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With a previous lecture, we have created this figure here of our nano ribbon of graphene.

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And this was, of course, an exercise for plotting and handling lists and arrays.

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But also it was supposed to teach you something about our Hamiltonian already because previously, for

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our infinitely large graphene where the units had only two atoms, we realized that our Hamilton matrix

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has a size of two by two.

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So now, since our unit cell is much larger, so where is it?

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Here is the unit.

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So we have to know, sir, we have eight times four, which is 32 atoms, so the size of our matrix

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will be 32 times 32.

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So how do we set up these elements of the Hamilton matrix?

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Well, we know that we have to describe all the hopping.

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So we have to describe the paths from one atom to another, for example, here from the atom, which

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hops to the blue one and then to the green one as well.

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So now, since these atoms here are not anymore the same basis atoms, there will be new elements in

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the hopping matrix, of course.

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So when you look at it carefully, we can now label our atoms going from top from bottom to top.

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So one two three four five, six, seven, eight and so on and so on.

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And so then this element here, for example, would describe the hopping from this purple atom to this

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blue atom.

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And then this element here two three two three element would describe the hopping from blue to red.

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This is why I wrote him the arm.

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Then the next one would be red to green.

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And of course, we also have to hop things along the opposite directions.

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And also, we have to consider the hopping not only to the left, but also to the right to the neighboring

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cell.

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So you see, we have the sequence here of different toppings purple, blue, blue, red, red, green,

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green, purple.

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And if you look at it more carefully, you will see, for example, that purple hopping to blue has

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two hopping terms one in this direction, the other one along this direction.

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Then the next one blue rat will just have a hopping along the y direction and then red.

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The green would again have two helpings along this direction.

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So these hopping paths are the same as these paths.

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So this means we actually only have two different hopping elements.

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That's always yeah.

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So first, we have these two and then we have this one.

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So they are always changing.

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So our matrix will just basically consider or consist out of zeros and only in the off diagonal.

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We will have elements and there will just be two different elements.

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So the first one will be this one describing the two hopping paths, for example, from purple to blue,

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but also from red to green.

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And then the other element will be B are so blue to red, but also green to purple.

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And this will have this shape here of exponential function, eye times and then the dot product of the

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K vector and the path.

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And so since the path is along Y with a length of A0, this will just be eight times A0.

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Quite.

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So then for the other of diagonal, the hopping path would just be exactly along the opposite direction.

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So this means these elements are basically the same, but just with a different sign and the exponent.

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OK.

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So the first thing that I'm interested in or that I had to think about was how can we cleverly design

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such a matrix?

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I mean, of course, we could say we make a loop or make it two loops and then we say, OK, if we are,

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add this element.

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Then we had this term here to the matrix.

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But it's actually possible to do this in a very clever way.

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And we can do this by using the diagonal matrix function.

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So the the command is and diak for diagonal, and then we can write, for example, one two three.

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And this will give us a diagonal matrix one two three.

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So instead, what we want, we want to have these numbers on the off diagonal, and this can be achieved

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with an option called or just written down.

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K is equal to one.

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So this means these numbers will not be shifted and it's not on the off diagonal.

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And you see our matrix.

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There's now a four by four matrix.

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So how can we now construct a matrix with both of diagonals filled?

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Well, we can just write like this and then here we write minus one.

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And you see, since we are adding these two matrices, we now have such a matrix exactly like here.

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Just the terms are wrong at the moment.

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So this we have to take care of.

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So let us now begin to really construct the Hamiltonian and then loop also over the key points.

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So at this point, we know technically how we can construct such a matrix.

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But we first need to define the elements here which you have written down analytically.

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So my idea is to construct an array which contains all of these values here.

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So better to say these values here and also these values and then constructs the matrix using this type

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of commands and.

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The nice thing is we only have to find two of these helpings.

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So, for example, this one and this one.

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And then we can use the Python command and pitot tile because tile is a really nice thing.

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You can then provide an array.

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And, for example, can write one comma two.

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And when you then write as the option, some number of 10, for example, then the result will be one,

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Komatsu repeated 10 times.

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This is exactly what we want to have.

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So we're not exactly, of course, but what we want to have is we want to repeat it several times.

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So first of all, we should define how many atoms do we want to consider at the moment, I would say

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32, because then it corresponds to this narrow ribbon, but we can increase the number later on.

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And here we write that we repeat this, of course, atoms divided by two times.

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So I write here the integer division just to make sure that this is really an integer.

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And now we must provide basically the correct toppings.

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So, you know, we can just do this and override toppings down here.

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So we just have to look, what did I write down here?

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So we wrote and he thought he exp than the imaginary unit, which is one j in Python a zero.

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And then this vector, or no, it's not a vector, does the dot product of two vectors, so it will

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be empty Times Square root of three over two times K X plus one half K.

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And then we have another term, which is pretty similar, actually just has a different X sign for the

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X coordinate.

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So like this, I think and then the other term will be this one here.

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So this will just be we copy this a zero times K-Y.

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So we run this, then we get, yeah, such a nice matrix.

