All right. I guess we can get started. Welcome back. It's the 15th lecture of the condensed matter, of course.
When last we left off, we were talking about trying to approximately solve the shortage equation for electrons in solids.
In the method we were talking about using was the nearly free electron model, nearly free electrons.
The general idea being that we're going to write our Hamiltonian as some bear,
just regular non interacting electron plane wave Hamiltonian plus some potential.
So the eigen states of this in a plane wave k h r squared k squared over two m just your regular electron Hamiltonian.
But then here we have some periodic potential which we're going to eventually consider to be weak, hence the name nearly free weak potential.
And the idea that we took from scattering theory is that the matrix element between some K and K prime from this periodic potential has to be zero.
If K minus K prime is not an element of the reciprocal lattice and we'll call it this of g, f, k, prime minus k is an element of reciprocal lattice.
So then we treated the the potential V as a weak perturbation and in second order perturbation theory, second order third.
We found that the energy of the plane way of K starts out as the bear energy
of the plane with k plus v sub zero zero for m mode of the potential lattice.
An overall shift of all of the plane wave energies that's uninteresting.
People sometimes just drop it all together because it's just a constant. But the interesting pieces are some overall reciprocal lattice vectors.
These are B squared over energy of K minus energy, not of K plus g.
So this is our second order perturbation theory expression.
And the thing that one has to worry about is that the denominator can be either zero or close to zero and that happens when.
So this diverges when we have K absolute K being equal to absolute k plus g so that the energy of can and
plus g are the same and that is equal to the same condition as K and K plus g being on on busy boundary,
boron zone boundary. So in that case, we have a problem.
Our second order perturbation theory doesn't work because of this degeneracy.
And what we're supposed to do is we're supposed to use degenerate perturbation theory of degenerate theory,
resisting all urges to make jokes about the word degenerate, but probably haven't made hundreds of times before.
The general rules of degenerate perturbation theory are that you should take the states that are causing you trouble.
The states that are giving you the divergence separate those out from all the other states in the
system and just diagnose the Hamiltonian with this within the space of those problematic states.
So for us, our the states that are causing us trouble are a states K and K plus G for some particular G.
So we'll write our trial wave function, actually. Maybe we'll be a little bit more general and assume, assume, k and K plus g assume.
These are near a zone boundary near a boundary.
So it may not be an actual divergence, it can just be a near divergence.
It's still bad enough to cause us to not trust our second order perturbation theory anymore and require us to use degenerate perturbation theory,
even though if it's not a you know, it's not a zero in the denominator to something close to zero in the denominator.
The general scheme is to write a trial wave function, which is going to be some some of the problematic cats.
So K and plus G may take some linear, some of these too.
And in the usual way, we will try to find the best linear combination of these two cats in order to get our lowest energy state within this space.
So we've done this a couple of times before.
The way one does that is you write down a Hamiltonian, an effective Hamiltonian equation sum over age and Phi Alpha equals E Phi.
And this should look familiar from what we've did when we solved the tight binding model for the covalent for the covalent bond here.
This H&M is a two by two matrix where an an m are chosen from K and K plus G.
This is how you how you do the general perturbation theory. You just restrict.
The space of states to the space of degenerates, states who are causing trouble in the second order perturbation theory.
All right. So the only thing we need to do now in order to follow this prescription is we need to write out the matrix elements.
So, for example, we have K, H, K, that's one of the four matrix elements we can write that is k,
h, not K plus k, v, k, since our Hamiltonian is made up of the H, not part in the V part.
This part is easy. This is just E not of K and this part is V, not just the overall constant energy shift.
The second subscript is zero because k minus K prime is zero.
Similarly plus g, h k plus g is equal to E, not of k plus g plus v not again.
The more interesting term is the term the off diagonal terms of the two by two matrix k plus g, h, k.
So let's write out the pieces of this. We have first K plus g, h, not K, and the second piece is K plus g, v k.
Well, the first part, this is easy because this would be H, not K is in Oregon, State of H not so we can just pull out an E,
not sub K and get K plus g k, but this is zero because these are two orthogonal plane waves.
So at zero the first term is zero. The second term is we've defined that as being the sub g,
so we're allowed to scatter from K to K plus G via the scattering potential G and up way up at the top.
