nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L2.5: Current Operator
========================================
 

[Slide 1] Welcome back to Unit 2 of our course. This is the fifth lecture. Now in the last few lectures, we developed these NEGF equations which are like the central result and in the next few lectures we'll be seeing how we apply these to different examples. And in Unit 3 we'll continue with more difficult examples, but the basic results are what we have developed so far. Now, this lecture is a bit of a detour where I want to talk about a more general concept of a current operator, which may be relevant to many problems of great present interest, but you don't, we won't really be making use of it in the coming lectures. Same principle you could jump from 2.4 to 2.6. 

[Slide 2] Now, let me explain what we'll be doing. The idea is as you know I try to get across this concept of inflow and outflow. That there is the sigmas which give you the outflow and the sigma in which gives you the inflow. And the way we got to this inflow and outflow is by looking at these dN/dt of psi dagger psi because that's like dN/dt and the semi classical picture that is guiding you is this idea that if you have currents flowing in from different terminals then the sum of all the currents is equal to the dN/dt and the q you need if you want the currents in terms of the charge flowing, whereas if we just want to know how many electrons are flowing then you don't need the q. So usually you find dN/dt and then you multiply by q. On the other hand, these days there's a lot of interest in knowing the flow of other entities, other attributes of the electron like spin and charge is a just a scalar number. Same for all electrons and we just put it outside. Spin is actually represented by a matrix, and I've used this symbol Y to denote any quantity that we might be interested in. It could be spin, it could be momentum, whatever it is and, of course, you have to figure out the right matrix to use to represent it, but in general for every observable property, there will be a corresponding matrix and psi dagger Y psi will give you the expectation value of that quantity, the average value of that quantity. And we want to know what is the d/dt of that quantity. So that will then give you an idea of flow of Y. Now, if you, we can proceed from here by noting that this quantity is just a number as we discussed before psi dagger any matrix in between psi and times psi. Overall it's a 1 by 1 quantity. It's just a number so you could write it as the trace of that number. It kind of looks silly but you know why I'm doing it. We went through this once before. We put a trace there and once you are traced you can move things around. So you can put, take the psi dagger from here and put it at the end. So we have trace of Y psi psi dagger. Now let's assume that this property Y that we are interested in has nothing to do with time. So it can be pulled out of the d/dt. So you could write it in this way, trace Y d/dt of psi psi dagger. So if you look at that quantity, this is what we could use to deduce a general current operator lop and the idea would be we do this once and for all. It doesn't matter what the quantity I is but once you have an expression for lop, you could go ahead and find the current of that I of that quantity Y. So you want to know the spin current then Y is the spin. So what you do is take the current operator multiply it by the appropriate matrix that represents spin and trace it. If you wanted to know the flow of momentum, well find the right matrix put it there, multiply it and you've got the flow of momentum, for example. On the other hand if you just wanted a flow of charge, that's fine too. Then Y is just q. That's just a number. So q times identify matrix and you could pull it out like we have been doing before. So, but what we want to do in this lecture is obtain a general expression for this lop, this I operator which in principle you could use to find the current of any quantity of interest. 

