nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L2.9: Dephasing
========================================
 

[Slide 1] Welcome back to Unit 2 of our course. This is the ninth lecture. 

[Slide 2] Now, when we discussed this NEGF equations I mentioned that there are these Sigmas related to the physical contacts but there's also a Sigma, the Sigma 0. That's not a physical contact, but it's about the interactions. And, in this lecture, that's what I want to talk about. Now, in respect to the origin of the Sigmas, we have set up this picture that tells you the inflow and the outflow that's associated with various Sigmas because basically each Sigma kind of represents. There's a Sigma which represents the outflow and a Sigma in which represents inflow. The difference, though, is that, when it's a physical contact then the inflow can be related to the outflow through this Fermi function. But this is a relation that does not apply to the interaction contact or the dephasing contact. And this dephasing contact when you Sigma 0 and sigma 0 IN don't have any necessary connection that you could write down in terms of any Fermi function because it doesn't have a Fermi function. But there's one requirement  it must satisfy, and that is that the inflow and the outflow must be balanced because you see, at the physical contact, they're not equal because there you could have a net flow of 100 electrons per second. So you could have 1000 coming in, 900 leaving with a net flow in of 100. That's fine. But this contact, you see, there is no physical contact to collect electrons. So, if 1000 are coming in, 1000 must be leaving just as well. Otherwise, you wouldn't have a steady state situation which means that, when you write down the Sigma 0 and Sigma 0 in they ought to satisfy this relation, namely, that the inflow is equal to the outflow. So let me go through examples of how we could choose various kinds of Sigmas so that this relation would be satisfied. 

[Slide 3] So one example would be the following that supposing we use a Sigma 0 that is equal to a number times the retarded Green's function. Okay. So this is a matrix. That's a matrix. This is a number. And, correspondingly, we choose the Sigma IN as d0 times this number matrix that Gn that we have defined. Again, we use the same d0 in both places. Now, of course, because I've introduced this relation, when I try to calculate the Green's function it will actually require an iterative process. Why? Because the Green's function depends on Sigma 0, and the Sigma 0 in turn depends on the Green's function. So I have to solve the two kind of self-consistently and the same here. Gn depends on sigma 0 IN. The sigma 0 IN depends on Gn. So, again, you have to solve them self-consistently. But the point I want to show is that, as long as you solve them correctly, this basic relationship will be automatically satisfied. And so current would be conserved. You would not be losing current into this dephasing contact which is, of course, the basic requirement because, if that happens, you don't even have a consistent theory to work with. Now, to see that note that, if Sigma 0 is d0 GR, then the corresponding Gamma which is what I need here and Gamma 0 is i Sigma 0 minus Sigma 0 dagger which is d0 i GR minus GR dagger which is GA. And the point is that quantity is A. So Gamma 0 is d0 times A. So, if you want trace of Gamma 0Gn, the Gamma 0 is d0 times A, so it becomes d0 times Trace of AGn. Here, sigma 0 IN is d0 Gn. So, when we look at the other quantity, sigma 0 INA, this one is d0 Gn. So, when I put it, in I get d0 trace of GnA. So you'll notice those are equal as long as I've used the same number here as here because this is trace of AGn. This is trace of GnA. And trace, as you know, you can reverse the matrices, move them around cyclically and trace stays the same. So this is a choice that will conserve electrons. So it wouldn't give you loss of electrons into this contact. But physically what effect, what does it lead to? What effect does it describe? 00:05:26,496 --> 00:05:28,006 

[Slide 4] Well, let's do this problem. We take this old problem we had, namely, the wire with a barrier which is kind of like a resistance, and you have the interface resistances. So, as you know, if you look at the electrochemical potential or the occupation of the levels inside the channel, what you expect to see is it's 1 at this end, 0 at the other. And wherever there is a barrier, there is a drop in the occupation. This is something we discussed at length in Unit 3 of Part A. But, basically, as I mentioned, that any time there is a barrier, that's where the occupation drops. And semi-classically you expect that. If you think of the electrons as cars on a highway, a barrier is kind of like a construction zone. As soon as you cross it, suddenly the road is empty. So it's kind of similar as this drop here. So that's what you expect semi-classically if you're thinking of electrons as particles or like cars. But, when you do the wave theory, you get these oscillations. This reflects the standing waves, interferences. That's what you get if you did a coherent quantum theory, coherent NEGF. But, as I explained in the last lecture, it's important to include these random potentials because that's what destroys this phase information. And what I claim is that this choice of Sigma, namely, Sigma 0 equals d0 times a Green's function, sigma 0 IN is d0 times Gn, that includes phase relaxation. And, indeed, if you do that and do the same calculation, what you find is the oscillations are subdued, I mean kind of suppressed. And you get a quantum curve that almost looks like the semiclassical with a little oscillation right around that barrier. This is the kind of thing people have seen. People can actually measure how this electrochemical potential varies around a defect, so on. And this is the kind of thing they've seen. Now, why does this work? How does it give rise to -- how does it include this effect of random potentials? Well, you could think of it this way, that you see you have E psi equals H psi inside your channel, and now you have a random potential which adds to the H, kind of like a UR times psi. And let us say we treat that UR times psi as a source term as if this UR times psi is what kind of goes back, goes out somewhere and then comes back in again as an s. Remember s was what came in from the contacts. So let's say we write the s as UR times psi. Then we could write ss dagger as UR psi, psi dagger UR. And ss dagger is what you call Sigma 0. That's what you have been doing. ss dagger is like the Sigma. Psi psi dagger is like Gn. And then you can see you have this relation, sigma 0 IN is d0 Gn. And what is d0? 00:08:40,236 --> 00:08:44,716 It is basically then kind of the RMS value of this random potential. Note that this d0 has the units of energy squared. That's because Sigma is like H. It is energy. It is electron volts. On the other end, the G is the number per unit energy so it's like per electron volt. And so this constant in between has the units of electron volts squared. And what it really represents is kind of this RMS value of the random potential that we have. And, of course, in this model, the assumption is this is a number. So this potential is assumed to be the same everywhere. So it's like this entire thing has a certain random potential but not like there's any difference between the potential here and there and so on. And that is why, I guess because it's like a uniform potential, it actually doesn't lead to any loss of momentum. Notice that the entire drop is still here. There is no drop here as in this region this phase relaxation but no momentum relaxation because, if there was momentum relaxation, there would also be a drop in the potential. And, after this, I'll show you an example where there is a drop in the potential. But for this choice of Sigmas, there is no drop in the potential. What you get is pure phase relaxation without momentum relaxation. Now, what I'm told is that people who do large-scale numerical computations, they don't like this one very much because you have to keep the entire matrix. They would rather have models where you need only the diagonal elements of the matrix. And I'll show you a choice after this where you do just that, actually; that is, you only keep the diagonal elements of GR when you construct the Sigma 0 and, similarly, the diagonal elements. And so, in a practical sense, of course, that saves you memory because you don't have to remember the entire 1000 by 1000 matrix, just the 1000 diagonal elements. But, in terms of the physics, though, while this method gives you pure phase relaxation, the other one gives you momentum relaxation, as well. And let me show you with an example. 