Oops I scrolled it off the top. Yeah.
There it is on the top that the sub g of came on is k prime is an element of g, so we call this thing the G.
So so what we have now is also so this thing is going to be overall this is going to be the sub g and we can do the other last matrix element h,
k plus g, which using the same argument is the sub minus G or the sub g star.
It's complex conjugate of its of its partner up here. So now we have our Hamiltonian R two by two matrix we can write down which is e not of k plus
v not v sub g here star here v sub g down here and e not tables g plus v not so far so good.
Everyone happy with this? More or less this go by too quickly.
Yeah. Happy. Okay, good. All right.
So for homework problems at four, I think you're going to actually solve this in general for a K near the Bronson on boundary,
which makes the algebra a little bit more complicated.
But what we're going to do here is we're going to simplify our algebra a little bit and assume K is right on the brown zone boundary.
Assume K and K plus G are on busy boundary, on busy boundary just to make the algebra a little bit easier.
So that means that E not of K equals E not of k plus g at the same energy.
So right on the degeneracy point, in which case this is a very simple matrix,
has two elements which are the same along the diagonal and then complex conjugate matrices,
elements on the off diagonal and the energies are then e not of k plus v not plus or minus absolute v.
And that's her answer. So what is this? What does this mean?
So maybe draw it here. So we have here's K, here's energy E and before we add the perturbation, we have a nice parabola.
We have the usual. Drop this. So this is the usual parabola which is e not of k equals h where squared k squared over two m.
So that's the free electron parabola. Then we can put down the periodic potential.
It has some r on boundary for this.
So it has a lattice constant,
a spiralling zone boundaries at pi array and minus pi over six at more browns on boundaries at minus two pi over a and two pi over a here.
So here at this point there's bronze on boundary minus pi over a and pi over.
You can scatter between them by this reciprocal lattice vector g equals to pi over a can get you from minus primary to pi over a.
So in second order of perturbation theory, you would discover that if you if you analyse this point here,
at this point here, you would get a divergence in second order perturbation theory.
And we would have to use degenerate perturbation theory in order to figure out what happens to these points.
And what we just derived is, in fact,
they split in energy in the way it's going to look is that this is going to shift down and then this will shift up a little bit so that,
in fact, you open up a little gap between the things coming in from the left and the things coming in from right.
And its size is too absolute big here.
And the same thing happens over on this side like this.
Opens up a gap of size to Fiji.
And the same thing would happen up here where you can scatter from bronze on bounded to bronze on boundary by g equals four pi
over a and here a gap would open up at the bronze on boundary here and the gap would open up with the bronze on boundary here.
Okay. So since I've drawn that fairly badly, I was prepared for this and I have a slide.
So here's the same picture and extended zones scheme. This is the it starts with the bear parabola, the the nice free electron parabola.
And then when you add the periodic potential, the major effect of the periodic potential is near the bronze on boundary where it opens up this gap.
You notice that the states inside the bronze zone boundary get pushed down in energy by
a little bit and they get pushed down more if they're closer to the bronze on boundary.
Because as you're close to the bronze and boundary, if you think in second order perturbation theory,
that's where the divergence is getting worse and worse and worse. It doesn't actually diverge.
But still, the perturbation is becoming more and more important as you get closer to the bronze on boundary.
So the gap on things get pushed down more as you get close to the bronze and boundary.
And then if you're just outside the bronze zone boundary, things get pushed up and you open up this little gap at the bronze on boundary.
Same thing over here. Now we're perfectly entitled to view the same thing in reduced zone scheme where we take this piece of the
spectrum and we shifted over by two pi over A So it's over here and we take this piece of this fixed spectrum.
We shifted over by two pi over eight in the other direction until it's over here and we view everything within the first bronze zone.
But now we have two bands within the first bronze zone. So we have a lowest band and we have a higher band.
And this looks a little bit like what we found when we studied the tight binding model a couple weeks back.
We had a four when we had two orbitals per unit cell.
There were two bands at two values of of two iron states at each possible value of K, which is what we have here.
And the structure looks more or less similar. We have a low band, a high band, a gap opening at the bronze zone boundary.
However, we're really coming at this from a very complimentary direction.
When we did a tight binding model, we started by taking a bunch of orbitals and then we weakly coupled them
together and allowed the electrons to hop from one orbital to the next to next.
