What are the spins of Goldstone bosons in Condensed matter systems In Condensed Matter systems, acoustic phonons and magnons are two famous examples of the bosonic quanta of Goldstone modes. 
Question What are their spins? Has it been measured in experiments? 

Some additions on $24.02.20$ in response to @ACuriousMind 's comment
The concept of spin in such nonrelativistic systems appears to be tricky. Spin is not dictated by the transformation of fields under Galilean symmetry but by Lorentz transformation. Correct me if I am wrong.
About the polarizations of phonons, I know that there are three independent polarizations for a given wave vector $\textbf{k}$. Unlike photons, there is no constraint that causes phonon polarization vector to be transverse to the wave vector $\textbf{k}$.
 A: Acoustic phonons are spin-0 bosons. Magnons are spin-1 bosons (in the presence of the spin-rotation symmetry). Because to excite a phonon, you only need to inject the energy, which is of spin-0. But to excite a magnon, you need to flip an electron spin, which will change the angular momentum by 1.
The spin of these Goldstone modes can be measured by scattering experiments. For example, in the inelastic neutron scattering (INS), neutrons are sent into the material to interact with nuclei or local spins. If the neutron is scattered by the nucleus, the nucleus will be kicked away from its equilibrium position. This will create a phonon (lattice vibration) excitation. If the neutron is scattered by the local spin (which is the electronic spin in the localized atomic orbital) due to the spin exchange interaction, the local spin will be flipped. This will create a magnon (spin wave) excitation. As the total angular momentum is conserved in these scattering processes, so by comparing the spin state of the incident and the scattered neutrons, we will be able to determine the spin of both phonons and magnons. Suppose the incident neutron is fully polarized to the up-spin state, then if the scattered neutron is still in the up-spin state, then we know no angular momentum has been transferred, so the excitation created in this scattering process will be of spin-0. Otherwise, if the scattered neutron flips to the down-spin state, then we know a spin-1 excitation should have been created. The phonon and the magnon have quite different energy scales and dispersions, which is quite easy to distinguish in the INS spectrum. So in principle, one can determine the spin of these Goldstone modes in the experiment. I do not know any reference, but I believe this experiment should have been done, and the conclusion should be that the phonon carries spin-0 and the magnon carries spin-1.

Appendix: an explicit demonstration for the spin of a magnon.
Consider a spin-1/2 system described by ferromagnetic Heisenberg model
$$H=-J\sum_{\langle i j\rangle}\mathbf{S}_i\cdot \mathbf{S}_j.$$
The ground states are degenerated ferromagnetic product states. For example, one choice is
$$|0\rangle=\otimes_i|\downarrow\rangle_i.$$
The spin wave excitation is created by $|\mathbf{k}\rangle=a_\mathbf{k}^\dagger|0\rangle$, where the magnon creation operator $a_\mathbf{k}^\dagger$ is given by
$$a_\mathbf{k}^\dagger=\sum_i e^{i\mathbf{k}\cdot\mathbf{r}_i}\,S_i^+,$$
where $S_i^+=S_i^x+\mathrm{i}S_i^y$ is the spin raising operator. It is easy to check that the operator $a_\mathbf{k}^\dagger$ transforms like a spin-1 (triplet) representation of the SU(2) spin rotation group generated by the total spin operators $\mathbf{S}=\sum_i\mathbf{S}_i$. Therefore, the magnon carries spin-1.
