nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 3.6: What a Probe Measures
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[Slide 1] Welcome back to Unit 3 of our course. This is the sixth lecture. 

[Slide 2] Now in the last lecture, we talked about this channel with a localized scatterer, where electrons were transmitted with the probability T. And what we discussed is that in that case, the quasi-Fermi levels would drop abruptly across the scatterer because as you cross the scatterer, suddenly the density of electrons goes down. And that's what is reflected in this quasi-Fermi level. Now quasi-Fermi levels, of course, are, again, deep concepts that you have to think carefully in terms of these deep concepts of statistical mechanics to get it right. And so often people are uncomfortable about it. And they say, well, maybe we don't need to talk of quasi-Fermi levels. Let's talk in terms of what you actually measure. So they say let's not talk about what's the quasi-Fermi level is inside that channel. Instead, let's say we put a little scanning probe up here. You see like a scanning, tunneling probe, and ask the question what potential would that probe float to? Now how would you answer that question? Well, how is this probe potential related to the quasi-Fermi levels? Well, you could think of it like this. You've got one group of states, the right moving states, with the quasi-Fermi level mu plus. Another group of states, the left moving states, with the quasi-Fermi level of mu minus, and these are things inside the channel. And then you have this probe, whose potential is mu p. That's this accessible quantity from outside, which this volt meter measures. That's this mu p. Now this region here, this little gap where electrons can tunnel through, you could represent that with a little conductance. And the thing is we have used two separate conductances because in general the probe -- electrons in the probe may be better coupled to right moving states than to left moving states. Normally, that wouldn't happen. Normally, you would couple equally. So those could be equal. But more generally, they could be unequal. So I've left it as g plus and g minus. So how would we determine the pro potential? The answer is that you see this is a volt meter. It doesn't draw any current. So no current flows out of here. So what that means is what Kirchhoff's law requires is that whatever current flows here must-- this current must be the exact negative of that. So if you have one micron coming out this way, you must have one micron going back this way. So what you can do is write this Kirchhoff's law for the current in this arm, which would be mu plus minus mu p times this conductance. And you could write the current in that arm, which would be mu minus, minus mu p times that conductance, and then say that the sum of the two must be zero. And if you do that, then with a little algebra, you'd get an expression for what the probe potential must be. And it will come out as a weighted combination, a weighted average of mu plus and mu minus. So these factors in front, they add up to 1. So if one of them is alpha, the other is 1 minus alpha. So if you had a probe which coupled very well to the positives, to the right moving states, then this would be a high conductance. That would be a low conductance. In that case, alpha would be 1. And that would be zero. On the other hand, if you had a probe that coupled better to the left moving carriers, then this would be zero, and that would be 1. So what would this probe measure? Well, if alpha is 1, it would measure the top curve, like so. And it would say measure 1, and then you see it drops to T. And so you'd say the voltage drop is 1 minus T. And the current is proportional to T. And so the resistance is like 1 minus T over T because that's the voltage drop. That's the current kind of. On the other hand, if alpha was zero, then you'd be measuring mu minus, which means you'd be looking at the bottom curve, this one. So you'd say it's 1 minus T, drops to zero. Again, you'd say that drop is still 1 minus T. And more generally, more commonly, of course, what would happen is alpha would be half, and that would also be half. What you would be measuring is the average, which is this black curve. And there also the drop is still 1 minus T. So in any case, you basically come to that conclusion, as long as the probe always had those fixed alpha as you move it along, as long as the alpha didn't keep changing as we were moving. Now this gives you a fairly good way intuitively of understanding what a probe measures relative to the quasi-Fermi levels. And it works well as long as this is a weakly coupled probe, or what people call a non-invasive probe. That -- what that means is it's so weakly coupled, it doesn't change what is going on in the channel. It just senses it. On the other hand, you see back in the '90s, when mesoscopic physics was developing, late '80s and '90s, scanning probes were not that common. 

