nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L2.7: Transmission
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[Slide 1] Welcome back to Unit 2 of our course. This is the seventh lecture. 

[Slide 2] Now in the past lectures when we are setting up these NEGF equations. One of the things I stressed is this picture of inflow and outflow. That is, there is this inflow and this outflow, and if you want the current at some contact, you can look at the difference of the two. Now what we'll be doing in this lecture, is I'll be giving you an alternative expression for current, which we'll obtain starting from this, and reason I haven't quite done this before is that this expression is not general. It only applies under a certain, limited set of circumstances, and I'll try to explain what it is, but it leads to this important concept of transmission, which is widely in the mesoscopic literature and in nano electronics, and so I wanted to connect with that. Now the first point is that in these contacts, which are described by a Fermi function, there is this basic requirement, which we discussed earlier, that sigma in should be related to gamma, which is the antihermitian part of sigma, by sigma in equals gamma times F. Similarly, for contact 2. If this is true, then you see, you could take out expression for current that we had before, which comes from subtracting outflow from inflow. You get that expression. You could replace the sigma in with gamma F and thereby get this expression for current. It's trace of gamma 1 times AF1 minus GF. Now this, though, is not generally true. This is the one that is generally true. This is only true when sigma in 1 is equal to gamma 1 times F1, and when is this true? Well, whenever your contact is actually described by a Fermi function. So, for example, I have stressed the importance of including this sigma zero, and later in this unit, we'll talk about it more. Which describes the interaction with the surroundings, and that, in general, would not be described by a Fermi function, but let's, for a moment forget that and assume this is true, so that the current is given by this expression, and we'll now proceed to do a little algebra to put it in form of this transmission formalism. 

[Slide 3] So, we have this AF1 minus Gn, and we need to simplify that. So the first thing is we start with the expression for Gn, which is GR sigma NGA, and sigma in has two parts. The sigma in due to contact one, sigma in due to contact two, and, of course, you are ignoring any other, this dephasing contacts. That is, we are assuming the contacts are described by this Fermi function, and that's all, and there are no other contacts. So we write it this way. A is GR gamma GA. That's an identity we discussed earlier, and the gamma is gamma 1 plus gamma 2. So if you want to write A times F1, you'll have gamma 1 F1 plus gamma 2 F1. What we need is the difference between the two. So when we subtract this from this, you'll notice the first term cancels out, and the second term gives you GR gamma 2GA times F1 minus F2. That's this. So now we can write the expression for current as trace of gamma 1, GR gamma 2 GA times F1 minus F2. Now I should remind you once that we are generally writing this current through a range of energy dE. So this is like current per unit energy, and if you actually wanted the total current, you should integrate over all energy. All right? But, in general, I won't be writing this integral, and you should remember what we are writing is the current per unit energy, along with the given energy. Now this is then what is called the transmission from contact one to contact two, between the two contacts. Now this is actually more general. You could apply to multi-terminal devices also. 

[Slide 4] That is, if you have a device with many terminals, but they're all described by this rule. Namely, that sigma in is gamma times F, and I should mention in this context that there is a widely used method for including dephasing due to Buttiker where you instead of these sigma 0s, what you imagine is as if there's another contact there, but you assume that that contact is also described by its own Fermi function, whereas what we'll be using is the NEGF method where would do not assume that these are described by a Fermi function. I believe that the NEGF method is more versatile, what the Buttiker method is widely used, and there you usually think of it as another terminal, but described by a Fermi function, and the expression we will now be obtaining applies as long as your contacts are described by Fermi functions. So if this is true, now we can start from our usual expression for current. That's obtained by subtracting outflow from inflow, and this is an expression that holds in general, but what we'll be doing after this assumes this relation, and so is not generally applicable. So if you use this, I guess, instead of sigma, you have a gamma times F, so you get AFP minus GN. How do you get AFP mine us Gn? Well, if we look at Gn, it's GR times gamma 1 F1, gamma 2 F2, et cetera. If you look at AFP, well, A is GR, gamma 1 plus gamma 2, et cetera. If you multiply by FP, you get FP everywhere. So if you look at these two terms, AFP and Gn and we need the difference, you'll notice each term is similar. They all involve GR, gamma 1, GA. GR, gamma 2, GA. But in this expression, every one of these Fs is FP, but here, it's F1, F2, et cetera, so when I subtract the two, you see, what I get is GR, gamma QGA, FP minus FQ, where q runs from one, two, and however many terminals you have. So if you put q equals 1, you get the difference of these two terms. If you q equals 2, it's the difference of those two, and so on. Now if you take that expression and put it back in here, you'll get this general expression for current in a multi-terminal device, and those of you that are familiar with the Buttiker equations, which we briefly discussed in part A, will notice the similarity. That is, what appears here is what in the Landauer-Buttiker formalism would have been called this transmission Pq. Okay? And its reason it's called transmission is that this quantity can be interpreted in terms of the probability that an electron can transmit from one contact to another, and the rest of this lecture, I'll try to explain this point. 

