Welcome back. It's now the 14th lecture of the condensed matter course.
More or less, we've finished talking about scattering.
There's a couple of little bits and pieces of scattering that I'm going to try to pick up later, maybe at the end of some other lectures.
Instead, I'd like to move on and use what we've learned when we study scattering in order to go back and study other types of waves in crystals,
the type of waves we were interested in originally where electron waves and vibrational waves are phonons.
So let me remind you a few things we had learned about electron waves in one dimension,
and then we're going to try to generalise up to higher dimensions from what we've learned.
So let's recall the type finding model type, binding model for electrons, binding model in one day, in one day.
And we're going to consider the case of two orbitals with two orbitals per unit cell orbits per unit cell.
And that can be two atoms, one orbital in each atom, or it could be two orbitals on the same atom, one atom per unit cell.
And we can, we can draw out the dispersion in reduced zone scheme.
Here's K, here's energy, here's pi over a, here's minus pi over a that's the barren zone, the big green zone.
We can write, draw the dispersion kind of like this and like this two modes at each wave vector K in the bronze zone.
Or we can unfold that picture in the extended zone scheme.
Looking like this. Here's K, here's pi over a year minus pi over a year to pi over A and here's minus two pi over a.
So we have the same picture for the first mode and then for the second mode, we're going to shift things over, shift the thing on the left,
over to the right by two pi over eight to make it look like this and shift the
piece on the right over by two pi over a to the left to make it look like this.
So we have what we called the first boron zone here.
First we're on zone busy here, and then these pieces here where the second bronze zone, second bronze zone.
And so it's useful. There's chemicals there right in the middle.
It's useful to try to generalise this concept to three dimensions or higher dimensions.
So let's let's do that. So generally, the Britain zone for one zone, since we're thinking about electrons in three dimensions,
is going to be always defined as a unit cell in reciprocal space.
And we're going to have to figure out what we mean by first bronze owned, second bronze on and so forth.
So it's general and bronze on general, general and bronze zone.
We define as follows. So we'll start with the first bronze on points in space.
That are closer to. That are. Closer to.
That are. Closer to. K equals zero, k equals zero.
Then do any other any other reciprocal lattice point receptor lat point are the first boron zone are the first boron zone first B's.
So all these points in the first boron zone are closer to the point of k equals zero, then to the other reciprocal lattice points.
Here's the reciprocal last point here. At two minus two pi over eight,
here's another reciprocal lattice point here at two pi over a and everything between minus pi over and pi over a
constitutes the first boron zone because it's closer to K equals zero than it is to any other reciprocal lattice point.
This definition might look a little familiar to you because in fact it's the same definition as the big nearside cell.
So this equals the big inner site cell sites.
Cell of the cake was zero point of k equals zero.
In reciprocal lattice. In reciprocal lattice.
Does that sound familiar? I hope all points that are closer to a given reciprocal lattice point than to any all points that are
closer to a given lattice point than to any other lattice point constitute its bigger site cell.
So all points that are closer to K equals zero than to any other reciprocal lattice
point constitute the vendor site cell of the cake or zero point in reciprocal space.
Good. Yes. Okay, good. So how about the second foreign zone?
Points where k equals zero is the second close.
This is the second closest. Recep lat point.
What point are the second brands on?
Are the. Second.
Second to B.C. So, for example, if we pick pick a point here.
X call point X, the first closest reciprocal lattice point to this point X is here.
The second closest reciprocal last point to the point x is equal zero.
Therefore, this point x is in the second burn zone. Good.
Okay, that's the definition. So a couple of notes about these definitions.
First of all, one bronze zone boundary.
Two opposite to opposite opposite busy boundary.
Is a reciprocal lattice factor is a reciprocal factor.
Sep lad back.
Let's see. So here's a bronze. On boundary is the boundary between the first bronze on the second bronze zone and minus pi over a.
Here's another one. It's the opposite one from minus pile raised plus pile ray.
Distance between them is two pi over a, which is indeed a reciprocal lattice vector.
Similarly, here's a boundary between the second Barone zone and out here actually would be the third bronze zone.
