Context 1 The spin $s$ of a relativistic particle of mass $m$ can be read off from the eigenvalue $s(s+1)$ of the operator $- \frac{W_\mu W^\mu}{m^2}$ in the rest frame of the particle where $W^\mu=\frac{1}{2}\epsilon^{\mu\nu\sigma\rho}J_{\nu\sigma}p_{\rho}$ where $J_{\mu\nu}$ and $p_\mu$ are respectively the generators of Lorentz transformations and spacetime translations respectively. In addition. one also has to find out the little group that leaves a given choice of $p_\mu$ invariant to find the allowed spin projections in any given direction.

Question 1 What is the theoretical origin of spin for a nonrelativistic quasiparticle such as phonons, magnons etc? Are their spins deduced from experiments and then put in the wavefunction by hand? Is there any theoretical reasoning for why their spin is what it is?

Context 2 For the massive (massless) relativistic particles, there is a one-to-one correspondence between the spin degrees freedom (helicity) and independent states of polarization. For example, photons have two helicity states which are in one-to-one correspondence with left and right circular states of polarization. However, the acoustic phonons of crystals have 0 spin but 3 states of polarization (2 transverse and 1 longitudinal).

Question 2 Does it mean that the spin degrees of freedom of phonons and its polarization states have no connection with each other?

  • 1
    $\begingroup$ What do you think spin is? $\endgroup$
    – Blazej
    Dec 30, 2017 at 11:40
  • $\begingroup$ @Blazej updated the post in response to your question. $\endgroup$
    – SRS
    Jan 2, 2018 at 4:38
  • $\begingroup$ The answers to this question might be helpful. In essence they state that the spin for transverse phonons is not necessarily zero. It is one for isotropic media and mostly ill defined for crystals. $\endgroup$
    – Crimson
    Jan 26, 2018 at 12:50

1 Answer 1


Answer to Question 1: The spin of a quasi-particle or collective mode in Condensed matter is purely determined by how the excitation transforms under rotation. In other words, it is basically the same as how it is determined in relativistic systems. In the usual case, the excitation is written in terms of field operators of a linked to a known quantity (displacement, spin, electron charge, orbital momentum, etc.). From there you examine how the operators transforms under rotations to determine the spin.

For phonons, which are displacements of atoms nuclei, you have a field which transforms as a vector, hence you obtain spin-1. The phonon hamiltonian is given by a sum of harmonic oscillators[1], in almost exact equivalence to that of light.


Where $k$ is the wave-vector, and $\lambda$ is the polarization. Note that the creation and annihilation operators of the harmonic oscillator are written in terms of [the position and momentum operators which are vectors in 3D systems][2]. Hence the phonon transforms as a vector, giving it spin-1.

For plasmons, which are harmonic modes of the charge density, you get spin-0 because the charge density is a scalar quantity and transforms under rotations accordingly[3]. The plasmon operator is given by $b_q\propto\sum_k c^{\dagger}_{k} c_{k+q}$

Magnons, which are oscillations of the spin-structure, are defined as vector fields, and written in terms of electron-spin operators, so they have spin-1[4].

You can do the same exercise for many condensed matter excitations (skyrmions, Majorana particles, etc. [5]).

Answer to Question 2: I believe the premise of this question is incorrect, acoustic phonons do not have spin-0, they have spin-1. If you are only looking at the longitudinal modes (which don't exist for light by the way), then sure you can call them spin-0.

Since you mentioned phonons, let me finish with one important remark on why spin of collective excitations is not necessarily a useful quantity in condensed matter. In almost all condensed matter systems you have a broken rotation symmetry. Usually, this symmetry is broken due to the lattice of atoms, and it means that the spin of collective excitations may not even be a meaningful quantity to discuss. The symmetry group of lattices is actually much more complicated than the vacuum, and unless you are looking at liquids, full rotational symmetry is broken.

To be explicit, if you have a phonon with a wavevector $k$ that is not at some high symmetry point, then the distinction between longitudinal and transverse becomes ill-defined. In fact, this is why the spin of phonons is largely irrelevant, in most of the Brillouin zone you cannot distinguish between polarizations in the normal way as you would do for light in vacuum. However, as you go to smaller and smaller $k$ where the lattice blurs into a continuum, then the concept of spin returns.

So I would conclude: You can determine the spin of collective excitations in condensed matter by looking at how they transform under rotations, or equivalently through how their operators are defined in terms of nuclei and electrons. However, because of the broken rotational symmetry of lattice, as opposed to the vacuum, spin is not very useful in condensed matter systems.

  1. http://www.phys.ufl.edu/~pjh/teaching/phz7427/7427notes/ch4.pdf
  2. http://www.phys.ufl.edu/~pjh/teaching/phz7427/7427notes/ch1.pdf
  3. Theory of Quantum Liquids by Nozieres and Pines
  4. http://www.phys.ufl.edu/~pjh/teaching/phz7427/7427notes/ch3.pdf
  5. Quantum Field Theory Approach to Condensed Matter Physics by Marino
  • $\begingroup$ physics.stackexchange.com/questions/306142/… The answer here (by Everett You) says that acoustic phonons are spin-0 bosons. What do you think about this? @user157879 $\endgroup$
    – SRS
    Jan 29, 2018 at 15:28
  • $\begingroup$ The longitudinal acoustic mode is spin 0 for sure. But there are transverse modes as well. $\endgroup$
    – KF Gauss
    Jan 29, 2018 at 15:45

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