Here we did the opposite. We started with a plane wave and we weakly perturbed it with a periodic potential.
But either way you look at it, you get the same physics of separate bands gaps at the Baron's own boundary.
If you look at it from the nearly free electron model, though, really this picture you can see in this picture,
it's it's actually just a free electron parabola with small perturbations at at the bronze on boundary.
So this parabola is sort of continued in the second band. So this this piece of the parabolas are reflected back and continue in the second band.
They don't quite touch because of the the gap opening up, but the curvature of this band near the bottom is actually given by the free electron mass.
So the thing that causes the curvature at the bottom of your parabola that's dependent on the free electron mass, not on anything else.
So we really should view this this bottom of the band, which looks like a parabola as just being essentially free electrons.
Okay, so a question that's fairly important at this point is maybe I'll put it over here is why the gap?
Why the gap? Mind the gap.
Why the gap? Okay. So in order to understand why it is that this gap is opening up at the bronze on boundary in a different language,
it's useful to, to, to give an example.
So examine a particular potential.
I'll choose V of X and we're going to do this in one dimension. Actually, this is one dimensional.
So I should have mentioned that this isn't one dimension. This is more this is a more general equation which hold in any dimension.
We're going to do this example in one dimension also. So VAX will choose it to b to v twiddle cosine to pi over a times x.
That's a nice periodic potential. Maybe I'll even draw that periodic potential.
Everyone probably knows what it looks like, but I'll draw it anyway. So here's our axis x.
Maybe we'll put a here minus a out here and a cosine potential.
Looks like this. And looks like. This.
So this is V of X and the lattice constant is a, the periodicity is a,
and the reason I've chosen this particular simple cosine potential is because it has nice Fourier modes.
V sub two pi over a equals v sub minus two pi over a equals v twiddle and all other all other v sub g are equal to zero.
So it only has two four modes which are non zero and they're both given by V twiddle.
All right. So now let's think about what this is going to do.
What this can do is it can scatter by v sub g only for g is two pi over a.
So in this picture it means it can scatter from this point to this point four by g.
These are the two points in the general perturbation theory. They can mix with each other.
This point of k equals minus pi over a and k close plus pi over a.
So the scattering we have to worry about is from k equals minus pi over a and to k equals plus pi over a,
being scattered back and forth by this potential.
So it can scatter back and forth between those two way vectors, which are right on the Bruns on boundary.
So in real space, if I think about x k representations of these of these cats,
these things are either the I pi over a x and you know, the minus I pi over a x.
I'll use that notation if we then diagonals this Hamiltonian in that in that's in this space.
What we'll get is we'll get our two eigenvalues or our two eigen functions are the CY Plus
will be the I pi over a x plus even minus I pi over a x and its energy e plus is is pi
over a not plus v twiddle and the other I can state sine minus is either the i pi over x
minus the minus I pi over a x and its energy e minus is e not pi over a minus b total.
So one of these linear combinations of these of these plane waves is getting pushed up in energy.
One of them is getting pushed down in energy by V-twin.
Also, again, we have these two plane ways, one going left, one going right.
They're getting mixed together by the by this by this scattering potential, by this periodic potential.
And one of the two I can say is has been pushed up in energy.
One of them has been pushed down in energy. Why is that? Well, let's see if we can figure that out.
Let's take five plus square it so we get the probability density.
So this is the density, the problem probability density prob density cy plus squared which is proposed passing all to cosine squared of pi over a
x and sine minus squared will be proportional to science squared of pi over a x in this plot those on this same figure.
So let's see cy plus looks like this cosine squared of the cy plus squared and then cy minus looks like this.
Sy minus squared looks like that so they have their amplify their probability
density is in different places along the in the positional space they're different
places along the axis and because of this the CI plus squared CS I'm putting
that in quotes c is mostly positive v v of x so you see where CI plus is great,
where it's where it's very large. That's where V of x is positive.
And when CI plus is small is where V of x is negative.
So that means CI plus e plus is pushed up and energy is pushed up up because it's experiencing mostly the positive potential.
Whereas in comparison CI minus squared sees mostly negative v v of x because CI
minus is large where v of x happens to be negative and so it's pushed down.
