This post deals with the nature of the CuO2 bands (within the compound \(La_2CuO_4 \) ). The 3 bands that one can construct from the \(O-p_x, O-p_y, Cu-d_{x^2-y^2}\) orbitals are somewhat interesting in their own right, and they also provide a format in which to understand how bands acquire "mixed character" and what it means to say a band is 70% Cu / 30% oxygen or something like that.
This first video derives the LCAO (linear combination of atomic orbitals) nearest-neighbor matrix. Maybe later I can do another one that looks at and interprets the eigenvectors. Feel free to do that on your own also. What eigenvectors do you get at k=0 and how would you interpret them in terms of, e.g., percent of Cu content? This is just exploratory, so feel free to make assumptions about the magnitude of t (maybe around 1 to 6 eV?) and about the diagonal terms, Ed and Ep.
Hmmm... I see now that my k=o suggestion has some problems and limitations. Why is that? What happens at k=0? Something kind of interesting, but a bit special compared to everywhere else in the BZ.
Here is another idea. There might be some specific zone-boundary spots that would be interesting to explore. Places where something is pi/a; or both things are pi/a. That might be more interesting and better. I think for this compound it is the zone-boundary regions that are most important and interesting.
That is pretty cool how the eigenvectors can tell us the ratio of copper to oxygen there are in the solid.
ReplyDeleteI understood most of this video, I did not understand why some of the t’s equations for CE have e^(-i*a*k). How did you know if the e^(-i*a*k) belonged on the first t or second t?
I do notice the difference on kx and ky. It was because it belonged to the axis of x and y.
Actually at k=0 there are 3 (eigen) states. One with energy Ed and 2 with Ep. Does that make sense? The eigen-vectors are 100, 010 and 001. For k not zero the eigenvectors get more interesting and reveal a more complex identity for the states.
ReplyDeleteCalculating t is an okay thing to do when you have time, but doesn't really tell you much. It will be a scalar number with units of energy. Try using 2 eV. Looking at eigenvectors (and energies) for specific cases is a good way to gain insight. Try something like t= 1 eV and Ed=Ep=0. Then you can try t= 1eV and Ed= 4 eV or so, Ep=0.
ReplyDeleteIt is the eigenvectors of the upper band that matter. That is the one that is 1/2 full. For those cases, how much oxygen nature does the upper band have at key zone edge points?
(set a =1 and solve a matrix of pure numbers at specific k points. The zero of energy for the diagonal elements is arbitrary so when Ep=Ed you can just make that your choice of zero.)
actually t is pretty difficult to calculate because it is hard to know what to use for the Bohr-radius length scale in multi-electron atoms. Also, the potential in the integrand is a bit of a mystery to me. Otherwise it is pretty straightforward - an integral over a wave-function like dx^2-y^2 and another like p_x, displaced by some length a, multiplied times some radial potential.
DeleteI would recommend giving up on t for now. It is a time sink with limited reward. The structure of the eigenvectors is more interesting i think.
Those are really interesting results!! Can anyone say what those tell you? Which eigenvector is most important and what information is it trying to communicate to us?
ReplyDeleteCan someone put into words the one or two most interesting and salient features??
Really interesting results. What do these eigen-vectors tell us? What is the most important band? What are the salient aspects of its eigenvector? How is that different for different k?
ReplyDeleteIt seems to me that the most important eigenvectors would be those that represent copper having strong coupling interactions with both the Px and Py oxygens. I'm interested as to what significance the signs on the eigenvectors have.
ReplyDeleteI'm not quite sure how to tell which band is which in Johns calculations, but I would think that the highest energy band is the most interesting to look at. The lower energy bands are filled so the conducting properties of the material wouldn't be shown by looking at those bands.
This is interesting, not really what I would have expected. I'm not sure if I am analyzing this correctly, but I would think that the most important eigenvector should probably be the one that corresponds to the lowest energy. Then, I think the most important vector would be the first. They seem to generally correspond to bands mixed between the Cu and the second O, distributed equally if $E_d = E_p$ and with more weight on the O function if $E_d$ is higher. I was expecting (or maybe just hoping for) some more pure bands, and that really only happens with the first oxygen and almost uniformly at the energy $E_p$.
ReplyDeleteRegardless, I find it strange that there appears to be such a dissimilarity in representation between the two oxygen wave-functions in the vectors, which to me doesn't seem to be much correlated to the direction of k.
Let me know if I'm on the right track with any of this. Also, earlier you said that the eigenvectors of the upper band are the most important. I am wondering, is the band determined/defined by the eigenvectors or the eigenvalues or both? Thanks.
"I would think that the most important eigenvector should probably be the one that corresponds to the lowest energy"
DeleteI thought that too at first, but since we care about the upper band, than maybe we should be looking at the eigenvector with the highest energy eigenvalue. With that in mind, we see that, in the case of \( E_d = 4 \) , the Copper coefficient is clearly the dominant one.