[Slide 3] So how do you go about it? Well, same as before we want to know the d/dt of psi psi dagger. Previously I think we were doing psi dagger psi and now we're doing psi psi dagger, and we want to make use of this modified Schrodinger equation. So the way you do it is, again, use the chain rule IH bar d psi dt times psi dagger, psi times IH bar d/dt of psi dagger and for IH bar d psi dt you replace with from this equation and there are 3 terms so we do it 1 by 1. As I said before if we put all 3 in, the algebra looks messy but it's linear so first put in H and see what you get, put in sigma and then put in the s. So when you put in the H, IH bar d/dt of psi that gives you H psi. So you have H psi psi dagger. This one gives you H psi whole thing dagger conjugate transpose and you pick up the minus sign because the I when you take conjugate transpose becomes minus I. Next you, so this is H psi psi dagger and this one is psi but this is where you have to note that H psi when you take the dagger is psi dagger H dagger and the minus sign. Now, these are the time varying quantities and the time dependent Schrodinger equation, but since you have psi psi dagger you could replace it with just the time independent quantities because their time dependent part of it will just cancel out to phase vectors of exponential minus IET over H bar. And so you get H psi psi dagger minus psi psi dagger H and now you can replace psi psi dagger with Gn over 2 pi so that you H Gn minus GnH over 2 pi. So that's what I have put there. So, the first term H gives you this. So what we now need to do is get the corresponding term for sigma and then the s. So how do you do the sigma? Well, it looks exactly the same. Instead of H you'll have sigma. The only difference is that here when you got a H dagger we replaced it with H because H is Hermitian. So H is equal to H dagger. In the case of sigma, we don't do that. And so what happens is when you do the same algebra for sigma that's when you pick up a sigma dagger here. So here it is Gn minus GnH, now it is sigma Gn minus Gn sigma dagger, but otherwise much the same algebra. Next we will look at the s term. And now the algebra is a little different, but what we do is instead of IH bar d/dt psi we put s. So you have s psi dagger and then you have psi times instead of this you put s dagger, psi s dagger, and the minus sign again is because when you take conjugate transpose you pick up a negative sign so that's what you get. And, again, we can go back to the time independent quantities because s psi dagger, s is the exponential minus IET over H bar the dagger has exponential plus IET over H bar they just cancel out. So you can write that. Now, you use the expression that we have been using from the time independent Schrodinger equation that psi is equal to retarded Green's function times s and psi dagger is equal to s dagger times the advanced Green's function. This is exactly what we have done before. Put that in and then s dagger becomes sigma n over 2 pi giving you this expression. So this you can now put back in here. So now you have all 3 terms corresponding to the 3 of them. 

[Slide 4] Now, you can write down the I operator, which is just d/dt of psi psi dagger and so you'd have to divide by the IH. Now what happened was there was a 2 pi here and the IH bar from there so it became this IH over here. Now this is the term that you could visualize pictorially because if you think about it you see this was something we had before. This was what we had interpreted as outflow term, there's the inflow term, but now we have an additional term here. So we have something like this that when you look at the outflow term that's what gives you this one. The inflow term that's what gives you this one if you put 1 that's for contact 1. If you put 2, that's for contact 2. It's kind of like before but now there's an additional term which you kind of interpret as the generation of the quantity Y inside the channel and this is the tricky point and I want to talk about it a little more and that is usually I said earlier that ordinarily the sum of all the currents has to be 0. So steady state this thing is 0 and all these just add up to 0, but this is not true if we are talking of an entity like spin or momentum which actually is being generated inside the channel. What I mean by that is supposing you have a channel so the electron comes in with some spin but through its interactions changes its spin. Then it's almost like spin is being generated inside and then the currents, so what is 0 is the sum of the current plus the rate of generation of that quantity and that is what this represents. 

[Slide 5] So, if you are talking about entity Y which is like charge so that you can pull it out from the trace, then you are just taking the trace of this matrix and then you'll get whatever we had before. Because if we just take the trace of the current operator. So let's say we take the trace of this. Well, that's 0 because trace of GnH is the same as trace of HGn, but as soon as they are putting the Y in there then you'll see the trace of YGnH is not the same as the trace of YHGn at least not in general. And so you could get a term here which you should interpret as the rate of generation of Y of whatever that entity Y is, but otherwise you will have these same terms like before looking like a, you know, an outflow and inflow and if you take the trace of this quantity you'll get the result we had in previous lectures. That is if you look here, for example, if you take the trace of this it doesn't matter whether the sigma n is the second term or the first term because I'm taking a trace and then what I have is GR minus GA, I times GR minus GA that's just A. And so this term just becomes trace of sigma n A. Similarly here it doesn't matter that this Gn is the second term here and the first term here because once I take trace I can move them around and then sigma 1 minus sigma 1 dagger times I just becomes gamma 1. So this is the old result we had. This is what we will largely be using in the coming lectures, but what I wanted to give you in this lecture was an idea of this general expression. The general current operator which you could in principle use to find the current of any entity like spin which I have represented here as Y. So that's basically what we did in this short lecture was try to show that what we have been doing so far that's kind of a special case of a more general current operator that you could use for any quantity Y but that Y is just a number that you can pull out that you don't have to put inside and take the, and multiply and take a trace. If you can pull it out, then you get the picture that will largely be using and charge, of course, is in that category because every electron just carries a charge of q. So now we can go on to the remaining lectures where we will be using NEGF equations and illustrating them with various 1D examples. Thank you.