And so it's a bronze zone boundary, the distance from it to the opposite Barone zone boundary over here at plus
two pi over a is four pi over A and that is also a reciprocal lattice factor.
That is generally a true statement. Okay. There's a similar so one, a sort of caveat, a corollary of this is that these boundaries.
Or maybe, maybe it's not a corollary, but it's almost equivalent statement.
These boundaries are points.
Where? Absolute value of K equals absolute value of k plus g for some g for some reciprocal lattice vector g.
Okay. So for example, you need, you need to have K and K plus G and see what's K and k k posted having the same magnitude for,
for G being some reciprocal lattice vector.
So if you take minus pi over a, you add a reciprocal lattice vector to pi over eight to it you get plus pi over a those have the same magnitude.
So you're on the bronze on boundary. Similarly, you take minus two pi over a at a reciprocal lattice vector to it.
For pi over eight you get two plus two pi over a that has the same magnitude.
So you're on a bronze on boundary. The second statement, which is important about bronze zones,
is each DC bronze zone has the same the same area or volume in three dimensions, same area and and represents.
Represents each k point. Each crystal momentum.
Each crystal wave vector, I guess. Wave vector k once.
That was our general idea of constructing the bronze zone in the first place,
that every physically different wave is represented exactly once in the bronze zone.
The second bronze zone is equivalent to the first bronze zone.
We've just removed the pieces around into different places, but each wave vector K is represented exactly once inside each bronze zone.
Incidentally, is it obvious from the second bronze zone that you would define?
The third bronze zone analogously points where the archaic zero is the third closest reciprocal at this point would be the third bronze zone.
So you have the first, second, third. You can have as many bronze zones as you want.
The final the final statement, which is useful, I think this is a homework assignment on the third homework set that the number of K states,
number of K states, we even prove this in lecture in each busy B.C. equals the number of units cells.
Unit cells. In system. In system.
Does that sound familiar? We prove this in one dimension. I think for homework, you're supposed to prove it in three dimensions.
It is also generally a true statement. So actually, let me do an example of this in two dimensions to make this more clear.
So this we can start with a square lattice and real space.
The reciprocal lattice of a square lattice is also a square lattice. So I've drawn a square last is supposed to be the reciprocal lattice.
So it's a square lattice. Again, I've coloured some of the points differently just to make them easy to talk about.
The point of k equals zero. I've coloured black in the centre and so first we're going to try to construct the
first boron zone that is the inner site cell of this black point in the centre.
So how do you construct the bigger site?
So you put down perpendicular by sectors to the nearest points, perpendicular bisected to the red points here we'll make this point.
That's a pretty good bisect here. Then this this plane, this perpendicular by sector,
these two also giving all the perpendicular bisect is to the red points and then you colour in the area that is the first boron zone.
Okay. Now the distance across the first boron zone is two pi over a which is a reciprocal lattice vector the way it's supposed to be.
That distance there from perpendicular, bisected or perpendicular by sector is a reciprocal lattice vector.
Now let's look for the second brian zone.
How do we do that? It's a very similar construction. You just start putting down more perpendicular by sectors.
So here the second closest sets of points are these blue points here.
So let's put down perpendicular bisected to those like this, like this, like this, like this.
So these are the perpendicular bisected to the blue points out here. We'll colour in the walled off area blue and this is now the second branch zone.
Now it's this check that affects the definition.
If you take a point like here in the second Barone zone, the first closest reciprocal lattice point is this red point here.
The second closest reciprocal lattice point is k equals zero.
Therefore, this point here somewhere in this blue region is in the second Brian zone.
Good. Okay. Now it may not be obvious that the distance across the bronze zone from brown
zone boundaries to bronze on boundary here is a reciprocal lattice vector,
but I'll show you that it is this is a reciprocal lattice vector from zero to the blue point.
You just shift it like that. It's indeed the distance across from beyond zone boundary to bronze on boundary so we can keep going on looking for the
third baron zone and to construct the third bronze zone we have to take put perpendicular by sectors to these green points.
So let's do that, put down those green lines which are perpendicular, bisected to the green points and you fill in the area now green.
And let's check that it satisfies the definition.
If you pick a point here in the green area, the first closest reciprocal lattice point is this red point here.