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I know it's not the matrix yet.

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It's a vector which we will then position in the matrix.

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So one thing that's important also to realize is that if we want to construct our 32 by 32 matrix,

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the off diagonal were only has half a length of 31.

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So this means we only have to take the first to the second last element.

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So you see, the last element has now been dropped.

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So this I would call off diagonal one.

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And then, of course, we have to do the same thing also for off diagonal, too, which is in fact pretty

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similar.

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It just has a different sign for Derby.

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Oh, by the way, here I forgot to write a minus sign in the latest court.

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Yeah, no, it's correct.

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So what we have to do is we have to just make here at minus sign, minus sign and here also a minus

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sign, and then it should work already.

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All right.

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And now we can think about how did we calculate the barn structure before and we can copy this.

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So let me copy all of this code from this pen structure here, because we will now do a similar thing.

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So

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we will first define a number of key points.

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We should probably go to a bit of a lower value here than we defined the list, the K X list, which,

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you know, maybe we could even leaf as it is.

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And then for the K y list here, I have to just tell you something.

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So of course, it would be a good idea to do it as it is.

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So we have a list and the K list, but it turns out that's the the solution will not depend on.

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Quite.

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So that's really hard to understand, to be honest, but it's basically due to the fact that our system

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is only periodic along one direction along the X direction.

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So this is also why, yeah, this solution will not depend on K-Y.

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So we can basically take any value for KYC, and I will just take zero.

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So if you don't believe me, by the way, you can of course check this.

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You can take a different value K-Y, and it should give you exactly the same result in every case.

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So then we have our empty energy list for the solution, which is good.

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This loop we can delete and now we can just copy and paste our off diagonals here into that loop, because

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the off diagonals will, of course, change every time.

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So just make this look a bit more nicely.

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And yeah, I think that's pretty good.

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Yeah.

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Maybe let's do it like this, it's easier.

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OK, so here we have to find the

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the off diagonal elements.

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And here we will define the Hamilton matrix, where it just depends on K X.

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So basically, what we have to right now is not something like this, but basically this going on here.

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But of course, with the updated lists, which will be off diagonal one and off diagonal to.

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All right, so now we have our Hamilton matrix for every key point.

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And now we solve for the energy.

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So let me think if this is correct, but it looks good, to be honest.

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So I think we can give it a shot.

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So I save the file and I run the cell, and that's already a good sign when it didn't give any error

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message.

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So let's look at the energy list.

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So it's, of course, a very large array where we loop over the key points.

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So let's, for example, look at the first key point.

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And there we will have several eigenvalues that corresponds to the energies.

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So now the next thing that we are going to do is we are going to plot this.

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So the first thing is we must reshape R8.

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If you don't remember, we had to do this also for the other band structure.

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This was basically because I have constructed here a list that we have then filled, but we must have

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an array, so we must update our list to an array and then we must reshape this thing to make it look

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more like.

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Yeah.

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So it took basically to satisfy the syntax for the plotting commands.

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So let me copy this and we paste it here.

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So our new energy list will be read shape and then we have to write here.

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And three, the energy list and the new shape will be key points comma.

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So we don't need two times two key points here because we only loop over K X and here we need the number

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of bands or the number of atoms.

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So let's update the list and let's plot.

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And this is pretty easy.

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Actually, we just right Pulte towards plot and then K X list, coma energy list.

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And for example, we could write all the, you know, all the key points and we only want to plot here

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to first band and color.

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Let's do it like.

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So this will be the first band, of course.

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We can also do another band, then 14, for example, which will look differently.

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So how can we plot all the bands at once?

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Well, we could, for example, just loop over the index for I in range of atoms.

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We plot this and here I wrote I.

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And you see, this is our new band structure where the x axis will show the K value.

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So in this case, k x, let's just use K because we only have one K value and then the y axis will be

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the energy as previously.

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And now we see our band structure here, so it looks kind of different compared to this one.

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But still, you see a bit of a similarity.

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You still see these this valley here.

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You still also see the touching point where to direct point is.

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So it looks a bit similar to this plot here.

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But the thing is, since we only have one quick turn now, everything from the other direction is basically

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projected onto K X equals zero.

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So, yeah, that sounds maybe a bit difficult.

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But all I want to say is this three dimensional plot here gets basically projected onto one plane,

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and then it will look exactly like this one here.

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So this will be the band structure of another ribbon of graphene.

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Now, of course, we can play with the parametres a bit.

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We can increase the number of key points, for example 201, and we can increase the number of atoms

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where it's really important to have higher multiples of four.

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So maybe like me to do 180 because we have to run all the cells now and you see we get even more bands,

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but we basically still get the same band structure and we can even continue this go to 200.

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And you see, now we have an almost solid area here where all the states are occupied by electrons and

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then only in the wide ranges, there are no electrons.

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And this is really, I think, an interesting fact that no matter how thick your nano stripe is or you're

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not a ribbon, you will always give a very similar bend structure.

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But the density of the states here in these areas, they will be much more homogeneous.

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The more bands you considered, the more atoms you consider.