Okay. Let's get started. Welcome back.
This is the seventh lecture of the condensed matter, of course.
In the last lecture, we introduced a rather important,
simple model of vibrations in one dimension the anatomic harmonic chain, chain of masses and springs.
Every spring is the same. Every mass is the same extremely simple model, but introduce a lot of very important concepts to us.
We solve for all of the normal modes of the chain.
When we quantised the system, we discovered the meaning of phonons with the quantisation of modes where we introduce the concept of crystal momentum.
The idea that if you shift away vector by two pi over a, you get back the same exact physical wave that you started with.
We introduced the concept of a bronze zone arranged in space,
usually taken from minus pi over a two pi over a where all of the wave vectors are physically different waves.
And if you go outside the broken zone, you just start repeating the waves that you have already described.
And we also manage to calculate the heat capacity of this chain. Exactly.
And we compared it to the Devi theory and the Einstein theory that we had looked at previously.
Now, one of the simplifying assumptions we used going into this model was that every single mass and every single spring in this chain were the same.
And in real materials you typically had different types of atoms.
For example, in sodium chloride you'll have a sodium atom and a chlorine atom, and these will be different.
So maybe a better model of a typical material would look something like this.
You'd have a light coloured atom and a dark coloured atom and drawing everything in one dimension again.
And then a light coloured atom and a dark coloured atom and a light coloured atom, dark coloured and so forth.
Now let me introduce some nomenclature that's going to be useful.
The idea of a unit cell is the repeated, repeated motif.
So, for example, we can take our unit cell to be this box.
And if you repeat this box over and over and you stack the boxes together with no space in between, you will reconstruct the entire chain.
Another important word that we introduced last time is lattice constant, which we defined as well.
We usually call it a and we define it as the distance between distance between we in equivalent atoms.
So for example, Gladys Constantine. If you like this, we'll call it a dislike for Adam to dislike called Adam.
You may also notice that that lattice constant also happens to be the size of the nasal.
Right like this this distance here is also a or if I drove properly it was.
Now one side comment is that the the unit cell is actually not unique.
Let me draw another picture of this chain. I could have drawn the initial sell equally well like this.
So that's something we should probably even write down because it's important in itself, not unique.
And if I had chosen this as a unit cell, I could stack duplications of that together and build up the entire chain just as well.
Now,
what we're going to do today in a real calculation is we're going to try to calculate the normal modes of a train of a chain that looks like this.
And the reason we're doing this is not just to add complexity to a problem we already solve last time.
In fact, by having different types of atoms in our chain, we're going to see that some fundamentally new and different things occur.
So that's why it's why we're doing it. So let me draw the chain.
This is known as the alternating chain, frequently known as the altering chain, sometimes known as the diatomic chain.
And it looks kind of like this. There's a dark atom, then a light atom.
Dark atom and a light atom and dark atom and light atom.
And they're connected together with springs. And so forth.
And we have a lattice constant A which is the distance between identical.
We see the difference between light and dark. And to make clear enough, you see that.
Okay, so a is a distance between equivalent atoms.
So between light atom and light atom. And now we have a choice as to what model we want to write down.
And there's two common choices that people make in solving these two simple problems, but they are similar.
One choice is is frequently made is to make the mass of the light atom, the mass of the dark atom different.
We're not going to solve that. You can actually solve that for homework. That one's a typical choice.
But we're going to solve instead is we're going to make the two different springs this spring Kappa one and this Spring Kappa two different.
And we're going to let them alternate back and forth. Kappa one, capita Kappa one and two and so forth.
And hence the name altering chain, because the springs alternate between Kappa One and Kappa to the physics of these two chains.
Whether we're letting the masses alternate or the springs alternate is extremely, extremely similar.
The reason we're doing this one is because it's algebraically just slightly easier to keep track of than the one you going to do for homework,
which is slightly algebraically more complicated. So let's write down some coordinates.
So for example, let's let the light coloured atoms have positions x, n and the dark coloured atoms have position y n.
So for example, that's like this one, the x one.
This one will be y one. It's going to be x2.
This will be y two and so forth.
