 You can follow along with this presentation using printed slides from the Nanohub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. So this morning we talked about this quantum transport and what I'll be talking about now, though again in a way we'll be going back to the semi-classical picture because much of what I want to illustrate is how this elastic resistor model you can use to understand all kinds of different physical phenomena. And what I'll be talking about is thermoelectricity and heat flow. And for that again one could include quantum effects is just that basic things you can understand fairly well without bringing in any quantum effects really. So we can go back to the semi-classical picture. But before I start on that let me just say a word about just summarizing what we did in the morning and that was that as you know the basic thing I was trying to express is how in semi-classical picture just as the Boltzmann equation is the central thing. I mean it's the anything you describe is basically Boltzmann is what captures all the physics. The point is to understand it and try to use it. Even the elastic resistor then is a good way to understand a lot of phenomena. The kind of quantum version in my mind is this NEGF. So in a way what Boltzmann did and this was his crowning achievement of this 19th century in a way is he took Newton's law and he put a scattering and all that into it and turned it into Boltzmann equation. You see Newton's law which describes mechanics, reversible processes. To that he included all these irreversible scattering processes and thereby came up with this equation. And as I said what he did was so subtle that in his time it was very controversial and even now you see people writing in a paper in physical letters about what exactly he did and the meaning of it and the implications of it. But the point is that just as Boltzmann needed to do something with Newton's law to get all that in, once you replace Newton's law with Schrodinger equation quantum mechanics you again have to do something similar. This is just as Newton's law wasn't enough, Schrodinger by itself isn't quite enough. You have to do something else. And NEGF is one way I know which essentially does that in a consistent way that is comparable to Boltzmann in exactly the same way. And those were the equations of the Boltzmann of the NEGF formalism and what it includes of course that is different from Boltzmann is that it includes interference effects essentially. And as I said if you are trying to learn this then two good examples to go through are first the one dimensional example that I mentioned about various scatterers. See how you get resonances out of the quantum. If you want to include defacing processes into the NEGF then you can actually turn the quantum calculation into the classical calculation by destroying the interference. So that's usually a nice example to go through. The other example I say if you are kind of comfortable with the 1D then I'd say let's try the 2D. And a good 2D problem is this Hall effect problem. Just a 2D conductor you can first calculate conductance without any magnetic fields and see what happens in a magnetic field. And one of the interesting things that actually happens in a magnetic field which I won't go into is that at high fields you see that Hall resistance you'll see show those very sharp plateaus, very nice plateaus. And that was the first example probably where this quantum of conductance played a very important role in the sense that Hall resistance actually is looks like H over Q square times an integer. And one way to understand it in the context of what we are talking about is that what happens at high fields is that the electrons that are going in one direction tend to have their wave functions on this side of the sample while the electrons that are going in the other direction tend to have their wave function on that side of the sample. So as a result you have almost a very good ballistic conductor because you see it's very hard to back scatter if that other thing is in a way out there on the other side essentially. So something non-trivial happens that is usually in a conductor you expect the density of states to be more or less uniform spread out all over. But in this high magnetic field regime actually you have what you call these edge states. These are things of course you know there are lots of papers about it and trying to understand all this and as I said the power of any gf is how it allows you to calculate a lot of this even if you don't quite follow what have happened. And so for example in the any gf equations I showed you there was something I called density of states. There's a matrix there just as there's an electron density matrix as a density of states and you could look at its diagonal elements and actually plot out what the density of states is at different points in the sample. So for example here I've shown a calculation at high magnetic fields. So what is over there is this grayscale plot. This axis is energy that axis is the width of the conductor. And if you see a white spot it means there's a lot of states there high density of states. So what you're seeing is the spatial distribution of the density of states. How was this calculated again with any gf put in h sigma etc and using those equations it's calculated. And you see those where those white spots in the middle. Those are what are called Landau levels actually at high energies, at high magnetic fields you have discrete levels at different energies. But the interesting thing is if you look there's these white streaks. I don't know if they come out here well, you do not see this very well. Right at near the edges there are these white streaks which are basically the edge states actually and not see I see that well on my screen not as well up there though. But right around the edges there should be these white things. And the way people think about it is that well electrons in a magnetic field they want to turn as you know in a magnetic field they want to go around like this. Why? Because the understanding is that dp dt as you know when you do semi classical dynamics dp dt is equal to force. So it's like q v cross v and if your magnetic field is perpendicular then it tends to turn you. And this you can show that as a result you want to and if you're in the middle you'll tend to go around in circles and those are the usual Landau levels etc. But when you're near the edge you'll tend to do this. Hit that, do this. This is what people call skipping orbits. And that's how it will sort of picture where the edge states came from. Whereas ones that are going this way tend to do this. So anybody going, so all the north bound people stick to one side, south bound people stick to the other side. So makes it a perfect ballistic conductor in a way. See? And that result is if your Fermi energy happens to be there, the actual resistance is given by this without any back scattering almost. Whereas usually with ballistic conductors they're kind of semi ballistic. In the sense that you get q square over h times m within a factor of maybe 5% or 10% because there's always some back scattering. But in what made this quantum hall very special when it was discovered in the early 80s was that this resistance they measured was like quantized to like the seventh place in decimal. It's very accurate. Something you never expect in a solid. Usually to make measurements that accurate you have to go to vacuum. So that was what was really surprising and unique about it. And I think that was probably the first experiment where this quantum of conductance kind of played an important role. These are fundamental constants which have been known a long time. But this combination was not quite as well recognized in general. There's one experiment where it actually played an important role. And again, so they said if you were trying to learn any GF and this is a good test problem, once you have done the 1D thing. Because 2D has more details in it, you have to worry about. So what I wanted to talk more about then today in this lecture is about the heat