E minus is pushed down down in energy.
So let's sort of think about this a little bit more carefully.
So what's going on is we have this space which has a left going wave and all right, going wave.
That's the space of the degenerate perturbation theory.
And with the variational principle, what we're supposed to do is we're supposed to put them together in a way that will,
you know, to get the ground state.
We're supposed to put them together in a way that will minimise the total energy and the way you minimise the total energy to get the ground state
as you put them together such that you get a sign minus cosine minus has its maximum where v of x is negative and that pushes down its energy.
Therefore it's in it pushes down as much as possible.
Therefore it's the ground state, eigen state and conversely, C plus has pushed up its energy as much as possible,
as high as possible by putting its density where the potential is positive.
So is that clear why that's going on? All right, good.
So do you ever read the Examiner reports people read these reports even though notice they're really nasty?
I think it's an unwritten requirement. The examining reports have to be really nasty.
I've never been on the examination committee yet.
One day my time will come and I'll probably read the rule, the unwritten rule, somewhere on some tablet that says you must be nasty.
Okay, so there was a question that came up about this nearly free electron model came up about four years ago and it really threw a lot of people.
So I'm going to go through the physics of what was involved in there, which isn't too much more complicated than what we did.
But but it's something that, you know, a lot of people found rather unexpected.
So I'm going to explain more or less what what was going on in that problem.
So when you run into it, someday we'll trouble you. So we're going to consider a square two dimensional lattice.
And I'm drawing reciprocal space. This is a reciprocal lattice. So a square 2D lattice, square 2D lattice.
And the reciprocal space of a square lattice is also square.
So I'm going to draw the bronze zone. So this is K equals zero here in the centre and the first bronze zone is this square here.
And this point here in the corner is pie. Over a pie.
Over a cake.
Now, the first thing this question asked you to do is find out what happens when you add a period of potential.
What happens to a point here on the bronze on boundary?
Okay. So that's basically exactly what we just did. That point there, due to the periodic potential, it can mix in with this point over here.
These two points are separated by a reciprocal lattice factor. This is a reciprocal lattice vector to pi over two, pi over eight.
So these two can mix together via scattering. They have the same energy because they're the same distance from k equals zero.
They have the same amplitude of their K. So.
So this condition holds x equals e, k plus g.
So they mixed together. And it's exactly the same calculation we just did.
It's just, you know, it's really a two dimensional model, but we're just considering this cut through two dimensions.
There will be a gap opening up at the bronze on boundary at this point.
At this point. So that much is simple or should be simple. I hope it's exactly like what we did in two dimensions, in one dimension.
But now in two dimensions. The thing that was complicated is what happens if you consider this point up here in the corner.
So let's call this .0.1. So now what we have to do is we have to ask where did the where did the divergences come in second order perturbation theory?
Well, this point, one can scatter by a reciprocal lattice vector to this point two,
but it can also scatter by reciprocal lattice vector to this point three.
And this is also a reciprocal lattice vector to this point four.
And in fact, four can kind of scatter the three. Working together to two can scatter to three.
They can all scatter to each other by reciprocal aspect is those vectors in case space are all reciprocal as vectors
and they're all allowed to mix to each other and they all have the same energy before you add the perturbation.
So when you do your degenerate perturbation theory, you'll find a bunch of these things, the bunch of these denominators that are all zero.
If you're considering this point scattering here, there'll be a denominator zero.
Scattering here, scattering here, all zero. So the way you handle this is you have to write your trial wave function.
That includes trial. That includes all of these wave vectors, cy one, k one plus five, two, k two plus five, three, k three plus five for k four.
All of these things have to be included in your trial wave function and then your Hamiltonian is then a four by four matrix,
which includes all possibilities of how you scatter back and forth between all of those four points.
Okay. So that was what was what was asked on the in the in the exam.
And a lot of people got thrown by it, even though it's not really that much more complicated.
Okay, good. Happy with that? All right.
Let's do it. Let's do a real example here. So in three D, in real example, real life, we have things like FCC lattices.
This is as we discussed a couple days ago, this is the branch zone of the FCC lattice.
The gamma point is appointed k equals zero in the middle.
The X point is that the barren zone boundary in the middle of this square face here in the direction the l point is in the bronze on boundary,
in the middle of this hexagonal phase, for example. And the material we're going to think about is the material.
