This is a. The Six Lecture of the condensed matter course.
In the last lecture, we discussed bonding, ionic bonding and covalent bonding,
and there were three other types of bonding that were on a list that we were going to cover eventually.
Van der Waals bonding, metallic bonding and hydrogen bonding. And for various reasons, I'm going to push those off until later.
We'll try to pick them up where they fit in. Instead, what we're going to do is we're going to go back and reconsider the motion of atoms.
We're going to simplify our life an awful lot today, and we're only going to consider the motion of atoms in one dimension.
So for the purpose of today, we live in a one dimensional world. We're going to summarise everything we know about bonding.
Very simply, we're going to imagine we have two atoms some distance apart.
It's called a distance X and there will be some potential V of X, which represents the force between the two atoms and the potential V of x.
As we discussed last time, it probably looks something like this v of x.
Something like that. So it has a attractive bonding force.
And then if the atoms get too close, the potential shuts off to infinity.
Okay.
So what we're usually going to do is we're going to expand around the bottom of this minimum in a quadratic way and approximate it as a parabola.
So we'll write V of x is some V not v not is the bottom v not here.
Plus it's called this distance here. From here to here, let's call that x equilibrium.
That's the the bottom of the well. In other words, the distance at which the atoms would most like two sets the minimum of the energy.
So we'll give a quadratic term here, x minus X equilibrium squared plus dot, dot, dot.
Okay. So the the energy is minimum. If the two atoms are separated by this equilibrium distance and then it's some sort of parabola,
if you if the atoms are either farther apart or closer together.
Okay. Now, we should be a little bit cautious in doing this,
because occasionally by approximating something as a parabola, you throw the baby out with the bathwater.
And in particular, if you're worried about thermal expansion, it's very important to keep the dot,
dot, dot terms, the fact that in fact the potential is not a parabola.
So let's think about that for a second. How does thermal expansion happen?
Well, if you're at low temperature, the atoms, you sort of think of a like a particle in the bottom of a potential.
Well, but what we're really talking about is the distance between the two, the two atoms.
But you can sort of think about that distance oscillating back and forth, just like it was a particle in the bottom of a well.
So it oscillates back and forth. The atoms get farther apart and closer together.
They oscillate back and forth and pretty much the average distance stays at equilibrium.
But if we give the atoms some higher amount of energy here, a higher temperature,
then the atoms can oscillate in to here, but out all the way to here.
Okay, because the potential is steeper on the inside than it is on the outside.
Generally the atoms will be able to make it far.
This x max here and this X-Men.
X-Men will differ a different amount from x equilibrium in particular, x max plus X-Men over two will be greater than x equilibrium at t equals zero.
So the average distance that the atoms are from each other when they start oscillating will start to increase.
And this comes from the fact that when the when the atoms oscillate, they can push in a little bit,
but the potential is really steep, so they can't push in that much.
But then when they oscillate out, the potential is softer so they can go much farther distance out.
So this is what gives you thermal expansion,
the particular form of the potential function that's softer as the atoms go away from each other and gets very,
very steep when the atoms get close to each other.
But as long as we're not considering things like thermal expansion, it's okay to just truncate our potential at quadratic order.
And we have a pure Hooke's law type spring between our atoms.
So far so good. Everyone happy with that? Okay.
So what we're going to do with the rest of the lecture is we're going to take a very simple model of atomic vibration,
actually extremely simple model of atomic vibration, which is known as the one atomic, one atomic harmonic chain.
Harmonic chain, which is potentially the most important model we're going to study all year,
not only because it happens to show up on the final exams very frequently.
So one atomic means that there is only one type of atom.
Harmonic means that it's going to be just simple springs between the atoms and chains mean we're going to have a lot of these atoms.
And the reason it's a very simple model, but the reason it's so important is because it introduces a lot of ideas.
It will come back over and over again throughout the term.
So this is what it looks like.
We have a bunch of atoms and again, we're living in one dimension, the lined up in a row, each atom has some mass m all of them identical.
