Is the formation of crystal due to internal symmetry or spacetime symmetry?

Question

In the contexts of field theory, we have internal symmetries and spacetime symmetries. Referring to crystal, people would say it is due to space translation symmetry. However, I don't think the formation of crystal breaks the local spacetime symmetry.

To be more solid, lets consider the energy of a crystal $$U_{\mathrm{tot}}\left(u_i\right)=\frac{1}{2} \sum_{i j} u_i K_{i j} u_j$$

We may have a specific configuration of atoms with a specific spacing with each other, as our ground state.

Now, if we move $u->u+\delta u$, i.e. move ALL the atoms by a distance, the configuration changes, yet this is still one of the ground states of the Hamiltonian.

The paragraph above exactly states the essence of spontaneous symmetry breaking, which typically leaves a large degeneracy of ground state when it is a continuous symmetry.

So in fact, the symmetry we consider here is actually an internal symmetry.(think $u$ as field) Therefore I would say the formation of crystal is due to internal translation symmetry breaking (of field). Is this picture correct? If so, does it mean all symmetry breakings are due to internal symmetries? But what do spacetime symmetries do if this is the case?

Answer 1

Your argument essentially shows that the presence of the crystal does not break translation invariance at a fundamental level. That's true, but not really what people mean when they say a crystal breaks spatial translations/rotations to a discrete subgroup.

Usually the degrees of freedom that we care about in condensed matter are the electrons in the crystal. (Sometimes, also other excitations like phonons and photons). The nuclei are -- to a good approximation -- fixed in place. So when we think about translating a degree of freedom, we are thinking about moving the electrons relative to a fixed lattice of nuclei by an amount $\delta u$. This translation will only be a symmetry of the system for certain, discrete, translations.