nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L3.2. Quantum Point Contact
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[Slide 1]  Welcome back to Unit 3 of our course on quantum transport, and this is Lecture 2. Now, as I explained in the introductory lecture, in Unit 2 we introduced this method for handling quantum transport, this general method. And what we are doing in this unit is looking at different examples to get a better feeling for how to apply the method. And the first example we'll look at is this very important example. It's about quantum point contacts. 

[Slide 2] Now, what's a quantum point contact? Well, the point contact means like what I've shown here schematically that current flows from left to right. You have this applied voltage, but they have to squeeze through this narrow region because this black regions-- what you have done is by putting a potential on these gates, you have depleted the conductor underneath it. And so all the electrons have to squeeze through this region, and so the conductance is largely determined by this narrow region. And since this narrow region is fairly short, what you measure is the ballistic conductance of a region whose width is this W, which is this little white region through which the electrons have to squeeze through. Now, as we discussed in Part A, what you'd have expected is that this conductance, this ballistic conductance would vary linearly with the width because ballistic conductance doesn't depend on the length of a conductor, but it sure depends on the cross section. So, this is what you'd have expected, and there were many experiments in the 1970's, I guess, where people had seen this, and this is generally what's referred to as a Sharvin resistance. But what had a major effect on this field was a series of experiments starting from 1988, which showed that in small conductors what you'd see is a conductence that doesn't change linearly like this but rather it goes in steps. So, this is one of the experiments from one of the groups, which appeared in physical review letters, and around the same time similar results also appeared from a different group in a different journal. And I've just shown this one. What it shows is the resistance versus the gate voltage. So what the gate voltage does is changes the width of this narrow region. So, this axis you could view as W. It's the width of your ballistic conductor. And you can see this resistance shows these steps, and when you subtract out the series resistance of the surroundings and invert it, you get the conductance, and what you see is this conductance goes in steps. And it can be expressed as this quantum of conductance q squared over h times 2 for spin because usually these moldes or channels they always come in pairs. Anytime there's an up spin, there's a down spin that goes with it. That's the two. And then M is the number of channels. So this was a seminal experiment because this quantum of conductance, I guess since then has become very-- is an integral part of our general discussion of nanoelectronics and mesoscopic physics because this idea that conductance is this quantum of conductance times the number of channels. And experimentally what you're seeing in this ballistic case is these steps. So this is a very important experiment, a very important physical phenomena. Of course, in this context I'm just using it as an example where you can try out the method for quantum transport that we have discussed because whenever you have a new method, in general the principle is apply it to a problem where the results are well known and make sure you get the correct results before you go off and do problems where you don't know the answers. And so I always suggest this as a very good example to start from because if your method works and you have set it up right, then you should be able to get this quantized conductance. So how do you do this? 

[Slide 3] Well, if this was a one dimensional problem, then you could do what we discussed in the last unit. That is, set up this one dimensional tight- binding model with a diagonal element epsilon and the nearest neighbor coupling is t, and so the h, the Hamiltonian would look like epsilons on the diagonal and t's on the lower and upper diagonals. And if you want the sigmas, well for the one dimensional problem, we had discussed in Unit 2 that the sigma looks like t e to the power ika. It is non-zero only for the boundary points. So at this point, it's non-zero so it's t e to the power ika. Rest of it is all zero. Here, also it's t to the power ika for the first point, rest of it is all zeros. And what's k? Well, that's related to the energy, and you get it from this dispersion relation for the leads. This e episilon plus 2t cosine ka, and you can invert it to get an expression for k. So, this is how you could write down the H's and the sigmas and once you have them, you can use the general equations that we talked about. This is coherent transport so you could just calculate the transmission. Now, if inside you happen to have a barrier. Let's say it's not a uniform conductor, but there's a barrier there, well you could include that as well because what it means is that the diagonal element for that point would have an additional U. That's the barrier. So what that means is to the H, you'd add another matrix like this which has a u at the second point, for example. Because here I'm kind of showing it as if the device has three points so the U is in the middle. So, the point is that numerically the process is quite straightforward. You could have any potential inside. You can easily include that in the H, and you can write down the two sigmas and then you can calculate anything you want. But the thing is, this is a 1D example. What we need to do is the 2D version of it because 1D will not capture the physics of this. 

