So far, we have considered the best structure of graphene, and we've considered graphene to be a periodic
lattice, which extends to infinity.
So it is not really like in this figure here.
Well, let us and at some point, but we are really thinking that every rat atom has three blue neighbors
and every blue atom has three rat neighbors.
Also, these at the edge here.
So the letters continues in all directions to infinity.
However, we know that this is not the reality, of course.
So in reality, we have a finite sample and we have edges of these samples.
And so we are going to look at how this band structure.
Also, this one here, how they change when we consider an actual finite sample, and this is called
considering so-called nano ribbons of graphene.
And so what we're going to do first is we're going to construct the lattice.
We're going to construct another ribbon.
So what we're doing here is we are basically cutting the infinitely large piece of graphene at the,
you know, it cuts parallel to the x axis.
So for example, here at the top and here at the bottom and then along the negative and positive x direction.
So here and here we are still considering it to be periodic and infinite.
So by doing so, we can actually consider an edge here and we kind of look at what happens at these
edges.
So the first thing that we are going to do is we are going to recreate this figure and here's how this
works.
So when you look at the figure and you think about that, this figure continues periodically to the
left and to the right.
Then once again, we have to consider our unit.
So previously we had only two basis atoms.
These were, for example, this purple and this green one here, because we could then repeat the setup
of two atoms and could construct the whole graphene letters from this.
However, now since our letters ends here and here, this is not true anymore.
And the reality is, since no atoms can be above here and below here, let us sell our D.
You know, the units, all of the letters must be really, really large and must contain many atoms.
So in fact, a unit cell contains a whole cut of the graphene from top to bottom.
And then we have to look OK.
What are then the atoms that we need for the unit so well, we always need these two pairs of atoms.
So maybe it's easiest to show you here on the left.
So we have this green, red, blue, purple, green, red, blue, purple and so on and so on until
we are at the bottom green, red, blue, purple.
So this whole set of atoms will be our unit cell, and this will be the first thing that we are going
to construct here.
And you see that for constructing the unit.
So we have four different atoms that we can then periodically to place to construct the whole unit itself.
So we're going to do this.
I will write coordinates eight, which will describe, I think, the purple ones we write and pitot
array.
And then we do it in a bit of a similar way, as previously we write all the times, letters, lengths
and then times and pitot array.
So this will be the vector, which we are just placing this later on.
And this will be zero come up and keep the square root of three.
And now we must specify what are the possible values for?
I sort of writing for I in the list, which is given by range minus and Max two and Max.
So if I run this, there is a problem because I have reloaded the notebook, so Nampai is not imported
yet.
So let me run the whole notebook, then it should work.
So now it's calculating all the things that we calculated before takes, of course, a bit of time because
we are constructing once again the structure of graphene, which was a bit heavy on the memory.
So here it is.
And the cuts and then here we are finished, and now the error message is gone.
So let's look at the letters that we have here.
OK, so now we have seven ATM positions.
And I think this corresponds to these seven purple atoms that are on this line here.
So the next thing that we're going to do is we are going to add the other atoms and we are beginning
with the next one.
So I will write basically the same thing and now we just have to shift from one atom to another.
So maybe it's more clever to first concept of blue and red, actually.
So from going from blue to red, we have to add here another vector, which is and pitot array and then
zero comma a.
And this should be sufficient.
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
x one comma y one as equal to chords he dots transpose.
So this is pretty similar to what we did before then we can do the same thing for the second set of
coordinates and then we can just plop this Pulte daughter's guitar.
Yeah.
And as I said, it's probably easier to start with red and blue and not with purple, as I first said.
So let's make this blue, and let's make this one red and change the index.
So we are now having the blue and the red ones from here.
And of course, now we need the other ones.
So let's continue.
I would say, yeah, let's copy this one.
Coordinates see, it's the same thing.
Now we have to, of course, add another factor.
So by the way, the letters length times, zero comma square root of three would be the distance from
here to here.
And then the other vector and p zero eight zero would be the vector from here to here.
So now we have to come up with the next set of atoms, and now I want to go for the green atoms.
So we have to add to your vector from here to here, of course.
So this will be lattice length and dan and p dot.
All right.
So this is really just geometry.
So of course, I did this before several times.
You know, when you do it the first time, then maybe this is a bit fast, but you can always pause
and think about why is this vector here?
Lattice length times one half comma square with three over two.
And now we're adding these atoms here.
And the color will be green.
Oh, yeah, forgot to chase this here.
So let's see what's the problem?
Coordinates see is not defined.
Oh, I forgot to run the self.
Okay.
Um, this looks.
Yeah, it looks wrong at the first glance, but you see here the x axis only covers a small distance,
so it's OK.
I think maybe we should change something here and right, Keelty told X.
These dots set aspects and then we can use the options equal and data limitation.
So I think once again, a typo.
Let's see what's going on.
He axes said, Oh, here.
As think, yeah, OK, now it looks much better.
And now we are only missing two purple atoms.
So yeah, let me.
So basically, I think nowhere just need to add two vectors.
And yeah, this one.
And let's see if this works already.
No, we have to, of course, go into the negative direction here and now we can at the last set of
coordinates.
So this takes a bit to build the units cell about building, then the whole nanoribbons will be very
fast because we only have to shift and this unit cells, so we almost finished.
And the color will be purple.
Oh, once again, here's a typo.
OK.
Yeah.
No, it works.
So you see, we have these four sets of coordinates which basically determine the the thickness of our
nano ribbon.
And here we have plotted all of these atoms.
So at the moment, we have four times one two three four five six seven eight atoms as the thickness
of the nano ribbon.
And now, as I said, we just have to take this plot.
I will copy this to a new cell and then we will just basically shift this multiple times along a vector.
And we do this in a clever way, of course, by using a loop for Iron Range.
And you know, we have to specify how often do we want to shift it?
Of course, in reality would be infinite times.
But let's use some number here 12, for example, and then we just have to shift the X coordinate.
And this will be eight times the lattice length.
And of course, we have to do this for all four sets of coordinates.
And here I got this.
And let me at the top here as well.
And now it works.
So we just have looped here over an index and have shifted the unit cell, and now we have reproduced
the figure of the NaNoWriMo.
And in the next lecture, we are going to define the Hamiltonian, which is a bit difficult.
You see here I wrote down already some math.
And then we are going to solve for the eigenvalues of this Hamiltonian to come up with the new band
structure.