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.

  • $\begingroup$ I am not so familiar with the multi-dimensional quantum harmonic oscillator. But since for the 1D quantum harmonic oscillator the relation $\omega^2=k/m$ holds, and for the classical multi-dimensional oscillator a corresponding matrix relation $\Omega^2=\sqrt{M}^{-1}K\sqrt{M}^{-1}$ holds (where $\Omega^2$ is symmetric and hence diagonalizable to the Eigenfrequencies), I suspect it goes something like that for quantum as well. Hope that helps. $\endgroup$
    – oliver
    May 4, 2022 at 4:53
  • 1
    $\begingroup$ Second quantization is an additional layer that might obscure the basics that one knows from mechanics: one is solving a system of coupled linear oscillators by transforming it to normal coordinates. I strongly suggest trying to solve the classical problem for a linear chain of balls connected by springs. $\endgroup$
    – Roger V.
    May 4, 2022 at 7:49

1 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.

  • $\begingroup$ It's not that you can't extract a mass for Bloch electrons, it's just that this mass depends on $\vec k$. In particular, this effect leads to Bloch oscillations, where an electron in crystal, subjected to an external linearly-changing electric field, goes from the bottom of a band (where $m^*>0$) to its top (where $m^*<0$) and then back, and then again forward and so on. $\endgroup$
    – Ruslan
    May 6, 2022 at 9:44
  • $\begingroup$ I don't think the concept of effective mass is needed to explain Bloch oscillations at all. In fact, at some $\vec{k}$, you'll end up with the second derivative of E being zero --- and hence an infinite effective mass. Considering that the electron continues to accelerate with an infinite mass, that doesn't seem particularly meaningful. So, yes, you can generally take the reciprocal of the second derivative and call it "mass", but it only acts like "mass" in the sense of Newton's second law if the band is (approximately) parabolic. $\endgroup$
    – lnmaurer
    May 6, 2022 at 19:32

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