This is the fourth lecture of the cancer matter, of course,
where we left off last time we were talking about the free electron or the Sommerfeld Theory of Metal's free electron.
Sommerfeld. Theory of metals.
And when Sommerfeld was doing is more or less following through this idea that
a metal is just a gas of electrons and he's trying to apply kinetic theory.
The only thing he was doing differently is he was respecting Fermi statistics.
He was keeping track of the fact that you can't put two electrons in one eigen state.
And last lecture we derived that the Fermi wave vector is related to the density of
electrons by three pi squared and to the one third where n is the electron density.
And unfortunately, in the last lecture I made an error.
I mistakenly call this thing the Fermi momentum.
It's actually the Fermi wave vector and I even wrote it on the board incorrectly.
H five times the way vector k is the Fermi momentum.
Sorry about that. From the Fermi wave vector.
We can get the Fermi energy in the usual way.
You have a squared KF squared over two m which we can substitute in our expression for
the Fermi wave vector h bar squared over two and three pi squared and to the two thirds.
This is a rather important relationship that we'll use again.
But what's important to realise here is the bigger the density of electrons you have, the bigger the Fermi energy.
And in a typical metal like iron ore lead, the density of electrons is really big,
a couple of electrons per atom and you have a whole lot of atoms, the very high density of atoms, an atom, every couple of angstroms.
So the Fermi energy gets to be enormous on the order of 80,000 Kelvin, or even bigger sometimes.
Now in in this lecture, what we're going to aim to calculate is something that we discussed earlier, the heat capacity.
And from experiments, we know the heat capacity from metals at low temperature takes this form of tube plus plus t,
whereas this tube term comes from vibrations or debye theory.
We discussed that already, and this gamma t term is special to metals and in fact is the heat capacity of the electrons.
So it's this gamma t term that we're going to be interested in today.
Now, before we actually do this, we need to do a little bit of preparatory algebra in particular.
We're going to need to take some over eigen states and put it into a more workable form.
And the eigen states in this case are going to be, you know, plane waves, even the I, k r and again, since we're writing this is exponentials,
what we're implicitly doing is we're putting the thing in a periodic box for von Karman, boundary conditions,
periodic in all directions, so we can work with exponential plane waves instead of signs and cosines.
What we'd like to do is we'd like to take some of our iron seats and we would like to convert it into an integral over energy,
g of energy where G is now a density of states.
This is very similar to what we did when we did the buy theory.
However, we can do something slightly different here, slightly different.
We're going to remove a factor of the volume from the density of states.
So this is now density of states per unit volume.
And we do this because it's conventional to do so and it's convenient to do so.
And if you didn't notice it before, conventional and convenient come from the same word.
So it is both of those things. It's just happens to be handy to do so.
So we're going to do it. At any rate, the definition of this density of states is that G of D is the number of states per unit volume.
In this case the volume.
With energies. Energies?
Uh, between Epsilon and epsilon.
Plus the epsilon. Very similar to what we had for Dubai Theory before, when we were thinking about 20 states per frequency.
Now, the general idea again, is that we're going to take the sum of all the eigen states and we're actually
going to convert that some of our individual states to an integral over energies,
times the number of states at each energy, just a different way of writing it that makes your life a lot easier.
All right. So what are we going to do here to get from the sum into the integral?
Well, first thing we're going to do, some of our eigen states is really a sum of a K,
but it actually has a factor of two out front because there are two spins per k spins.
Electron can be spin up or it can be spin down with the same way vector.
Then we're going to do the same manipulation we did with the by theory.
Leave the factor of two out front is going to replace the sum over K with an integral d3k over two pi cubed.
This is the way sums get converted into integrals and we'll make that replacement many times this year.
A sum over K becomes the volume times integral d3 k over two pi cubed and then.
Since we're thinking about isotropic system, we can convert the integral over three Cartesian directions into spherical polar coordinates.
So we have to be over two pi cubed and then we have an integral zero to infinity.
For pi k squared decay where the four pi k squared is the usual for pi as the directions on the sphere.
So the usual spherical polar coordinates. And this is a pretty good result.
But really we'd like to write this in terms of energy is not in terms of wave vectors.
So we'll use epsilon as h bar squared K squared over to M or I guess we can write
that as k is square root of two m over H for our times epsilon to the one half.