The second closest is the blue point out here. The third closest is the cable zero point in the centre.
So this is in the third Brian zone.
Now also the distance from Brown Zone boundary to brown zone boundary across the system is four pi over a, which is also a reciprocal lattice vector.
Now, another thing that you've been promised from these definitions is that each Barone zone has the same area,
looks like the kind of might and that they should each represent each k point once.
So the bronze zones represent every possible point. Exactly.
Wants to see this. What we're going to do is we're going to take the pieces of this puzzle and with some very crude animation,
we're going to move them by reciprocal lattice vectors and show that they're actually the same shape.
So let's see how that works.
So I'm going to take this red section, the first boron zone, and I'm going to get that exact same shape of the first bronze zone,
same area, same shape, exactly by moving these pieces of the second bronze zone over.
So this piece, I'm going to move by a reciprocal lattice vector like that, two pi over a over.
Then I'll take this piece, I'll move it to pi over a over this way and I'll take the top piece,
move it to pi over eight down and the other one to pi over a up.
And it fills exactly the first boron zone, so. Well, okay.
My animation skills are not not perfect, which you see, the other one is even worse.
But. But you get the picture that it's supposed to fill the first bronze on exactly one side we by shifting things
by two pi over a we actually didn't change the waves at all because if you shift by two pi over eight,
you're getting back exactly the same crystal momentum, the same physical wave by doing this.
So the second bronze zone is effectively exactly equivalent to the first bronze zone.
Now let's do the same thing for the third bronze zone. We'll shift this over by two pi over a this over by t by over a okay.
Now it's really going to get bad, but but you get the idea that it's supposed to is supposed to fit perfectly.
And had I been better with my animation skills, it would have fit perfectly.
Okay. So that's the general general idea of O'Brien's answers.
A couple facts about Brian Jones in 3D.
So in 3D, the lattices that we're interested in right down a list of the lattices we're interested in are mainly in.
And we have simple cubic, simple cubic.
We have BTC and we have FCC. So the corresponding reciprocal lattice is very simple.
That is well, a simple cubic lattice has has a reciprocal lattice which is also simple cubic.
The VC lattice. This is something I mentioned and I think it's a revision homework problem.
The reciprocal lattice of a BC lattice is FCC and the reciprocal lattice of an FCC lattice is C.
So the shape of the first Barone zone, first being B.C. shape is well for a simple cubic lattice.
The Fichtner site cell is a cube. So the first Barone zone for the simple cubic lattice is a cube.
For the BC lattice, the reciprocal lattice is FCC.
So we get something that's the shape of the bigger site cell for FCC, which was a funny shape.
It was a truncated octahedron. I'll show your picture in a second. And for FCC lattice, the reciprocal lattice goes back.
And so you get the bigger site cell for BBC here.
Actually, I said this. This one is the truncated octahedron.
Now this one is a truncated octahedron. And this is the let me just show it on the slide.
Okay. So this is the first Bruns on an FCC lattice.
This one's a chunk. This one is a truncated octahedron. This one is the truncated the dodecahedron.
This one is 12 sides. This one has 14 sides. All right. Whatever. These are the shapes of the of the of the bronze zone.
So in this case, the FCC lattice has a reciprocal lattice, which is boxy.
And this is the Vicknair site.
So the BC lattice, and this is the bigger site cell of the SCC lattice, which is also the first boron zone of the BC lattice.
Now you'll notice that these pictures are drawn in K, space, K, X, Y and Z.
And you'll also notice that there are some letters dropped on to these pictures.
These letters are there because people typically, when they describe points on in the bronze zone, they usually say,
we're talking about the Gamma Point or the X point or the L point, and that's just shorthand nomenclature that you don't need to know.
But it's worth knowing that they have these when they say X point, they mean some particular point in the bronze zone.
So the point in the middle of a square face is called the X Point.
This point, this point at this point, the point in the middle of one of the hexagonal faces is called the L point.
The point in the centre is always called gamma. You don't need to know these things.
All right. So now that we we've we have these pictures of the bronze zone for, you know,
for our various lattices, we can start thinking about the excitation spectra,
either the electronic expectation spectra or the phone on excitation spectra for various different crystals.