And now the way we solve problems like this, these mass and spring problems, is we start by writing down Newton's equations of motion.
So, well, okay,
one more definition is the Delta X and delta Y are the deviations from equilibrium emissions from equilibrium position from from equilibrium.
Okay. And we're going to write down Newton's equations in terms of these Delta X and delta Y.
So maybe actually, maybe I'll put it on another board over here.
So mass delta x double dot.
Mass times acceleration equals. We have to be a little bit careful here.
XA atom, a light atom on the right, it has a Kappa two capital.
And then on the right of an x, Adam is a y atom with the same number.
So Delta Y and minus delta x n.
And then on his left is a Kappa one, plus Kappa one, and on its left is a delta y atom with the index one lower right is Delta X.
And does everyone agree with that? That on the left of x two is y one.
So I have two. The capital one spring has the lower index.
Yes. Good, happy, good. And then delta Y and double that.
Okay. So on the right of a Y is an X with the index one higher and that's a Kappa one spring.
So Delta X and plus one minus Delta Y n and on the left, to the left of y one is x one.
So it's an x with the same index, delta, x and minus Delta.
Why? And I think I got that right. Okay.
So those are our equations of motion. And the way we're going to solve this is the same way we did last time.
We will write down a wave on that wave onwards.
So there is some disagreement on what the word Anzacs actually meant in German.
And I actually saw someone stop me after a lecture, a German speaker said, No, no, no, you were right about what that means.
And then someone else said, No, no, no, you're wrong. And then I asked a couple of German speakers and they all disagreed.
And then actually someone pointed out that that is actually an English word.
It is in the OED. You can look it up. It means it means mathematical.
Yes. So so I'm going to say it's in there.
It really is. Okay. So anyway, we are going to write down our wave on that.
So Delta X and is sum and is the I omega t minus i k and a and delta y n is oops I called x rather and a y in the eye omega t minus high k and a.
And what we really mean as last time is we mean to actually take the real part of this whole expression.
And because of that, we're allowed to choose omega a greater than or equal to zero.
But K can be either sign, either sign to represent either left going or right going waves.
Now, as with last time, we realise that if you take K and you shifted to K plans to pi over A, you get back exactly the same wave.
And the reason for that is because if you look at the exponential factor, even minus i k plus two pi over eight times n is.
Times a times, eight times and is equal to exactly the same thing as in the IK and a and a.
So it matches that those two are exactly the same.
So if you shift k by the two pi over a, you get back exactly the same way form rather important fact we found out last time.
Just a couple other things we're going to use periodic boundary conditions,
periodic boundaries because we use a system of length L equals N times A, so this means we have n unit cells in its cells.
And if we have a system of size l that means the case must be on a periodic boundary condition, must be two pi over l times p where p is an integer.
So the spacing between different l different possible values of k are two pi over l.
So now if we want to count the number of different k's.
We did exactly the same calculation last time, different case.
That equals the range of possible case. That's two pi over a before we start repeating,
since you can shift by two pi over a and get the same way back and the spacing between k's is two pi over l that gives us l over a or n.
So the number of different k's is equal to the number of unit cells.
In the system. A rather general rule that we will use.
Many times. Okay. So given our wave on that up there, we can solve this set of Newton's equations.
It's by plugging in and doing a little bit of algebra.
And then I can see that the algebra I'm about to do is a little bit messy, but bear with me.
So here goes. First of all, plug in for Delta x n.
So two derivatives gives me minus omega squared m then the wave is a x in the I omega t minus i k and a and that equals
well k equals kappa two times delta y n which is a y in the i omega t minus i k and a and then let's do the other capital.
That's the other term over here, the other Y term.
So that's plus kappa one e to the I omega t minus i k and then it's and minus one a is an index and minus one and then must be the two x terms.
So we have minus kappa one plus Kappa 2ay here is in die why a x in the i omega t minus i k and a everyone happy with that.
Did I do that correctly? No one reacts, seems to work.
Okay. And then the other equation is minus omega squared and a y in the eye, omega t minus i, k and a equals.
And let's do the two x terms first.
So we have kappa 1ax in the I omega t minus i k and then the coefficient and in the second equation of the index of the x with kappa one is one.