Silicon Carbide. Silicon carbide.
So actually, it's another important industrial material. It has this structure here is FCC with a basis.
So the yellows are silicones and the and the blues are carbon.
The silicon can be taken at 000 and the carbon at one quarter, one quarter on quarter.
We've seen this structure before. This is a zinc blend structure. Zinc sulphide, gallium arsenide takes the same structure.
Silicon carbide takes this structure as a very common structure. Okay, now.
All right, so here let's see if we can we can understand how this picture,
this complicated picture of all the eigen states for electrons can come out of nearly free electron models.
So let's look down at the lowest energy, the lowest energy. Here we have the gamma point.
That's a cake with zero. Gamma is another name for cable zero.
And as you go out in the bronze zone towards the point X, you see this thing that looks kind of like a parabola.
And if you back up that oops, back up back here.
So it looks kind of like this parabola here. And that's exactly what you're seeing here.
This is a free electron parabola coming out from K equals zero.
And when you get to the bronze zone boundary, there's a gap opening up.
And this is the second half of the parabola going up as back, back, back there.
Here it is. So here's the second half of the parabola up here, just like the second half of the parabola is here.
And he could go off in another direction.
So going from Gamma out to the L point, again, you have a parabola and there's a slightly bigger gap in the L direction.
And here's the second half of the parabola here. So it's basically free electron model.
And if you measure the curvature down at the bottom of the band and your gamma equals zero, you get basically the mass of the free electron.
It's not exactly the mass of the free electron, but it's it's pretty close to the mass of the free electron within 30% or 20% or something.
So. So it really tells you that what's going on near the bottom there is basically free plane way physics just perturbed
slightly by the scattering potential that at the X point scatters you across the branch zone and opens up this gap.
Is that clear so far? All right.
So when we're on the subject of really nasty exam questions, about six years ago, there was an exam question about this.
Then not a single student got right and the examiner went berserk on that and ranted about the awful students and so forth.
But but, you know, that's just, you know, what that's required of them to do.
But actually, the question was actually quite difficult.
And I think it was I'm going to explain to you what the question was about.
It's something that if you hadn't seen it before, it would be very, very hard to figure it out.
But I guess the expectation was you were supposed to be taught it, and I don't think anyone taught it.
So now I'm going to teach it. So if it comes up, you'll all know it.
So the question was about comparing something like this to something like silicon.
So this is silicon here. It looks almost exactly like silicon carbide.
The only difference is instead of carbon at one quarter, one quarter, one quarter, you replace those carbons with another silicon.
So it's all silicon silicon in 000 in silicon and one quarter, one quarter on quarter.
We've seen this structure before, too.
This is the same structure as diamond silicon and diamond structure are are equivalent, same, same crystal structure.
And actually, if you look at the two two energy eigen energy spectra here of K across the bottom and energy up top,
they they're actually silicon and silicon carbide they look almost well, I wouldn't say identical, but extremely similar.
A lot of the features in one are there in the other as well.
So for example, here you have the same parabola,
but you'll notice there's one one sort of shocking difference here that there's no gap at at the bronze on boundary here at this point.
The X points here, the parabola is come together and they they don't they don't split at all.
And so what one is supposed to deduce is, in fact, that there's no no back scattering at the X Point at all.
For some reason in silicon, you do not get scattering at the X Point.
So it's now we're going to try to figure out why that is.
So, first of all, for for the chocolate bar,
the scattering at the X Point goes from an X Point on one side of the barren zone to the opposite side of the barren zone.
What are the Miller Indices associated with that reciprocal lattice vector?
What? Not. No, this is really good.
This 1001 is a very good guess, but it's not right.
Does anyone want another guess? No, no, no, no, no, no.
So 001 is the right direction. Or I guess it's in in the X direction or 100, if you prefer.
If you're doing the X direction,
it would be that would be pointing in the right direction because the x is in the x direction and the minus x would be in the opposite direction.
So be in the 100 direction that you're scattering or the 001 if you're going the other way.
So that's the right direction, but it's not 100. And the reason it's not 100 is because 100 is not a reciprocal lattice vector for FCC lattice.
And you'll remember back to this picture here, they remember that 100 is a reciprocal lattice for the simple cubic lattice.
