nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L2.3: Quantum Model ======================================== [Slide 1] Welcome back to Unit Two of our course [Slide 2] this is the third lecture. Now if you remember in the second lecture we set up a little semiclassical model, to guide us through the quantum model that we'll now be talking about. So if you remember, we are looking at a small range of energies, and there is inflow and outflow from the two contacts; which you could write in the form of a differential equation like this, and that steady-state there is no dN/dt but there' is an I1 and I2. Those are the currents that you are interested in. Now the question is how could we get a similar picture in the quantum model, that is out of Schroedinger equation; because Schroedinger equation as it stands does not have this inflow and outflow. Now, the first point, as I have mentioned in the introduction is that in order to see this inflow and outflow you need to go to the time dependent Schroedinger equation, and by and large you're talking of steady-state so there is not much explicit time dependence that we'll actually be discussing, but in terms of understanding this current occasionally and the next two or three lectures we'll have to go back to the time dependent one rather than that time independent one, and as I mentioned the connection between them is this time dependent one. The psi tilda is -- has this exponential minus iEt over h bar; and the way you get from here to here it is by noting when you take the time derivative of that quantity it's just like multiplying with mine iE over h-bar, and so effectively just multiply by E and that's how you get this equation, okay. Now what we'll show, is that is if you add a term like this that corresponds to this outflow, and if you add a term like this, that responds to inflow. And where do you get these terms? Well the direct way would be from the boundary conditions, that is you say that the Schroedinger equation kind of applies everywhere, but we restrict our attention to this region, and apply boundary conditions on the two ends; and those boundary conditions effectively give rise to terms like this. But how you do that from the boundary conditions that something we'll discuss later in the course as we look at different examples. For the moment, all we want to do is understand how a term like this and a term like this, gives rise to inflow and outflow. [Slide 3] Now in order to understand that of course the first thing is you have to know how to connect this number of electrons to the wave function, and as I mentioned in the introduction the connection is that N is this quantity psi dagger psi, 00:03:17,370 --> 00:03:22,040 and in order to understand dN/dt the rate at which the number of electrons changes what we need is the d/dt of psi dagger psi. And what the Schroedinger equation, the time dependent Schroedinger equation gives us is the d/dt psi. So the question is how could we use this to find out that quantity? Well the way you can do it is began, the I guess what you call it the chain rule. So you have the d/dt of a product of two things, so take the d/dt of the first thing, multiplied by the second thing, and then the first thing times the d/dt of the second thing. Now we want to replace this ih-bar d/dt psi from Schroedinger equation, and here there is three terms so we can do it one by one. So if you put in all three the algebra you wrote will kind of look messy. So first, put in the first one, see what we get, second one see what we get, move on so let's first look at the first h psi if you put that in what you get is this is psi dagger ih-bar d/dt psi you get h psi here you have psi but here now need the ih-bar d/dt of psi dagger right, as opposed to ih-bar d/dt of psi. Which means you kind of need to take the conjugate transpose of both sides. Now when you take the, -- that is just dagger so you have h psi here, so you got the h psi dagger, and the reason you pick up a minus sign is because, when you do this conjugate transpose of course that only affects the matrixes, the transpose part, but the conjugate part affects this i and makes it a minus i and that's why you pick up that minus sign there, okay. Next what you could do is then note that h psi dagger is really psi dagger h dagger. so you have psi dagger h dagger psi, and and then psi dagger h psi. So you pull the psi dagger out in front, a psi at the end, and inside you have h minus h dagger. Now one property would be Hamiltonian is that it is Hermitian. What that means is it's equal to its conjugate transpose. So H dagger is equal to H, and so this term drops I'll so basically then that first term doesn't give you anything as far as d/dt of psi dagger psi is concerned; and that makes sense because you see what H psi the E psi equals H psi describes as an isolated channel, and isolated channel that can scan disappear, go anywhere, so there should be no dN/dt. In fact, that's why fundamentally when you're doing a theory of this isolated channel the, --- you expect the Hamiltonian to be Hermitian, Hamiltonian meaning this H to be Hermitian and that's what is reflected here. Now when you look at the second term this sigma, the one that is supposed to cause outflow, the point is that one is not Hermitian. It is not equal to its conjugate transpose, but otherwise you see that term looks just like that, this is H psi and this is sigma psi, which means we can run through the same algebra, and here we would get a sigma minus sigma dagger. But now we wouldn't be able to do cross it out and put zero. So what we'll get when we look at this term is and additional done here which would be like psi dagger sigma minus sigma dagger psi. So that would be one of the flow terms there. Now what about the s? Well now you can do the same algebra but now is to know ih-bar d/dt psi you put in s so each psi dagger and here you put in the s here it is psi and for ih-bar d/dt psi dagger you put in s dagger again with a minus sign because then you a conjugate transpose you pick up a minus sign there. So overall then ih-bar d/dt of this number of electrons is given by that term and that term [Slide 4] and what we see is how this is kind of like outflow and this is inflow. Now moving on then, so what we have is this now the Sigma minus sigma dagger as I said this quantity sigma is not Hermitian. So it's not equal to its conjugate transpose, and the difference, you see, sigma minus eight times i is what we'll call gamma. So as I said, this is if Sigma where Hermitian this would have