And in particular that would give us k is the same fact.
Well, it's a one half times the same factor. A square root of two M over h four times epsilon to the minus one half the epsilon.
And then if we plug these things into here, what we then get is.
Okay, so we now have to volume.
I'll pull out the four pi and we have the two pi cubed downstairs.
Then we have an integral zero to infinity, the epsilon.
And then putting in those factors for the case, square decay, we get one half.
There's three of these factors, two M square over, two M over par cubed and then epsilon to the one half.
So this looks almost like what we want.
It's almost the integral of g of e d.
So we then identify, well, okay, so I'll just write it out again.
So this thing is integral zero to infinity g of e d e where we define g of e to then the.
Two m to three half hour h four cubed times one over two pi squared times epsilon two one half.
So density stays proportional to epsilon two one half.
And this is a perfectly good answer, but it's actually convenient and therefore conventional to.
To convert this factor of two over four cubed into something that looks a little nicer.
And the way we do that is by using this equation here. So that equation there and you should maybe write it over here.
If I take that equation to the three halves power I get f to the three halves
equals h bar cubed over two m to the three halves times three pi squared.
And did I do that right? I think I did that right. Okay.
And then you'll notice that I can turn this around or make it upside down to m to
three halves over a cubed is then three pi squared and over the f to the three halves.
And this factor here is this factor here.
So plugging that in we get g of of e is then what is three pi squared then?
See, I know this is a lot of algebra.
It's a big algebra day because it's a monday three has one over two pi squared are epsilon two one half and then cancelling a few things.
We get our final result. Do you have a epsilon is three half's density over e f times energy over e f to the one half.
Don't make any mistakes. Does that look right? Anyone object? Look good.
All right. This is going to be something fairly useful.
And in particular, it's useful to look at the density of states at the Fermi Energy,
which is just three halfs density over F, which I think is something that you're asked to derive in their homework as well.
The first homework set, and I'll give you a quick hint that I think there's a easier way to get there than what I just did.
But, you know, if you can't figure it out, you can just follow this.
But there's there's there's a cheaper way. But this one, this is sort of the more you know.
It is the more direct route the other way. So sneakier. Anyway, see if you can figure it out.
Okay. At any rate, now we're going to try to use this result figure.
We know the density of states here in a volume.
We're going to try to use this to figure out the heat capacity of the of the electrons at low temperature.
Now, there is always more than one way to do something.
There is the right way. And there's the cheating way.
And what I am going to do is I'm actually going to explain how the right way is done and then we're going to cheat.
And the reason we're going to cheat is because the right way is algebraically
really horrible and it's really hard to get any intuition just by doing algebra.
We just did enough algebra and I promise you, doing it the right way is is three times more so.
Furthermore, the actual calculation is so algebraically complicated that you'll never be asked it on any exam an Oxford.
And that is. I mean, I can't 100% guarantee it, but I can 99% guarantee it.
So. So. In fact.
So because of that, we're going to just do it the cheating way, which gives you the intuition for what's going on and avoids a lot of algebra.
But it's worth knowing at least how you would go about it if you really want to be honest.
So if you really want to be honest, what you would do first is you would write an equation for the number of electrons in the system,
and we wrote this equation before last time. It's the sum over all eigen states of the probability that each eigen state is filled,
and that probability of an alien state being filled is the Fermi function of beta,
the inverse temperature times, energy of the eigen state, minus metre, the chemical potential.
And being that we just derived the density of states,
we can rewrite that as an integral while volume times the integral from zero to
infinity of the density to states per unit volume times the Fermi function.
So it's exactly the same, the same expression, the epsilon, exactly the same expression, except instead of writing some of our eigen states,
you integrate of overall energies the number of states of each energy, and of course you're always integrating the probability that a state is filled.
Is everyone good with this? Yes. Yeah. Okay, good.
Now, this equation here, you can think of it in two different ways.
One way is, if you could fix the chemical potential and you knew the temperature, it would tell you how many electrons you have.
But more often than not, it goes the other way around.
You know, the number of electrons you have in your system because you know how many atoms you have or something like that,
and you know the temperature and it enables you to figure out the chemical potential.
So it's sort of an inverse relationship, you know, this, this.