If you define the planes by the 100 mil index, you get these plains here, but the SCC lattice has some additional lattice points in the middle.
So 100 is does not get all the last points. Therefore it is not a reciprocal lattice vector for the FCC.
Lattice 200 is the smallest reciprocal lattice vector in that direction.
So here is scattering from x to minus x. I'll even write that down.
So x to minus x scattering is to zero zero for silicon.
Well, actually for any two FCC lattice x two minus Xs to zero zero.
Okay, so we've gotten that far. Now, the question is, why is it that we don't get any scattering at 2009 index is anyone's guess of that.
I have one more of these left. What is it?
I can't hear who the master. The the mass.
Yeah, well, they're both silicones, so. Yeah, so. So that doesn't.
That doesn't do it for you any other. It's not an excuse.
The rest of it is. It is FCC latest FCC list with a basis.
So I'm not going to give you give it to you for that, but it's close to that.
So maybe I'd give you a half of this, but I don't know how to do that. So. So it's not PR, it's not just an FCC loudness.
And that's the point that you have to keep track of the lattice and the basis to find out if they're scattering.
So let's think back to I mean, is the FCC lattice which allows the silicon the silicon is a lattice plus of eight times a basis.
So if we think about what we learned about scattering from from x ray scattering, remember that s is s lattice times s basis, right?
And S is what gives you the amplitude of scattering, right?
If we square as you get the amplitude of scattering this work for x rays and it will work for electrons as well.
So s lattice that enforces selection rules and force selection rules and to zero zero actually satisfies the selection rule for an FCC lattice.
All even are all odd for FCC to zero zero is all even.
So we're okay with that. So that's not going to that's not going to vanish.
But let's look at s basis as basis you'll recall is some over atoms alpha in units l of e f sub alpha the form factor it is a i g that are alpha.
And if we do this for silicon, we'll get f silicon, and then it will be one for the point at 000.
And then we'll get either the two pi,
I mean the Miller indices of the scattering to zero zero times the position of the lattice point one quarter of 1.2 one quarter of that.
Does this look familiar from our x ray scattering adventures? Yeah.
Okay, so you'll notice that this thing here is actually minus one, so as a basis is zero.
So there's no back scattering in silicon in the 200 direction because you know, I'll give you I'll give this to you anyway because oops,
because I don't want to keep myself because I'll eat it and I've been gaining weight.
So okay so so you so you don't get that back scattering that in the 2005200 because there is what this is telling you is that
there is exact destructive interference between the scattering of the two silicones in the unit cell with the both scatter,
but then you get exactly destructive interference by them. And so in fact, you get no net scattering at the end of the day.
And this this physics is actually it occurs for x rays in silicon.
There's no scattering of x rays for silicon in the to zero is the direction it occurs for electrons.
There's no back scattering of the x point for electrons in silicon and also occurs for phonons.
Let's see, here's the phone on spectrum. For silicon.
This looks a whole lot like the phone on spectrum for Diamond, not coincidentally, because it's the same crystal structure.
So here's the silicon photon spectrum it has.
So we went we had this a previously there's six phonons at every K vector because
there are two atoms per primitive unit cell and each one moves in three directions.
So there's six photons one, two, three, four, five, six, and three of them are acoustic coming down here.
Now, you'll notice if you look at this point here at the X Point in this picture,
silicon carbide, a gap has opened up in this picture, a gap has not opened up.
And it's coming from essentially the same physics, even with phonon waves that in the 200 direction, there's not back scattering.
Therefore you don't have an opening of a gap,
whereas in Silicon Carbide you don't have perfect cancellation because this term would have f silicon, this term would have f carbon.
Those don't equal each other. So you would still have back scattering and you have a gap opening up at the branch on boundary.
Okay. So I've realised this is, this is a little bit complicated but it was on an exam and hopefully,
you know if it comes up on an exam again people will be able to get it right.
But I do admit that if you've never seen that before, it's something that's pretty hard to figure out.
So. All right. So having basically discussed everything that we need to know about the nearly free electron model,
we can now move on to talking about generally band theory, band structure and theory of electrons.
So these are complicated pictures here.
There's silicon carbide again with it's complicated picture of all this I can say that every every K so let's I'm.