And then there's a spring constant, all of the spring constants identical, identical between the two atoms, Kappa, Kappa Kappa and the Kappa.
The spring constant comes from the expansion of the of the bonding potential, the harmonic expansion that we used up above.
Okay. The equilibrium distance between the two atoms.
X equilibrium here. Let's call it a.
For the purpose of of argument generally put it over here.
This distance A is known as a lattice constant.
And generally we're going to use the word lattice constant frequently later in the term.
Lattice constant generally generally means distance between between identical atoms.
And in this case, all of our atoms are identical. So it's just the distance between the atoms here.
Let's define some positions here. So let's call this 1x1.
Maybe this one will be x2x3 that.
So generally say that x of an is position of Adam and.
And let's let xe and the superscript zero be the equilibrium position.
Position of Adam and. So that's you've imagined letting the chain come to rest and you measure the positions they're all spaced by a distance.
A So X and zero is just we can let it be.
If we let the zero s atoms start at position zero, then we can just set zero to be n times a each atom separated by a distance a from the next.
And that quantity we're actually interested in is the deviation from the equilibrium position, which we'll call Delta X.
So delta x sub n is x, sub n minus x and not.
And what I'm going to try to do is we're going to try to figure out the vibrations of this chain using completely classical physics to begin with.
And we'll worry about quantum physics later.
So you've probably done a couple of spring and math problems probably in your first year, and it's pretty straightforward.
What you're supposed to do is you're supposed to write down Newton's equations for all of the masses.
So that's pretty easy. We just have an ethical summary. So F on the atom and is mass times the acceleration delta x double dot.
And so what is the force on an X on on the mass.
Well okay, it has a force from the atom to its right, so it's Kappa Hooke's Law,
Delta X and plus one minus Delta X and and then it has a force from the atom on its left, Kappa Delta X and minus one.
Minus Delta X and. I guess we can simplify that a little bit by writing.
It is Kappa Delta X and plus one plus delta x and minus one.
Minus two. Delta x and. And.
What we'd like to solve for is we would like to solve for the normal modes.
Want, want normal modes of this chain.
Normal mode to remind you means all atoms oscillate.
At. At a common frequency. At common frequency.
And if you remember how you did this in in your first year courses,
it ended up being an eigenvalue problem with sort of a matrix, the dimension of the number of of masses you have.
Now, we might have an infinite number of masses here to deal with if we have a very long chain.
So it looks like an infinitely large eigenvalue problem and that might be sound a little bit frightening,
but it turns out that solving problems like this is really easy and we're going to do it again.
We're going to use the same kind of trick over and over this year, and the trick is to guess the answer.
And fortunately the guesses are all the same. So it's easy to guess. The guess is you use what is known as a wave on dots.
On dots is a German word that means something like Guess so.
We're going to guess that the solutions are wave forms. Someone who speaks German is probably saying, no, that's not what it means.
Is that what you're saying? Yeah, I have no idea.
What does it mean? Rule of thumb or something? It means work.
Okay. I stand corrected. Thank you. Well, in physics, we frequently we use it to mean.
To mean. Guess what? So I don't know where it came from.
All right. So so the trick is we are going to guess that that the that the the form of the of the oscillations are simple waves.
So we'll write down a for a wave form delta X is some constant a you know,
the I omega t minus i k times x and not k here is the wave vector and omega is the frequency.
Now, this might be a little bit confusing because what we've written down is we've written down something complex,
whereas we know the positions we're interested in are actually real.
But this is just like when you study circuits in your first year, you're trying to think about currents that are oscillating,
and instead of writing signs and cosines, you write down some sort of complex expression.
And what you really mean is that you're supposed to take the real part at the end of the day.
And that's what we really mean here is at the end of the day, you should take the real part.
And the reason we do this is because it's always easier to work with exponentials than it is to work with science and cosines.
Now, because we're going to take the real part,
we could put an overall minus sign up in the exponent and still get the same answer so we can fix that.