And so you can figure out this in principle, although it's algebraically messy to do so,
but in principle it would allow you to figure out the chemical potential given that, you know,
the number of particles and you know the temperature, once you had the chemical potential,
you could write an expression for the energy in the system, integral again integrating over all states.
But now you integrate the energy times the probability that a state is filled.
Okay. So instead of just counting the particles, you count the particles times their energy to get the total energy in the system.
So you find the chemical potential first. You then find the energy.
Once you know the chemical potential. Then therefore, you would know the energy as a function of temperature.
You can differentiate that to get the heat capacity. So in principle, from this kind of argument, you could get the heat capacity.
If you could do these integrals. The problem is that these integrals are really nasty, and that's why we're not going to do it this way.
Instead, we're going to make some assumptions which aren't quite right, but to see what the assumptions are.
Let me first draw a diagram. This is the Fermi function again, which we drew last time and app.
Zero temperature and if goes from one to right here.
F 1 to 0. So this is this is t equals zero.
And then finite. The Fermi function smears out a little bit like this is greater than zero.
Okay, this all looks familiar, I hope. Okay. Incidentally, I believe this full calculation.
I think you actually did it last year in your stat neck course, or at least the lecturer did it.
And you probably remember that it was pretty awful.
And it's hard to actually remember anything about the intuition of what's going on, if the algebra is really awful.
So we're going to try to do this in a way that is going to give you the intuition a lot better.
So one thing we're going to assume, which is not quite right, but it's pretty close to right,
is that the chemical potential doesn't actually change as a function of temperature.
It does change the function of temperature because of this equation, but it only changes a little bit.
Why is it only changes a little bit? Well, if you look at the Fermi function here.
You can imagine as you raise the temperature, what's happening here is some of the states here that we're filled now moved to here.
So there were some electrons here. They empty out and they fill these states up here.
Now, if the number of states that empty out and the number of states that are filled are equal to each other,
the chemical potential doesn't have to move at all. You would keep in constant keeping the chemical potential constant.
Now, in truth, you have to adjust the chemical potential a little bit as a function of temperature to keep the total number of particles fixed.
But you do not have to adjust it a lot. Basically, the chemical potential stays almost exactly the same.
So we're going to make this assumption. We're going to write it. Many here assume the View is in-depth, independent of tea.
It's not quite right, but it's not too bad an approximation once we have that.
Well, actually, once we had that, we could take this equation here and then say, okay, let's plug in new fix as a function of temperature.
And then we could calculate the energy as a function of temperature here.
Assuming new is fixed, differentiate it and get the heat capacity.
But even that's too complicated because this integral is just really nasty.
So we're not even going to do that. We're going to do something even simpler.
So what we're going to do is we're going to write the energy at some temperature.
T is the energy at zero plus two very approximate things.
Approximate thing one is number of electrons that can get excited, that can get get excited.
And thing two is times amount of energy.
Of energy each absorbs.
So you imagine starting at zero temperature and then you turn on the temperature and then there's some number of electrons can get excited.
Each one absorbs a total amount of energy, some amount of energy.
And the product of these two roughly gives you the total amount of energy that you've increased your system by.
Sound reasonable? Yeah. Yes. Yes. Is it reasonable?
It's roughly true. You know, it's roughly. Right. Okay. Bear with me.
Okay. So now all we have to do is we have to figure out. What are these two factors?
So the number of electrons that can get excited. So this is this is sort of the interesting piece here.
The number of electrons that are excited is roughly number of electrons within within KB t of f y.
So you have to be so this range here is about T and you have to be within this range of F in order to get excited.
Because if you are further down here somewhere, you can't get excite.
It's hard to get excited, right? So it's you can't get it. They're supposed to be funny of.
An electron in a state down here. I can't get excited at all because all the states that we get excited into are already filled, so it can't move.
It's just stuck in that state and it's going to be frozen there unless you turn the temperature huge.
So it could jump all the way up to here. Okay. So it's the number of electrons within cavity of F and how many is that?
Well, it's basically the density of states per unit volume at the Fermi Energy Times, the volume.
So this gives you the total number of states per unit energy at the Fermi Surface Times can t so we're taking the number of states
per unit volume at this energy and we're multiplying it by this range to give you the total number of states within KB t of F.