And so everywhere in the bronze zone there will be different like in the reduced zone scheme,
you'll have different excitation modes at each possible value of K in the in the bronze zone.
So let's, let's do an example being this Valentine's Day happy Valentine's Day everyone is we're going to do diamond
for s all right so well we do diamond every year so just happens to be a Valentine's Day this year.
So Diamond, you'll recall, is an FCC lattice with two atom basis, one carbon and 000 and one carbon at one quarter, one quarter, one quarter.
This is the shape of the bronze zone as we had in the previous slide.
Now the first question and opportunities, wind, chocolate, how many phonon modes should there be at each k vector in the first boron zone?
How many phone unloads? No, no, no, no.
Oh, my gosh. So there's there's two atoms in the.
I'm going to eat this one myself. All right. I'm going to gain weight this year.
Now, there's two atoms in the primitive unit, so each atom can move in three possible directions.
So that means you have six modes. Yeah. So. Let's see if we have a picture of it.
Yeah. So here's here's a picture of the diamond phonon spectrum.
So the way it's drawn is that you take cuts through the barren zone along certain directions that you're interested in.
So, for example, this line. So here's the frequency vertical and then the K vector is horizontal, but this given to you in a cut.
So this is cutting from Gamma to X along the 100 direction gamma two X along 100 direction along this dotted line here.
And then from here to here is Gamma two l along the one, one, one direction from here to here, along that dotted line.
This one is a little more complicated. It's X to gamma along the 110 direction.
So that actually goes from this X through this K into the gamma in the centre of the next brown zone over.
So that was a little complicated, but. All right. Now you'll notice that if you pick a point in the centre here, you will count exactly six modes.
One, two, three, four, five, six. If you count over here, you'll discover there's only four modes.
Why is that? Well, the reason is because two of them are degenerate. And you'll see these two modes coming in to this point here.
And then they have the same energy all the way down from here to here.
And then two modes here are degenerate up here, and they have the same energy or same frequency all the way over here.
And the reason for the degeneracy is because you're looking along a particularly symmetric direction where
vibrations transverse in this direction and vibrations are transverse in that direction look identical.
Whereas along this direction, it's a very symmetric direction.
So all directions of vibration actually have different energies.
You can see a couple of interesting things in this. In this figure, the gamma point is K equals zero.
You can see the acoustic modes coming down to 22k equals zero linearly.
And these are the optical modes up here at at high frequency.
And one of these modes is longitudinal. Two of them are transverse.
Now we can ask the same questions about an electronic excitation spectrum.
And you when you calculate an electronic excitation spectrum in this type binding approximation,
you have to decide how many orbitals you're going to consider on each on each atom.
So for example, in that picture up there, we considered two orbitals per unit cell.
So given that a typical type of calculation for carbon, you don't want to keep all the orbitals out to infinity.
So generally you just keep the orbitals you're interested in.
What is usually interested in is the orbitals where there's some sort of interesting action going on that the, the one, the one orbitals.
There was a core orbitals that just completely filled. Nothing interesting going on there.
The orbitals that are three three has three p and so forth.
Those are just completely empty, totally uninteresting. The ones that are interesting are the ones they're sort of partially filled, partially empty.
Those are two S and the 32p orbitals. So given that we're considering the three S and and the 32p orbitals,
how many excitation modes eigenvalues should we find at each k point in the bronze zone?
One more chance. We're considering four orbitals per atom.
What was the answer? Eight. Did you say? Yeah. Okay, so almost.
I got that. All right. Eight. Yeah. So there's two orbitals in your account.
What you need to do is you need to count the total number of orbitals per primitive unit
cell and that will give you the total number of excitation modes per per k vector.
Now this diagram, this is an excitation spectrum, an electronic excitation spectrum of diamond.
These things are frequently known as spaghetti diagrams for obvious reasons,
because it looks like some sort of complicated ball of spaghetti that you can't make any sense of.
But but it's actually it's fairly simple. Well, not that not that complicated.
What's again, you're looking at cuts through the barren zone. So this is a line between the L and Gamma Point along this lot dotted line.
This is a line from Gamma to X along this dotted line, then X to K, K to gamma.