We started talking about about band theory when we talked about the tight binding model.
Maybe let's draw a picture of some bands I like over here.
So here is energy. Here's K in this direction. So I'll draw the bronze zone.
There's a bronze zone from minus pi over a two pi over a and then you might have some bands that look like this,
and then a higher band that looks like this and maybe some higher bands up here as well.
Now, some things that we mentioned when we talked about the tight binding model is that if you have a filled band,
filled band and a gap plus gap, you have an insulator equals insulator.
Repeating this because it's fairly important. So if we fill a band, we fill, say, this band.
And a gap. So the next band, we have an insulator. And the reason this is an insulator is because there's basically no freedom of where
to put your electrons unless you overcome the gap to the next to the next band,
which could be a lot of a lot of energy required to get up to the next band.
You can't rearrange the electrons at all. They can't absorb any heat.
They can't change. They can't change their momentum because all the momentum states are already filled.
So it's basically just completely inert. So I'll write that down. It's inert.
And if you have two bands that are completely filled, again, you'll have an a gap to the next band.
It will also be an insulator. A little bit of nomenclature, which is useful is the highest fill band.
Highest filled band is usually called the Valence Band.
Valence Band. So here this would be Valence Band and the lowest empty band.
The lowest empty band.
Is known as the Conduction Band. Which would also be a really good name for a rock group.
So if anyone makes a rock group, I'm suggesting a name induction band.
Okay. So another important statement this insulator was small gap with a small gap.
Gap is known as a semiconductor is small here is sort of less than about four electron volts equals.
This is known as a semiconductor.
And the reason for this nomenclature is that if the gap is sufficiently small, then at zero temperature this would still be a really good insulator.
But at room temperature 300 Kelvin, if the gap is less than about four electron volts,
then a few electrons will still be able to get thermally excited from the filled band into the empty band.
So you'll have a couple electrons running around free,
just thermally occupying the conduction band and they will carry some amount of current that you'll,
you'll get a couple of electrons running around up here and that will allow them to carry some small amount of,
of current, some small amount of conductivity. So it's a semiconductor conducts sort of semi conducts poorly, but it conducts similarly.
It can absorb small amounts of heat, but not a lot of heat because only a very few electrons can be rearranged.
Okay, that makes sense. Yeah. Okay. So actually, so this picture of silicon carbide.
Silicon Carbide, these bands here are all filled. The Fermi energy or the chemical potential lies in the middle of this gap.
So you would say that it's an insulator or semiconductor.
And if you sort of look at the difference in energy between this point, the lowest point of the conduction band and this point here,
which is the highest band point of the Valence band, it's a little less than 40.
So we say silicon carbide is a semiconductor. Okay.
There's something else that we mentioned when we talked about tight binding model is that counting electrons
is counting electrons and states is rather important to figuring out if you have a field band or not.
If you have any unit cells in your system, then you have empty states.
K States. But then you have times to spin states.
Spin states per k parquet means that two mn electrons will fill a band will fill band.
And this to get to an electrons you need two electrons per unit cell so that you expect that if you have any,
any, even number, even number of electrons per unit cell number of electrons per unit, cell per unit.
So you would expect that you can fill can fill integer number of bands, integer number of bands.
So let's take the case of silicon carbide, for example. So silicon carbide.
Silicon Carbide. So silicon carbide both have valence four.
So silicon and carbon both. Valence for both valence for.
Valence equals four, which means we should count four electrons for each of them.
If you wanted to count all of the so why am I only counting four instead of all of their electrons?
The reason is because it's only we really generally only count the electrons in the furthest out shell,
because only the farther in cells are completely inert. They just weigh buried at much, much lower energies.
But if you wanted to count their filled shells as well, you would discover that the filled shells are all even as well.
So. So we would still have an even number of electrons both for silicon and carbon.
So the primitive unit cell here, primitive cell in this picture, it has four electrons.
Four, sorry, eight electrons. Total, total, one silicon atom, one carbon atom, each carrying four electrons.
So we should fill four bands. Four bands.
I think I asked the same question about Diamond about a week ago or more.
And in fact, if you count the number of bands here which are below the Fermi Energy,
there are four of them one, two, three, four, and they're all filled.
We do have this little bit of a confusion here that if you count here, it would look like there's only three.