flow, that's the second, the topics down here. What we talked about in the morning was what's up there, this thing. And the basic fact, experimental fact is I guess let me. Explain this. Is that if you take a piece of semiconductor, then one experiment that is very easy to do, I guess even I have done this. Is that if you put down two probes, one hot and one cold, then there's a current that flows, even without any voltage, right? Or if you don't allow the current to flow, then the voltage would be developed, right? So this is what the hot point probe experiment. So in this picture, I could say that if this side was hot and this side was cold, then a current would flow in a certain direction. And the interesting point is that the direction of that current is different in n-type semiconductors and p-type semiconductors, okay? The question is why? And what I claim is that we don't need anything new to understand this. That's all right here. If you just take this, and the way we said it was the Fermi functions in the two contacts are different because they have applied a voltage. We had two different chemical potentials. But supposing we consider the situation when the chemical potential is the same. I have not applied any battery, any voltage across it. Just the temperatures are different. Of course, that would make the Fermi functions different. Because you see the Fermi function is E minus mu over KT. Usually the reason F1 is different from F2 is because mu1 is different from mu2. But let's say no voltages. All the mu's are the same, but that's fine. The Fs are still different. There's a KT1 and a KT2, the two different things. And that's why you have a current. And what I claim is that just from this point of view, you'd understand all the basic things of thermoelectricity, the standard stuff, right? That's what I want to get across. So the picture I had drawn before, you know, you had a density of states here. And you have a mu. And this is mu1 and this is mu2. So when I draw the Fermi function, the way I want to draw it is both sides have the same mu, but they have different temperatures. So this is the hot side. So I'll draw the Fermi function kind of broadened out like this. So this is F1. This is a hot side. And the F2, let me draw it, it's very cold. So I'll draw it very sharp like this. And for this discussion, let's assume the density of states is somewhere here. And what I claim is that this equation basically will tell you, the term will give you the thermoelectric current. So which way should the thermoelectric current be? Well, it's F1 minus F2. So if I look up here, I see that F2 is essentially zero because it's cold. Whereas here, F1 is not quite zero. And so electrons will flow in this direction. So the way the current, you can figure out what the direction of the current is. So remember this one now has no voltages across it. We're just looking at the short circuit current now, no voltages. There's two different temperatures and the electron will go through this. And so electrons will go this way, which means electrons will come in here, electrons will go out this way. And of course, the direction of the current is the opposite of the direction of the electron flow. If you actually have to write down the conventional current, the one that you measure, it would be the other way here. That's it. So this is how it looks like, usually. Now that's in an n-type semiconductor. Now if you take a p-type semiconductor, then the thing is it's reversed. Why is that? Well, p-type means that instead of something up here, it could kind of go down like this, let's say. And the point is that if you look below mu, actually it is that F that is bigger. This one is smaller. You see, when you are looking up here, this is zero. And that's like, say, 0.2, or some number like that. And so it is going this way. Here, this is one. Whereas this is more like, say, 0.8, so it's less than one. So when I look at F1 minus F2, I'll find that electrons are going the other way. So this is the interesting point that is kind of difference between putting a temperature difference and putting a mu difference. You see, when mu's are different, F1 is always bigger than F2. If mu1 is higher than mu2. And so current will always flow in the direction of higher, from higher mu to lower mu. With temperature though, it doesn't have to flow that way. It's like at energies above mu, you'll tend to go this way. At energies below mu, you'll actually go from cold to hot, rather than hot to cold. Whereas above mu, it will be hot. And if you had a constant density of states that did not change, that was more or less the same above mu or below mu, you wouldn't see any net currents. That was much above as below, it all cancel out. But if you had an n-type conductor, so you had only density of states on one side, you'll see one direction, p-type another direction. Graphene, graphene, what would happen with graphene? Graphene, right, so the density of states would look like this. So if your mu is somewhere here, if your mu were here, it is perfectly symmetric, it's completely. If you were here, you'd probably see something, because there's more on this side than on this side, so it wouldn't quite cancel. And same reversed on the other side. So the point I wanted to make here is that, so when you're trying to understand this thermoelectric effect and where it comes from, that's it. I mean, it just follows from this, there's no new principles involved. It's just this f1 minus f2, nothing. We don't need to discuss this anything else new. Now, usually when you talk about thermoelectric effects, sometimes people say that, well, the reason this is reversed is because in the case of the valence band, it's actually holes that are moving. But I feel that that's really not a convincing explanation. Because finally, in all solids, what moves is electrons. And holes is, at best, it's supposed to be a conceptual convenience, usually. And a physically measured effect should not depend on what I find convenient or what you find convenient, this is a physical effect, you see? And the point I'm trying to make is that the reason it flows in one direction is just this, f1 minus f2, this is it, really. So, but one point I should make is that if you looked at the Hall effect. No, because one of the things in this discussion I've tried to do is, we have said that all that matters is really density of states. We don't have to distinguish between conduction band and valence band, necessarily. Because usually people say, well, those are two very different things. You treat electrons separately, holes separately, right? But what I'm saying is not really. You can just treat the entire density of states all together. But let me just qualify that with a little bit, since right before this we talked about the Hall effect. Because usually the way you tell the difference between n and p-type conductors, there's these two standard ways of doing it. One is something called, one is this Hall effect. And the other is what we just discussed, this thermoelectric effect. These are the two standard ways of telling the difference between n and p, because just from conductivity you can tell usually, which is way. Because you just measure conductivity, you don't know which way. Now, thermoelectric effect, all I'm saying is it is really all from f1-f2. That's it. But for Hall effect though, there is actually a real difference. It goes like this. But if you looked at the density of states, when you look at, there's something here. So this is E, this is like a n-type material. And this is like a p-type material. And we are what you're drawing here is a E-P relation. And here the E sort of looks like EC plus P square over 2M. Now when you're describing an EK relation in a valence band, so it looks more like E equals EV minus P square over 2M. It is going downwards. So in a valence band when you write a EK relation, you write something like this. And the difference between n-type and p-type comes about because when it looks like this and you put a magnetic field, let's say an electron going this way curves upwards. In a material like