So it's basically just taking cuts to the branch zone.
And at each K point, there are eight modes one, two, three, four, five, six, seven, eight and some in some directions.
It looks like there's only six. Let's see, for example, this direction.
One, two, three, four, five, six. Because two of them have become degenerate here.
See, these two came together. Oh, no, hang on. These two came together here to make only one.
That's because are actually two modes with the same energy. So you have to be a little bit careful about that.
Four Now carbon has a valence of four, which means each carbon donates four electrons and that means with two carbons per primitive unit cell,
we have donated eight electrons total. So you fill the bottom four bands with both spins.
So the bottom four bands are filled and there's a gap in the upper four bands.
And in fact, as we'll discuss in a couple of lectures later on,
that means what we have is we have a really good insulator filled bands, then a big gap and then a bunch of empty bands.
It's also what makes Diamond nice and transparent, so it makes Diamond beautiful, appropriately enough, for Valentine's Day.
All right. So let's move on from here.
I think that's enough of diamond for today.
So one of the things that's fairly important about this, this story is that with our tight binding spectra,
we always have the principle that if you take K to some K plus G,
you get back exactly the same wave and you have the same principle for vibrations as well.
Now, you might ask whether this are the principle that if when you shift K by a reciprocal lattice vector,
whether that is actually special to the tight binding approximation we use or whether this is actually a general statement,
let's remind ourselves why it is that we found this for an exit, for r for a tight binding model, basically the tight binding wave function.
This is a slightly different language than I wrote it before. But you can you can write it as either the I KKR times electrons sitting on
lattice point R is this we used to call this thing phi and we gave this a no.
I think it's more or less I guess the difference is now we're doing it in three dimensions rather than one dimension.
That's why it looks a little different. But this is basically the type binding wave function that we wrote down before and
this thing is obviously invariant undertaking K to K plus g because either the I k,
either the Igot R when you shift k by g g that R equals one because g is in the reciprocal lattice and R is in the direct lattice.
So if cake is shifted by g, it doesn't change this factor here and you get back the same wave.
So but you might ask whether this is just a property of the tight binding model or if this is actually something that's general.
And this is actually a really important question and it's a question that came up very shortly after the discovery of the shortening equation scoring.
I discovered his equation in 1926, in 1928, that people were already worried about this issue.
And it was resolved very famously by Bloch, Felix Bloch, and what's known as Bloch's Theorem.
Block's theorem. All right. Okay. Which is frequently.
Referred to as the underpinning of all of material science.
It's an underpinning of all of semiconductor physics.
It's an extremely important principle which may not seem so important until we talk about some of its implications.
But let me state the theorem first, then we'll prove the theorem. Then we'll talk about some of the implications.
So the statement is an electron in a periodic potential.
In periodic potential potential.
Periodic vrr, I guess.
Periodic means V of our equals vrr plus v of capital r with capital being a lot of spectre has eigen states of the form as eigen states.
Of the form. Okay.
There's a lot of symbols here for a second size sub K of R equals e to the i k
dot are times you superscript alpha some k of our where where you is periodic.
So it's the same if you shift it from you, you've r equals your R plus r and ks.
And KS. Can be chosen in the first bronze zone, can be chosen in first bronze zone.
In other words, K is crystal momentum. K is crystal wave.
Vector. Crystal wave vector. This form is known as this form of SCI.
OSSI is known as a modified plain wave is known as as a modified plain wave.
A modified plain wave. Modified plain wave.
That that should be fairly obvious. It's a plane wave, the iCar, and it's modified by being multiplied by a periodic function.
You. That's the modification of the plane wave.
The index. Alpha. Alpha is the band index.
And the band index is meant to describe.
You know, you get a different wave function if you're talking about this point here or this point here at the same k vector,
at the same K vector, you can have more than one eigen state.
So the energy's eigen energies. R e superscript alpha of k so e one and e two depending on which which band we're talking about,
which which which energy level, the lowest or the next lowest or the next next lowest and so forth.
Okay. So this is a pretty big bunch of statements we've made here,
but it's fairly important because it tells you about the general structure of iron states in a periodic potential.