And we take that factor and we multiply it by the amount of energy each electron absorbs,
which has got to be roughly cavity, cavity, times cavity here.
So some electron from down here within cavity of the Fermi surface got excited up to here by absorbing about cavity of energy.
And then, you know, to be to try to be a little bit more honest about the fact that we're cheating, completely cheating here.
Well, I am going to add a fudge factor.
So E total is a T equals zero plus those factors there which we just derive.
So it's V G E F times CPT squared.
And then the fudge factor is we'll call gamma will over two and gamma twiddle is just our admission of guilt that we didn't do a real,
real calculation. We and we know we're going to get the answer wrong by some factor of water one.
So Gamma Tweedle could be, you know, two. It could be a half.
It could be pi, it could be two pi. But it's not going to be 100. It's not going to be a thousand.
It's not going to be 0.01. Okay. So it is some number of order one, which is our admission of guilt that we didn't actually do the real calculation.
Okay, once we have this expression, we can of course differentiate it.
C.V. The ETI, which is this is why I put the factor of two in so it goes away when you differentiate that clever the K.B. and then G,
add the F and you're left with CPT.
And in fact, we already derive g e f over there.
So it's plug it in three halves and over e f and we're also going to use maybe I'll put it here convenient,
conventional with the density times, the volume is the number of particles.
So when I multiply this factor by write it out here.
So I multiply three halves and over e f here by the the v and the small n give you a big n
and we'll get this gamma twiddle factor times three halves big n kb times cavity over e f.
Okay. So that's our final result.
And actually, if you did the calculation really honestly, you would discover that gamma tweedle is is pi squared over three.
And I'm sure you're not going to be held responsible for knowing that.
But just for our general edification, that's what the number actually happens to be.
If you really want to see how to do this calculation, you can go back to your stat,
make notes from last year or you can, you know, it's a lot of a lot of books and so forth.
And I hate it when people say, you know, it's in a lot of books, but trust me, you know, the algebra doesn't really teach you much.
If you really want to see how it's done, you can you can work through it.
Sommerfeld did work through it. He got the right answer. But we we're not going to be held responsible for it.
Anyway, a couple of important things about this result. First of all, it is linear in temperature.
And the reason it's linear in temperature comes fundamentally from the fact that only electrons near the Fermi surface can absorb energy.
When temperature goes to zero,
the range of electrons that can get excited near the Fermi surface drops to zero and you lose your heat capacity altogether.
And in fact this is what we wanted experimentally. We wanted to get a heat capacity that is linear in temperature.
So that's the first good thing about this. And another thing that's really nice about this expression is you'll recognise this piece here,
this piece here as the classical result classical results for a monotonic gas three has and KB sounds familiar right and then
is multiplied by this factor here K over F which is what it's tiny it's room temperature 300 kelvin over 80,000 Kelvin.
Tiny, tiny tiny amount. So you have only again coming from the fact that.
Only a few electrons are participating in the heat capacity.
So this result in Sommerfeld went and he said, okay, this is my new prediction for what the heat capacity of a firm gas should be.
And so let's compare it to experiment so well.
Okay, so let's write it as gamma t all of those constants get absorbed into the overall gamma.
And he's going to well, if he if you if you want a real theory of of how much heat capacity a particular metal should have,
you need to know what the density of the electrons in the metal is so you can calculate EAF.
But he did the usual thing and said, okay, let's assume one electron per atom the same way we had done before.
And if you do that, then you write out gamma assay gamma for the experiment divided by gamma from the theory and with the theory,
assuming one electron per atom for lithium, we get 2.3 for sodium, we get 1.3 for potassium, 1.2 copper, 1.5.
This is a pretty good agreement. It's extremely good agreement considering that the classical theory,
what you really had done, the three has NCBI result is too big by a factor of 100.
And all of a sudden, we're getting results, which are really pretty close to the right answer.
And the reason we're getting results that are close to the right answer is because we're treating Fermi statistics more honestly now.
Okay, well, we didn't treat them at all before. And now Sommerfeld said, you put it in the Fermi statistics, you get the heat capacity, right.
