With previous lectures, we have considered a finite lattice of graphene, a so-called nano ribbon.
And this is the corresponding band structure which gives you the relation of the energy of an electron
with respect to the momentum.
And we have seen that this pen structure is essentially the same as the other band structure from before
of the infinitely extended graphene.
It's just projected on a single plane.
So the relation of energy to the momentum of the electrons is pretty similar in this case.
But now we are going to apply a magnetic field perpendicular to the graphene, and we will see how this
changes the band structure and the relation energy of K.
So maybe a bit of background about this.
So if you apply a very small magnetic field, then the band structure should not really change that
much.
And the only thing that will happen then in an experiment is that you will see that electrons that move
through the graphene will be deflected along the perpendicular direction due to the Lorentz force.
This effect is called the hall effect.
And what happens if you increase the magnetic field to very large values of maybe a few?
Tesla even is that the band structure will drastically change.
All of a sudden, this is called Landau Quantization.
We will see that here.
We will find several very flat bands and very flat energy levels.
And also in the experiments, they observe a quantized version of the hall effect, where all of this
all of a sudden, the number of deflected electrons will increase in steps.
So before I lose too many words about this, let us program and let us see what is going to happen here.
So here we need even more physics backgrounds and to be honest, it's a bit difficult here.
I just give you the essentials and then we are trying to program this.
So what's happening here is we are applying a magnetic field along the Z direction, so out of plane.
And maybe you know this from electrodynamics that we have to introduce now a vector potential which
has fulfilled this relation to curl or the rotation of a vector potential must be equal to B.
And so this is like some indirect definition of A.
And there exist several choices for a which all give the same magnetic field and we really have to freedom
to select one of these choices.
And I have considered Tier two following choice minus B times y times the unit vector along the X direction.
So this vector field only has an X component, and it has the value minus B times Y.
And if you now calculate to curl and you will see that the magnetic field will be exactly zero zero
B. So just a magnetic field along the z direction.
Now how do we implement the magnetic field or the vector potential in the Hamilton matrix?
This is done by so-called parallel substitution, but not really that important.
The only thing that's going to happen is that we will get an additional factor for all the hopping terms.
So additionally, to such terms here, like exponential function and then the hopping, we will get
another factor which will have to shape exponential function times I.
And then a line integral where we have to scale product of a.
And here this will be along the direction of the hopping path.
And this is a mathematical exercise that you may want to do or may not want to do.
It's not really that important for this cause.
When you solve this, you will see that for all of our toppings, the solution will be exponential function
minus i.e. times Delta X times b times the average value of Y of the two atoms that correspond to the
path.
So basically, if we look at a hopping path, for example, here from purple to blue, then we have
a certain Delta X and we have an average value of the Y coordinates.
And this determines this additional factor.
Also, we can see that are hopping along the y direction, for example, from blue to red, also from
green to purple will not have an additional face factor because the Delta X, where is it here?
The Delta X is zero in this case.
This is because we have chosen our potential to be along the X direction and then it is perpendicular
to the Y direction, of course.
So this dot product here will be zero for these particular topics.
So let's go ahead and implement this in our Hamilton matrix.
So here I went to the top of our notebook here and just copied these three cells because we will want
to calculate exactly the same thing.
So you see, it gives the same results and we will just want to modify the hopping terms.
So basically these off diagonal vectors here, that's the only thing we have to modify.
So we have to multiply a face factor to half of the toppings.
So all the toppings that have an X component or an X direction will have a modified topping.
And the other ones which only go along y or minus Y, they will not have a modified hopping.
And of course, as always, there are many ways how you can do this, how you can implement this.
And I found it to be most easily when we divide these off diagonal this into three sub lists corresponding
to the three individual toppings.
So this one, that one and this one, because then we can address all three toppings individually.
So let me do this.
I will just write that the first off diagonal will be called off diagonal one eight and the command
will be empty tile, then the array.
And let me just make this all in one line so that we save a bit of space here.
So we will then have just the first topping and then that's it.
So we will to lead this one, though, after just be careful with the brackets.
Um, let me see.
I think this is not correct at the moment.
Yes, so I have to close the array first.
And then this one's for empty tile.
Yes.
OK.
So I think now is correct.
And then the last thing we can do for of diagonal 1b, which will be the other hopping, which will
just have a modified X component.
And of course, we can do the same thing for of diagnosi, which will just be eight times eight times
k y.
So like this?
Yes.
OK.
And the same thing.
We're also going to do for the other things which have just a different sign in the exponent.
So let me copy this to two two, and that's change to sign here for the exponents.
Now what we only have to do is we have to multiply here the modified hopping.
So of course, only for those that have an X components or not for this one, but for this one and for
this one.
And for this, I will introduce an array of the modified hopping as well a remote, for example.
Right.
And here also Ara mods, right?
And here our mod left and our mods left.
And now we only must define what are these arrays.
So Ara modified right equals and pre modified left equals.
And the good thing is, when you look here, these these arrays, the modified hopping, they do not
depend on art.
So we are sorry.
They do not depend on K.
So we do not have to put them into the loop.
We only have to calculate the ones in the beginning.
And essentially they are just what is written down here.
