nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L3.3. Self-Energy
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[Slide 1] Welcome back to Unit 3 of our course on quantum transport. This is Lecture 3. as you know, in this method of quantum transport that we are discussing, very important piece is to write down the self-energy for the contacts and in unit 2 we saw how you could do that straightforwardly in 1-D problems. Now, in the last lecture we discussed a 2-D problem whereby using the method of basis transformations you could apply this 1-D technique to this 2-D problem but this approach doesn't always work and so what we need is a more general method and so what we'll be doing in this lecture is to develop this general expression for the self-energy that you can always use. Now- 

[Slide 2] before we get into the general method let me just quickly just recap what we had done in unit 2 for the 1-D problem. You see, what we did was we said let's say we have a 1-D wire described by the Schrodinger equation E psi equals H psi and then you enforce these boundary conditions, that is you assume there's a wave incident from the left contact and there's a reflected wave and transmitted wave and by eliminating, by applying those boundary conditions and eliminating the points outside the boundaries, you can get these additional terms, this one that represents outflow and the one that represents inflow. And what we showed is that the expression for sigma has this form where it is zero for all points except the boundary and at the boundary it's t e to the power ika and the source term again is zero for all points except the first point where the wave is incident. And this B is the amplitude of this incident wave, which is related to the Fermi function. So using these results we obtained these expressions for sigma and sigma in. That is, how to write down these self-energies for the contacts. Now what we want to do is a more general, 2-D version of this and the way we'll do this is- 

[Slide 3] think of it like we have a device and it's connected to 2 contacts and for starters lets assume the device is not even connected to the contacts so device is just by itself and is described by Hamiltonian H, so that's this middle point, E minus H times, if there's a wave function psi it will be E minus H times psi equals zero. What you are assuming is that to start with the device is empty. And then we have the 2 contacts this H described by H1 and H2, and we are assuming that this contact is kept filled by a source, by a source term from somewhere so that electrons here continually go out but then new ones come in. And so E minus H1 times phi, phi is the wave function in this contact, is equal to the source term in this contact and E minus H2 this is zero because this contact is empty. Now, what would this minus i eta? Well, eta is assumed to be a small, imaginary number and what it ensures is that the wave functions in these contacts decay eventually so that left to itself they all go to zero, that is usually what happens is if a state has an energy epsilon then the wave function goes as exponential of minus i epsilon t over h-bar. But because of this minus i eta, you have this additional minus i eta, which gives this decaying part of the exponential. So what that means is usually if you put an electron in a state epsilon it stays there forever because you just have the wave function looks like this, the phase changes with time but psi star psi does not change. But now we have this imaginary part, which gives this decay, so what that means is if we wait long enough the probability will become zero, which is why, in this contact for example the wave function is zero because left to itself it just decays. The middle one I haven't put in a i eta, so technically the wave function didn't need to be zero but let's assume it's empty to start with, no electrons in there, and in the left contact that's where we are assuming there is some wave function. Why? Because external sources continually put electrons in. That's this s1. So, in this contact also electrons leak out but because new ones come in you have a certain steady state number of electrons in there. So that's how we are setting up the problem and for the moment there is no connection between the 3 regions. And now what we do is introduce this connection to the, between the 3 regions, which is this tau 1 and tau 1 dagger. Overall this thing is hermition , so if you a tau here you must have tau dagger there. Similarly, tau 2 is the connection of the middle region to this contact 2 and the tau 2 dagger here. And when you connect it, of course, you still have the source term which wants to keep it at phi, but because of the connection it will change from phi, there will be an additional part, which I have written with this symbol, the chi 1, and then there's the psi inside the device and there is the chi 2 in this contact. So you see, originally all you had was electrons in this contact, which were kept there by new ones coming in continually. And now that you have connected them what happens is you actually have electrons inside the device and the electrons in the contact are changed a little bit. There is some electrons in this contact, that's the chi 2, and in this contact it has changed from what it used to be. So these 3, then, would be our basic equations, okay? Remember, these are all matrices and one point where, just to simplify the notation and being a little sloppy is that you see this H is a matrix and this H1 is a matrix, the tau is a matrix and so on, so whereas this E and this eta, these are numbers. So, I should really take E plus i eta and multiply it by the identity matrix which, of the same dimension as H1. Similarly here, it should be E i, where i is an identity matrix with the same dimension as H, right? But just to avoid clutter I'm not writing that i, so wherever you see this E plus i eta, it's implied that it is multiplied by an identity matrix of the appropriate size. Okay? 

[Slide 4] Okay, so we have these 3 equations, so what we'll do first is write them out 1 by 1. So look at the last equation, so if you write it out you get minus tau 2 dagger psi plus E plus i eta minus H2 times chi 2 is equal to zero. So you can solve this to write this chi 2 because it's equal to the inverse of this times this quantity. So that's this expression. Chi 2 is equal to this inverse times tau 2 dagger times psi. So that's what I've written there. So next we'll look at the first equation. So that first equation is E plus i eta minus H1 times phi 1, E plus i eta minus H1 times chi 1. This is kind of what was there before, we connected, and this is the additional part after we connected and then minus tau 1 dagger psi is equal to S 1. Now we note that you see this is equal to that because, you see, before you connected things that's what you had, so you can cancel that out. So what remains is this part and, again, you can solve for chi 1 and it will look like chi 1 is equal to inverse of this times tau 1 dagger psi. So what you have obtained now is expressions for these additional terms in the contacts and related them to the psi inside the device. And now we write out the middle equation, which looks like this. It's tau 1 times phi 1 plus chi 1 E minus H times psi, tau 2 times chi 2. That's equal to zero. So, what we now want to do is use these 2 expressions to replace chi 1 and chi 2. And when we do that- 