Okay. Now he went on and said, in fact, with this newfound understanding of what's going on and Fermi statistics being important,
we can fix some of the problems with due to theory. You know, one of the things that you recall from Judith Theory we call Duda,
one of the things we calculated which we actually got pretty close to, right, was the thermal conductivity thermal con.
We had an expression, this kinetic theory expression capture the thermal conductivity one third density
KV Heat capacity per per electron that it was b squared times tao scattering time.
And drew the in this expression. Drew, they used used these things that we don't like anymore.
The TV is three halves KB the classical result and he also used the classical result for v squared eight kb t kb t over pi m.
So the combination CV times v squared which actually enters in the thermal conductivity, it has the value 12 over a pi and c k.
B. T over m. Right.
Okay. Now, Sommerfeld said both of those results are wrong, so let's use the right results.
So his result was CV is well,
he has the pi squared over three and then three halves can be the classical result and then kb t over e f and then he said,
okay, and this squared is not not given by the classical kinetic theory result,
but should instead be given by v f squared, the Fermi velocity squared.
And this makes things a little bit simpler. We can write f is one half mass times v f squared.
But those together we get some cancellation. Again, if we look at the factor CV times v squared, we get pi squared times kb t over m.
So in fact, what we discover, it's not too far from the to projection.
In fact, we're over Sommerfeld. Sommerfeld.
Is is what? It's 12 over five cubed or something. It's about a half.
So it gets pretty close to the same result. And I get the right pulverised cubed.
Yeah, I think so.
So it's pretty you get pretty close to the same result, which is good because we like the answer for the thermal conductivity that we got in.
Due to theory, it satisfies this vitamin France law that was known to exist experimentally.
And so we didn't want to ruin that. And in fact, Sommerfeld theory doesn't ruin it.
Two mistakes, the heat capacity and the velocity.
Cancel each other out. Exactly. Well, not exactly, but pretty close to.
Exactly. But. There were other things that Drew Taggart completely wrong.
One of them was the Peltier coefficient. Peltier, which let's see, what was Peltier coefficient.
I think. Think I left it over here. Yeah, okay.
It was a coefficient. Pi is kd times t over three times minus E and in due to theory, this came out 100 times too big, more or less.
But now if we plug in the new value of V, which is 100 times smaller, where all of a sudden getting things, they start to look right.
Okay, so this is this is good. So Sommerfeld was pretty happy with this result.
But unfortunately, as with all as with all things, he introduced a new problem.
What's the new problem? Well, okay, let's go back and remember the conductivity expression any squared tao over m.
That's still the same result.
In in Sommerfeld theory give you the same prediction and better be the same prediction because we have the ratio of, you know,
we wanted to get the right ratio of thermal conductivity to electrical conductivity and we didn't change the thermal conductivity much.
So we better not change the electrical conductivity much either so more or less up to maybe a factor of 12 or pi cubed.
We expect that the conductivity should be given by this expression.
Probably this expression is a good one to stick with and we don't we don't know how,
but we can measure sigma measure this measure sigma to get to our get this get Tao.
And then from Tao we can calculate the mean free path.
Lambda mean pass. Scattering lines, which would be the Times Tao.
Now in Sommerfeld theory we would probably replace this with VRF Times Tao because the the electrons
are moving around at speeds close to VF and the problem is that the F is close to the speed of light.
Well, not closely, but it's 1%. The speed of light is extremely fast, which means the mean path is extremely long.
So lambda is huge, unreasonably big.
How big? Well, at room temperature at tea room at tea room room lambda can be say 100 angstroms may not sound enormous,
but at low t lambda can be a millimetre.
And again, that may not sound huge to you, but you have to think about how many things the electron has to go past before it scatters.
Well, every angstrom there is another atom, or every two angstrom there is another atom that the electron could bump into.
So here it goes by 100 of them. Here it goes by a million of them before it bumps into something.
So what can it have bumped into? It could have into the nucleus. It could have bumped into the core electrons.
It could have bumped into the other free electrons in the free electron gas that are running around.
And for some reason it does not bump into any of them. Really, really strange.
And this is something that we're not going to answer until much later in the term when we study band theory of solids.
So for now, the unreasonably long memory path is just a puzzle that we're going to have to deal with.
So this was something that troubled Sommerfeld back then, troubles a lot of people back then.
But, you know, Sommerfeld was brave and he decided what else?