So let's first prove this theorem and then we'll talk about why this theorem is so important
and why it underlies so much of material science and semiconductor physics and so forth.
So first, okay, so this is sort of a quasi proof. It's actually fairly rigorous, but we'll we'll not fill in the details.
But the quasi proof goes like like this.
First we'll consider a Hamiltonian, which is h, not just the free electron Hamiltonian plus or potential V, which is periodic.
So V is periodic and H not is the usual p squared over two m.
Now, if we throw away the periodic potential, we would know the iron states.
Those are just plain wave eigen states.
Plane waves are e not k k where enought you know of k is just the squared k squared over two m we usually have.
So so far so good. We know we have the plane waves, then we have to add the periodic potential.
Now, what can the periodic potential do? The periodic potential can scatter you from some K to some K prime via a matrix element of this form.
Now we know a lot about matrix elements of this form. From what we learned in scattering theory, this thing here,
this matrix element from a plane wave K to a plane wave k prime via a periodic potential
v is the for you transform for a transform a vrr evaluated at k minus k prime.
And that is the following it is zero.
If K minus K prime came out as k prime is not a reciprocal lattice vector is not equal to g and we'll call it v sub g, f k minus k prime equals g.
I mean it's called k prime minus cable's g that.
And this is just Allawi's condition.
You can scatter from a wave vector K to a wave vector k prime only if the difference between and K prime is a reciprocal lattice vector.
We learned this when we studied scattering of X-rays and neutrons and so forth.
So this gets us almost to the proof. So we can almost conclude the result here.
Why? Well, the point here is that the Hamiltonian with the scattering potential is blocked diagonal.
No, no pun intended. BLOCK, diagonal.
BLOCK of the K diagonal. What I mean by this is that you can scatter from K to a k prime if k k prime are
separated by G and then from k prime you can scatter to another k double prime.
If those are separated by a g and you can scatter to another k prime and so forth and so on and so forth.
But you can never change the crystal momentum.
All of the ks that you can get to by scattering by the via the periodic potential, all of them differ from each other by reciprocal lattice vectors.
So the Hamiltonian breaks up and if you go into these sectors, if you like.
And so the wave function must be of the form, cy k must be of the form,
some over g of some coefficients of column AMG plus k times either the i k plus g dotted with are.
Now why should it have to be of this form? Well, you start with some of the iCar and then you mix into that.
The Hamiltonian mixes into that different plane waves whose wave vector K differs by a reciprocal reliance vector g.
And you can mix in as many of those as you want and they all get some coefficients a.
But at the end of the day, the wave function must be a sum only of things that have the same crystal momentum k plus some reciprocal lattice vector g.
There's no way that I potentially can take you can change your crystal momentum.
That's the important realisation. Good. Yeah.
Happy? Good. All right. So from here we're pretty much done because you can just factor out the, you know,
the I k dot are and then you have some over g of a of g plus k into the I,
g r and this piece here is the is the periodic function u u of R which and you can tell this thing must be periodic because if you take e to the I,
g g dotted with R plus r, move it by a direct space lattice vector.
This thing is the same as the i k that are.
So this thing is periodic. As as claimed.
Okay. Furthermore, we can also check what happens if you take and you shift it by a reciprocal lattice vector g.
So let's let's actually take this form of K up here and let's shift K by G.
So I have K plus g, I actually I've already used the letter G, so let's call this a G prime shifted by K to Kate plus g prime.
So we'll write it as some over g a sub g plus k plus g prime e to the i r k plus g, prime plus g.
But the sum is over a dummy index so we can define a dummy index.
It's called g twiddle equals g g twiddle of vector equals g plus g prime vector.
And then if you substitute that in, that becomes sum over g twiddle vector a of g twiddle of vector plus k into the i k r k plus g
twiddle vector dot r which is exactly the same expression that we started with up up there.
So if you shift K by reciprocal lattice vector G, you get back exactly the same wave function,
which was what we were trying to answer in the first place.
Is it special that when you shift K by reciprocal last vector you get back the exactly the same wave function?
Is that special to the tight binding model? It's not. It's a property of the shortening equation in the periodic potential.
Okay. All right. So the implications of this theorem, which is viewed as so important, is one implications.