[Slide 5] so we are using those 2 expressions, putting it in here, and when you do that what you'd get is E psi minus H psi, that's E psi and H psi, and then this tau 1 chi 1, instead of chi you put that in so you get sigma 1 psi. Similarly, tau 2 chi 2, that gives you sigma 2 psi. And, this tau 1 phi 1, remember phi 1 was the initial wave function in this contact before you connected anything, and so tau 1 times that, that's the source term S1. So S1 is equal to tau 1 phi 1 and the sigma 1 is tau 1 times this inverse times tau 1 dagger. Sigma 2 is tau 2 times this inverse times tau 2 dagger. So that's sigma 2. So we have now obtained a general expression, you see, for the source term it's equal to tau 1 times the wave function that existed before you connected it. We have obtained an expression for the sigmas, what does it look like? Well, it's tau 1 times this inverse times tau dagger and this inverse, this quantity is what's called the surface Green's function, which we will talk more about shortly. So that's what I'll write as g1. It's the property of contact 1. Note that it only involves the Hamiltonian of contact 1. Similarly, this is the surface Green's function for contact 2. It involves only the Hamiltonian for contact 2. So what we have managed to do is we have obtained a general expression now that looks just like what you had done for 1D in unit 2, that is E psi equals H psi, that's the usual Schrodinger equation but now we have put in these boundary conditions and come up with the outflow term and the inflow term and we have obtained expressions for sigma 1. It looks like tau 1, g1, tau 1 dagger. Sigma 2 is tau 2, g2, tau 2 dagger and we have obtained an expression for the source term. And this then is like that, generalization of this 1-D result that we had before. 

[Slide 6] So we now have an expression for the sigmas and the expression for the s and I'd just like to make a little point about the connection between them. So sigma 1 is tau 1 G1 tau 1 dagger. Now if you look at gamma, you know, the broadening function, which is related to the anti-Hermitian part of sigma, then you could write it in this way. See, the sigma 1 minus sigma 1 dagger and sigma 1 is tau g1 tau 1 dagger, so you could write it as tau 1 and then i g1 minus g1 dagger and then tau 1 dagger. Right, but this requires a little matrix algebra. You should try this out and you have to remember that when you have the dagger of a product, like AB, you have to reverse it, it's B dagger A dagger. So to see how you get from this step to this step using sigma 1 you'd have to make use of that. And if you do that you'll get this expression and then you notice that i times g1 minus g1 dagger, this g is the surface Green's function of contact 1 but this is like the spectral function or the local density of states on contact 1 and so you'd get tau 1 a1 tau 1 dagger. So that's this gamma. Now look at the source term. If you remember, we have this expression for this S1 and the sigma in is s1 s1 dagger. So you can put it in. s1 is tau 1 phi 1 and s1 dagger is phi 1 dagger tau 1 dagger and notice that what's here is like the electron density operated in contact 1. Remember phi was the, this wave function that you had in contact 1 before you connected it. So this is like what you had in contact 1, the electron density, before you connected things. And so, that you could write as gn for contact 1. Now, if the contact happens to be an equilibrium then this gn for contact 1 is related to its spectral function by the Fermi function. So it's like density of states times Fermi function is electron density. And in that case, you see sigma n is tau 1 a1 tau 1 dagger times f1 while gamma is tau 1 a1 tau 1 dagger. And so you get sigma n is gamma times f. And this is the general result for contacts in equilibrium that we have mentioned before in unit 2. And here it comes out naturally but point to remember is this part of it is only true if the contact is in equilibrium. If it's not then this isn't true but this expression you could still use. 

[Slide 7] Now the next point I wanted to make quickly is about the surface Green's function because given these expressions that we just went through- you see, it's now straightforward in any problem to figure out the sigmas because the tau tau dagger are given, what you need is the surface Green's function. And in the next lecture actually we'll talk about how you find the surface Green's function. Now, 1 thing about the size of these matrices, now what tau describes is how the device is connected to the contact. Now, the contact is, of course, usually a large thing but the connection, immediate connection is usually just to the surface so, for example here, the surface has like 9 points, let's say 9 basis functions. And the conductor has 3 points, so your tau would be like something that connects 3 points to 9 points, 3 by 9. Tau 1 dagger would be like the conjugate transposed, so it's like 9 by 3. So in order for this multiplication to make sense the g1 would have to be 9 by 9 because this, of course, must match that and this must match that. Now, this is slightly tricky because you might say, well, H is a very big matrix. It's not 9 by 9 because H is the matrix for this whole thing so the way I've drawn has got 9 points this way and 9 points this way so it's like 81 points. So you'd say well this is really an 81 by 81 matrix, not a 9 by 9. Well the point is that for this calculation we only need the surface. So what that means is in general you could take this E plus i eta minus H1, which is say 81 by 81 matrix or something much bigger because this goes on to infinity. It could actually be very big but what you want to know is just this 9 by 9 part here that belongs to the surface. You don't really need the rest of it. So for our purpose, when you want to calculate self-energies what you really need is just this surface greens function, all right? And on the other end it's not straightforward to write down because you have to invert this whole thing. You cannot take the H and just take this part of it and invert it because that won't give you the right answer. You have to take the whole big H, invert it, but then look at only this part and how you do that is what we'll talk about in the next lecture, okay? 

[Slide 8] So that's what we'll get into. Thank you.