You know, we did pretty well over here, understanding the heat capacity, the thermal conductivity and Peltier coefficient.
What else can we calculate that we might be able to get right?
Not worrying about this particular little problem.
And so now we're going to take a little bit of a out of order detour and discuss a little bit of magnetism.
Now, the last couple of lectures of the year, week seven, are entirely about magnetism.
So this is a little bit out of order,
but I think it fits in here well because it really is based on exactly the same same business that we just we just went through.
And the particular type of magnetism we're going to study is what's known as Pauli power magnetism,
apparent magnetism of free electrons, of free electrons.
Now, Howley, of course, we all know him. He was the exclusion principle guy.
His also people said he was the most arrogant man who ever lived.
You know, he used to tell his students that it was okay for them to make a mistake because he never made a mistake himself.
But it was okay for his students to make a mistake. So he was sorry.
That was typical of him. But he was a very great scientist and he actually did this calculation before Sommerfeld did.
He did his work. So a lot of this maybe should be called Pauli theory, not Sommerfeld theory, at any rate, per magnetism.
What is that? So what one does is one applies a magnetic field to your system and you measure
the magnetisation that comes out the proportionality constant sky over mu not you,
not here is just the usual constant, the permeability permeability and this wrong permeability.
Is that right? Maybe that's right. Anyway, that's a constant.
It shows up on your data sheet, you know, some fundamental constant.
The sky here is the susceptibility is known as susceptibility.
Susceptibility. So this equation defines chi power magnetism.
Power magnetism. Means chi is greater than zero.
So in other words,
you apply a magnetic field to the physical system and it develops a magnetisation in the same direction as the field that you applied.
So how are we going to address this?
Well, first thing we have to do is we have to write a Hamiltonian for our electrons or I p squared over to m the usual kinetic term,
plus a coupling of the electron spins to a magnetic field.
You may have seen this in the atomic physics course last term.
So Sigma here is the power of spin operators, spin up operators, and it has importantly, it has eigenvalues, x plus or minus one half.
B is the magnetic field new here? Is the border magnets on board magnets on a fundamental constant is each bar over two m and numerically,
it is useful to keep in mind that that number is somewhere around a Tesla, a kelvin per Tesla of Energy.
G Here is the g factor. And for electrons, the g factor is typically, well, a free electron out in space.
The Z factor is two, so we're just going to use two. And that's going to prevent us from having to write G and g,
twiddle and get confused because we're already using G for density of states because it's conventional, therefore convenient.
Okay. Anyway. If you're at this point, you might be wondering if you took the relativity course last term.
Why is it that I wrote P-square over to him?
If we have magnetic fields where you might have expected that instead you should have p plus e a squared over to m.
Where A is the vector potential? Does this look familiar from?
Yeah. Okay. So the EAA that shows up in the kinetic energy.
This is the piece that makes the electrons curve. If you leave it out, the electrons don't curve.
Okay, we're going to leave it out. The reason we're leaving out is twofold.
First of all, because it's hard to treat.
It is actually a rather complicated calculation to deal with how much the electrons curve and what that does to the magnetisation.
But the better reason to leave it out is because the influence of this term as a term is actually less than the influence of the coupling to the spin.
The to the spin is actually more important than the coupling to the actual physical motion of the electron.
So we're going to treat this term. We're going to throw out this term. Okay.
If you really want to know the answer, in fact,
what this term will do is it will change the answer by a third in the it's all we'll get if you put this in.
It actually has the opposite sign.
As the result we'll get here and it's only a third is big so will make the final result of two thirds the final result.
It's not important. We're going to ignore this for now. We're going to keep this. All right.
So ignoring the term, just keeping the the spin term here, what we have is that the energy.
For an electron with wave vector k and spin up is r e k not plus movie b,
whereas energy spin down is e cannot minus UVB where e cannot is h bar squared k squared over two m the usual free electron energy.
So the idea is that when you apply a magnetic field, the up spin electrons become more expensive, the down spin electrons become less expensive.
And so what's going to happen is some of the other spin electrons are going to flip
over to try to become down spin electrons because that would lower their energy.
However, they can't all flip over because a lot of the states are already filled and they can't flip over into states that are already filled.
So you're going to get some of them flipping over, but not a lot of them.