One are an excitation are all excitations can be described all excitations.
Can be described be described in one bronze zone in one B.C., because whenever you go outside of the bronze zone,
you can just zip by a reciprocal lattice vector to get back into the first back into that bronze zone.
So that's an important statement. The second important statement, maybe more important is that a periodic potential,
a periodic potential does not change, does not scatter crystal momentum.
And this is something that I mentioned before when we discussed the type finding model that the crystal momentum is conserved.
The crystal momentum is always conserved in the fact that we have this very strong periodic potential doesn't ruin that.
The reason this is so important, remember, we had this puzzle when we studied the Sommerfeld model that you have this mean free path of electrons,
which is enormously long, hundreds of atoms or thousands of atoms long.
And we couldn't understand it because there's a big nucleus that the atom can hit every few angstroms.
It could scatter off of tons and tons of nuclei. The point here is that the you don't have pure plane waves.
What you have is modified plane waves.
You have these plane waves times a periodic function, but once you modify the plane wave by multiplying it by this periodic function,
that is perfectly good plane wave to go all the way across the system without scattering at all.
So if you take a, you know, an electron, you put in a little wave packet of this modified plane wave,
it would travel clear across the system as long as it doesn't hit any impurities or or nasty things like that.
But it could go all the way across the system without scattering at all.
So this sort of resolves our our initial issue of why it is that electrons can travel so far without without scattering.
Okay. And where did this this result come from?
It's important to realise that this important result really came from Louise condition.
It came from the fact that a periodic potential only scatters you by reciprocal wave vectors.
All right, so the one thing that this doesn't do for us.
Is it does not manage to. It does not manage to help us actually solve the shorthand equations.
So solving solving Schrödinger equation is still hard and we've already discussed one way to do it.
One way to do it is tight binding. And by type binding, what we're really doing is we start with.
Start with atomic orbitals. With atomic orbitals. And allow week hopping, allow week hopping.
But that might sound a little non generic because maybe the hopping is strong from one place to another.
So with useful too,
it's actually very useful to think about it from a very complementary viewpoint where instead of starting with a type binding model,
we start with plane waves, start with with plane waves, plane waves and add week periodic potential, the air.
So it's entirely the opposite way of looking at things.
This approach, approach to two is known as the nearly free electron model, nearly free electron model,
and is a very useful model for understanding a lot of the physics of semiconductors and band structure in solids.
So let's let's go about that. Let's start it. We probably won't finish it today, but the so how do we do this?
We start with our H0 on k are plane waves without the periodic potential.
So is zero k on k where is zero is the usual h bar squared k squared over two m?
It's a good place to start. And then let's assume this is weak, assume weak.
V All right.
So what do you do when you have a weak perturbation to a Hamiltonian? Well, you use perturbation theory, which you presumably learned last year.
So you remember what happens to the energies in perturbation theory.
So let's start with first order of perturbation theory. First order, order, purge the energy this probably look familiar the energy of the wave.
The Eigen State K is the bare energy without the perturbation plus the first order of perturbation theory term which is k,
v, k and this matrix element here when I scrolled it off the top of the board.
No, maybe I didn't, maybe I did. I think I did. We defined it as the sub zero because k minus k is zero and this v sub zero is the same.
Same for all K for all k.
So this just gives an overall constant energy shift of the the eigen state.
So the the I can say t to get shifted up in energy or they get shifted down in energy.
This is sort of the the zero thorium mode of this potentials where a constant potential is added on top of any fluctuating potentials,
things that change in space. So the zero three mode of the of the potential just shifts up and down the overall energy of K.
Does this expression look familiar first or probation theory. You just take the exponent.
Yeah. Okay, good. So this is not interesting. And as a matter of fact, people frequently drop V zero because it's so uninteresting.
They just get rid of it. I'll try to keep it, but maybe I'll forget it by mistake at some point.
So second order is more interesting.
So it's second order you have ic equals e k not that's the they are ps plus v zero.
That's the first order piece. And then you have the second order piece which is sum of k prime not equal to k.
Then we have k prime v k squared over e sub k not minus e k prime not.
Is that the right order? Yeah. Does that look familiar from second order of probation theory?