Omega is always greater than or equal to zero without any loss of generality.
Because if you just change the sign of everything in the exponent, once you take the real part, you end up getting the same result.
But we must keep track of the fact that K can have either sine k either sine.
Which corresponds to either a left going wave or a right going wave.
The other thing, I guess we should probably substitute in here the value of X and zero, which is end times A.
Okay. So then all we have to do is we have to take this wave and that's plug it in to Newton's equations up there and see what we get.
All right. Well, if we plug it in on the left hand side,
we get minus M omega squared times A in the I omega T minus i k and a right to derivatives brings down minus omega squared.
And then on the right hand side, we get a kappa a kappa a there's an eye omega t which is common to all the terms.
And then we get either the minus I k and plus one a plus either minus i k and minus one a minus to in the i.
K and a. Good.
Now we can cancel out a whole bunch of things from this equation to simplify our life.
So we get minus m omega squared equals kappa in the i k a plus minus i k a minus two.
Then we can use a trig identity.
This thing is actually two times a cosine and I actually move the m to the other side as well and we get omega squared is kappa
over m two minus two cosine k another trig identity that we can replace one minus cosine k a as sine squared of k over two.
So that gives us a total of four with. Yeah. Two on the four K over time sine squared.
Okay. Over two and then I'll just take the square root of this whole equation.
We get the final result. Omega is two square root of Kappa over m absolute value of sine k over two.
And there we have it. We've, we solve for the frequency of the normal modes of our chain.
Given the way. Vector K. Now, it's probably worth plotting the answer.
Just go. So what is it look like?
So here we have omega of k vertically and we have K horizontally.
Like this and we'll put some points on let's make this point pie over a maybe this point over here is minus pi over a
and we have the sine absolute value of sine and it looks like this absolute value of sine kind of looks like this.
That. Okay, well, this must be the same height.
And the height of this, uh, curve here is to square root of k k over M, and it has its peak at pi over A and at minus pi.
Okay, so this curve is known as a dispersion curve.
Dispersion. Dispersion.
Term just means omega as a function of K.
Now, somewhere in this picture, we should expect that there should be sound waves.
We're talking about oscillations of of some solid.
I mean, we have this we have our picture of a solid, which is just atoms stuck together in springs and somewhere in there.
There should be something that looks like sound waves. Where are the sound waves?
Well, sound wave is a very long wavelength phenomenon. The definition of sound is it has to be a long wavelength oscillation.
So we should really be looking down here at small K to look for sound waves, just a as a, you know,
a sort of pneumonic that the sound that you hear has wavelength somewhere between like a centimetre and several metres.
That's what you can you can hear depending on how good your ears are.
Whereas the spacing between atoms that we're talking about this pi over a a year one is so this the wavelength
associated with this pi over a wave vector is on the order of the distance between atoms with a appears here.
So two pi over the way vector will give you a number on the order of the spacing between atoms.
So most of this picture is very, very, very small wavelength compared to what we usually think of as sound.
Sound is for very, very small wavelengths right down near the middle, near K equals zero.
So let's actually try to figure out what the sound will be doing.
The way we figure out what sound is doing is to go to K very close to zero.
We expand this thing, we expand the sine and we end up getting what do we get?
Well, the one half cancels the two and we get square of Kappa over m r times a times absolute K And you remember the definition of the sound velocity
is that frequency should be sound velocity times absolute K And so we've derived sound velocity is square root of kappa over M times.
A There we have our sound velocity now.
If you took the fluids course last term, you'll remember that sound can also be defined in terms of things like compressed ability of a fluid,
and we should be able to get the same velocity of the sound from the formula that you derived last term, which is square root of one over density.
Mass density here. Mass density times the data, the compressed ability.
Okay. So what's the mass density of our chain? That's easy. It's just one mass per distance, eh?
And what's the compress ability? Well, first we have to figure out what the definition of compress ability is.
It's minus one over V, the volume, the pressure.
