A A good quantum number for Cnv symmetry?

[Amentia]

Hello,

I was wondering if it was possible to define good quantum numbers in solid state physics or chemistry when systems posses a discrete cylindrical symmetry Cnv. I know that in terms of angular momentum, L and L_z will be good quantum numbers for spherical symmetry, then only L_z is a good quantum number for cylindrical symmetry.

What about C2v, C3v, etc.? Is it possible to define an operator which commutes with the Hamiltonian and provide a quantum number similar to L_z?

I am sorry if this is common knowledge but I have not found the answer on the internet nor in the textbooks I know of.

Thanks for your help!

 

[TeethWhitener]

What about C2v, C3v, etc.? Is it possible to define an operator which commutes with the Hamiltonian and provide a quantum number similar to L_z?

Of course. A $C_{2v}$ molecule is symmetric under a $180^{\circ}$ rotation about the symmetry axis, so the Hamiltonian commutes with the $180^{\circ}$ rotation operator. Tinkham’s book on group theory in quantum mechanics is a good resource.

 

[Amentia]

Of course. A $C_{2v}$ molecule is symmetric under a $180^{\circ}$ rotation about the symmetry axis, so the Hamiltonian commutes with the $180^{\circ}$ rotation operator. Tinkham’s book on group theory in quantum mechanics is a good resource.

Thank you for your answer. When you use group theory and couple this discrete angular momentum with the spin projection, $L_{z}(180^{\circ})+S_{z}$, is there a way to define the eigenstates of the Hamiltonian? I am used to couple L+S for L=1 and S=1/2 in semiconductor physics to obtain J=3/2 and J=1/2 states that define heavy-hole, light-hole and split-off. In some textbooks, they take only J_z as a good quantum number to define some new set of states.

I have Tinkham's book but I don't remember seeing a derivation of this procedure. Are you recommending the book in general or thinking to a specific chapter?

 

[TeethWhitener]

I guess I’m confused about what you’re asking. Are you thinking about spin-orbit coupling in molecules vs solids? Or maybe spin-rotation coupling like we see with spin isomers of hydrogen?

 

[Amentia]

Sorry I tried to stay general as I was thinking it would be easier for people to answer but I guess it makes everything confusing.

I am thinking about nanostructures where you have a part of the wave function which is a Bloch state (like in regular solid state physics) and a second part which is an envelope function. So the Bloch state basis is already $J=L+S$ and then the envelope has also some angular momentum $L_{env}$.

In the book The k.p Method by Lok Lew Yan Voon and Morten Willatzen, they first assume a spherical symmetry and say they can now couple $J$ to $L_{env}$ and generate a new good quantum number $F$ by the usual composition of angular momentum rules with Clebsch-Gordan coefficients.

Then, they consider systems with cylindrical symmetry and say $F$ is not a good quantum number anymore but we can keep its projection $F_{z}$. Then they find that for a quantum wire for example you can write the wave function as $|\psi_{F_{z}}\rangle = \sum_{J_{z}}C_{J_{z},F_{z}}|\frac{3}{2},J_{z}\rangle\otimes|F_{z}-J_{z}\rangle$.

Actual nanostructures can have "discrete" cylindrical symmetry like $C_{2v}$ and I was wondering if their approach could be extended to find a good quantum number, basis states... If I write the rotation operator for $C_{2}$ in the cylindrical case, I would use $exp(i\pi F_{z}/\hbar)$ but $F_{z}$ is not supposed to be a good quantum number anymore.

I hope what I am trying to understand is better explained. I am sure it must be taught in group theory books somewhere as you hinted if it has been done before in other cases. But my first question is really: is it possible?

 

[TeethWhitener]

Ah ok, yes it's much clearer what you're after now. Unfortunately, I don't have anything really insightful to say about it. I imagine it will be analogous to the symmetry breaking seen with an atom in a crystal field. $F_z$ definitely won't be a good quantum number, though. I'll think more about it if I get a moment.