been zero. So if sigma were a number then gamma is kind of like that imaginary part of that number. Of course it's a matrix so its like the antihermitian part of the matrix, multiplied by this i, all right. Now with that what you could do is you could i dN/dt that is this is and so dN/dt is psi dagger psi and of course you have to divide by ih-bar, and then Sigma minus sigma and here you replace with gamma. So you then get an expression looking like this, and you get gamma, but note that the Sigma has to parts sigma one and Sigma two. Similarly the corresponding gammas also have to parts gamma one and gamma two so gamma one is i i sigma one minus i sigma one dagger gamma two is i Sigma two minus Sigma two dagger. So that's this quantity. So this then becomes that after dividing by ih-bar, and the last term we just have it as it is. Just divided by ih-bar. so this is then the expression for dN/dt. This is what else you the rate at which the number of electrons is changing, right. This is this inflow and outflow that I mentioned. Now and that you can see by comparing with our semiclassical picture, so this one is kind of like the new one plus new two times n. but as the second one is kind of like this the inflow term except that right now it's not quite clear how this, looks like, that we'll have to do a little more algebra to get there. that's what we do next but the first one I think you see the correspondence kind of the gammas are like that nu's now noted that again dN/dt that's what we equate with I one plus I two as I mention for steady-state flow, the dN/dt is actually zero, but when you look at the expression for dN/dt, you'll find certain terms that involve I I1, contact one and certain terms that involve contact two and that is how you identify which is i1 which terms give you I1 in which terms give you I2 and the point is the sum is zero it's a steady-state so what we'll do is instead of dN/dt we we write I1plus I2. [Slide 5] divided by Q so we have an expression like this I1 plus I2 divided by Q equal to this this term and that and that term as I said the first one corresponds to the outflow in the semiclassical picture the second one would require a little more discussion now note that from the previous slide I have now taken off the tilda we are now back to the time independent picture we're not talking about dN/dt anymore we're talking of steady state currents I one plus I two and the whole thing is zero but this part of involves contact one that part of it involves contact two now next find point is that you see in our modified Schroedinger equation psi depends on s1 and you could rewrite it in this form as we, as I, discussed in the introduction this I, capital I is like the identity matrix with the same size as H Sigma so you could invert that so psi equals two retarded Greene's function times s1 gr times this and you can take the dagger of that, the conjugate transpose that gives s1 dagger times the conjugate transpose up the retarded Greene's function which is called advanced Green's function that's that's this, So you have psi and you have psi dagger and you could use these to replace the psi and the psi dagger in there. so instead of the psi here you could write GR s1 instead of the psi dagger, we could write psi dagger GA so next we define you noticed you have a quantity here GA miners GR so what we can do is define a quantity A which is actually like the density of states that's not obvious though but like the density of states and this A is defined as i times GR minus GA and if you write this quantity as A then you see that term becomes s1 dagger A over h-bar s1. So this whole thing that we had before now looks like this and now I think you can see gradually the connection to this [Slide 6] inflow term. So, in other words we have I one plus I dual it is that comes we had before one involving gamma 1, one involving gamma two and we just considered a source term s1 and you saw the inflow do that of course as I mentioned in the introduction it is a little tricky to put two source terms into the original Schroedinger equation because if you put them directly in you will get interference terms that actually don't exist so what you should do is add the, --- consider only one source get your result and then add in a second source term. So overall the picture is your E psi equals H psi, then you have the Sigma one psi and the Sigma two psi, and that gives you the outflow that is, those are two terms, and then the s1 and s2 they give you inflow and that that these terms. and you can now kind of see the connection between the two. Although this connection will get clearer and a next lecture when we write down the energy of equations but here you will notice then there's gamma over h-bar is a little bit like this nu the rate at which electrons flow out and gamma has the dimensions of energy so when you divide by H the dimension is frequency, like per second just like that nu, and here you have this s1 times density of number of states this is also kind of like that the capital S1 is like this s1 dagger s1 and the density of states is like that A in here and as I said we'll develop it further next lecture; but now the point is that overall at steady-state I1 plus I2 is equal to 0 but if you look at the terms that involved one then you could identify it with I1 and if you look at the two that involve two, you can identify that with I2 and total is of course at up to 0, okay. [Slide 7] so to sum it up then what we discussed in this quantum model is how you start from the Schroedinger equation and then get these additional terms Sigma one, sigma two, s1 and s2 and if you look at the inflow and outflow by looking at this dN/dt and you can identify the inflow and outflow you get an outflow that would look like this psi dagger gamma psi that's like the nu one and, and for the inflow you have things like s dagger A s which is kind of like this s1 D zero so those are the inflows and those are the outflows so the main picture I wanted to convey in this lecture is how by adding the terms into the Schroedinger equation and as i said the way those terms come about is by considering boundary conditions which will get to when we look at examples but the point I'm trying to make is if you have problems like this they correspond to outflow and inflow much like that common sense semiclassical picture I mean this picture the semiclassical picture is to guide our understanding of these different terms and with this picture we'll go on In the next lecture to developethe setup of NEGF equations.