So it's end plus one a here and then we have plus kappa to a x the omega t minus I, k and A and then the two,
then the two the two y terms minus kappa one has kept to a y in the eye, omega t minus K.
And a notable thing about this equation is this the longest equation I'm going to write on the chalkboard all year long, and now it's done.
So we never have to do that again.
But it's not actually kind of complicated as it might look because there's a bunch of exponential factors which drop out the counts on both sides.
So we have two equations and we can simplify it into one matrix equation and write it this way minus omega squared times the vector x,
a y equals some big matrix times the vector x, a y, and the big matrix is what is it?
It's a minus kappa one plus Kappa to and then minus Kappa one plus Kappa two on the diagonal, Kappa one plus two.
And then the off diagonal has Tabata plus Kappa one.
Either the I k it's that part of the exponential didn't cancel it comes from the plus one and then the Kappa two plus Kappa one, the minus a.k.a.
Come coming from the minus one and the exponential. Does everyone good with that agree?
Okay. All right. So what is this? This is an eigenvalue equation or the eigenvalue is this guy over here minus seven mega squared.
And the way you solve an eigenvalue equation is by moving the omega squared to the diagonals over in this side,
then you set the determinant equal to zero.
You get the characteristic determinant or characteristic equation, or sometimes called the secular equation.
We do that the one step and omega squared minus capital one plus capita all squared
minus capital plus capital one in the i k a absolute value squared equals zero.
And we can solve that rather rapidly.
Omega squared equals capital one plus capital plus or minus absolute value capital plus capital one.
Okay. And that's our solution.
Now, one thing you may notice is that for each K, there are two possible solutions of Omega, the plus solution in the minus solution.
Right. So does that mean we'll write that down over here?
Each K has two normal modes.
Modes? We can call them omega plus and minus.
Okay. And how many cases do we have? The number of cases are equal to the number of unit cells, and each K has two times two normal modes.
So number of normal modes. Number of normal modes equals two times number of human cells.
Unit cells equals the number of masses.
And this we should have expected because we had that number of masses as the number of degrees of freedom we have.
So when we solve this problem, we should add exactly that many normal modes.
And indeed we do. So that's that's good. Good.
We're happy with that. All right. Good.
Now, so it is useful to actually plot this thing, and I'm going to sketch it out first and then we'll sort of justify why it looks the way it does.
So herring and fat. Here's K, here's omega.
Okay. Pi over A is minus pi over a, minus pi over a.
So that's the the brown zone outside of the brown zone. It's everything just repeats.
So we should have a mode that looks like this and then a higher frequency mode maybe looks like this.
Okay. And then you could actually reproduce this periodically outside.
I mean, you could you can make it go onwards as you like, because as you shift everything by two pi over a, you just get back the same wave.
But we're going to really focus on the bronze zone because within the bronze zone here, every wave you find is different from every other wave.
Now, it's useful here to probably look a little bit more carefully at this plot and see why it looks the way I've drawn it.
So let us look at some convenient points. Maybe at K equals zero.
That's a good point to look at. Well, what happens with cake zero?
So over here we have this absolute K to plus k one either the a.k.a that goes to just a K to plus k one.
So M omega squared is either two times k one plus k to or zero.
And so I've drawn that here, here zero at cake zero.
And here is the higher end. You want me to draw the higher one also.
So omega plus k equals zero is square root of two Kappa one plus Kappa two over m.
I get that right? I think so. So that point here, we'll give it a square root of two.
Kappa one plus Kappa two. Over. And so we have a long human with the high energy mode.
Now the low energy mode will come back to the high energy a moment, but the low energy,
low frequency mode is one that we should expect should have been there,
because we should expect that there should be sound waves somewhere in the system.
And when there are sound waves,
we know the sound waves should have a spectrum which is has frequency linear in wave vector and comes down to zero at zero wave vector.
Now to convince ourself that that's what's going on with this equation.
We have to actually do a little bit of more math and expand the dispersion curve near K equals zero.
So let's see how we do that. So let's take ak2 plus k one of the i.
K. That quantity there is.
Well, yeah. That thing there can be written as square root of capital squared plus capital one squared plus two capital one kappa two cosine of k.