this, an electron going this way will actually curve downwards. So that it's not a f1-f2 issue. It's a different thing. And that is again only because of this plus and this minus sign here. Because as I mentioned before, the way you think about it is this dp dt is equal to qv cross b. And I don't quite want to go through the algebra. But one of the things you know is that this causes an electron to go around in circles. You have probably seen this, it's called the cyclotron frequency. And the cyclotron frequency is given by this qb over m. And of course, in general, when you're interpreting this, this m is like the ratio of momentum to velocity. That's the thing. And so for example, as I mentioned in graphene, the mass actually increases with energy. And that's again experimentally observed that when you're doing this in graphene, as you have higher and higher carrier densities, the cyclotron frequency actually changes. Whereas usually in ordinary materials this is fixed. But actually it's very carrier density dependent in graphene because that mass is energy dependent. As you go up, the cyclotron frequency gets smaller. Now in this context, though, the important point is that if you think of this m as this ratio of momentum to velocity, and remember again velocity is like dE dp, you'll notice that with this minus sign it is almost as if the mass is negative. Because when you look at velocity, you see as you increase p the velocity is actually because the energy is going downwards, the sign is opposite. That's the point. And so there is a difference between this kind of a band and this kind of a band so that one type wants electrons to curve downwards and the other goes upwards and which is why the Hall effect is different. But when you talk about the thermoelectric effect, which is what we are discussing, there is no real difference between the two bands anyway. And one interesting point as a result I feel is kind of related to this is when you look at amorphous semiconductors. Now I made this point that a lot of things we are talking about in terms of density of states is kind of applicable even if you don't have a EK relation, even if you are talking about anything small where you don't really have an EK relation or an amorphous material. When you go to amorphous semiconductors, they still measure a thermoelectric effect that is quite well understood. There are n-type amorphous materials, p-type amorphous material. But often when you do the Hall effect measurement, you get something that doesn't agree with thermoelectric measurements at all. An n-type material shows a Hall effect that's had the wrong sign. Very hard to understand what happened. Hundreds of papers trying to understand Hall effect in amorphous materials. Why? Because there is no EK relation. There is no simple picture like this, etc. Anyway, so that's an aside. What I really want to talk about, what I want to get back to is this thermoelectric effect. And the point I wanted to make is that this thermoelectric effect is really just a consequence of this. And a few years back, I remember we had actually proposed that there was a lot of controversy in this field of molecular electronics about whether these molecules were conducting through the valence band or through the conduction band. In the way, the conduction is like below your Fermi energy or above your Fermi energy. And what we had proposed is a good way to tell the difference would be to measure the thermoelectric effect. And one question that came up, well, I know it's a molecule. Can you really have a thermo? Do you really have any thermo power? And the point is, because of this, you can see it really doesn't matter where this density of states comes from. And I think it's been about two or three years now that a group at Berkeley that performed a series of experiments actually measuring this thermoelectric effect in molecules. It's a very difficult experiment actually. And they show quite understandable results, what these are. Yes, please. The relationship between the same requirement as the chart, the chart signer change for both is possible and for very much the same. Yeah, you shouldn't bring in both of those, right? Yeah, so what I'm saying is, the way people usually explain it is by saying that the reason it has reversed is because the sign of Q has reversed. And what I'm saying is that no, what really moves inside a solid is always a negative charge. What really moves is always minus Q. And the reason there is a difference is because the M is different. So the thing is that electron with a negative M, you could conceptually convert and think as a whole with a positive M. But you shouldn't do both. I mean, don't make it a whole with a negative M, right? That's the point, right? What you're really saying is that gradient of that curve is different between the M and P, and that's where the negative comes from. Right, that's okay. If you, right, since the velocity is the DEDP, okay. So now you can easily get an expression for the thermoelectric coefficients, for example. So you could say that from that expression, we could go to a linearized thing, where you could say i is equal to G times V1 minus V2. This is what we did before. If you remember, we said let's use a Taylor series expansion on this F1 minus F2. And thereby write current as just the difference between the voltages. And the point is you could do something similar and write it as some coefficient times T1 minus T2. Now the way we got the G, if you remember, was, write it here. What we had done then was, say, we'll write 1 over Q, signal DE, and then for F1 minus F2, I said let's write it as del F, del mu, and then del F, del mu gets multiplied by mu1 minus mu2. And then we said that del F, del mu was like minus del F, del E. That was the way we got a conductance thing. Well, we could do something similar and get the temperature coefficient also. That is, you could say, well, just as you have this, you'd have something like this, del F, del T times T1 minus T2. That's it, right? And the thing is, here it was relatively easy to turn this into minus del F, del E. Here also you can do that, but it takes a little more thinking. Here this is very easy because when you look at the Fermi function, you find that it goes in as E minus mu. So it's quite clear that if you take the derivative with respect to mu or E, just a minus sign between them. That's it. But now what we are trying to do is take the derivative with respect to T, right? So the way I could do it, as you could say, well, del F, del E is like dF dx times derivative of x with respect to E, which would be 1 over kT. Whereas when I do del F, del T, it would be like dF dx. And then the derivative of x with respect to T, which is like E minus mu over kT square with a minus sign problem, the minus sign. So between them, what I can see is that del F, del T is equal to minus del F, del E times E minus mu over T. So this is algebra. So what I need is the derivative of this function with respect to temperature. And what I'm trying to do is relate it to the derivative of this function with respect to E. Since that we have seen before, it appears in the conductance expression anyway. And what I'm trying to do? Do you take a partial derivative or do you take a full derivative of this quantity? Because it can go kind of with the independent of temperature and so on. Right, so in this thing though, when you are calculating it, it's like you're holding the mu constant, right? So in my thinking, it's like when I write del F, del E. It means I'm changing E, but everything else is fixed. And when I say del F, del T, it means I'm changing T, but everything else is fixed. So the way it works is if I change E, X changes by a certain amount and that's why the F changes. And if I change T, again X changes and then the F changes. And that's why they're kind of related. Because finally, it is all about how F changes with X. And then the question is a little change in E, how much does it change X? Or a little change in T, how much does it change X? And that's what this part is. And yeah, this part is of course fixed, so that's