[Slide 5] And for this point, let us consider a simple example. We have a 1D wire, and if you took a course in quantum mechanics, one of the first things you'll learn is how to calculate the transmission through a potential barrier, and we are assuming a very sharp potential barrier, like a delta function that simplifies the algebra a little. So you have U0, a delta function at Z equals 0. You have an incident wave a reflected wave, and a transmitted wave, and one of the things you do in elementary quantum mechanics is you try to use the continuity of wave functions and their derivatives to calculate the reflection and transmission, and what I'll show you is that by using our expression, based on the NEGF method, you'll get the same answer that you would have got if you had done it the usual way, the way one does it in quantum mechanics. Now what we'll be doing, though is the problem not with the continuous, special variable Z, but a discrete version. You know, where you have a discrete lattice. So Z is actually an integer n times the lattice spacing A. So you still have incident wave and a reflected wave and a transmitted wave, and this is discrete lattice. Diagonal elements are epsilon. The connecting elements, the upper diagonal and lower diagonal elements are T. One tricky point, when we go from the continuous to its discrete, this delta function, note that this is U0 delta Z, what you should put here is U0 over A, the potential, because here, because it's a discrete lattice, your width of this is like A, and the definition of a delta function is something that's infinitely tall but also infinitely thin, such that the area under the curve is 1. In discrete lattice, of course, the thinnest you can get is A. So in order to have an area of 1, the height needs to be U0 over A, so that the area under the curve would be equal to 1. That's the delta function, and, of course, you want the area under the curve to be U0, because it's U0 times the delta function. Okay, now what we'll show then is we'll apply our method to calculate this transmission using the Green's Function Method, and we'll show you that you get the usual answer. That is, you would obtain for transmission through a scatterer, which is given by h bar v, the velocity, squared divided by this U0 square, plus h bar v square. So how do you do that using the Green's Function Method? 

[Slide 6] Well, the way it working is takes this discrete lattice and, in general, if I was doing this numerically, say I have 10 points in my lattice, so all of my matrixes would be 10 by 10, but for this problem, because it's a delta function, it just involves one point, so I could just take that point as my device and represent the rest the here, in terms of a sigma, rest of it there in terms of a sigma. So, you see, if my device had 10 points, then H would be a 10 by 10 matrix, and I'd be dealing with matrixes that were 10 by 10, and for that, usually, it would be kind of inconvenient to do it by hand, but on a computer that's nothing. Computers, you know, thousand by thousand matrixes, no big deal, really. And for numerical simulation, you just write down the matrixes and get the computer to evaluate it for you, but if you want to do it on paper, then it's good if the matrixes are 1 by 1, maybe 2 by 2, okay? And for this problem, we is use a 1 by 1. There's just one point, the diagonal element is ordinarily epsilon, but because there's a barrier of height U, it becomes epsilon plus U. What's sigma 1? We discussed in the last lecture, it's t to the power ika. Sigma 2? That's also t to the power ika. Remember, these are also 1 by 1. So usually sigma 1 is t to the power ika on one point, and 0 is everywhere else, but this is just a 1 by 1. There's only one point. So it's just t to the power ik. What's gamma? Well, that's I sigma minus sigma 1 dagger. And as you have seen, that gives you this 2t sine ka minus 2T sine ka, which you can write as h bar v over A, like we did in the last lecture. The gamma 2 is also h bar V over A. So now we are ready to evaluate the transmission. Gamma 1GR, gamma 2GA. We have gamma 1 and gamma 2. We need this Green's functions. So how do you get the Green's function? It's E minus H minus sigma inverse. So that's E minus H. H is epsilon plus U, minus sigma is sigma 1 and sigma 2. This is t to the power ik. That's also t to the power ik. So minus 2t to the power ik. Now E to the power IX is cosign X plus I sine X. So if I spell that out, I'll get 2T equals sine ka. plus I sine ka. Now you'll notice here that because you have the dispersion relation in the leads. E is equal to epsilon plus 2t cosine ka. It means this E minus epsilon minus 2t cosine ka, that just is 0. It cancels out. So what you are left with then is minus U, minus 2It sine ka, and 2t sine ka minus is h bar v over A. So you get minus U plus Ih bar v over, so that's the Green's function. 