And just by multiplying that out and if k is small, we're going to replace this cosine by one minus k squared over two.
So then what we have is I guess we can write it like this square root of this
whole thing of two plus kappa one squared that takes care of the one term.
When I put together this, this and this, I get capital one plus capital squared.
And then what's left over is twos council, capital one, capital K squared.
I get that. I want to leave that. Yeah.
Okay, then this conveniently factors out to give her a capital plus capital one times the square root of one plus capital one.
Capital squared over capital one's capital square.
And if you remember square root of one plus x, I get a sign wrong.
I got a sign wrong. That's minus. That's minus.
No one gets chocolate for that is one minus X over two.
So so then we have we can rewrite this as capital plus capital one minus capital one.
Capita K squared over two.
Capital one plus capital. Does that look right?
Did I miss anything? Yeah.
Capital kept to thank you. Cavity.
Yeah. Yeah. Good. Thank you. Whoever said that, I owe them chocolate, so.
Yeah. Okay. Okay, good. So I don't have one today. I was going to bring one, but I ate it, so.
All right, so if I plug that into into here,
we see that the Kappa one plus Kappa two with the with the minus solution cancels this this
capital one plus capital also will cancel this and we'll end up just getting m omega squared.
Omega squared equals capital one kappa two K squared over two,
Kappa one plus Kappa two or equivalently omega square root Capital One, Kappa to a squared over, uh, to Kappa one plus Kappa two.
I guess there's an m down there and then absolute k, all right.
And I realise that's a lot of math. We're not going to do too much math that heavy this year, but I have to do it once in a while.
Okay, so this we recognise as the sound velocity, so we just derived the sound velocity of this wave.
The sound velocity is the slope of this curve here.
Now usually actually I should use the proper nomenclature.
Usually people refer to this as the acoustic mode and they tend to call it the acoustic mode,
even all the way out to the zone boundary out here at very high K,
even though it's really only sound when it's when it's long wavelength or small K, they call the whole branch here the acoustic mode.
Now, you remember last time when we derived the sound velocity.
We also were able to derive the same sound velocity by using a hydrogen panic argument.
So let's see if we can do that again this time.
Remember from hydrodynamics we have the v sound should equal square root of one over the mass density times the compress ability.
The mass density is where there are two masses per lattice constant a and the compress ability while we derive last time,
the compress ability should be one over the spring, constant times a and the spring constant.
Well, here the the unit cell has two springs in it and the spring constant for two springs in series is Capital One,
capital over capital one plus capital. Does that sound familiar?
You remember that from first year. Okay. So if we plug this and this and this into here, in fact, we get exactly the same result.
V sound equals square root capital one kappa to a squared or to kappa one for step two.
So the hydrogen AMOC calculation gives you exactly the same result as we got by actually solving the system completely.
Now let's talk about this mode up here.
This mode is known in comparison to the acoustic mode.
This mode is known as the optical mode, optical mode, high frequency mode, occasional zero.
And we should probably I mean, I should probably tell you why the name optical mode.
So the word optical mode comes from experiments on unreal.
Real materials. When you shine light on the material and you induce vibrations by by shining light on it.
Now, if you think for a second,
what's the property of light that we know similar to sound light has frequency which is proportional to its wave vector.
The only big difference is that C is huge compared to sound velocity.
So if I drew the light dispersion curve on the same plot, it would be an extremely steep slope line like this.
Probably so steep I couldn't even. Right now, if you imagine shining light on a material and creating vibrations in quantum mechanics,
we should think about absorbing a photon and creating a phonon.
So imagine a process where you absorb a photon and create a phone on.
In order for that to happen, you have to conserve both energy and momentum.
So you have to match up both the frequency and the wave vector of the light with the frequency in the wave vector of a phone on.
The only place that can happen is right here. And the reason you're never going to match the acoustic mode because the velocity mismatch is so badly.
But with the optical modes, since the object optical has finite frequency, even a cable zero,
there is a point where the optical mode frequency matches life frequency and the optical
mode momentum matches the the light momentum or even crystal momentum in this case.