how you do it. And so just as this del F, del mu then became just minus del F, del E, which is what we had done before. This one then becomes minus del F, del E. But then there is this E minus mu over T in front of it. So the conductance here is like this first one, like we had done before. This would look like integral D E minus del F, del E. And then the conductance. Whereas this one will look like integral D E minus del F, del E. E minus mu over T, that's it. So you have this extra factor in there. And this of course includes this physics that I discussed. That this coefficient changes sign depending on whether E is greater than mu or less than mu. You know, that physics that I was trying to describe to you earlier. That's all in this math. Right here. Capture it immediately. And so you can see if you had a, I guess the G also should be there of course. So if you had a material whose density of states, which is reflected in this function, is all for positive E minus mu, you'll have a positive coefficient. If it's all for negative, you'll have a negative coefficient. So when you do the Taylor series expansion, so it has to be done around the temperature T1 or T2, right? So when you're writing this coefficient T2 over there, you have one option. Right, so whenever you're doing the Taylor series expansion, it is around the equilibrium point, right? And the understanding is that the T1, T2 difference is small enough that it shouldn't matter which one you use. But same with the mu1 and mu2. You could ask the same question, right? So when you write a T here, I guess it could be either T1 or T2. And point is it shouldn't make a difference. Because we are talking of small differences such that that doesn't matter. So when you need some relative devices, the normal temperature difference is about 200 Kelvin or something like that. So is the expansion still valid for that temperature then? Because we're doing Taylor series expansion for small things, right? So is the expansion still valid there? Yeah, I guess what is important, it would depend a lot on this specifics of this function, I think. That if you had material with a very sharp spike, then something may be valid otherwise it may, whereas if it is more uniform, it may be valid in a different condition. So that I think needs a more careful consideration. And my thing is that in doubt you should always just go back to this. This is the one to use really, right? But when people evaluate thermoelectric coefficients, what is the expression that they usually use? And actually with conductance also one can ask that when does the Taylor series expansion work in a way? That is in the sense that, yeah, because your question, you have this current versus voltage and you have something like that and we are trying to calculate that dI dV. Now you might say that, well, we use the Taylor series expansion and that only applies if the voltage you apply is less than 25 millivolts, right? Because much less than 25 millivolts. Why? Because in doing this expansion, I assume that mu1 minus mu2, it was a very small fraction of kT. That's what allows you to do the Taylor series expansion because this is in the denominator. So the change in x should be small. So basically it's the change in x that should be small. Now most resistors, of course, usually are linear far beyond 25 millivolts. Why is that? Now that has to do with the subtleties of, you know, earlier I said that when you have a real device, you could, how do you justify using the elastic scatterer model? It would be sort of like you have to think of the real thing as lots of little things in series. Where each little thing is the length of an elastic scattering length, let's say. And the point is, so if in an electrical resistor, you are trying to figure out whether something would be linear or not, whether your Taylor series expansion works or not, I would say that you shouldn't drop more than 25 millivolts across that one little thing. If you are thinking of a very big resistor, then you could easily put one volt across the whole thing. You buy a resistor from RadioShack, you could put one volt across it, it's still one kilo ohm or whatever you bought. It's perfectly linear. And then you say, well, how could this be working? After all, it's 25 millivolts. You're putting one volt, this theory doesn't hold. But the point is, the way you justify applying it to something that big is by saying that there's really lots of little things and I'm doing a theory on one of those little things. And then you say that okay, as long as you don't drop more than 25 millivolts over one inelastic length, you are okay. And so I think this is the same with the temperature. The point you raise I think is right, that there's an enormous temperature difference, but as long as you don't drop too much across an inelastic length, you're probably okay. That's how you justify it. Anyway, so this then leads to these coefficients that I just discussed. So that kind of slide sort of summarizes what I just said, that you could take that, start from that same expression, linearize it. You'd have something depending on voltage, something depending on temperature. And the first is just the conductance we have been talking about. But then the other coefficient, the Gs, that is what will have this e minus mu over qtn. And that's the standard expression for evaluating that quantity, actually. And usually in thermoelectrics, the measured thing is this C-beck coefficient. And that is, like if you leave it open-circuited, what voltage do you get? That is, what you do is you take, leave this open-circuited, so the current is zero. And then you say that, well, for a given temperature difference, how much voltage difference do I get? And that is actually the ratio of this and this. So these coefficients, the C-beck coefficient, the sort of line, the ratio of this voltage to the temperature, and that would come out as minus Gs over G under open-circuit conditions. So that's what is usually measured, which is smaller. But then at each small devices, you have different intermediate temperature teeth. So you have to expand at different temperature when they're adding up, right? So don't they give you a significant difference? Right, so that is why. So if you had a long series combination of lots of things, so you're putting, say this is zero and this end is 1 volt, and you're breaking it up little by little. And in one of these sections, let's say you have 0.01 volt, because you had hundreds of these. So the point I'm making is that when you calculate you're trying to find the conductivity of this bulk material. So what you'll do is you'll assume this much voltage and find what current flows, and you'll take the ratio and extract the conductivity. And that would be representative of this whole big sample. But on the other hand, if you say that, well, I put 1 volt, how can it be linear? Then I'm saying that's not quite right. It is this quantity that needs to be more comparable to KT. It's going to be much less than KT. When you look at the difference, you have the coefficient E minus mu over KT squared. Right, but that T is supposed to be the, again, if you're using a linear response theory, it is supposed to be a fixed temperature. That is true. So in a big sample, if it is varying, then you have to be careful at one end from the other, because just the average temperature is different. Get your point. So now, what I wanted to talk about next is, there's a related effect from this C-beck effect. That is, this is this thermoelectric effect. Here it is. You have a temperature difference. It drives a current. There is a related effect, and these two are connected. And ordinarily, it's hard to see why the two are connected exactly. And this is the other thing I'm saying. You should see it very clearly, in the discussion. And that is that if you have a material that shows a C-beck effect, so let's say you have something like this, and you put a temperature difference, and so a current flows, and we see a C-beck effect, you'll also see something called the Peltier effect. And that is that if instead of a temperature