But if you look more carefully, you'll notice this fact. There's two bands here which come together and have the same energy.
So this one is actually two bands which just happen to have the same energy here.
They have different energy, so you can count that there is actually four of them.
So in Silicon Carbide, all these four bands are filled and all these bands up here are empty.
Good, then? Yes. Yes. Okay, good. All right.
Now, in in contrast, we have metals have part filled bands.
Part filled bands, which is almost always true, are mostly mostly true either today or next time.
We'll discuss about why it's only mostly true. Mostly true for odd for odd number of electrons.
Odd number of electrons per unit cell, for example, if we had a picture like this.
So here's the bronze zone, here's a band, here's a band.
Suppose we have one electron per unit cell that would be enough to half fill the lowest band.
And so we have a part full band. This is one electron per unit cell part fell band here.
And we would call this thing a metal because since the band is only part filled,
we can make low energy excitations by taking an electron from the fill point to the unfilled point, arbitrarily low energy.
We can also change the momentum of these filled states by just taking some of the electrons from over here
and moving them over here at very low energy cost of just slightly shifting the entire the entire picture,
slightly shifting it to the right. Fill a couple more states over here and a few less fewer states over here.
And so as you're changing the momentum, you're actually adding current in that case.
So by changing momentum of all these electrons, you actually change the total electrical current in that system.
And so since it can carry current, we call it a metal.
However, metals can also occur. Metals also occur also occur with even number with even number of electrons per unit cell.
Also something I mentioned a couple of weeks ago when we discussed trying to model.
The way that happens is here's your boron zone. If you have a lowest band, it looks kind of like this.
And then you have a higher band that dips down like this. You and you had, say, two electrons per unit cell.
That would be just enough electrons to fill one band. But I mean, you could choose to fill this entire band if you wanted to, this entire lowest band,
but that would be energetically unfavourable compared to partially filling both of the bands.
So instead it's better to partially fill this and partially fill this instead of filling these higher energy states in in this lower band.
So that makes sense. Yes. Yes. Okay, good. So so in this case, it bands overlap.
All right. This if bands overlap overlap and it looks it looks a little unnatural when I draw it this way.
But if you look back to the silicon carbide band structure here,
you could certainly imagine your head that this minimum just dip down a little bit further and this maximum came up just a little bit further.
And if that were true, so that this mat minimum went below this Fermi Energy and this went above the Fermi energy,
then some of the electrons that was previously in this band that was filled would then go into this band which was empty to lower the energy.
Okay. So one thing that we that's really important to keep track of is how big are the gaps between bands?
So if the gaps between the bands are really huge, then you're not going to have this situation ever occur.
You know, if the bands are sufficiently far apart, then they're not going to overlap.
So how big are gaps? How big are gaps?
Big gaps between bands, gaps between bands.
Question Well, we had different ways of describing how we understand the band structure.
One way of describing it was from tight binding. In the tight binding picture.
Recall that a band, each band, each band comes from an atomic orbital is from an atomic, orbital, atomic, orbital.
With energy. With energy. With energy.
It's called Ace of I. And the bandwidth is from hopping, from hopping, hopping T.
So if the original atomic orbitals are very well spaced apart from each other and the hopping is small,
then the bands don't have a chance to overlap because initially start very far apart from each other.
And the hopping is what spreads the bands out from a single energy into a band.
So if the hopping is small and the energy spacing is large, then the bands won't overlap.
So that's one way of understanding it. That's sort of a natural way to understand why it is that there something like noble gases.
When they form crystals, they are insulators because the hopping between noble gases is essentially zero.
So these are these original topic orbitals, they're spaced a little bit,
but the hopping between that spreads out these orbitals into bands is negligible.
So they don't form very wide bands and so the bands never overlap.
So, so that's one way of understanding it. But in a nearly free model, nearly free the gaps at the Britain zone boundary at these boundary.
Our from from the periodic potential from potential vivax.
So if you know that the periodic potential is strong, you should expect gaps that are pretty big opening up at the bronze zone boundary.
Whereas if the potential is fairly weak, then you expect not to have big gaps in the bronze on boundary.
All right. I think maybe we have 2 minutes left.
I think maybe we should stop there and we'll pick up more on Bang Theory tomorrow.