But we're in one dimension, so that gets replaced by minus one over a length d length de force.
So if you think about one little piece of the chain, the length is one over a length is A and D length.
The force from Hooke's law is minus one over Kappa,
so the compress ability is just one over Kappa A and then if we plug in these two expressions into that, well, okay, let's do it here.
This is then using this mass dancing, this compressed ability.
It's one over a, mass over A and then of Kappa one over Kappa A, and that gives us,
in fact, the same result that we had over here, square root of Kappa over M times.
A So it agrees with the expression we would get from a hydrodynamic picture of what sound waves should be.
Okay, so everything all sorts of seems to work pretty well, but that is way down here in this regime,
whether where the dispersion is linear at higher k, it's not linear anymore.
It starts to curve. Now think back to what it was that Debye was saying.
Debye was saying he was going to just assume that the frequency of his sound modes was always linear in K,
and then he cut off the spectrum at some maximum frequency.
Well, it starts out linear. It doesn't stay linear curves, but indeed it does have a maximum frequency above some frequency.
There's no sound modes left. We found all of the normal modes of the system and this is all of them.
And they have only and they only have frequencies between zero and this frequency here.
So Debye was right in that respect that there really is a maximum sound wave frequency.
It's not well, we wouldn't call sound, but vibrational frequency of of this chain.
It's worth looking a little bit more closely at this point here, which is the highest frequency excitation, the highest frequency normal mode.
So that occurs at K equals PI over A.
And how are we going to look at that more closely? Well, let's let's write down what the wave form would be.
It's a heat of the eye, omega t minus ei pi over a times a and or it would be a in the eye omega t times minus one to the n,
which means that every alternate, even versus odd mass, is moving in the opposite direction.
And that's the highest frequency you can possibly get. Okay. At this point, it's worth seeing a movie.
So let me. Oh, here we go.
See if it works today. Oh, beautiful. Okay, so this is a program we can download from my website.
It was written by Mike Glaser. It works on Windows, it works under wine and Linux.
I think it probably works on on Mac, but I'm not sure about that.
So the it the purpose of it is to show you what oscillations of one of these chains actually looks like in detail.
So here we click anatomic chain and we're going to click longitudinal, which means, oops, my gosh, change the speed.
There we go. So you can change the speed of oscillation as well to make it easier to see.
So what we have here in this corner over here, you have this dispersion curve and you can change k back and forth from high K to low K.
So let's start with a very fairly low K and make the speed fairly fast so you can see what's going on here.
Okay. And you can see at fairly low k come on.
Yeah, you can see that this is sort of a long wavelength oscillation where everyone sloshes left,
then slashes to the right and back and forth and back and forth. Now as we increase K,
it oscillates faster and faster and the wavelength gets get smaller and smaller and until eventually it's oscillating really, really fast.
And I'm going to have to slow down the speed so you can actually see what's going on when you get to PI over a this point here,
you can see that every other mass is going in the opposite direction.
And if you think about it for a while, you'll realise that this is there's a good reason why this is the highest frequency that you can possibly get,
that you can't do any better than having every other mass opposing its neighbour,
that if you want to get really high compression or high frequency, this is the highest frequency you could possibly get.
Okay, so I'll. Leave that for you guys to play with.
I recommend messing around with it a little bit. Oh.
Are we going to have problem turning back on? All right, anyway.
Good. So the most important thing that we that we deduced in this calculation is that the dispersion is periodic in K.
It is proportional to assign and you'll see that the dispersion just keeps going
exactly like itself over and over with sort of a periodicity know from here to here,
you just keep replacing over and over. It looks exactly the same. This is not a coincidence.
There's a general law that if you have something that's periodic in real space,
real space like a chain which has a periodic periodicity, delta x equals a.
If you take our whole chain and you shifted over by A, you get back exactly the same chain again.
If you have a something that's periodic in real space. I should also say that real space is also known as direct space.
Direct space. That will imply that you have a dispersion, which is periodic in K and K space, also known as reciprocal space.
