oscillations?

Question

When you step through the procedure of deriving a phonon dispersion relationship for a given crystal structure (i.e. small oscillations from equilibrium, harmonic approximation, collective coordinate transformation, normal mode orthogonal transformation), you can rewrite your original interacting Hamiltonian in second quantization language like so, \begin{equation} \mathcal{H}=\sum_{q,\lambda}\hbar\omega_{q,\lambda}(b^\dagger_{q,\lambda}b_{q,\lambda}+\frac{1}{2}), \end{equation} where $q$ and $\lambda$ are the crystal wave number and branch respectively. The above procedure diagonalizes the Hamiltonian, and the space of states that is acted upon by the Hamiltonian operator is spanned by simple harmonic oscillator wave functions with respect to the collective coordinates. i.e. \begin{align} \psi_n(Q)&=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\frac{1}{\sqrt{2^nn!}}H_n(\sqrt{\frac{m\omega}{\hbar}}Q)e^{-\frac{m\omega}{2\hbar}Q^2},\\ \\ Q_q&=\sum_\alpha\hat{\eta} e^{-iq\cdot R_\alpha}, \end{align} meaning we can think of our phonons as decoupled oscillations in our new collective coordinates.

I think my question is pretty simple, but I can't find an answer anywhere: What is the mass of this oscillator?

My naïve guess would be that it has something to do with a derivative of the dispersion relationship similar to the effective mass of Bloch electrons, but I haven't seen a definitive answer.

Answer 1

My naïve guess would be that it has something to do with a derivative of the dispersion relationship similar to the effective mass of Bloch electrons, but I haven't seen a definitive answer.

This is a good guess, and if you try it you'll find that phonons have no (meaningful) mass, which may not be surprising since phonons are often considered analogous to photons, which are massless.

First, how would you get a mass? For a massive particle, you expect a "dispersion relation" like

$$E = \frac{\hbar^2k^2}{2m}$$

In other words, energy should be proportional to momentum squared, and as you note, you can extract the mass by taking the second derivative (as with electron effective mass).

In contrast, for a photon

$$E = \frac{\hbar}{c}k$$

Here, energy is proportional to momentum --- not momentum squared. If you try taking the second derivitive to extract the mass, it wont work. That shouldn't be a surprise because photons have no mass, so you cannot extract their mass by taking the second derivative of the dispersion relation!

Mass means a quadratic dispersion relation (ignoring relativistic effects). A non-quadratic dispersion relation means there isn't a mass.*

Now, consider the simplest phonon model (masses connected by springs in 1D). You'll find

$$E = \hbar \omega \propto \left|\sin\frac{ka}{2}\right|$$

In particular, at small $k$, $E \propto k$ --- just like for a photon. Just like for a photon, you can't define a mass. For larger $k$, the dispersion relation gets messier, but you still can't extract a meaningful mass. The dispersion relation isn't quadratic, and that means you still don't get a mass.

* Strictly speaking, Bloch electrons don't have quadratic dispersion relations either. However, if we're only interested in electrons near an extrema, and the dispersion relation is approximately quadratic at that extrema, then we can approximate the dispersion relation as quadratic and back out a mass.