difference, if you had kept the same temperature and just had applied a voltage across it, a current through this, you would actually cool one of the contacts, cool this contact, and you'd heat up this one. So it actually, and this is of course the basis of the Peltier effect, which is actually used to make practical cooling devices also. And the two are related in the sense that any material which has a good C-beck coefficient also has a good Peltier coefficient. So although these two sound like totally different things, and it's not even immediately clear why one had to go with the other. And what I'm going to explain is in this picture how you would understand that. So the argument goes something like this. For the Peltier effect, let's say we don't really, we don't have hot and cold, it's all the same temperature, but there's a voltage difference. So that you actually, let's say you're sending electrons in here. Now the point is, and so the electron is going at this energy. But then, once it comes out here, of course we discuss that all the dissipation appears in the contacts. It gets rid of this energy into the contacts, and then when it gets out of the contact the electron is roughly around mu 2, down here. So it goes through here, and this is the, it heats it up. But look at what happens here. It had to go up in the first place because what comes in from the contact is here and it actually had to go up in order to get through. So right now we are thinking of same temperature, but there is a difference in the mu 1 and mu 2, let's say. This is mu 2. So the important point is that if you had a material that shows you C-back coefficient, so C-back effect, so that the density of states is above then whenever a current flows you would actually be cooling one contact and heating the other one. In the past I said that with elastic resistors what wasn't clear was that this I squared R loss everyone knows there has to be a joule heating. Where does it happen? Well I said it happens in the contacts. And the point I am now trying to make is that with these n-type conductors when you look at it it is not all heating. It is like one is getting cooled the other is getting heated but then it is getting heated more than it is getting cooled. Why? Because in order for electrons to be flowing this way, since the temperature is the same mu 1 must be above mu 2. So here you would be cooling, picking up that much energy from this contact but then you will be dumping that much here. And the extra that you are dumping is like mu 1 minus mu 2. That is the joule heating. I mean overall of course you are heating it more than you are cooling and this minus that is basically the joule heat because every time an electron goes through from here to here basically what happened was an electron came in at mu 1 left at mu 2 and so it dissipated an amount of energy is mu 1 minus mu 2 which is q v. So every time an electron goes through it dissipates overall that much energy. Why? Because it comes in here and goes out there and it loses this extra energy q v. And so if electrons are going through at a certain rate you can show that the current the power would be v times i. I mean every electron that goes through dissipates that much. And how many electrons go through? That's divided by q. That's the number of electrons going through per second. I mean that's the current. If you divide it by the charge on an electron tells you how many go through per second. And every time one goes through it dissipates that much. So what is the energy dissipated per second? v times i. I mean that's the joule heating overall no question. But if you look carefully what you find is it's not heating everywhere it's like this end is getting cooled and that end is getting heated if you look carefully. Now of course you wouldn't have this. You'd have just the reverse if it was a p-type conductor. If all your density of states was down here then you'd kind of first be going down and then be going up. It would have reversed. And of course if you have density of states everywhere more or less uniform then you'd have none of it. All of it just cancels out. There would be some going above some going below etc. But the point I'm trying to make is when you look at it this way you can kind of see that anything that gave you a C-back effect would also give you a corresponding Peltier style effect. It would have this cooling and heating in it. So in the usual situation the argument was that if this energy is E and this is mu1 so usually we have been drawing the picture like this. And the argument is an electron goes through from here comes out here. No energy is dissipated. Where is it dissipated? Well eventually it has to lose this energy and come to the Fermi energy. And that is because I'm assuming this contact has just a constant density of states. So left to itself with enough static processes all the current will tend to be right around here. You know according to this del F del E function it will that if there is once it has come back to equilibrium the way it wants to it won't have all these hot electrons up here because when you look at the for occupation once the electron comes out here this contact really likes to have a Fermi function and an electron up here looks like an anomaly. I mean it looks like a big spike at an energy that shouldn't be there. And so all inelastic processes will quickly lower it down. That's what we're talking about. And at this end same story it's like it wants to be a Fermi function and electron left so there's a kind of spike downwards there. And that's an anomaly and again gets filled up from the electrons in here and whatever comes in here is somewhere here. And overall what happens is an electron comes in at mu one leaves at mu two and thereby gets rid of this energy Qv. And that's why the heating is V times I overall. That's the joule heating where we know that a battery with a voltage V delivering current I that's the heating. But in between you would say this so this is how I explain the normal resistor. The elastic resistor where's the heating at. And what we are now saying is that in a thermoelectric it's a little bit like the mu one is here and the mu two is here let's say and there is a current but then it is going through this energy and the only reason you have a current there is because of course there's a long tail of the Fermi function so there are electrons here that are going through that's how it is going and now you see you first have to cool it before you can heat it. That's the issue. This is the argument. And if you wanted to write down the heat current you could almost write it down in a way in a similar way that is what we had here is this I if I wanted to write the heat current I Q it would be so this is the what we wrote down here is the current the electrical current and the point is every time an electron goes from here to there the amount of charge transferred is Q but the heat that is taken out of here is E minus mu one so the heat current is like E minus mu and the charge is like Q so we had an expression for I if I wanted the rate at which electrons go through then I should have divided by a Q that would give me the rate at which electrons go through and then every time an electron goes through it removes that much heat so what I should do is put an E minus mu one over Q here if I am writing the heat current at contact one if I am writing the heat current at contact two I would write just the opposite problem the reason I this reverse is just a direction of heat flow how you define it I think I reversed it because the way this is chosen electrons are going out from here whereas here it is kind of coming in the way the reference is done but the point is that when you try to calculate heat current it almost follows easily from whatever we have discussed before the main point to note is every time an electron goes through the charge Q but it also transfers an amount of energy which is E minus mu and of course if you are doing linear response then usually you don't distinguish between mu one and mu two you just say that the heat current is