[Slide 5] So, in general, then, the point I want to make is previously I wrote Sigma 0 is a number times GR. You could do something more sophisticated where you say, okay. I look at the iG element of Sigma 0 and relate it to the iG element of GR and multiply it by Dij. So, in other words, previously, we were taking the entire matrix and multiplying by a single number. Now we're taking the entire matrix and multiplying each element by a separate number, Dij. And we use the same matrix here, sigma 0 IN, Dj, Dij times Gn. And the point is that, as long as the D matrix is symmetric, the Dij equals Dji, what I'll show is you'd still have a consistent theory because the trace of the Sigma nA will be equal to Trace of Gamma Gn which means inflow and outflow will be balanced and you won't be losing electrons. That's always a zero order check to make because, if you're losing electrons, you won't have a consistent theory. Well, the way it goes is, again, Sigma 0 is DGR, so Gamma 0 Dij, Aij. You could work that out. And so trace of Gamma 0 Gn would be like Gamma 0 Gn. I mean, when you write trace, what you're supposed to do is take the ij element of this multiplied by the ji element of that and sum overall inj. The trace means you're looking at the diagonal elements. So then, instead of Gamma 0, you put in D times A, so you get something like this. Sum over ij, Dij A and then Gn. You could do the same here. When you take Sigma 0 n times A, you'll need the trace, so that's like Sigma 0 nji Aij sum over inj. And then, if you put in the DGn, you'll get that. And the two will be the same as long as D is symmetric matrix. So, in general, you could use any D. Now, what I'll show you is one where we use a D for which has non-0 elements only on the diagonal. 

[Slide 6] And will see that is where the D looks something like this, some number, and then you have 1s on the diagonal. The rest are all 0s whereas what you used previously was D was like the same for every element, single D. So that's what we used. And, when we use that, we saw what we got, namely, pure phase relaxation. And so we got a curve that looks almost like semiclassical. Well, when we use this one what we get is, again, the oscillations have gone down, but now you see there's a big slope here. And these slopes again reflect this momentum relaxation in addition to the phase relaxation. And what is this D matrix? Well, it is basically then the average value of the potential at location i and the potential at location j. And previously we were assuming that it was basically the same potential everywhere which is why all these elements were 1. Now we are assuming that the potential at any point is totally uncorrelated with the potential at the next point and with the next point, which is why any off diagonal element is 0. Of course, the diagonal is 1 because it is correlated with the potential right there. It's the same thing. But there's no correlation from one point to another, and that kind of potential will destroy momentum because the potential is changing rapidly, will always cause strong reflections whereas here the potential is almost constant, doesn't cause any reflections, no loss of momentum. And the effect is very evident when you look at the potential profile in your channel. 

[Slide 7] So, to sum up, then, here what we tried to do in this lecture is that talk about this dephasing contact because so far we've been talking largely about the physical contacts. The dephasing contact, the key relation to keep in mind is, however you choose your Sigmas, this relationship is satisfied to make sure electrons are conserved. And this will be satisfied if we choose it in this way according to this prescription with the same D in both places as long as it's a symmetric matrix. So any symmetric matrix would work. And what I showed you was two examples: one where it was a full matrix all 1s and the other where it was a purely diagonal matrix which gives rise to momentum relaxation while the all 1s gives rise to phase relaxation. And, in general, you could have more sophisticated choices also, you know, where you actually have a tensor D with four indices, etc. You see, usually the way you get these Ds, the approach is you start from a microscopic theory. You say that, well, you're talking about phonon interaction, and phonons give rise to a potential like this. And this is the correlation of that potential. That's the usual approach. The approach we are taking is, well, we won't worry about the microscopic theory of where the potential comes from. But we'll say what are, I guess, acceptable choices for D so that you don't lose particles. And then, for a particular choice of D, what does it relax? Does it relax phase? Does it relax momentum? Both? Or, if you're talking spin transport, you might want to come up with a D that relaxes spin. And that one could take is we know the momentum relaxation time or the phase relaxation time or the spin relaxation time from experiment, and so we'll adjust our D appropriately to give us the right relaxation times and then go on to analyze devices as opposed to saying that we'll have to get the D from a microscopic theory of all the detailed scattering processes. What we are doing is we are saying we are not going into that, instead, treating the D as something you get from known experiments. Thank you. And, in the next lecture, we will I guess sum up this unit.