Is it true that isotropic biaxial strain does not lower C2 symmetry?

[Amentia]

Hello,

My question is simple. I have read that isotropic biaxial strain does not lower C2 symmetry, but no proof whatsoever was provided. I would like to know if it is actually true and have a solid proof. If someone can provide it, that would be wonderful. But also explaining me how to start so that I can prove it myself would be great! Of course I mean C2 symmetry is a rotation of pi along z axis and the strain is in the perpendicular plane xy, in case it would not be clear.

Thank you in advance for any help!

 

[TeethWhitener]

Try collapsing everything into the xy plane and converting to cylindrical coordinates.

 

[Amentia]

Thank you for your answer but I don't get it. What should I collapse?

For example, if I want to know the symmetry properties of a wave function in a system which possesses the C2 symmetry, I thought I could start by finding the eigenvalues and eigenvectors of the rotation operator for spin 3/2 and angle pi. And for me it is easy to calculate, no problem with that. Although when I get the eigenvectors I am not sure what I can conclude about the symmetry of the two sets of wave functions obtained.

When I add strain, this method does not look good, how do I know a priori that the total angular momentum is still a good quantum number? I do not think it is possible to start like this when there is a perturbation since I need to show what the perturbation is doing. With the example of biaxial strain, I am not even sure that the result is always the same, depending on the orientation of the strain relative to the crystallographic axes. Perhaps here your idea to use cylindrical coordinates is helpful but I do not see how.

 

[TeethWhitener]

If you have your wavefunction (or molecular coordinates, or whatever) in cylindrical coordinates, then the strain operator only affects $r$: $S\Psi(r,\theta,z)= \Psi(r+\Delta r,\theta,z)$, whereas the only nontrivial $C_2$ operator rotates the function about the z axis by $\pi$; that is, $C_2\Psi(r,\theta,z)=\Psi(r,\theta+\pi,z)$. If the function is $C_2$-symmetric, then $\Psi(r,\theta,z)=\Psi(r,\theta+\pi,z)$. So now just put all the info together to show that the strained function remains unchanged under the operation(s) of the $C_2$ group.

 

[Amentia]

So the point is to show that $S\psi$ is an eigenvector of $C_2$ if $\psi$ is an eigenvector of $C_2$? If this is the case, I think I understood your reasoning.

But in this case, how can I apply it to a function with a spin degree of freedom? I think we have $C_2\psi(r,\theta,z)\phi = \pm i \psi(r,\theta+\pi,z)\phi$ where $\phi$ is the spin function, or more generally, combining the two eigenvectors: $$C_2(\psi_{1}(r,\theta,z)\otimes|\uparrow\rangle+\psi_{2}(r,\theta,z)\otimes|\downarrow\rangle) =-i\psi_{1}(r,\theta+\pi,z)\otimes|\uparrow\rangle+i\psi_{2}(r,\theta+\pi,z)\otimes|\downarrow\rangle $$. Do you simply assume strain and spin do not live in the same space and then get an equation like: $C_{2}S\psi = \pm i S\psi$ instead of $C_{2}S\psi = S\psi$? Then just draw the same conclusion from this?

 

[TeethWhitener]

The spin and spatial functions do not live in the same space. The $C_2$ operator only operates on the spatial function. Or at least I’ve never heard of a situation where the spatial rotation operator can affect spin.

 

[Amentia]

If we are dealing with a wave function that has a spin degree of freedom, I do not think it is possible to just omit it. Normally, there is an equation telling us that the application of a unitary operator on a function is equivalent to the application of the inverse transformation on the coordinates. The question is what is the form of the operator?

Without spin, the equation is:
$$e^{-i\frac{i}{\hbar}\pi\hat{L}_{z}}\psi(r,\theta,z)=\psi(R_{z}(-\pi)[r,\theta,z])=\psi(r,\theta-\pi,z)$$

But with spin, I believe we need to change to the total angular momentum J=3/2 which includes spin-orbit interaction. This is what we usually do when we perform rotations in Hilbert space. I am thinking about the states in the valence band of semiconductors. They are decomposed in heavy-holes and light-holes with spin up and spin down. According to the choice of operator for $C_{2}$, the result will be different for the rotation of the contour. I am a bit struggling because I was stuck with the simplest case but in the end I would like to apply that to semiconductors in the envelope function approximation formalism where the wave function is given by a sum of products between an envelope function and a Bloch function. So we could have something like this:
$$\psi = f_{3/2}(r)|3/2,3/2\rangle+f_{-3/2}(r)|3/2,-3/2\rangle+f_{1/2}(r)|3/2,1/2\rangle+f_{-1/2}(r)|3/2,-1/2\rangle +...$$

At least it seems strain does not bring more difficulty to that! But it is a bit surprising since it is known that strain can actually couple the states in the valence band.

 

[TeethWhitener]

This is an interesting question, and not one I have an immediate answer to. Spin-orbit coupling is a term in the Hamiltonian, not the wave function, (as is the strain operator). It's possible to have a wavefunction that retains its symmetry when strained in the xy plane but not when spin-orbit coupling is applied (since that term mixes spin and spatial coordinates). So if both strain and spin-orbit are present, I can imagine the symmetry of the wavefunction might be lowered. I'll have to think more about it.

 

[Dr_Nate]

Hello,

My question is simple. I have read that isotropic biaxial strain does not lower C2 symmetry, but no proof whatsoever was provided. I would like to know if it is actually true and have a solid proof. If someone can provide it, that would be wonderful. But also explaining me how to start so that I can prove it myself would be great! Of course I mean C2 symmetry is a rotation of pi along z axis and the strain is in the perpendicular plane xy, in case it would not be clear.

Thank you in advance for any help!

This is an interesting question. Which field/sub-field are you reading this in?

 

[Amentia]

Hi, if I have to define the field, I would say semiconductor nanostructures although I do not know if this is the official name! The document where I have read this is not anything published that I could find in a book so far, but I can provide an example of an analysis that I cannot understand well based on my poor knowledge of group theory...

https://arxiv.org/pdf/1803.00340.pdf

In this article, on part II.A., there is some analysis of the effect of strain on symmetry. But even the simplest example without strain is still beyond my abilities in the application of character tables, so when they introduce a complex parameter beta to make a linear combination of irreducible representation that should take into account strain effects, I get it qualitatively but I am unable to understand or derive something like that by myself. I looked at some books on group theory and I have not found anything like this.

 

[Dr_Nate]

I think you'll need to know group theory to understand this.

When I was learning about the International Tables for Crystallography and the symmetries of crystals structures, I came across an idea that I didn't have the background or vocabulary to truly understand. It was very deep into the group theory of crystals. I believe (I could certainly be wrong) that a crystal can only smoothly transition from certain symmetries to other closely related symmetries, but a jumps from wildly different symmetries aren't allowed. I can certainly give reasons why I think this is true or almost obvious, but to prove it or understand it in group theory is beyond me. I believe it uses the vocabulary: groups and subgroups, and Hermann's theory.

I hope that points you in the right direction.

 

[Amentia]

Thank you for your answer. I think there are not many symmetry groups below C2 anyway. If you have nice explanations that do not make use of group theory, I would like to hear them. But I think I will also just study group theory a lot now, this self-isolation period might be a good moment to simply learn what I put aside before...