With a periodicity. Delta K equals two pi over a and hence the name reciprocal.
Because here the A's upstairs and here the A is is downstairs.
Now the periodic here's some some more words that are going to be useful. The periodic unit here is known as the Bruen zone for a one zone.
Maybe I'll write that definition a little bit because it's rather important.
The periodic unit in case space. In case space.
Equals the brand's own. And I apologise not only to German speakers but to French speakers.
Now that this word Bruyne, Louis de Bruyne is a word that English speakers have a terribly hard time pronouncing correctly,
and probably anyone who speaks French will cringe at how I butcher it.
But I'm not the only one. Every English speaker butchers this name.
I really apologise about that. I almost failed French in high school, so I've never gotten any better at it.
But I think it's something like brie wine. That's my best guess about how it's supposed to be.
Okay, now, why is it this is a very important principle that we have periodicity in case space?
Why is it we have periodicity in case space? Well, let's find out.
We have our waveform delta x n is a, b, i, omega T minus I, k and A,
and then let's take K and shift it by two pi over a shifted by this entire length.
From here to here, we'll take some particular K and move it by two pi over A and see why we get back the same thing.
Well, let's do that shift in the make a T and then we get minus i k plus two pi over a times
and a that equals a in the iomega t r minus i k and a times e to the minus I to pi.
And, and this thing here is one. So in fact, we're getting back exactly the same wave form that we started with.
The reason the dispersion curve looks the same at this point here and this point here is because they're describing exactly the same wave.
The wave is unchanged by shifting K by two pi over a.
This is an extremely important principle that will come back over and over and over.
Now, this might be something that you'll find disturbing to begin with.
Ah ha ha ha. Okay, now we have a problem here.
Okay, now I'll see if I can show you this. Come back. Okay.
Let me show you. Something else.
See if it comes back up eventually. Yes. Okay, good.
So you might ask, well, if K is the same as K plus two pi over a, what's the wavelength.
Right. I mean, we usually think of the wavelength as being two pi over a two pi over cake with the wavelength two pi over k because lambda.
So if k is the same which isn't physically the same wave here and here, which one's the wavelength?
Well, the way to understand that is from this picture here.
So. Delta x the x here is plotted on the vertical axis just to make it more clear.
So I plotted two waves here. One of them, the solid line is at K are actually mean.
The dashed line is K and the and the solid line is at K plus two pi over a OC.
The two curves agree at the position of the masses.
And we're only talking about the displacement of the masses. The waves doesn't mean anything in between the position of the masses.
The wave form is meant to describe what the masses are doing.
You can't ask what is the value of the wave in between the masses?
It doesn't have a value. The wave is meant to describe what the waves are doing, what the what the masses are doing.
So it doesn't actually mean anything whether you extend the the curve in the solid line way or the dash line way in between the two masses.
What only all that matters is what the masses you know, how much the individual masses are displaced.
This is a phenomenon known as aliasing. And if you get this in your head, you'll understand pretty well why it is at this point.
And this point are actually describing the same wave, even though they have different case.
Okay, good. All right. All right.
So. So a couple a little bit more nomenclature, that's fairly useful.
The real space lattice lattice.
Let me define it as all points. Points equivalent to x equals zero.
So in our picture there that would be x and equals m times a.
Because you can shift each point by an entire spring and you get back to something that looks exactly the same.
We also have the reciprocal space lattice, reciprocal space subspace lattice, which is all points in case space points in space.
Equivalent to K equals zero.
And for this particular model, it would then be let's call those points g sub m equals two pi m over a.
All those points are equivalent to archaic or zero shifted by two pi units of two pi over a.
Now it's actually something that is kind of interesting and will be important later on this year is the the exam X
and is always equal to one for all of the points in the direct lattice and all the points in the reciprocal lattice,
you can choose any combination of those points and we'll always give you one.
In fact, we're going to use that when we go to higher dimensions. We're going to use that property as a way to define the reciprocal lattice.