Okay. So I'm going to run out of room here pretty quickly.
So let's start with B equals zero. And here the whole calculation will do it A zero.
That's okay because T is much, much less than T, so it's pretty close to two equals zero.
And if we calculated the number of spin up electrons or the density of spin up
electrons as the number of spin up electrons divided by the volume at B equals zero,
it should be the same as the number of spin down electrons up and down to symmetric in that case v and we can write that is integral zero to infinity.
The E actually will cut off the integral at F.
We're going to count only the electrons that are the states that are filled over two and it's put in divided by two.
Because here we're writing expression for only the spin ups or the spin downs, not both of them, of g of E.
The density of states that we calculated was the total density of states of both spin ups and spin downs.
Okay, so let's let's draw here density of states g of E for spin offs here.
This is E, and you'll recall somewhere on the board, I think I scrolled it off the top.
Oh, well, maybe it's over here. Yeah.
Here it is. The GOP is proportional to each of the one half, so this thing looks like a parabola that way.
And then we can also plot gravity for the spin down.
It's going to look exactly the same for the spin downs like this, and they both get filled up.
So the Fermi Energy, if this is in zero magnetic field.
Yes. Now, when we add the magnetic field way up there, the spin up electrons are going to become more expensive.
These guys are going to get shifted up in energy by UVB and the spin down electrons are going to become less expensive.
So you're going to get shifted down in energy by maybe actually maybe I should raise some things here.
So some of the spin ups are going to get going to want to turn over, has to become spin downs to lower the energy.
So let's let's draw that.
What happens we do this so once we add so this is for B equals zero in this picture B equals zero over here let's try to draw a B,
not equal to zero over here. So here we have energy.
Here we have G for the spin ups of E, and here we have G for the spin down of E.
And what happens is that these guys got shifted up in energy this way, this distance here is movie times B and these guys got shifted down.
Like this. Buy this much movie. Okay.
And then we fill them both up to F. Here.
This is the F, this is F. So you see that some of the ones that were spin up, these guys here emptied out and filled these states here.
Is that clear how that happened? So these dates here got pushed up in energy.
Above the Fermi surface. So they emptied out whereas these states here.
Got pulled down in energy, so they filled up. Okay.
Is that clear? I know this. This this gets a little bit confusing.
Bear with me. So if we want to write an expression for the spin up density, we can write integral from zero to F.
Minus UVB. Because actually we only want to integrate up to here.
Because in financing only this region, not this region here, because once we had magnetic field, it's only this region here.
It's only the smaller region here that's filled up. These guys have now emptied D of g of E, I guess over to.
Whereas the number of spin downs is integral from zero to F plus maybe the g of e over to.
Because we want to integrate all the we have to here since those have been pulled down in energy that clear.
All right, so we're almost there. So now what I want is I want to calculate and down, minus and up, which is integral D from F,
minus B, B to F plus me b, b of G of E over two, and then we can G,
since this is all over only a little small sliver of energy,
we can use the rectangular rule to calculate that integral and we get one half g of f times two.
Newby cancel the two is we want to write out g of F times movie B and then finally the magnetisation, which is what we're looking for.
The Magnetisation is new B, each electron has a magnetisation of one.
It has a magnetic moment of one more magnets on. And then we have the density of spin downs minus the density of spin offs.
We'll give you the magnetisation and if you want to know where the sine comes out this way,
why it's down minus it's not up to minus downs is because the charge in the electron is negative,
meaning the spin of the electron actually points opposite its magnetisation, which is pretty confusing,
but that's what we have to deal with because the sign of the electron is is negative.
Okay, so we plug in and up minus n down. So we'll get a G, E, F, B squared times the magnetic field.
And you recall the definition of susceptibility chi over mu, not magnetic field.
So we identify the susceptibility, the pauli paramagnetic susceptibility is mu b squared mu not G at F.
And as a final result, again, we can take this result, compare it to the theory theory our chi experiment and do it for various different things.
For lithium, that ratio is about 2.5. For sodium, it's 1.8.
For potassium is 1.6, which is pretty good agreement.
Now, if we were thinking about classical particles, classical atomic gas.
Then in fact, when you apply the magnetic field, nothing would stop them all from flipping over.
They would just go right into their, you know, the the flipped down state because that would be lower energy.