[Slide 3] And the way people measured voltages was usually with conductors that were connected to the channel. So in other words, it wasn't a noninvasive probe, not this weak thing hovering a few angstroms above the surface, but more like actual conductor that was connected directly to the channel. And what that means is this conductor itself changed what you were trying to measure. It acted as a scatterer on its own. So what that means is these quasi-Fermi levels we have been discussing, none of it would be relevant anymore because those were, of course, calculated assuming these probes weren't there. And just the presence of the probes changes the entire distribution. Now in terms of interpreting experiments of that types, you see, these Buttiker equations prove to be very successful. This is what Buttiker suggested back in 1986, I believe. And what he said is that when you have these four, I guess, probes, mu1, mu2, mu3, and mu4, treat them all on an equal footing and write down the current at the four terminals in terms of the potentials at the fur terminals. So you've got four things here, four things here. So you've got this conductance matrix, a 4 by 4 matrix that connects them. Now what does G1-2 represent? Well, it represents this connection from 1 to 2. So how easily electrons go between those two; 1-3 would be 1 and 3, 1-4 would be like 1 and 4, and let's say we have a way of finding these. you could measure it experimentally or you could have a theoretical model for calculating those things. So if you have those, then you could calculate the current if I gave you all the potentials. But then you might say, well, I can't give you all these potentials. I kind of-- I could assume one of them to be ground, and then I could tell you what I applied here. But I have no idea what these are at because they just float. I don't know what it is. Yes, that's true. But you see, on those floating terminals, you do not know the potentials. But you know the current. The current is zero. So I3 and I4 are zero. And so the key observation here is that you got four currents, four potentials, eight quantities. But if I give you any four of those, you could find the other four. That's-- I mean it might take a little work. But the point is it's straightforward algebra. So while I could do something like this, let's take this 4 by 4 and kind of write it in a form so it kind of looks apparently 2 by 2. That is what I do is I take this I1 and I2 and give it a name, I1-2. I3 and I4, give it a name, I3-4, similarly, mu1-2, and mu3-4. And then this conductance matrix kind of looks like 2 by 2. But of course, it's still 4 by 4. It looks 2 by 2, but then each one of these elements, this A, B, C, and D, are 2 by 2 matrices. So now we say that I3 and I4 need to be zero. So what that means is what is I3-4? It's C times mu1-2 plus D times mu3-4. And that must be equal to zero. And now with a little matrix algebra, I could write mu3-4 as negative D inverse C times mu1-2. Note this is a 2 by 2 matrix. That's a 2 by 2 matrix. This is a 2 by 1 column vector. So when you multiply it all out, you get a 2 by 1 column vector. So it's a matrix. Now I can take that a put it back into the first equation because the first equation says I1-2 is equal to A times mu1-2. That's the first term. And then I have a B times mu3-4. But for mu3-4, I can put minus D inverse C. So finally, I1-2 is equal to A minus BD inverse C times mu 1-2. So what that means is if I gave you mu1 and mu2, you know how to calculate the current. We just find A minus BD inverse C. And if you want, you can calculate the mu3 and mu4 as well. So this is the procedure that is widely used. You see Buttiker's 1986 paper, that's very widely cited, lots of calculations that have been based on just doing this and allows you to compare with experiment. Now there's a few subtle points that I'd like to mention here. 

[Slide 4] One is that you see you have this 4 by 4 conductance matrix, and you might say, do these sixteen elements have any constraints, or could they be anything? And there's very-- a couple of very, very important constraints, which you could call the sum rules. And the sum rules come like this. One important constraint is that if I put the same potential on all 4 terminals, like 1, 1, 1, 1, or .5, .5, .5, .5, there should be no current. That we know from experience. You know if you have four contacts, and you put the same potential on all of them, no current flows. But in order to make sure that your equations know that, what must happen is if I put the same number here, I should get zero. The only way to ensure that is if every row adds up to zero, so these four things, this, this, this, and this, must add up to zero. The next row must add up to zero, the next row must add up to zero, and so on. So that's one of the sum rules. All rows must add up to zero. Now another constraint is that no matter what potential I put, the currents must all add up to zero. That's Kirchhoff's law. You see if you have this, know where all these currents are coming in total must be zero, at steady state. And that requires that each column must add up to zero. So those are the two important sum rules. Now if you had a two terminal device what that means is, that when you have a two terminal device, let's say that's your conductance G zero. Then the sum rule requires that must be minus G zero because the two must add up to zero. Similarly, the sum rule requires that must be minus G zero and that must be G zero. So when you have a two terminal device it's not like you have four conductance elements. We really just have one, G zero, which is why normally we don't even bother to write that matrix. We just write I equals G zero times mu1 minus mu2. But once you have multi-terminal devices, it's much more subtle. There's many more possibilities. In general, G1-2 doesn't even have to be equal to G2-1 if you have a system like in a magnetic field. In a magnetic field, G1-2 for positive magnetic field is equal to G2-1 for negative magnetic field. And of course, as I mentioned, the matrix elements should still obey the sum rules that I mentioned. One other little subtle point here, and that is that, I've mentioned before that conductances are usually obtained by averaging over energy. And usually I have often gone back and forth between the averaged quantity and the actual energy dependent one. The idea is you can talk about this and remember at the end of the day you average. In this case, though, it might actually make a difference. That is you might ask, I've got two possibilities. That is I want to do this four terminal device, and I've got a model for calculating these G's. Should I use the energy averaged quantities here, or is it like I should do it one energy at a time? For every energy put in the right value, calculate currents, and then average at the end of the day. And my belief is that wouldn't be right. What you should do is use the energy averaged quantities right here because if you're doing it energy by energy then at every energy you might calculate a different mu3 and a different mu4, which wouldn't be right because physically of course in this structure you have a single mu3. It's not like different energies have different-- flow through different potentials. So in general, when you're using this here, I belief these should already be the averaged quantities. That is that's what you put in, and then you do the algebra for figuring out the quantities of interest. 

[Slide 5] So that then we can move on to the next topic, which is, as I mentioned, that one way to make this discussion concrete in terms of potentials inside the channel is to talk about what is measured, and that's what we just discussed. Another way people sometimes talk about this is to say well, let's talk about the electrostatic potential. And the question is can we do that? That is just as we have a quasi-Fermi level that's changing, you'd think that there'd also be an electrostatic potential inside which would change, because of which the entire bands would also change the same way, for example. And this, of course, needs some discussion, and that's what we will do in the next lecture.