So sort of put it in the back of your your head.
So one thing that's actually really useful for us to do at this point is to count the total number of waves that we enumerated.
We solve for normal modes of the system. So how many of them are there?
How many? How many normal modes? Normal modes?
We have. Well, we know that if we're counting different normal modes, we only have to look within the broken zone.
Then if we go outside the bronze zone, we can always we're describing something that's equivalent to another point inside the bronze zone.
If we take this point here, for example, it's equivalent to this point here.
So we only have to describe things within the bronze zone. We have to figure out how many modes there are within the bronze zone.
So we'll do the usual thing. Use periodic boundary conditions, periodic boundaries.
So we're putting the springs in a in a in a big circle. Let's let the circle have a length l which is and so and masses around in a circle length l.
So k then has to be two pi over l times some integer p.
P is an element of integers. Usual way that happens when we put everything in a periodic box.
So the total number of modes, number of modes is then the total range of k's that we're looking at two pi over.
A So that's the entire range of K's we're looking at divided by the spacing between modes to PI over L.
So what we're doing is we're dividing up this broad zone into lots of little pieces.
Whose spacing is two pi over out. Okay.
And so that is L over A or N.
So the total number of normal modes that we solve for is end.
There are normal modes because there are end masses to begin with. This should not be surprising.
You probably found this when you study mass and spring problems before.
If you start with and masses you inevitably get and normal modes.
And this is actually exactly what Devi had guessed before,
that he wanted to have just the same number of modes as he had degrees of freedom in his system.
Okay, so so far everything we've done is completely classical so far.
So now we're going to move on to quantum mechanics. Quantum.
So in quantum mechanics, there's a very simple rule, which is that if you have a normal mode, a classical normal mode of frequency.
Frequency Omega. When you go to quantum mechanics, that becomes ion states with energy.
H bar east of NW equals H Bar Omega and plus one half.
So what is this? What does this end telling us? This NW is telling us you pick a particular normal mode that you're looking at.
And this RN is basically telling you the amplitude of that mode.
Then if you if you let it oscillate a lot, it has a lot of energy.
If you let it us a little bit, it has only a little bit of energy. So it's a difference between big and small end.
But the key thing is that the energy is quantised in integer units of this bar omega.
Now this is an important definition that one quantum.
A quantum. A quantum of vibration.
Of vibration. Is a phone on is known as.
A phone on. Is a phone on. A phone on. This is entirely analogous to a quantum of light being a photon and its energy.
Energy is h bar omega of K depending on which which K we're actually considering.
So I now I have to I have to rant for a second about my my normal pet peeve.
So my rant is that if you open up a lot of books,
they will tell you that a phonon is a quantum of vibrational energy or a photon is a quantum of light energy.
And I think that's actually a bad definition.
I think you shouldn't specify energy because it's true that a photon carries light, but it also carries angular momentum and carries momentum.
So why did you specify energy and not just and not just specify there's a quantum of light.
Similarly, with phone on here, a phone on carries carries energy, but it also carries momentum and other things.
So I'm just going to say a quantum of vibration is the phone on.
So it'd be a little bit more specific about what I mean here by by a phonon we pick a particular normal mode or a
particular oscillation frequency of particular wave vector K and then we say if it's in its ground state and equals zero,
we say there are no phonons filling that state.
If you excite rn up to one, we say there's one phonon in that mode.
If we excite and up to two, we say there's two phonons in that mode.
It's exactly similar to what you do with photons and light.
Now, of course,
these phonons have to be bosons because you can put phonon is Bose on is a boson because you can put more than one phonon in a given mode,
given k mode. And it's energy.
Energy of the phonons in a particular K mode is h bar omega k.
So this is the energy of the phonons in a particular K mode times the Bose factor.
There's another indication that we're talking about bosons here.
Bose factor. This factor plus one half.
So the bonus factor here tells you the expected number of phonons in the particular mode at a given at a given temperature.
Okay. Is everyone comfortable with this? It should look a lot like like what you did with photons earlier.
