Question: What is the relativistic (quantum) wave equation that governs that motion of phonons?

My Attempt(s): The phonon Hamiltonian is given by, $$\mathscr{H} = \dfrac{1}{2}\sum_n \left(p_n^2 + \omega_n^2 q_n^2 - \hbar\omega_n\right) = \displaystyle\sum_n \hbar\omega_n \hat{a}_n^\dagger \hat{a}_n$$

Now if I write $\mathscr{H}\Psi = i\hbar \dfrac{\partial\Psi}{\partial t}$ following the Schrödinger equation, then the integer spin of the phonons will not be considered and hence we need a relativistic version of the wave equation.

In this circumstances, if I try to write the Klein–Gordon equation (if phonon is spin $0$) or the Proca equation (if photon is spin $1$), then another problem should arise from this answer on SE Physics, which is saying that the spin of a phonon can't be defined in a general way:

Since the group of rotations is not a continuous group in real crystals, it is not possible to define spin in a meaningful way. It is only in an isotropic ideal medium that is possible to define spin for a phonon (quantized accoustic wave). Equivalently it is only possible to define a spin if the the wavelength of the phonon is long or if one is restricted to phonons in special directions. It is only in such case that one can say that longitudinal phonons have spin $0$ and transversal phonons have $1$. From this article A.T.Levine, "A note concerning the spin of the phonon" you can read in the conclusions:

... Thus, the spin of the phonon will be well-defined in a medium which is isotropic but for a real crystal it will be well-defined only along certain restricted directions of propagation. The precise effects of phonon spin should, in principle, be detected experimentally by observing its interaction with other fields, e.g. spin waves. In any case, it is a quantity of fundamental interest that must be considered in any program of quantization.

Under this case, how should one proceed to arrive at the correct relativistic wave equation?


By mentioning the name of Klein-Gordon equation and Proca equation, I didn't mean to say that phonon travels at the the speed of light. Rather I wanted to point out that we don't know the spin of phonon, from which we can construct a wave equation.

As phonon corresponds to an elastic wave (sound), we may perhaps need a relativistic version of the elastic/matter wave. And in order to construct that we have to know the exact spin of the underlying (quasi)field. Unfortunately, such kind of unique spin doesn't exist, as i mentioned earlier. So what to do? We need to need construct such a wave equation, which builds-in the fact that the phonons are bosons.


  1. Tiwari, S. C. "Do phonons carry spin?." arXiv preprint arXiv:1708.07407 (2017).
  2. Levine, A.T. A note concerning the spin of the phonon. Nuovo Cim 26, 190–193 (1962).
  • 1
    $\begingroup$ Why would you expect phonons to obey a relativistic wave equation? $\endgroup$
    – Andrew
    Jan 23, 2023 at 11:56
  • $\begingroup$ Let's consider a general case. There are some cases, where the phonon can move in comparable velocity of the light. Besides this, another reason I have mentioned in my question. In Schrödinger equation, spin of the particle is not included. So to keep the information that the phonon is a boson in it's wave equation, we need a relativistic version. $\endgroup$
    – SCh
    Jan 23, 2023 at 12:00
  • 2
    $\begingroup$ Hm. The wave equation for phonons should be a long-wavelength approximation (effective field theory) to the microscopic dynamics. If the underlying crystal structure is not rotationally invariant, then in general I'd expect the effective field theory not to be rotationally invariant either. Also the natural speed for propagation of sound waves is the speed of sound (which in general can depend on polarization), not the speed of light. $\endgroup$
    – Andrew
    Jan 23, 2023 at 12:25
  • $\begingroup$ The Levine quote mentions two different things: 1) isotropic materials and 2) specific directions in real crystals. Which are you interested in? Coming up with a relativistic version of linear elasticity will get you 1 but not 2. $\endgroup$
    – lnmaurer
    Jan 27, 2023 at 4:39

1 Answer 1


TL;DR: The same wave equation(s) that we obtain from Maxwell equations for the classical EM field.

$a_{\mathbf{k},\nu}$ and $a_{\mathbf{k},\nu}^\dagger$ in the Hamiltonian can be though of as the expansion coefficients of the field in terms of its eigenmodes, which are obtained by solving Maxwell equations with the appropriate boundary conditions. Thus, if, instead of creation/annihilation operators, we were to write the Hamiltonian in terms of the field operators $\mathbf{E}(\mathbf{x},t)$ and $\mathbf{B}(\mathbf{x},t)$, these would satisfy the same equation as the corresponding classical fields - ie.g., the usual wave equation (which is, of course, already relativistic.) The difference with the Klein-Gordon field is that a) KG is usually written for a scalar field, and b) KG has an additional mass term (inversely, one can say that wave equation is KG with zero mass $m=0$.)

The answer above is mistakenly addressing photons rather than phonons that appear in the title. The answer however would need only a few adjustments: instead of Maxwell equations, we would have (linear) elasticity equations, whereas instead of electric and magnetic fields we would have strain and displacement fields. The rest is the same - as always when we deal with an eigenmode expansion and second quantization.

What remains unanswered here is the relativistic form of elasticity equations, but this would be better answered, if disentangled from the quantum aspect.

Additional remarks
Photons are quantized elastic vibration - as I tried to explain in the answer above, quantization does nothing to change the equation, and we just as well talk about the elastic waves in crystal. We certainly can define phonon polarization (longitudinal and transversal), but it remains to decide, whether it can be called spin. Indeed, as noted in the OP, crystal has lower symmetry than free space, so we cannot claim existence of a conserved quantity that results from symmetry. In this sense phonons are like electromagnetic modes in a cavity/waveguide, where the symmetry might be broken by the shape of this cavity, and hence polarization is poorly defined. On the other hand, when we talk about a wave equation, we usually imply the long-wavelength limit, where the phonons can be described as the quantized waves in continuum elasticity theory. These have higher symmetry - they can be even isotropic, so spin can be defined. The problem thus reduces to knowing the symmetry of the continuum elasticity in relativistic case. I suggest posting this as a separate question - googling relativistic elasticity (even in combination with wave equation) produces multiple hits, so there might be an exert in this community able to give a clear answer to this.


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