Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

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.


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.

share|improve this answer
    
Thanks! This was exactly what I was looking for! I've never really understood the standard notation, but I guess I've heard things like how the T2g representation transforms like an effective l=1 angular momentum. An additional note: the acoustic phonon is the Nambu-Goldstone mode of broken continuous translation. Is there such a boson for the broken rotational symmetry of a crystal? I'm gonna guess it's still the phonon, but I'm not sure. –  ChickenGod Feb 27 '13 at 9:16
    
Oh, and I guess this hasn't been addressed: what quantity is conserved as a result of rotational symmetry in a crystal? –  ChickenGod Feb 27 '13 at 9:21
1  
@ChickenGod edited to address this. –  Emilio Pisanty Feb 27 '13 at 13:20
add comment

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(?)

share|improve this answer
    
This is related to the Einstein-de Haas effect. –  Emilio Pisanty Feb 26 '13 at 11:38
    
@EmilioPisanty Many thanks for the links, much appreciated. Your dicsussion is interesting, but you are focussing on the electronic wave functions, and I don't think this address the question, it does not complete the answer. –  JKL Feb 26 '13 at 12:49
    
Well, where else would the angular momentum be? The nuclear lattice does move - it carries phonons - but for it to have angular momentum you'd need the whole crystal to rotate. –  Emilio Pisanty Feb 26 '13 at 12:58
    
@EmilioPisanty That is quite right. That is how I interpreted the question, which might be the wrong way. However, in the light of our conversation, the atoms in the crystal lattice not only they vibrate linearly, they may also have rotational vibrational modes that can contribute to the total angular momentum of the crystal. What do you think about this? –  JKL Feb 26 '13 at 13:07
    
Yes, I would expect phonon modes analogous to structured light to be possible, but those will have angular momentum $L_z=m\hbar$ around the phonon wavevector $\mathbf{k}=k\hat{\mathbf{z}}$ and will not really reflect the properties and symmetries of the lattice. –  Emilio Pisanty Feb 26 '13 at 15:20
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.