Yes. Yeah. Okay, good. So now what we can do is we can use this information to actually calculate the quantum mechanical heat capacity of our chain.
Now, our chain is a rather simplified chain. It's a simplified model of what's going on.
It has you know, it has these a number of simplifications that, you know,
the springs are only between near his neighbours and the perfectly harmonic springs.
But given those simplifications, what we're going to end up with is an exact answer for the heat capacity.
No approximations at all. So the exact answer for the heat capacity, the total energy in the chain is the sum over all of the modes.
K equals minus pi over a two pi over a of h bar omega.
That mode those factor beta h power and make a k plus one half where the keys are taken in in steps of steps of two pi over l.
Now, the usual way that we have before, so many times we can replace that sum with an integral decay over two pi times the length of the system.
And we only have to integrate from minus pi over eight to pi over a because once we're outside the Brian zone,
we're just re describing some of the modes that we have already previously describe.
And if we count the total number of modes in the system, total number of modes, total number of modes.
Number of modes.
It would be well, okay, we can write it as l over two pi L over two pi integral four minus pi over a two pi over a decay of the number one.
So we're just going to integrate over all the number of modes called the total number modes that will
give us what L over A or the number of atoms in the system is again exactly similar to what divided.
He cut off his integration over modes such that he would have exactly and modes in the system no more, no less.
So we use we use our omega of K that we derived wherever it is.
Well, off the top of the board, I'll write it again. We use two square root of Kappa over m absolute value sign of K over two.
And if we plug that in to that, some of which we can turn into an integral,
we will get the exact amount of energy in that system at any given temperature and you can differentiate it to get the heat capacity.
The by used omega k equals v sound times absolute value of k which agrees pretty well
at small k but doesn't agree at large k but it was a pretty good approximation.
Einstein You can describe Einstein the same way. Einstein used all his oscillators omega k r at the same frequency omega zero.
That will give you the Einstein model basically saying there's an oscillators and they all have the same, same frequency at any rate.
This enables us to write down at least at least you can algebraically write it down
an exact expression for exactly the heat capacity of our chain quantum mechanically.
There's one more rather important concept that we need to introduce today, which is, again,
coming back to this idea that Kay and Kay Plus are a member of the reciprocal lattice two pi over a represent the same wave.
I mean, the fact that we we, you know,
we like to think of these phonons as being particles the same way we think of photons as being quantum mechanics
is great because you can think of things as particles and waves and waves and particles can go back and forth,
however, however you like.
So sometimes we like to think about these phonons as being waves, and sometimes we like to think of out of them as being particles and particles.
We like to think about them as, you know, some object that's moving and carries some amount of of momentum.
And in fact, sometimes in some models, these phonons can be more complicated models than this one.
But in your phone asking can scatter into things in their way.
They can scatter into other phonon scatter off of other phones, that can scatter off electrons, they can scatter off of light, for example.
So we'll discuss some of those things later in the term.
And the thing you might you might think that there would be a conserved momentum in the scattering process,
but how is it possible we can conserve momentum in a scattering process if K isn't
even well-defined up to this change that K and K plus g represent the same wave.
So what one does is one defines define what is known as crystal momentum.
Crystal Crystal. Well, okay, if it's K, it's the usual thing.
If it's K, it's way vector. If it's H marking its momentum, crystal momentum.
So I'll put in the H bars crystal momentum, which is defined as h bar k modulo two pi over a.
Now what's modulo mean modulo is a word which means up two additive terms of additive terms.
Of. So, for example, 12 mod five equals seven mod five modulo five because up to an additive term of five, 12 and seven are the same.
Two mod five is also the same. So in other words,
where we're confessing that momentum is not conserved up to this factor of two pi over a that you can take and give factors you know,
additive terms of two pi over a freely and this might disturb you because you might you know,
you might really like this thing called momentum conservation but we really need to
think back to why is it that we wanted momentum conservation in the first place?
