Crystal Angular Momentum

Question

In a crystal, we don't have full translational symmetry, but we still have discrete translations. This allows us to define "crystal momentum" that is conserved modulo a reciprocal lattice vector.

In a crystal, we don't have full rotational symmetry, but we still have discrete 2, 3, 4, and/or 6-fold rotational symmetry. Can we define a "crystal angular momentum"?

Also, unrelated question: if we have a semiconductor impurity state that resembles a hydrogen 1s state, 2p state, etc... what is its angular momentum?

Answer 1

If a crystal has a discrete group of point symmetries then the electronic eigenfunctions will be suitably invariant under that group. Formally, the symmetry requires that the eigenfunctions of a hamiltonian with symmetry group $G$ belong to the various representations of that group.

In the abstract, a representation of a group $G$ is a vector space (in this case a subspace of degenerate energy) $V$ and a "recipe" for unitarily transforming the vectors in $V$ with the transformations from $G$, in the form of a group homomorphism $R:G\rightarrow U(V)$. If one knows the structure of a group (in the form of its multiplication table) then there is a lot that can be said about its possible representations, which are typically denoted by some standard notation (e.g. $E$, $A$, $B$, etc.). The wavefunctions are then labelled by the representation type of the subspace they belong to.

To bring this back down to angular momentum, the subspaces with different $l$ are the different representation subspaces. The group in question is the rotation group $\text{SO} (3)$. It has an infinite family of representations of increasing finite dimension, and the index $l$ that labels them is precisely the angular momentum quantum number of those wavefunctions. In group theoretic terms, then, "having definite angular momentum" simply means "belonging to a suitable representation of $\text{SO} (3)$".

Thus a crystal with a point symmetry will have electronic eigenfunctions that do have "definite crystal angular momentum", in the sense that they belong to a certain representation of the point group.

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Added in response to comment:

Unfortunately, there is no physical quantity that corresponds to this symmetry. This is due to the general fact that discrete symmetries have no generators. While you can write rotations, for example, in the form $e^{i\mathbf{J}\cdot\hat{\mathbf{n}}\theta}$, where $\mathbf{J}$ is the generator, this is not really meaningful for discrete symmetries.

A good comparison for this is parity: if $\Pi$ commutes with $H$ then we say parity is conserved, in the sense that the transformation itself is a constant of the motion. For a more general discrete group $G$ (instead of $G=\{-1,1\}$ for parity) then the labels $+$ and $-$ are replaced by the group representation. Similarly the labels $l$ and $m$ correspond to the group representation and to the eigenvalues of some particular group transformation. Both are conserved under $H$, but there is no generator.

Answer 2

I am not sure I am clear what you mean by Crystal Angular Momentum, but here are some ideas.

One must distinguish between angular momentum due to rotation of an object as a solid system, and that of a system made of subsystems each having angular momentum - atoms in the crystal in this case.

Individual atoms in a crystal do have angular momentum, and this is how they get their magnetic dipole moment. In crystals that have magnetisation the majority of atomic angular momentum point in the same direction, hence such crystals display magnetic properties. But that relates to the total angular momentum of individual, "loose" so to speak, atoms of the crystal and not that the crystal itself rotates as a solid object. I hope this makes sense(?)