So let me start, it's, of course, in the narrator describing the hopping for all the individual hopping
paths.
So for example, this one, this one, this one, this one.
But of course, subdivided, divided by left and right direction.
So for a right direction, hopping this term will always be positive.
And this one will just be the average value of the Y coordinates.
So basically, what we have to consider here is this hopping, this hopping, this hopping, this hopping.
And when you take the average value of the Y coordinates, you will see that these points will be equally
spaced.
And this is what we can use now to program this in a very simple way.
We are just writing this in and p dot exponential.
And it's yeah, it's difficult times, one j times and P as Q R T U three L for two
times zero.
And so this basically we have to define first.
So the fields is equal to some number.
Let's I don't know, maybe let's use zero point one.
And so this is here to be feel.
This is the I.
And then we have square root three over two times a zero, which is, I think, the distance and the
X direction.
And then we must also provide an index I times and pitot square root three over two times letters length,
which is basically describing this average value of Y for the different hopping paths.
So the AI index will describe basically all the different toppings.
So this one, this one, this one, this one, this one, and so on.
OK, so now of course, we have to once again define what the value of AI is.
So I think I have to close the exponential function first.
Let's see.
No, I think, yeah, this one was incorrect.
Yeah, now I have to write for I in range zero comma atoms, and now the only thing that we must modify
is that only every second topping and every second atom will have a hopping to the right.
So it will actually not be every I, but it will be every second I.
Where is it?
Here it is.
So what we can write down instead is.
Like this?
And once again, since it's the off diagonal, we just go from the first to the second last element.
So maybe this is a bit complicated here, but I think if you think about it very carefully, you will
manage.
But maybe it's a good idea.
If you didn't understand to just post a video, look at the figure once again and think about how can
we express these modified hopping factors here for the different topics?
So now if we look at the left hopping, then we will have essentially the same thing just with a different
sign.
So let me add a different sign here.
So no, we almost at the end we have defined the race for the modified hopping along the right and along
the left.
And basically, what we have done here with this trick is we have basically modified is hoping for all
the different toppings.
But here's the important thing is that for these terms, we use this and Tyler commands where actually
I just realize now that I deleted accidentally too much.
So I forgot.
Um, let me see and pray.
Yeah, I forgot the zero here.
I deleted this one because this is the important point.
Every second term in this rate is zero.
So it doesn't matter if we multiply this one here.
So this is a bit of a trick to make it a bit more easy.
So let's restore the zero here that I deleted accidentally.
And I think also here.
Yeah, exactly.
And here, here and here, this is exactly as we did before.
I just accidentally deleted it when I updated these terms.
So now we have our off diagonal terms, and this means we only have to update the Hamiltonian.
So we right here one a
one b one C.
And the same, of course, four of diagonal to a b c.
And I hope that now everything should work because we are then solving and adding it to the energy list
and we reshape and let's have a look at the plots.
Yes, that looks really good.
So yeah, it seems that the beef was already quite large.
Let me go to a very small value here.
You run that again because first we have to check that for very small beef you values, we can restore
the old result and actually change this one.
You also to 160 so that we can compare it better.
So this is the original result for 160 atoms and zero magnetic fields.
And when we have just a small magnetic fields, then we will get a very similar results.
So now let's see what happens if we increase the fields.
So here only very small changes visible.
So this would not really, I would say, would not correspond to any land or quantization, and no quantum
hall effect would be measurable.
But you see already a bit of a split up here.
So now if we keep on increasing the value, then you see, oh no, it gets really interesting.
So this blog area here totally the forms and especially in the middle, you see very flat levels and
these are these so-called Landau levels to form when you apply large magnetic fields and that give rise
to the quantum hall effect.
So now let's go back to our value of zero point one.
And here you see, the quantization is enormously.
So you have these totally flat levels here, here, here, here and then in between, you have a few
states that bridge these gaps.
And these are in fact the so-called edge states that give rise to this quantum hall effect.
So what we can say is, and I wrote this down here and the headline that we have a quantum hall effect,
which basically means we have a metallic behavior due to etched channels.
So it turns out that if you would only have these flat Landau levels here, our sample would be an insulator
because there are these gaps and the electrons cannot bridge these gaps.
So a current cannot really flow in the sample.
And in fact, this is really what's happening in the inner part of the sample if we apply a large magnetic
field.
However, in the arch layers so pretty close to to these atoms here at the bottom end to top the electron
behavior changes, and they will form such edge states that bridge the gaps and they will give rise
to this very interesting conductivity phenomenon called quantum hall effect.
And in fact, this phenomenon has to be analyzed, of course, in much more detail to really be able
to understand it.
But I think even this lecture was very difficult because I had to introduce these modified toppings,
and it was quite difficult to program these.
And so I would say we stop at this point with having the band structure, with the Landau levels and
with the edge states bridging the gaps.
So if we would now continue to analyze these edge states in terms of their so-called topological properties,
we would at some point recover the results of these very clever men here, which got the Nobel Prize
in 2016 for their theoretical discoveries of the topological phase transitions and topical topological
phases of matter.
So they did not only do this for graphene, they did this in general for the quantum hall effect and
also for other systems.
But here we have basically combined the quantum hall effect and graphene.