[Slide 4] Now, for 2D, you need to start from a two dimensional lattice. Again, the diagonal elements are epsilon, the nearest neighbor coupling is t. If you want to include this point contact, what you could do is put in something like this where on these sites you could add a U, the same way we added a U in the 1D example. So that way, we can put a very big barrier here so that the electrons can't even get near it. And thereby, the electrons would have to squeeze through this point contact. So numerically, that's how you could include or you could make this structure look like the physical structure that you are trying to simulate for that to be part of the H. Now, you then need the contacts, and you have to be able to write down the sigmas. And that's what I'd like to explain next is how you write down the sigmas because what we discussed in Unit 2 was how to write down the sigmas for a 1D problem. Now, we have a 2D problem. How do I write down these sigmas and what you'll see is that in this case but using a clever trick, you can extend all results from 1D to write down the results for 2D. So let me explain how you do that. 

[Slide 5] So the way it would work is let's say just to be clear we take three points along the way, and you want to find out this sigma. So this is the actual device side, and it's connected to this side, and you want to write down this. So, what you could do is when you're trying to think, visualize this H matrix, it is convenient to think of this column as one unit. You see, this is a 2D problem, and a very good way to do the bookkeeping of how when you're setting up this Hh because in 1D it's relatively easily to set up the H. In 2D, the bookkeeping isn't completely obvious. In other words, it's not even unique. There are many different ways of doing it, but the one I found most convenient is to think of this column-- I've shown here three things-- there could have been ten things there. The column is one unit, which I call alpha. And the coupling of this unit to the next, that's what I call beta. So this alpha itself is a 3 by 3 matrix. So usually in 1D epsilon is a number, t is a number. But now this kind of looks 1D, but the point is instead of epsilon, I actually have a 3 by 3 matrix there. And the beta is a 3 by 3 matrix. And this is a very good way to visualize what's happening here. How do you write alpha? Well, alpha is basically this 1D chain. So alpha would look something like this. Epsilons on the diagonal and t's connecting it to the nearest neighbor. So t's, so that's it. So this is alpha. How do you write the beta? Beta is the connection from one column to another. And there you notice the connection is from point 1 to point 1, point 2 to point 2, point 3 to point 3, and so beta looks diagonal. It's just t's along the diagonal. It connects one to one, two to two, and three to three. Now comes the important point, and that is that what you can do is do a basis transformation so that in the new basis alpha becomes diagonal because this is not a diagonal matrix. You got off diagonal terms, but you can always diagonalize the matrix by using the eigenvectors of alpha as your basis. So, the general procedure is that if given an alpha, you can find the new alpha, which I have indicated with the tilde by finding an appropriate transformation matrix V. So if you're doing it numerically, there's a command in metlab or in any other software that you're using, which will allow you to do that, which allows you to diagonalize this matrix. It will tell you what this transformation matrix should be. By hand, analytically if you want to find it, that's also straightforward. It's something you can look up in a linear algebra book if you want to see how to find V's. But once you have that then, you could transform this alpha from this off diagonal-- get rid of these off diagonal terms so that it looks diagonal. Now, one point though is that you see in our model we got two matrixes, alphas and betas. And usually often what happens is if you diagonalize one of them, the other one may not remain diagonal. And so doesn't help because, of course, whenever you're solving a problem all matrices need to be in the same basis. But here the advantage is because this beta is like an identity matrix, it's the same t everywhere, so what that means is when I change basis, the identity matrix always stays identity. And so in the new basis, we still have exactly the same beta. And so in the new basis, alpha is diagonal and so is beta. And what that means is that you can visualize in this new basis, which I'll call the eigenmode basis-- you can visualize the problem in terms of this new alpha, the alpha tilde, the beta and they're all diagonal, and what that means is it's like separate 1D wires. So this wire has a diagonal element of epsilon 1, and the connection is t. The next wire has a diagonal element epsilon 2. The connection is t. Epsilon 3 connection is t, but the point is there's no connection in this direction, unlike the original problem. The original problem had connections going this way. Now, there's no connection in this direction at all and so the original 2D problem has now decoupled into a set of parallel 1D problems. And what that means is, of course, that since there are parallel 1D problems, I can do them one at a time. 