I'm not sure how we could quantify the amount of "oxygen nature" present. If I try to think about it geometrically, it makes sense to me that the higher energy should exhibit more "copper nature", since the \( d \) orbital extends from the \( x-y \) plane much more than the \( p_x \) and \( p_y \) orbitals. Maybe I should stay away from that interpretation, however...
DeleteWould it really be a simple percentage? Should we look at how the coefficients oscillate from one point in k-space to another? My guess is that we don't care as much about how they oscillate, going by the assumption that \( E_d = 4eV , E_p = 0eV \) is closer to reality than \( E_d = E_p = 0 \), since the Oxygen coefficients seem to oscillate way more than that of Copper for \( E_d = 4eV \).
Patrick: you said "Regardless, I find it strange that there appears to be such a dissimilarity in representation between the two oxygen wave-functions in the vectors, which to me doesn't seem to be much correlated to the direction of k." Can you explain that a bit? It seems to me like the Oxygen components are highly dependent on location in k-space, so I'm trying to see how you're interpreting the data.
Out at dinner I realized my mistake here...the highest energy is definitely the one to be interested in because that would be the conduction band...I'll revise more when I get home
DeleteAaron: You're right, I was mistaken about which eigenvector I should be paying the most attention here. The highest energy fits the k-space directionality exactly like I would have assumed it should...probably should've been my first clue that I was thinking about it the wrong way.
DeleteIt seems to me that the most important eigenvectors would be those that represent copper having coupling interactions with both the Px and Py oxygens. I think this is because this coupling leads to a lower energy state like Patrick said.
ReplyDeleteThe professor asked us to find the most important band, but I'm not quite sure how to tell which band is which in Johns calculations, does anyone have some insight into that? I would think that the highest energy band is the most interesting to look at. because it would shed light onto the the nature of the materials conductivity. The lower energy bands are filled so the conducting properties of the material wouldn't be shown by looking at those bands.
On a second glance I think I was wrong. I think each eigenvector represents a different band and thus the highest energy eigenvector would shed light on the conduction band.
DeleteActually, it looks like the points with "less coupling" correspond to lower upper energies. Looking at the \( E = 4eV \) data, you can see the k-points where there are more off-diagonal 0 elements in the matrix, the highest energy eigenvalue is lower than in the k-points where there are more non-zero off-diagonal elements.
DeleteI feel like I'm not being clear....
I agree with your correction. Does what I said about the "coupling" make sense though?
DeleteI think I have an idea about how we can quantify these percentages. Since the eigenvectors are normalized, the squares of the coefficients all add up to 1. So for \( E_d = 4eV , \vec{k} = \pi (-1,-1) \), our eigenvector [0.88807383 0.32505758 -0.32505758] implies that, the upper band ( \( E_\vec{k} = 5.46410162 \) ) is \( 0.8881^2 = 0.7886 = 78.9 \% \) Copper, \( 0.3251^2 + 0.3251^2 = 0.2112 = 21.1 \% \) Oxygen.
ReplyDeleteThis comment has been removed by the author.
DeleteWhich is consistent with what Zack guessed before!!
Delete" 70% Cu / 30% oxygen or something like that "
By three bands, do you mean the components from Copper and the Oxygens individually? If so, my follow-up question would be if you have an idea about what that would imply physically.
ReplyDeleteIf you mean the bands given by each eigenvalue, then my follow-up question is: If the highest energy is the upper band, and the lowest energy is the lower band, what does \( E_\vec{k} = 0 \) correspond to? Is it just a limit, maybe? I'm just wondering out loud, if anyone can answer these.
Also, thanks a billion for writing that python script. We wouldn't have gotten this far without it.
ReplyDeleteWhat do you think having a flat band would do? I've done a little bit of reading and I think it will give topological media. Meaning that this materials conductive properties would be different at the surface of the material. Is there a way we could do a Density of States Calculation with this material? I think that if we did a DOS calculation it would be very singular which is also a sign of a topological material.
ReplyDeleteHere are some abstracts I was reading about flat bands in topological matter...
ReplyDeletehttp://arxiv.org/abs/1110.4469
http://journals.aps.org/prb/abstract/10.1103/PhysRevB.83.220503
By three bands we mean the 3 eigenvalues. A band is all the energies that one obtains, as a function of k, for one of the eigenvalues.
ReplyDeleteThe three bands come from 3 atomic orbitals. Mathematically, starting with 3 atomic orbitals makes the matrix 3x3 which leads to the 3 eigenvalues and thus 3 bands.
Here's a finer-grained visualization of the text file I sent.
ReplyDeletehttp://i.imgur.com/OoMaoCk.png