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So far, we have considered the best structure of graphene, and we've considered graphene to be a periodic

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lattice, which extends to infinity.

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So it is not really like in this figure here.

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Well, let us and at some point, but we are really thinking that every rat atom has three blue neighbors

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and every blue atom has three rat neighbors.

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Also, these at the edge here.

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So the letters continues in all directions to infinity.

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However, we know that this is not the reality, of course.

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So in reality, we have a finite sample and we have edges of these samples.

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And so we are going to look at how this band structure.

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Also, this one here, how they change when we consider an actual finite sample, and this is called

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considering so-called nano ribbons of graphene.

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And so what we're going to do first is we're going to construct the lattice.

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We're going to construct another ribbon.

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So what we're doing here is we are basically cutting the infinitely large piece of graphene at the,

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you know, it cuts parallel to the x axis.

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So for example, here at the top and here at the bottom and then along the negative and positive x direction.

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So here and here we are still considering it to be periodic and infinite.

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So by doing so, we can actually consider an edge here and we kind of look at what happens at these

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edges.

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So the first thing that we are going to do is we are going to recreate this figure and here's how this

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works.

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So when you look at the figure and you think about that, this figure continues periodically to the

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left and to the right.

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Then once again, we have to consider our unit.

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So previously we had only two basis atoms.

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These were, for example, this purple and this green one here, because we could then repeat the setup

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of two atoms and could construct the whole graphene letters from this.

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However, now since our letters ends here and here, this is not true anymore.

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And the reality is, since no atoms can be above here and below here, let us sell our D.

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You know, the units, all of the letters must be really, really large and must contain many atoms.

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So in fact, a unit cell contains a whole cut of the graphene from top to bottom.

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And then we have to look OK.

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What are then the atoms that we need for the unit so well, we always need these two pairs of atoms.

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So maybe it's easiest to show you here on the left.

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So we have this green, red, blue, purple, green, red, blue, purple and so on and so on until

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we are at the bottom green, red, blue, purple.

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So this whole set of atoms will be our unit cell, and this will be the first thing that we are going

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to construct here.

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And you see that for constructing the unit.

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So we have four different atoms that we can then periodically to place to construct the whole unit itself.

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So we're going to do this.

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I will write coordinates eight, which will describe, I think, the purple ones we write and pitot

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array.

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And then we do it in a bit of a similar way, as previously we write all the times, letters, lengths

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and then times and pitot array.

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So this will be the vector, which we are just placing this later on.

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And this will be zero come up and keep the square root of three.

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And now we must specify what are the possible values for?

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I sort of writing for I in the list, which is given by range minus and Max two and Max.

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So if I run this, there is a problem because I have reloaded the notebook, so Nampai is not imported

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yet.

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So let me run the whole notebook, then it should work.

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So now it's calculating all the things that we calculated before takes, of course, a bit of time because

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we are constructing once again the structure of graphene, which was a bit heavy on the memory.

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So here it is.

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And the cuts and then here we are finished, and now the error message is gone.

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So let's look at the letters that we have here.

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OK, so now we have seven ATM positions.

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And I think this corresponds to these seven purple atoms that are on this line here.

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So the next thing that we're going to do is we are going to add the other atoms and we are beginning

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with the next one.

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So I will write basically the same thing and now we just have to shift from one atom to another.

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So maybe it's more clever to first concept of blue and red, actually.

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So from going from blue to red, we have to add here another vector, which is and pitot array and then

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zero comma a.

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And this should be sufficient.

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By the way, we can already have a look at this, how we're how it's how it's going so we can just write

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x one comma y one as equal to chords he dots transpose.

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So this is pretty similar to what we did before then we can do the same thing for the second set of

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coordinates and then we can just plop this Pulte daughter's guitar.

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Yeah.

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And as I said, it's probably easier to start with red and blue and not with purple, as I first said.

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So let's make this blue, and let's make this one red and change the index.

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So we are now having the blue and the red ones from here.

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And of course, now we need the other ones.

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So let's continue.

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I would say, yeah, let's copy this one.

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Coordinates see, it's the same thing.

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Now we have to, of course, add another factor.

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So by the way, the letters length times, zero comma square root of three would be the distance from

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here to here.

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And then the other vector and p zero eight zero would be the vector from here to here.

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So now we have to come up with the next set of atoms, and now I want to go for the green atoms.

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So we have to add to your vector from here to here, of course.

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So this will be lattice length and dan and p dot.

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All right.

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So this is really just geometry.

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So of course, I did this before several times.

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You know, when you do it the first time, then maybe this is a bit fast, but you can always pause

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and think about why is this vector here?

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Lattice length times one half comma square with three over two.

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And now we're adding these atoms here.

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And the color will be green.

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Oh, yeah, forgot to chase this here.

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So let's see what's the problem?

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Coordinates see is not defined.

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Oh, I forgot to run the self.

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Okay.

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Um, this looks.

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Yeah, it looks wrong at the first glance, but you see here the x axis only covers a small distance,

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so it's OK.

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I think maybe we should change something here and right, Keelty told X.

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These dots set aspects and then we can use the options equal and data limitation.

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So I think once again, a typo.

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Let's see what's going on.

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He axes said, Oh, here.

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As think, yeah, OK, now it looks much better.

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And now we are only missing two purple atoms.

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So yeah, let me.

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So basically, I think nowhere just need to add two vectors.

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And yeah, this one.

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And let's see if this works already.

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No, we have to, of course, go into the negative direction here and now we can at the last set of

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coordinates.

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So this takes a bit to build the units cell about building, then the whole nanoribbons will be very

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fast because we only have to shift and this unit cells, so we almost finished.

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And the color will be purple.

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Oh, once again, here's a typo.

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OK.

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Yeah.

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No, it works.

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So you see, we have these four sets of coordinates which basically determine the the thickness of our

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nano ribbon.

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And here we have plotted all of these atoms.

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So at the moment, we have four times one two three four five six seven eight atoms as the thickness

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of the nano ribbon.

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And now, as I said, we just have to take this plot.

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I will copy this to a new cell and then we will just basically shift this multiple times along a vector.

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And we do this in a clever way, of course, by using a loop for Iron Range.

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And you know, we have to specify how often do we want to shift it?

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Of course, in reality would be infinite times.

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But let's use some number here 12, for example, and then we just have to shift the X coordinate.

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And this will be eight times the lattice length.

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And of course, we have to do this for all four sets of coordinates.

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And here I got this.

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And let me at the top here as well.

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And now it works.

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So we just have looped here over an index and have shifted the unit cell, and now we have reproduced

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the figure of the NaNoWriMo.

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And in the next lecture, we are going to define the Hamiltonian, which is a bit difficult.

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You see here I wrote down already some math.

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And then we are going to solve for the eigenvalues of this Hamiltonian to come up with the new band

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structure.