[Slide 6] So what I can do next is, I can write down the sigma for these 1D problem using what we had learned in Unit 2. Remember what we learned in unit 2 is that sigma is given by t to the power ika for the boundary elements, and it's 0 for the rest. So we just focus on the boundary elements so for 1D you'd have got a t e to the power ika. Now what happens is you've got a whole bunch of things in parallel so you get t to the power ik1a for wire 1, t e to the power ik2a for wire 2, t e to the power ik3a for wire 3 and so on. And how do you find this case? Where they're related by this dispersion relation when e is epsilon 1 plus 2t cosine k1a. Or epsilon 2 plus 2t cosine k2a, etc. So you can determine k1a or k2a from the old formula except that each wire gives you a different k. They all have the same energy, e, but the k's are different because the epsilons are different. So the point is in this basis you have managed to write down the sigma. So does that mean we need to solve the problem in this new basis? Not quite. Once we have this, you can always transform it back because just as you can go from the original lattice to the eigenmode, you can always go backwards, take the eigenmode representation and transform back to the lattice representation and the method is here if you had a V dagger and a V, here you'd go V and V dagger, right? This is the standard principles of basis transformation in linear algebra. So here we have the sigma in the new basis, in the eigenmode basis, so you could just transform it back, and you'd have the sigma in the lattice basis where we want to work. So, what we have done is by using this very clever method of basis transformations, we have managed to use the 1D results that we learned in Unit 2 to do a 2D problem because we took the 2D problem and doing a basis transformation decoupled it into a bunch of 1D problems. And this is conceptually a very important thing because that's essentially what you mean by modes, that when you have a conductor with a cross section which is not a 1D conductor, a 2D conductor, a 3D conductor, you can think of it as lots of modes in parallel, which are all decoupled so the mathematical basis for that is this diagonalization of these alphas. 

[Slide 7] Now, you could use this method now to numerically simulate a structure like this. And if you're doing it on a computer, that's all there is to it. So you set up the H. You write the sigmas, and then you use the formulas we have discussed earlier. Now, if you want to understand it, then this one is not so straightforward. What you can do is think of a couple of related problems where you can see the answers in a relatively straightforward way. You can kind of see why you'd get this quantized conductance. The point I'm making is if you just set this up on a computer and you have done it right, you should get this quantized conductance straight out of the simulation, and that's a very good way to check your simulation because as I said, you don't get steps by accident. It means you have done it right. On the other hand, if you want to understand why it looks like steps, then it's easier to think of related problems like this. That is, supposing instead of this gradual constriction we think of something like this where it's uniform with everywhere, and equal to the narrowest region. Now you might say, "Well, that's not the real structure." The real structure has to widen so that in these regions they are very highly conducting, and you can actually assume that the chemical potentials are constant. You actually need these wide regions in general. But for doing this problem, you could assume that you have this narrow region connected to regions which are not widened but is exactly the same. And the reason this works or gives the results-- the same results as you get from the wide case is that both have this important property that when an electron comes from here in this uniform structure there is no reflection because it's a uniform structure. The electron just goes straight out. And the real structure where it broadens, they're also-- there is not much reflection unless the energy happens to have certain specific values. So in general, electrons when they try to come out in the original structure, they don't have much reflection either. So it's that reflection-less contact that's important in order to get nice clean steps. And so if you assume a structure where there's no reflection because it's uniform, you get the right results. And often that's a much easier problem to simulate then the real one. Of course, as I said, if we just went ahead and simulated the real one, you'd get the steps. Now, how do you understand this uniform structure? 

[Slide 8] Well, in this case the way you can think is I got multiple wires, multiple 1D wires so what is the transmission of one of these wires? Well, the idea is if we looked at the first wire, you'd get a transmission, which would be 1 over a range of energies of 4t, and where the center of it is epsilon 1. Similar, the next one would be the center around epsilon 2 and the width would be 4t again. Epsilon 3, and the width would be 4t again. And where does the 4t come from? That's the width of the dispersion relation. E = epsilon + 2t cosine ka. So each one of these then has a band where the transmission is one because it's a ballistic conductor. This band is again one. That band is one. But the bands are offset. And so when you add these up because all of these are in parallel, so when you add them up, you get these steps like so. Incidentally, I write here t whereas here I've written t0. It's really meant to be the same thing. Now, what is the actual transmission that determines the measured conductance? Well, at low temperatures, it is determined by the transmission around the Fermi energy or this chemical potential. So if your mu happens to be here, you have that value. And as you change the width of your conductor, what would happen is these epsilon 1, epsilon 2 would move around and so you may be here at some width and then if you make it a little narrower, this edge might move up so that effectively you could be at this value. So, as these edges move, you'd be going from one step-- one integer value to the next. So you could actually set up your computer to calculate this transmission at a given energy as you change the width of your conductor because as you change the width of your conductor, the epsilons change, and as the epsilons change, the transmission at a given energy changes. And that is kind of what you actually measure experimentally. So, this would be how you could try to understand what you should be getting out of the computer once you set this up numerically. Well, that kind of brings us to the end of this example, this quantum point contact. And as I mentioned, what allowed us to do this was this clever idea of basis transformations, which worked only because when we diagonalized alpha, beta remained diagonal, and the reason was beta had this very special form so when you transform basis, it doesn't change. But this property is not true in general. So in general, if you try to diagonalize alpha, beta won't stay diagonal. And so we need a more general method that we can use in any problem. And that's what we'll be doing in the next two lectures. So we'll be setting up I guess the general method for finding the self-energy sigma. Thank you.