E minus mu that's it you don't really distinguish between those two because there is of course a difference between them difference between in the sense that the amount of cooling and the amount of heating and the joule heating from the battery but you say that well the joule heating is proportional to the current squared and so for linear response you can kind of ignore it and so you just said Iq one equals Iq two for the linear response formulas so for a very linear system so current flows around the formula so for that the heat current is zero if the density of states above and below are almost the same that is true which is why probably semiconductors usually make better materials for all this right than metals than metals metals also have some c-back coefficient some failure but semiconductor but commercial devices are usually p-type materials and n-type materials in series because if it's very whichever then dfpx will be like data household basically by e f or e f actually there has to be an in order to get a non-zero coefficient there has to be an asymmetry between e less than mu and e greater than mu otherwise as long as it's all symmetric you won't get any c-back or failure any of that this is also a way to define the heat current with the device it's saying this is where I have come from in the room and what it's maybe hard to define a mu in the device right this is why I say that this elastic resistor this concept it helps say all those things a little more clearly because usually when you think of a long device it is harder to understand what mu means and all that whereas here what we are saying is think of this channel connected to two contacts and the contacts are big things so mus are well defined things right and so the thing is you take this electron at this energy but then the contact when electrons come in it comes in at mu so they kind of have to absorb all this energy from this contact in order to rise up in the long run and that's why it gets cold that's the view so this is why I say this elastic resistor because it separates out these two things it helps clarify a lot of points and the justification to some extent is that after all this the expressions we are getting is exactly the standard thermoelectric coefficients in that sense I'm not telling you anything new these are all things you could find in textbooks it's just that they wouldn't come out quite this easily that's the point usually again you'd have to go to Boltzmann equation and of course our whole thing is consistent and based on Boltzmann equation it's just this whole thing of elastic resistor helps you see a lot of things quickly that's this heat current so that's what I've written up there that if you're writing the heat current iq it would be exactly what we had before but now you have added the e minus mu over q in there and this one also again you could now do a Taylor series expansion something voltage and another one temperature and then you'd get a gp and a gk and what you can see is that you see in the last slide we had taken the we written i equals g v1 minus v2 plus gs d1 minus d2 and now with iq we can do something similar at a gp v1 minus d2 plus q d1 minus d2 you could take this current expand it the same way and you'd get all those coefficients and one interesting thing you'll notice is that this gp and the gs are related it is almost basically the same coefficient and that just comes out of doing the Taylor series and all you get almost the same coefficient and one is related to the Peltier effect other is related to the Seebeck effect which kind of tells you why they always go together they're basically the same expression any material that has a good Seebeck coefficient will also tend to have a good Peltier coefficient much for the same reason and again none of this is not that anything that's not known in the literature and thermodynamics is all consistent with exactly what you'll see there that's the on-solid relation that's the on-solid relation actually this particular this relation between the Peltier and the Seebeck that's called this Kelvin relation but the fact that these coefficients are similar that's a special case of this general on-solid relations on-solid relations which says that the flux due to a particular force and the corresponding this flux due to this corresponding force they would have to be related that's the on-solid relation g1, 2 and g2, 1 in a general sense so it'll be consistent with that yes it's just that usually on-solid relations in town will take a lot more work it's a lot more subtle to understand and derive you know this actually took many years lots of discussion to get there whereas here all I'm saying is it follows pretty easily for the elastic resistor that's all now as one last example of this viewpoint what I wanted to talk about next is so far we have always been talking about electron flow and what I wanted to show is just with a relatively simple extension you could talk about phonon flow as well and this is something I believe Professor Fisher will be talking more at length tomorrow about his talk is about Landau's approach to phonon flow in a way we'll talk more about that but while you're on the subject I wanted to show how you could get that pretty easily very quickly so the expression we had was d e then we had g of e of 1 minus f2 g of e was like q squared over h and Landau this is what we had called this conductance function and I guess you could cancel out one of the q's but we had written d e conductance function we had a bunch of different expressions one of them was this one and I think in this context this is most useful which is why I'm writing it so now the point is supposing we want to write down what is the heat current due to the flow of phonons and the thing is that whatever discussion we had about electrons with this elastic resistor you could have exactly the same discussion about the flow of phonons only thing now is that these two contacts you don't talk about the Fermi function but you talk about the Bose function n1 and n2 which tell you how phonons are distributed in energy inside this contact you know the electrons have this extrusion principle so at equilibrium they are distributed according to this Fermi function and the Bose function looks a little different let me just write that h of r omega so kind of similar with the this is plus that's minus of course that makes the function look very different temperature temperature so just as here the driving force was the difference between the two F's and that difference usually comes from mu but could come from T in the case of phonons we don't quite have a mu so the only driving force is the T is it kind of density difference diffusion? yeah this is the point in general as a general principle I say that what drives flow is not so much the density difference as this chemical electrochemical potential difference that's the point of sometimes I ask this question that if you had a material a very high density of states here and a very low density of states here then when they come to equilibrium the point is the mu will be the same across and so there will be lots of electrons here very few electrons there it's not like the number of electrons is getting equalized it's the same concept that you learned earlier on that what drives heat flow is not how much heat this has and how much heat that has so it's a similar issue here so what drives it is always this temperature difference and that is why what should appear here should always be f1-f2 or n1-n2 in this case I've got two reservoirs with two different Bose functions coming from two different temperatures and that's what it is that's it and when I'm trying to write this iq then I'd put a dh bar omega ok and the thing is here we were talking about the rate at which charge flows of course if I take this q and divide it here it will tell me the rate at which electrons are crossing over per second and so and then I can multiply it by h bar omega to get the rate at which energy is flowing so my point is the correct expression here should look something like h bar omega and then this