Now, so that is an area of where why people call this the optical mode.
Whenever you have interactions with light and vibrations, it's inevitably the optical mode.
But I should be a little bit honest that that process I described to you by which you absorb a photon and you emit a phonon actually does not occur.
And the reason it doesn't occur is because there is a conserved quantum number,
the spin of the photon, which I mean, photons have spin phonons do not.
So if that process were to occur, you would violate angular momentum conservation.
So that is, you know, definitely bad. You don't want to violate angular momentum conservation.
So, in fact, no such process exists by which a single photon is absorbed and a single phonon is created.
However, there are more complicated processes that can occur involving photons and phonons.
For example, absorb two photons and emit some photons.
That's okay, because the angular momentum of the two photons can cancel and you can still have
conservation and is still the same principle that is very hard to conserve,
both energy and momentum in any sort of interaction between phonons and photons,
unless you're getting optical modes into the game because you need to get high frequencies into the game somehow with phonons.
And the only way to get high frequencies with small wave vectors in Phonons is to use the optical mode.
So that's where the name comes from, actually. More generally, let me give more general nomenclature here.
Acoustic mode. Generally, acoustic mode is any mode where Omega is proportional to K at small K and optical mode.
Optical mode is any mode where omega goes to a constant, not equal to zero at small K.
Okay, so let's think about this, this small K regime a little bit more seriously.
So we found the eigen value is the frequencies, but let's look at the eigenvectors.
So the matrix were actually diagonals and is that big matrix there?
And let's look at it at take zero. So at zero the matrix we're interested in is of the form, I guess it's minus Kappa one plus Kappa two times.
What is it. One minus one, minus one one.
Something like that. I get the minuses in the wrong place and I think it's right.
Okay, so it looks something like this because when when the wave vector goes to zero, the exponential, the okay both go to one.
And so you have a matrix of the form one, one minus one, minus one one.
If I pull out minus one plus kappa two. So the eigenvectors we have two possible eigenvectors one is for the omega equals zero solution.
We have an eigenvector x, a y equals one one.
Whereas for the the optical mode solution, the other eigenvector the high energy high frequency eigenvector we have x a y equals one minus one.
So what is this telling us?
What this is telling us is that for the acoustic mode near k equals zero, the atoms are actually moving with their neighbours.
They both are. The atoms in the unit cell are moving in the same direction at the same time they're moving in concert they are.
Whereas for the optical mode, the higher frequency mode, the two atoms in the unit cell are moving opposite each other.
And it is actually quite natural to explain then why it is that the optical mode is so much higher in frequency,
because an optical mode you're compressing the springs maximally,
whereas with the acoustic mode you are hardly compressing the springs at all because everyone is moving in the same direction.
These sort of trends occur more generally.
So General, if if there are m if there are m atoms in the unit cell.
Atoms in units cell. We will have emojis that each k one is acoustic.
That's the mode where everyone moves in the same direction at k equals zero, and then the remaining and minus one of them are optical,
meaning they don't all move in the same direction and k zero in d dimensions.
The dimensions there are d times m modes at each k at each k.
These are optic are acoustic and the remaining D times and minus one are optical.
Okay. Why is this? Well, we're in the total count of the modes is a total count of the number of degrees of freedom.
If there are m, there is always the same number of ks per as as the number of unit cells in the system.
Each each value of each unit cell has m masses in it and they can move in d dimension.
So it is total of D.M. degrees of freedom per unit cell.
So we expect the M modes for each K.
Now of them there can be d of them there acoustic and that corresponds to all of the masses moving in any of the three possible directions.
So we know what these things are. We have discussed them before. One of them is longitudinal and two of them are transverse.
So for each K, one of them is moving in the direction of K and then two of them are moving perpendicular to K in three dimensions.
Okay. There's one more point on this picture that we should probably look at more carefully,
which is these interesting points here near which are at the so bronze zone boundary.
Pi over a when we put it over here. So consider consider these boundary.
These boundary. K equals pi over a.
So what do we have then? Well, then this factor capital plus in the I k a kappa one is actually absolute k two minus k one.