If you took the symmetry in relativity course last term, you'll remember that momentum conservation comes from a symmetry of space,
and the symmetry of space that it came from was translational symmetry of space.
And so you can translate, you know, your attention in space to anywhere you want.
And all the laws of physics remain the same in our model here.
We don't have that anymore because we're only allowed to translate by a our translation group is smaller than in, you know,
in out in the vacuum of space here we were only allowed to translate by a and because of that
we get a less strong conservation law that instead of having momentum completely conserved,
we only have momentum conserved up to two pi over a week.
So that's what's gone gone on here. It's very similar to this idea of North as Theorem.
It's not exactly north this theorem into Did people tell you about Amy North?
There she was. She was amazing mathematician from the early part of the 1900s.
Northeast Theorem, strictly speaking, only applies to continuous symmetries.
And this is a discrete symmetry, but it's very, very similar.
And the reason we're getting less strong momentum conservation than we had in, you know,
in in when you think about regular particles in space is because we don't have the same translational symmetry of space.
We only have a discrete translational symmetry. Okay, we have a few minutes left.
So I'm going to I'm going to add in subject that I promised you I would address, and that subject is hydrogen bonding.
So this is a totally different subject. I'm just picking this up because it's only could take us 5 minutes.
Hydrogen bonds. The general idea of a hydrogen bond is that hydrogen is a very special element because it's basically just a proton and an electron.
So you might imagine having a big atom like so here's a big atom like fluorine,
and then it bonds to a hydrogen atom, more or less ironically, but maybe slightly covalently.
So the hydrogen atom is sitting out here, it's a proton and its electron is more or less sitting over here.
Its electron has been sucked to the the fluorine atom, and what's been left behind is a bigger proton.
So this is our hydrogen nucleus, and what's left is a better proton.
A proton. And that's extremely special because it's the only time in chemistry where you ever get a better proton.
It's not screened by some electrons around it.
So whenever you have chemical bonding, usually some of the electrons are moved around, but there's always some shell left over,
except in the case of hydrogen, where the one electron can be removed from it and you're left with a proton.
And that's why hydrogen bonds in very special ways.
So what can happen then is if you have another atom over here, say, oxygen or maybe another fluorine atom,
this bearer proton will then make a very strong polarisation of this fluorine atom and attract it.
And this is the hydrogen bond, h bond, which is the bonding of the hydrogen to another atom via this strong dipole force.
So let me show you a classic picture of this, how this shows up.
So ice is sort of the classic example of that where you have hydrogen sitting on the surface of oxygen to make your H2O.
But the oxygen has essentially removed the electrons from the hydrogen to a very large extent,
leaving behind a bare hydrogen, their nucleus of a proton,
that their proton polarises this other electron over here and attract this other electron here on this oxygen and attracts the oxygen,
making a very weak dotted bond to this big red oxygen sitting over here.
And this you know, it's a weaker bond. It's a much weaker bond and a covalent bond or an ionic bond.
But it's strong enough to freeze ice, you know, below below zero.
So it's a you know, it's a strong enough bond to do some interesting things.
Another classic case where hydrogen bonds enter is in biology and particularly in DNA.
If you took the biology segment last term, you'll recognise this as being DNA if you didn't.
Well, now you're seeing it. This is DNA and you'll see it.
So DNA is made up this this double strand.
And right down the middle of the double strand where the two strands zipped together, there's a hydrogen bonded to an oxygen.
This bond is the hydrogen bond. The hydrogen is really bonded to the nitrogen over here.
And it has a weak bond to the oxygen over here. And you'll see that there's a similar hydrogen bond from a hydrogen here to a nitrogen here.
Similarly, hydrogen bonds right down here and hydrogen bonds down here, right here.
So it's a hydrogen bond that holds together the DNA right down down the zipping.
So. So when you pull the strands of DNA apart, you're breaking hydrogen bonds.
So I think that's all pretty much you need to know about hydrogen bonds.
And I guess we will finish. We'll pick up again tomorrow.