m lambda over lambda plus l that's just the only thing is here this was all electronic quantities now I should be looking at phononic quantities but otherwise we are more or less done that's it so the omega is the frequency omega is the frequency of the phonon because of the phonon so going from here to here then all that I did was there was a you remember let me put it this way so the overall expression was that I started with was something like this and all I did was took the q out replaced it with h bar omega because every time a phonon flows it carries that much energy just as every time an electron flows it carries that much charge so that's all I did I might have said it a little more complicated but this is all I did really then the f1 minus f2 replaced with n1 minus f2 so lambda here is the phonon mean free path however phonons are getting scattered and often those mean free paths depend on boundary scattering all kinds of things so those things you would have to understand a lot about the phonon system to know what to use there and so on now this one then again you could do this linearization so you could try to get an expression for the thermal conductivity so you could try to write it that way and then you would have a thermal conductivity of this thing so in order to do that of course you need to you can do the same Taylor series expansion so it becomes t1 minus t2 and then del n del t that same thing again now in this case we will notice when I take del n del t I suppose what I should do is like before dn dx and then how x changes with t which would be like minus h power omega over kt square dn dx of course is like e to the power x minus 1 whole squared I think that's about right so if I put this in here then so for del n del t what I am getting is I mean this is dn dx but then of course this whole thing is there I think you should get and note that the x of course is h bar omega over kt so I think this is the expression I have on the this basically what you get is sort of summarized up there that so you will see like integral over dh bar omega with little renaming of variables you know you could put a divided by kt there so it would become like dx and multiply by kt here and here make this kt square put a k square there and put k square there move it around a little bit and that's how you would get that second expression that way it's like k square t over h in front and then there's integral dx m lambda over l plus lambda and then you have this x square e to the power x over e to the power x minus this is of course a different function from the dfd that we have been using all along because now we are talking phonons I mean do this you get that function interestingly though that function doesn't look too different from dfd looks almost the same that's what I have tried to plot here if you take that x square e to the power x over e to the power x minus 1 square you get that blue curve whereas when you take dfd and actually multiply by 4 you get that red curve so it's kind of similar but not exactly the same curve of course you don't expect it to be the same curve actually and in a way actually you know I was never sure when I first thinking about it how anyone would get this whole theory of thermal conductivity from this approach because this Bose function of course I knew that at x equals 0 it blows up you know this e to the power x minus 1 it kind of blows up and I didn't quite see how you would get a sensible thing out of this it seemed like something else needed but when you calculate this the thing is you pick up one h bar omega from this energy and then from this derivative this another one so that finally the function you get has this x square in it x square e to the power x over e to the power x minus 1 square goes to 0 it doesn't blow up because what you can show is the denominator for small x is like x square just cancels out so it's a perfectly well behaved function there's no problems at all as far as this is concerned from small x when you write the conductance equation we have the term q square by h this thing we are having a case called q by h can we think of this term as almost like a quantum of thermal conductance essentially almost right as far as phonons goes it is kind of like that I'd say yeah so I hope it's difficult to do both functions how can it be this tradition if it blows up it has to overall influence to convert somehow right because verbi function is a number between 0 and 1 it's a probability you can understand it but here it blows up at x to 0 it even takes minus value if x goes to 0 if x goes to 0 this blows up and this is all related to that whole thing about the Bose condensation and things of that sort right so the Bose function is very different with different physics in it and of course the Fermi function it applies to electrons with exclusion principles you can have no more than one electron in something which is where the Fermi function goes from 1 to 0 so you either have it's either 1 or 0 and in between means that sometimes it's full sometimes it's empty whereas with photons or phonons they don't have that exclusion principle and millions of them can be in the same state laser of course puts lots of them into one particular omega etc but at equilibrium this is the distribution that people believe where you are so you could argue that for a given H bar omega there would be the half H bar omega energy that goes with it that was the argument so when you look at it this way then it looks very similar to the conductance formulas the point I was making is q square over H it's this k square t over H and then there is this internal thing with m lambda over L plus lambda now I have a question so most functions that actually mean that's real that H bar so what does that actually mean so the average number of phonons or photons in a particular frequency mode so one of the first times it was used probably was suppose if you are counting the number of the black body radiation for example from something you see that so the various places you would see actually the Bose and Fermi both of those follow from a much more general principle of statistical mechanics which is that the general principle of statistical mechanics is that if you have a whole bunch of states then the occupation of these states is proportional to e to the power minus e over kt the energies involved now with Fermi with exclusion principle things that obey exclusion principle you say that there's only two states there's a 0 or 1 and so you say that the probability of being here let's say is 1 probability of being here is e to the power minus e over kt and of course overall probabilities must add up to 1 so let us say we divide it by z I mean z is some constant which we figure out so that the sum is 1 and from that you can get the Fermi function I'm not going into it but this standard thing now if you apply the same principle to something with lots of states like this 0, 1, 2 and no upper limit and add up all of these then you would get the Bose function so in that sense both are derived from one general principle you could think of it the general principle being that the difference is with electrons anytime you have a state you can either occupy it or not occupy it that's your option nothing else you can't put two electrons in there with bosons it's like once you have a state you can occupy it you can put two in there you can put three in there you can have four in there all these possibilities and then you find the average so I guess under high temperature x is small so you get some simple answer there can you get some simple expression for G at room temperature what kind of argument for the conductance itself oh I see because lot of the h bar omegas involved are probably comparable or less than kT then that might because usually when you are calculating the energy in a particular mode this is the number of phonons in a particular mode if you want the energy then you have to multiply by h bar omega so what happens is if you want the energy in a particular mode you write h bar omega over e to the power x minus 1 which would be like x divided