And so the frequency is omega squared is one over mk1 plus kappa one plus capital plus
or minus absolute kappa one minus Kappa to a cap of two minus capital one does matter.
So absolute value. And that means that we have two possible solutions.
Omega equals either square root of two kappa one over M or square root of two Kappa two over M.
So we'll mark them here. So this one we'll call Square Root of two Kappa one over M, and this one will call the square root of two Kappa two over M.
And here I've assumed we have Kappa one is greater than Kappa two otherwise.
We are around the higher end. You want to squeeze on top and these modes actually never cross each other.
You can convince yourself by looking at the form of of the of the frequency these two modes can never actually cross.
So what is going on at this at this brand's own boundary, at the brand's own boundary, the wave form delta x, delta y equals x,
a, y, you know, the i omega t either minus i k and a this e to the minus i k and a becomes minus one to the n.
And that tells us that what's going on is that alternate unit cells are moving out of phase with each other.
They're moving in the opposite direction as each other. So now is see if we can get this to work.
Uh oh. No, no, no.
Stop, stop, stop, stop, stop. Stop.
Know, uh. Oh, dear.
Does this happen? Only on Mondays. Here we go. Screen up. Okay.
Oh, but that's. Oh, wow. Amazing. Okay, so we can still do this while it's going on.
Okay, so this program with is the thing written by by Mike Glaser and you can click alternating
chain and longitudinal longitude means that everything oscillates in the single line.
You can change the K here by changing the slider and you can see that the the dispersion is plotted down here is plotted from
zero to pi instead of zero instead of minus pi over zero to pi instead of minus pi over a two pi over a like I've drawn here.
So it is the same picture. So first let's start by taking K very, very small.
O And also you have these little clickers up here which will change the masses on the springs that's making the masses the same,
and C one and C two, or we have what we call Kappa one and Kappa to the two spring constants.
So we have two different spring constants, Kappa one, a cap of two.
And here's the dispersion, the acoustic mode down here, the optical mode up here, here we've clicked acoustic.
So we're seeing actually the acoustic mode here. And you see that in acoustic mode is very low frequency.
And basically all of the let's move this up a little bit so it happens a little faster.
You can see that basically all the masses are moving in concert with each other, sort of slow sloshing back and forth.
Hydrodynamic oscillations you can turn viewed as one big fluid.
Everything moves together. Now if we go to K equals zero, it becomes zero frequency.
But we can click on Optic here and we'll get the optical mode. You see, the optical mode is each mass moving opposite the other mass in the unit cell.
And you can see that this is something that should be high frequency because it's compressing the springs maximally.
Now if we then change the K vector to the brand zone boundary, what we have here is,
is as predicted, that the alternate units cells are moving opposite each other.
So this unit cell, these two masses are moving opposite of these two masses.
And you can see actually why it is that only one of the two spring constants enters in the
frequency because the spring between these two masses actually isn't getting compressed at all.
It's only the spring between the unit cells that are getting compressed and not the spring within the unit cell.
It's getting compressed now if you switch than from looking at the optical mode at the zone boundary to the acoustic mode of the zone boundary,
it looks extremely similar, but it's a lower frequency here.
But you'll notice this the other spring, this now getting compressed is still you know,
you have one unit cell moving in one direction or another unit cell moving in the opposite direction.
But now it's the opposite spring, the key to spring, which is being compressed rather than the Kappa one spring being compressed.
Okay, so I highly recommend that people download this program and mess around with it.
It's got a lot of other fun features. This is what we can do on a boring Saturday night or something.
Okay, so let me redraw this picture here.
Maybe maybe stop this for a second. Okay.
Redraw this picture here. All right.
I don't know why it's still still doing that, but. Okay. All right, good.
Me redraw this picture, except I'm going to draw slightly differently.
Here's K again, here's Omega and I'm going to put Pi over here and minus pi over here and then two pi over here and then minus two pi over here.
I'm going to draw the acoustic mode just like I had it before. And then the optical mode, what we had before was something that looked like this.
But I am perfectly allowed to take this piece here and shift it by two pi over a and plot it instead here.
Okay. So I did research at this piece and I shifted it by two pi rate to move it over here.
