I am trying to understand crystallography and the space groups of crystals, but I have one major question bugging me. The book I am using adresses different lattice symmetries and applications of group theory. More specifically, the electron energy bands are characterised by the irreducible representations of the point group of the Bravais lattice. By basis I mean the atoms associated with a lattice point.

Is this characterisation altered or does it break down in any way when a non-trivial basis is used?

I would suspect the answer to be along the lines: "The form of the bands are changed, but the characterisation is conserved...", but this is only guesswork.

  • $\begingroup$ Are you asking a question about like, finding good quantum numbers for such systems? The point groups are obviously altered, but you can still have an underlying Bravais lattice. The species / unit cells however will have different (or even just the trival) point groups - hence, distinct space groups. $\endgroup$ – daydreamer Oct 14 '20 at 15:43
  • $\begingroup$ The energy bands of a crystal can be characterised by the irreducible representations of the space group of the Bravais lattice. My question is then, is this classification still valid with a non-trivial basis? I will edit my question to make it more clear $\endgroup$ – B. Brekke Oct 14 '20 at 16:21
  • $\begingroup$ I see no other reason but extreme luck/coincidence for the presence of non-trivial basis not altering the wave functions symmetries in general cases. If you put distinct basis, the point group changes. Find the new group, compute its invariants and proceed in the usual Wigner-lesque way to find the finite-dimensional irreps, or just the nicest irreps possible. Group Theory always gives you good quantum numbers $\endgroup$ – daydreamer Oct 14 '20 at 16:40
  • $\begingroup$ Although I'm not very experienced, I have doubt in this answer. The reason is that compound materials are mentioned several times in the discussion of specific simple lattice symmetries in my book. Also, if I were to find a new symmetry respecting the basis, I would very often end up with only the trivial symmetry group rendering the method quite useless. Maybe I didn't understand your comment, however, I am still not sure of what is going on $\endgroup$ – B. Brekke Oct 14 '20 at 17:03

By adding atoms to the crystal basis, the space group will generally change and so will the symmetry classification of electronic states $|\psi_{nk}\rangle$. To be a bit more complete, let's look at exactly how states are classified and labeled.

If we have a symmorphic space group $\mathbb{S}$, then for a given $k$ we can define its little cogroup $\mathbb{G}_k$ (group of the wavevector) by $$ \mathbb{G}_k = \{\:(R|0)\in \mathbb{S} \:\:\:|\:\:\: Rk\equiv k \:\} $$ where $Rk\equiv k$ means equivalence up to a reciprocal lattice vector. In a crystal, electronic states $|\psi_{nk}\rangle$ are classified by the irreducible representations of the little cogroup of $k$. Each high symmetry line in the Brillouin Zone has a single little cogroup associated with it and the irreps will then label the bands along that line.


FCC with one atom basis

The basic FCC lattic has space group $Fm\bar{3}m$ (#225). If we look at the high symmetry line $\Delta:\Gamma\to X$, we can find (from this source) that the little cogroup is $4mm$ ($C_{4v}$). Then looking at the irreps here, we see there are 5 of them including one 2D irrep

FCC with two atom basis

A zincblende crystal is FCC but has a two atom basis (like CdTe). It has space group $F\bar{4}3m$ (#216). Looking at the same high symmetry line $\Delta:\Gamma\to X$, the little cogroup in this case is $2mm$ ($C_{2v}$) (as found here). $C_{2v}$ has only 4 irreps and they are all 1D. So the symmetry classification is clearly different.


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