[Slide 7] So if you put all that in then, your transmission involves these four terms, and we just calculated GR. So if you write it out, the gamma 1 is h bar V over A. We just calculated GR. That's gamma 2, and GA is just a conjugate of that. This is 1 by 1, so I don't need to worry about transpose, but that's the conjugate. When you multiply it out, you get h bar v over A square, U square, h bar v over. Now remember that U is related to the strength of the function U0 by U0 over A. So if you put that in here, then all these As will cancel out, and then you will have this result, and this is a standard result. You could have obtained directly by solving Schrodinger equation with a delta function potential, but the bottom line of all this is that this t bar, which came from our Green Function Method, you see, it can be interpreted as this transmission to a barrier, as if your device is some kind of barrier, and what you're calculating is the transmission from left to right. Now I should point out though that this transmission probability here, this transmission here, is actually, in general, it is the number of channels in your conductor times the probability. So let's say you have 10 channels, and the average probability is of an electron to get through from left to right is 0.2. Then this quantity would have been ten times 0.2, which is like 2. Here because we are using a one-dimensional wire, of course, this quantity itself is the transmission probability. But in general, if you had many channels in the wire, so if it's a two-dimensional wire, then you see that number, could, in principal, what you calculate here could even be bigger than 1. 

[Slide 8] Okay? Now so if you plot this out then. So it's like, you have this one-dimensional wire, what we had seen in Unit One is the EK relation which was obtained by assuming periodic boundary conditions, and what we have now calculated is the transmission as a function of energy. Now if U is 0, which means there's nothing here, no barriers, then, of course, you have a ballistic wire, and in that case, the transmission is 1. One across the band. That is, from the bottom of the band to the top of the band. Once you go outside the band, it's 0, because then there's no states. The transmission is exactly 1 here, and that comes from this expression. If you put U0 equals 0, that's it. It's 1. On the other hand, if you put a U that is equal to 2 times t then you can see the transmission goes down. And it goes down by different amounts at different points. Why? Because how much it goes down depends on the velocity, and the velocity varies along the EK plot, because the energy is epsilon plus 2t equals sine ka, so the velocity goes as sine ka. So the velocity is 0 here, where K is 0. It's also 0 here where KA is equal to pi. It's a maximum right in the middle where ka is pi over 2. So the maximum value of this velocity is like, 2t. So that's why right around there, that's where the transmission is the biggest, because that's where h bar v is biggest. So it all follows from that expression. So this then gives you a feeling for how to connect what we did. What we did based on this Green's Function Method to the scattering theory and this transmission formalism that is widely used. 

[Slide 9] So to sum up here then, basic point is that when you calculate current, this is a very general expression that is generally applicable, but then you can write the current in this form, in terms of a transmission easily, if sigma in is related to gamma through a Fermi function, and that holds if your contacts are described by a Fermi function. May not hold for, let's say, this interaction sigma. Because there you cannot assume, in general, it's a Fermi function. So if you assume that, then you can write it in this form, and this is a form that is widely used. Now, in general, of course, as I mentioned in part A, you can always write the current in terms of F1 minus F2 times something. So in principle, I could have used this expression to find the current, divide it by F1 minus F2, and found an effective transmission. That you could always do. It is just that if these things are not described by a Fermi function. I won't have a simple expression like this for transmission. So what I should do is still, if I really want current, I should use this, and then if I really want to find a transmission, so I can compare with other things, then I should take whatever current I get and divide it by F1 minus F, and remember, here I've explicitly shown the integral over energy. By and large, I have not been doing that. Here, for example, this is current per unit energy, and it's understood that we should have an integral over energy on this one as well, okay? And so this is what I tried to illustrate with this 1D example, a simple 1D example with a barrier. This is a very instructive one to go through because, you see, we have the general equations. That we did in the first part of this unit. What we now want to do is look at different examples to get a feeling for how to apply the method. How to apply these equations, and we have kind of done the simplest of these, which is the one barrier. In the next lecture, we'll do two barriers. Thank you.