And then I can take this piece here and shift at two pi over a this way.
And put it here. Okay. So now I'm going to erase.
I know people hate when you do a race of transforms, going to raise the peace in the middle.
And I leave these pieces here. So this piece and I have this piece.
Okay. So I just took this piece and moved it by two pi over this way, and this piece moved by two pi over three that way.
And I got this picture here. Now, this way of drawing things is known as the extended zone scheme.
Extended Zone Scheme.
And it has a nice advantage that there is only one K, there's only one mode at each frequency at each wave vector.
K So whereas over here we have two modes at each wave vector.
K Here we only have one, but we have used twice as much range of K and that's sometimes convenient to do just some nomenclature which is useful.
This range here is known as the first boron zone.
Boron Zone. And this region here is known as the second grand zone, along with this region here, second brand zone.
So we took the optical mode and we moved it out of the first boron zone and put it into the second run zone for the extended zone scheme.
This way of drawing things over here in comparison to the extended zone scheme is known as
the reduced zone scheme because everything has been reduced into a single bronze zone.
Now, why is it that it's so convenient to to spread everything out so that there is only one mode at each wave vector?
Well, a case where this becomes extremely useful to do is the following case.
Suppose consider Kappa one very close to Kappa two.
So in that case, what's going to happen is that these.
This little gap here is going to get extremely small. So let's see what that happens, what that looks like.
So here's K, here's pi over a year to pi over a year minus pi over A and here is minus two pi over eight.
So it's going to look like it's kind of this.
But like this. Like that is a very small gap at the bronze zone boundary here.
Because Capital One and Capital are very close to each other. Now, what happens when Capital One actually equals capital?
Well, then what we have is the gap closes. It becomes a single, single connected line.
But more importantly, we recover the monetary McCain Right.
If Capital One equals capital, then every mass and every every spring is the same as every other spring.
Every mass is the same as every other spring. So we get back the modern atomic chain.
But the lattice constant has changed.
The new lattice constant is only half as big as the old lattice constant,
because the white and black masses now actually are the same, the same object now.
So we don't have to make a unit.
So with two things, we now make a unit cell with only one thing in it. Right.
So the bronze on the bronze on the back.
Now ranges from K is goes from minus pi over a prime to pi over a prime.
Okay. And that is the same as two pi over a minus two pi over a two to pi over a.
Okay. So this entire range here, one, if the two masses become the same, this entire range here becomes the first branch zone.
First because of monotonic chain. My atomic chain.
And then if you make Capital One capital slightly unequal, you open a small gap at the new own boundaries at PI over a.
But really what's underlying is a single mode for the atomic chain from here to here and here to here, just a single connected node.
And you open up make when you make capital one in capital slightly different, you open up a small gap at the bronze zone,
at a bronze on boundary of the new smaller branch zone associated with the new larger unit self.
Happy. Okay. In the last few minutes, I want to discuss a completely different topic,
something we didn't we didn't get to last time, which was or the time before, which is van der Waals bonding.
Van der Waals. You may remember this guy from the van der Waals equation of state that you probably studied in stap mac van der Waals bonds,
also known as molecular bonds. Molecular or fluctuating dipole.
Fluctuating dipole bonds.
Bonds. And these are Vancouver's bonds are actually quite different from either ionic or covalent or
hydrogen or metallic bonds in the sense that these bonds occur for inactive chemical species,
inactive species. And what I mean by inactive species is that no electron is transferred between atoms, no electron is shared between atoms.
Things like noble gases are classic cases of Andreev bonding.
The noble gases are filled shell. It doesn't share its electrons.
It doesn't donate electrons. It does not covalently bond.
It does not, ironically bond. But it can still van der Waals bond.
Now, another example of an inactive species are molecules such as nitrogen to nitrogen by itself as an atom is very active.
But when you put two nitrogen atoms together, they form a nitrogen molecule, which is extremely inactive.
You can think of it as being filled shell of a filled molecular shell, and it also does not share.
Once you have the molecule, the molecule is very stable.
It doesn't share its electrons with anything else and it doesn't donate its electrons to anything else.
So nonetheless, even though you're not donating electrons and moving electrons around, you can still running out of space.