And the only reason they don't all flip over is because of Fermi statistics.
Fermi statistics prevents them all from flipping over because some of the states are already filled.
And so you get a finite susceptibility, whereas the classical calculation would would predict an infinite susceptibility.
So this is pretty good. We're getting results that are within a factor of two or three of the actual experiments.
So this is a more or less all we have to say about Sommerfeld theory.
So summarise a couple of things. Our successes, successes of free electron free electron theory theory, which means drew up by Summerfield.
Duda plus Sommerfeld. Well, okay.
A couple of successes. We got the heat capacity, right? We got the conductivity good.
The thermal conductivity good. The ratio of those to were good.
The Peltier coefficient is in the right ballpark, the susceptibility.
And there's actually many, many other things that you can calculate that you'll get right.
So the free electron picture is actually pretty good.
However, there are still some problems that we're going to have to deal with.
One is that, as I mentioned earlier in this lecture, Lambda seems too big.
Too big. That's a big problem. Having a mean free path of a millimetre just seems completely unreasonable.
We used that the density of electrons should be one electron per atom.
Then we had some intuition why we should do that?
Because, well, you know, there's a bunch of the electrons are bound in core orbitals and maybe they didn't count.
They don't run around, they just stay fixed. But there are some other atoms, you know, that work perfectly well for sodium and potassium.
But what about for carbon of carbon? Mixed diamond. It has four electrons in its valence, orbitals for electrons in its outermost shell.
And in fact, it's an insulator. There are no electrons running around free, so why not?
What's going on there? Why do we count one electron per atom? Why do you not count some electrons at all?
So what happened to those other electrons? What about the sign of the hall effect?
Sign of our h our hall? The hall coefficient is always supposed to have the same sign and due to theory.
Also in Sommerfeld there is always the same sign. And we measured experimentally.
Well, we didn't measure, but it was measured experimentally that in fact the sign comes out wrong sometimes.
Another thing that's kind of interesting I didn't mention before optical properties.
Are different, maybe y different.
Y different. What I mean by this is that different metals look different.
So, you know, gold looks kind of gold is silver. This kind of silver is copper the kind of copper it is.
Hence the name of a tin looks kind of, you know, lighter and then looks kind of darker.
They look differently. They have different optical properties. They reflect different colours.
Why is that? In the free electron theory, they should all look exactly the same.
Another thing we didn't address is magnetism.
Magnetism in particular things like iron are ferromagnetic coming from the name for iron that you can have magnetisation not equal to zero.
Even when even when b equals zero. You probably study this in and you're like your magnetism.
Quite. Why is that? Real picture would never get that.
And finally, what about Coulomb interactions?
Coulomb interactions? When we think about the electrons running around in the solid, you know, they have a huge Fermi energy, you know, 80,000 Kelvin.
But if you think about the the energy scale of the Coulomb interactions,
it's just as big the interaction of the electron with the nucleus nuclei close by, also 80,000 kelvin, also a huge number.
The interaction of the electron with other electrons that are running by it, also 80,000 Kelvin, we threw it out completely.
We just treated these electrons as if they are free gas, no interactions whatsoever.
We only treated the fact they have Fermi statistics to a large extent.
All of these problems come from the same thing. They were neglecting the same thing over and over and over again.
And we are going to have to take more seriously one particular item.
And that one particular item is that materials have microscopic structures, detailed microscopic structures.
Atoms are stuck together in a particular way, frequently a particular periodic way, and that completely changes their properties.
And that's what we're going to have to deal with for much of the remainder of the term.
All of these things to a large extent will be sorted out once we deal honestly with the fact that the atoms are arranged in some particular way.
The first thing we're going to have to do is we're going to have to understand why it is that the atoms actually stick together to begin with.
And so starting the next lecture, we're going to discuss a bit of chemistry and chemical bonding.
Just a really quick thing. How many people had A-level chemistry?
Oh, that's pretty good. How many did not?
Okay, you're the lucky ones, because you're going to have to unlearn a lot of the things that, you know, maybe not like.
It's always good to learn. Learn things, even chemistry.
But but you may have to unlearn some of the things that you learned previously, because we're going to look at it from a more physics perspective.
And I will see you Thursday, Thursday, Thursday, Thursday.