Good. All right, now, what do we know about this second order term?
Well, again, we have lousy condition for the upstairs matrix element that in order for that matrix element to be non-zero,
you must have K prime minus K must equal a reciprocal lattice vector g.
So this term here can be rewritten as some over g of reciprocal.
Last factor is g of v sub g squared where g is k prime minus k divided by e sub k not minus e sub k plus g not.
Good luck. Happy with that? Somewhat.
Okay. So this is our expression for the second order shift in energy of the plane waves that we started with.
But we have to be a little bit careful about expressions that look like this.
And the reason we have to be careful is because you might have a situation where the two terms in the denominator are either equal to each other,
in which case you get zero downstairs or very close to each other, in which case you get something very small downstairs.
And in either case, the secondary term would blow up, in which case you have a divergence.
And that means the second order perturbation theory is not correct.
So you must be careful.
Be careful when he sub k.
That's just right. Is SMK not approximately equal to E sub k plus g?
Not now. When does that happen? Well, e sub k is actually just proportional to k squared.
So that will happen when absolute k equals absolute k plus g.
Well where's that. You'll remember somewhere here.
There it is. It's the condition for being on a bronze on boundary.
When you're on bronze on boundary e sub K and it's a K plus g are actually the same.
Let's actually draw a picture of that. So what we have is we have a this is our bare spectrum K and E, the bare spectrum is just a parabola.
So this is e sub k is h bar squared not k squared over two m and here is pi over a here is minus pi over a.
And the energy of these two, these are separated by a reciprocal lattice vector to pi over a and their energy is identical.
So this would give us a divergence in second order perturbation theory.
Similarly, if we go to minus two pi over a and two pi over a,
these are separated by a reciprocal lattice vector for pi over a and their energies are identical.
So they will give you a divergence in second order perturbation theory.
So the place where second order perturbation theory fails is exactly when you're on a bronze zone boundary.
So what we're going to need to do is we are going to need to are probably going to have to do this next time.
But what we're going to do to fix this problem, fix the problem by using using degenerate perturbation theory,
degenerate perturbation theory, effort theory, which hopefully you've learned something about already.
So we're not going to have time to launch into that today.
So I'm going to actually take the opportunity to insert one piece of information or maybe two
pieces of information about scattering that I promised you we would talk about some time,
and that is scattering off of. So maybe I'll put that over here.
Scattering off of x rays and neutron scattering x ray and neutron scattering.
This is different subject just inserted in this last 5 minutes.
Neutron scattering from.
From liquids. Liquids and amorphous solids.
Amorphous. Solids.
So in liquids, in amorphous solids, we do not have periodic structures.
We do not have periodic arrangement of atoms. But nonetheless, you get a lot of information out of the neutron X-ray scattering.
If you think way back to when we did Fermi's Golden Rule, that the scattering amplitude is still the 48 transform of the periodic of the potential.
But it's not a periodic potential, it's Fourier transform of the non periodic potential.
But does that mean there's no information in it? No, there's still information in it.
It's just that you don't get sharp peaks anymore. What you get is here.
There it is. So this is X-ray scattering on liquid aluminium.
It's heated up to some high temperature. So it's so it's a liquid.
There's a function of a scattering angle or the reciprocal wave vector.
That's the way vector along this axis.
The amplitude of scattering it has peaks. And the peaks represent roughly the distance between between atoms.
It's one over the distance between atoms. The same way the peaks represent the the periodicity of the unit cell when it becomes crystalline.
Now what happens is as you cooled down a liquid and it gets more and more crystalline, that becomes more and more local order.
And these peaks get sharper and sharper and sharper.
And finally, when the thing actually crystallises, they become a delta function peaks the way we expected them to for for a crystal.
But if it's a liquid, it's you can sort of think of a liquid as locally being like a solid.
Each atom has a couple of neighbours, but if you sort of look farther away, it's not ordered any more.
It sort of loses its order. But if you sort of look just in some small region, it probably has, you know, it's organised over a small region.
So you get these sort of broad peaks in in the scattering spectrum.
Okay. I think maybe I won't try to cover any further issues today and we'll start again on Wednesday, I think.
Have a good weekend.