You can still form a bond.
The way the bond occurs is as follows Imagine that you have, say, a nice, noble gas atom over here and maybe another nice noble gas atom over here.
And I suppose at some moment this atom has some dipole vector p want a polarisation?
Now if you remember from your in m a distance away, there will be an electric field due to that dipole moment.
In this case pointing in that direction and e will be equal to minus p one over four pi epsilon, not our cubed.
And if it was, if I had drawn the angles differently, I conveniently chose R to be perpendicular to the moment there.
But if I didn't make them perpendicular, there'd be some cosines and signs and stuff like that, which I'm leaving out now.
When this atom experiences this electric field, it develops a polarisation, it gets an induced polarisation.
P two which is chi some constant times.
E This is this constant here is known as the polarised ability, polarised ability or electric susceptibility, electric susceptible.
And you can calculate this for simple atoms like hydrogen, you can calculate the susceptibility in perturbation theory.
And maybe if you had done that as an exercise, as an exercise in the book, they asked you to do that using quantum mechanics anyway.
Now what we have is we have two dipole moments and you remember that the energy of two dipole moments is minus P,
one P, two over four pi epsilon, not cubed.
So the total energy is then just plugging these things all together.
We get minus p one squared chi over four pi epsilon not r cubed all squared.
And in particular this is negative and goes out to the sixth.
So the force goes as one over to the seventh.
Now that is more or less how the calculation goes that you have a polarisation,
one atom induces a polarisation in the other atom and the polarisation is a track.
But you might say, well, wait a second, this whole argument was predicated on the statement that the polarisation is not equal to zero to begin with,
but in a spirit, the symmetric atom like like helium, the expectation of the polarisation is zero.
Why is that? Well, you remember that the the polarisation of an atom, here's the nucleus,
here's the electron electrons running around the atom in some sort of spherical orbit.
So the, the position are of the electron is related to the polarisation just by minus eight times R or something like that.
So it's really, we're just saying that the average position of the electron is actually at zero,
but you'll notice that what actually comes into this equation is not the polarisation or the dipole moment P, but it's P squared.
So P squared is not equal to zero so that you can get a nonzero dipole or attractive force, even though the average dipole moment is a zero.
If you like to think about fluctuations in quantum mechanics as being some sort of dynamical fluctuation or fluctuation in time,
what we really mean is that it's a sort of a course picture,
a crude picture of what quantum mechanics is, but you can sort of think of it as at some moment in time,
the electron is on one side of the nucleus, so there's a dipole moment.
The other atom responds to that and oriented dipole moment in the opposite direction than the attract.
And then at some later time, there's a question. Yeah.
If there's an angle between. You're saying that if there's an angle between you two, there would be cosine theta and so forth and things like that.
Yeah, but the force will still be attractive. You can.
And actually there's, there's an exercise in the book that works you through it.
And you can and you can get all the angles right. And it will always come out attractive.
Good question, but not worth the chocolate bar. So but it's still a good question.
Thank you. Anyway, the.
So it's basically you get a little bit of fluctuation on one atom.
It causes a fluctuation in the other atom and they attract.
And no matter which direction the fluctuation is, the responding fluctuations in the opposite direction and they always attract.
So you can still get a nonzero van of force. Now the van de Vos force is much weaker than covalent ionic or hydrogen bond,
but it's still strong enough to cause important effects such as, you know, it holds together things like argon at low temperature.
And when argon becomes a solid a more interesting case of where Vandervoort forces show up is with this guy.
If you've ever been to a tropical climate, this is a gecko, a little lizard.
They also they sell car insurance in the United States. I don't I don't know why that's true.
This is the mascot of the Gecko Geico Gecko Company.
Anyway. Geckos are amazing little creatures because they can crawl up almost completely flat glass walls with no trouble at all.
And the thing that actually sticks them to the wall turns out to be Van der Waals forces.
They have you know,
they have a very flat foot and they stick their foot onto the glass and there's enough van der Waals force between their foot and the glass.
And they're fairly like little creatures that they can actually walk up the wall because of that.
All right. I apologise. I went over. I will see you on Wednesday.