by e to the power x minus 1 times kT and this one doesn't blow up either as it is and so if you actually looked at the energy in a particular mode that wouldn't blow up somehow just this you have this millions of them with not much energy but the point you were making was that when x is small then I guess this becomes 1 and so the energy per mode is kT which is this equipartition theorem but when x is large it cuts off so when you actually look at the energy per mode it stays fixed up to a point and then drops off that's the and the low frequency end is kT and in the long run it drops off now this k square T over h that actually has the dimensions of I think watts per Kelvin so I think it comes to about I forget 80 picowatts per Kelvin or so right this has this you can put in the numbers I think this is what you have to get just as when we talk about conductance you know I made this point about when you look at conductance it's like q squared over h and then there is m lambda over L plus lambda so if you wanted conductivity you would look at q squared over h m over a m lambda over a so last day I was trying to say that when you are trying to estimate the conductivity I think we went through an example of how you would estimate the conductivity of say copper and the way you could do it is you could say well this is the quantum of conductance standard thing real important point is how many channels do you have per unit area when you are thinking of a big conductor and then there is the mean free path so from this point of view what determines the conductivity of something is that around the chemical around the Fermi energy this chemical potential how many channels do you have per unit area and then there is also the mean free path and this is then what I identify as a conductivity and for phonons also then we are having a similar expression you see it is like k squared t over h and again you could ask the same question what is number of modes per unit area what is the mean free path same issues now what determines this number of modes with electrons I said that the way you can think about it which is not immediately obvious when I define modes as d times v but with some discussion you can see it is that it is like how many wavelengths fit into a cross section and the big difference between metals and semiconductors is that metals the Fermi energy is such that the wavelength is almost the atomic distance so basically how many wavelengths can you fit in a cross section is almost like asking how many atoms do you have in a cross section you see whereas in semiconductors the de Broglie wavelength is more like tens of nanometers and so the number of modes is a whole lot less and that is why semiconductors conduct a whole lot less because mean free paths if anything semiconductors have much better mean free paths so it is really the M that cuts it down so much now the thing is with phonons the number of modes is almost like this electrons in metals it is almost like you got if you look at the phonon wavelengths those are the highest ones are what matters of course is the wavelength of a phonon of course changes with energy and what matters most is again average energy which is of the order of kt because the peak of that curve you know if you look at how far it spreads out it is a few kt and so if you look at the corresponding wavelengths of phonons it is again atomic distances so the point is when you look at the number of modes involved in thermal conductivity usually it would come to about this number of atoms so what I mean by that is this number here m over a that number would be roughly one for every you know three angstroms by three angstroms so 0.3 nanometers by 0.3 nanometers if it was one nanometer by one nanometer you would have put in probably 10 to the how many per unit area so you would have put in about 10 to the 18th per square meter if it was one nanometer by one nanometer with 0.3 you would probably put 10 to the 19th per square meter and the mean free paths for phonons they vary widely you know the best thermoelectrics of course thermoelectrics are materials where you really want to cut down the thermal conductivity because you are trying to get in the way there and those are usually the best ones are those with very low mean free paths you know you try to use materials where the phonons have very low mean free paths because of various grain boundaries or other reasons inside and so then this can be say even one nanometer or it can be much longer 100 nanometers or so and so a lot of the difference in thermal conductivity that you see between materials is because of this big difference in lambdas not so much because this changes widely I mean there isn't quite the equivalent of metal semiconductors that I have seen it is more I mean they are all basically like metals as far as modes go this is the argument I have went through oh by the way just to complete that if you put in this 80 picowatt per Kelvin and go through this you will get numbers of you know typically like what you see for thermal conductivity of materials it would be about right now so the argument with that one would be that you are trying to find the wavelength let's say and so with phonons omega is equal to velocity times k so this is the velocity of sound times k and the k is like 2 pi over wavelength and let's put h bar h bar so and let's say we are looking for a phonon whose energy is kt saying that that would be a typical one that is then equal to h c s over lambda so the corresponding mean flip so by the way this lambda is de Broglie wavelength nothing to do with the mean free path totally different thing lambda de Broglie is equal to h c s over kt now typical sound velocities are kilometers per second 3 kilometers per second that's the numbers I have so when I put those in this is the kind I tend to get wavelengths that are nanometers or less than a nanometer yeah they move so much forward say this again they move so much slower than electrons although the c s is I see that's what is making the wavelength short is anti electron wavelength different from the phonon this commands a good conductor for p and so you are saying the wavelengths are the same the de Broglie wavelength de Broglie wavelength it's going on wavelength in a level it's the same wavelength twice they are quite similar which is why as far as the number of modes go phonons seem to have much the same number as electrons in metals as far as the number of channels go because of the number of wavelengths that's what I seem to be getting right this is why c t is so low you have a lot of local conductivity thermal conductivity m over electron is divided by m over so that's always going to be small right that does look like a problem that is true right from what I have seen but this needs a lot more thought I have not given all this careful thought yeah mostly acoustic mostly acoustic for that they are very large so not considering zone boundary yeah though a lot of the zone boundary ones are also involved so you know you draw this omega k picture so acoustic phonons once you are we can here are probably way down here but what conducts heat I think is a lot of these also from that if you go to the boundary then you will be getting less and less but around here this is still comparable to an atomic distance because if you were here then the wavelength I think would be atomic distance if you are at the zone boundary but it's getting comparable a lot of these because there's a lot more modes density of modes goes up in energy also but then it gets cut out by the that function the Bose function but that's why I said kt may be a good number to go with but it could be less than kt all that needs careful discussion but the thing that is different I notice is that while with electrical conduction there is this orders of magnitude difference between a metal and an insulator there is no such comparable thing here which is why thermal conductivities don't really differ by major orders of magnitude it is within us relatively short wind a small window really that's what I have seen ok thank you yeah let me stop here