nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L4.6. Spin Hamiltonian
========================================
 >> 

[Slide 1] Welcome back to unit 4 of our course on quantum transport. This is the sixth lecture. 

[Slide 2] Now, in the past we had used a Hamiltonian like this, you know, with a conduction band edge and effective mass and discussed how you could translate this into a tight binding model, how you could choose your parameters of that model, this epsilon and the t, the onsite and the nearest neighbor elements, such that they reproduce the correct dispersion relation that you expect from here. Now, later in lecture 6 in unit 3 we talked about how to include a vector potential and there we saw, again, you could use the same model but an onsite element remains the same but these nearest neighbor couplings need to pick up a phase factor that is related to these vector potentials. And, again, if you do that, it will give you the correct dispersion relations. Correct meaning the dispersion relations for small K. You cannot mimic it exactly for all K but for small Ks it will match. 

[Slide 3] Now, what we have done in the last lecture of this unit is we included spins and so we picked up this I and we picked up these spin-dependent terms, the ones that involve sigmas so these are operators but the operators is, again, is 2 by 2 matrix operator, right? So you have these Is and sigmas and the question is how do we translate this into a tight binding model like this? And the basic result is that, yes, you could still visualize it much the same way but here this onsite element was a number, epsilon. Now the onsite element will, itself, be a 2 by 2 matrix, alpha. Similarly, nearest neighbor coupling was a t to the power I phi x. This will now be, again a 2 by 2 matrix, beta x and this will be beta y and what you get here, what appears here will be the conjugate transpose of that. So here it was the conjugate, now it's the matrix and the overall thing has to be Hermition so this will be the conjugate transpose of that, et cetera. So this is, I guess, what we'll be trying to show and our objective is to come up with the expressions for these alphas and betas such that it corresponds to this Hamiltonian that we obtained in the last lecture. Now, this is relatively easy to do if we forget this lower one, which means if we just look at the I part of it then it's relatively straightforward so let's first discuss that and then we'll go on to these, okay. So we first dropped that. In that case, actually, the answer is very simple. It's like this, the alpha is just epsilon times I. So here it was a number, now this is a 2 by 2 matrix but it's basically epsilon times a 2 by 2 identity matrix. Similarly, beta x is just the number here times I. Beta x dagger, that's this conjugate transpose of this, of course, you only need to take the conjugate, the transpose of I is still I so you get that. So it's just that number times I and same with beta y and beta y dagger. So that's the result. Let me quickly explain why that works. 

[Slide 4] Okay, now the basic principle, as I had mentioned before, is to choose your parameters such that the dispersion relation that you calculate using this tight binding method, you know, which we discussed in unit 1, so that that dispersion relation corresponds to what we expect from that continuum model. And for the continuum model it's straightforward to obtain the dispersion relation, just replace B with H-bar K, right? So p x becomes this H-bar k, H-bar square k square and I guess I've taken the H-bar from here, divided it in there and so from here you can get there just with B equals H-Bar K. For the lattice model what you are supposed to do is do this summation over all nearest neighbors, so you start there, you got 1 where m is equal to n when you do the summation, that's the first term, and then you have the terms corresponding to the 4 nearest neighbors, right? This is what we did, I guess, in unit 1. And then you can combine these using these trig identities to get these cosine functions. And how do you make this correspond to that? Well, the idea is this is a cosine function, that's a square function, they, of course, won't match all the way but for small k you can replace cosine theta with 1 minus half theta squared and using that you could make the two correspond. So that was the approach that we had used before. Okay, now what happens is we have an I going with this. So it is like this operator but then times a 2 x 2 matrix, the identity matrix here. So if you calculate, if you look at this dispersion relation then you'd also have kind of like an I here, right? The same I carries over and the way to include this into the discrete model is put an I on all our parameters so make the epsilon into epsilon I, put the t to the power I, phi x into that t to the power minus I phi x into, just multiply it with I and so overall then everything becomes, I mean, just picks up an I, which then goes in there. And so if this corresponded to this before, now including the Is also they will correspond. 

[Slide 5] So, you can see relatively straight forwardly that for this Hamiltonian with the extra I there, all we do is take what we had before and multiply everything by I. So each one is now a 2 by 2 matrix and we have this picture here with a alpha for this onsite and beta is for nearest neighbor, okay? Now we come to the harder part of the problem, which is how to include these terms. Now, here actually the first one is relatively simple. This is a 2 by 2 matrix, I mean, actually it's a, sigma dot B and you could just include it there because this does not involve the momentum. This does not involve this momentum part of it, if you just add it there you'll see that in the dispersion relation you'll have added this term in there. So all you need to do is add it to the diagonal term. This one is harder because this one actually involves momentum p and the way to include this is if you write it out so z cross p so in a sigma dot z cross p so it would be like sigma x times the x component of that. And that would be like py because z cross p and the x component of that would be p y, which you could then write as H-Bar k y. So that's how you get sigma xky. Similarly, sigma y times the y component of that, which would then involve p x, which gives you the k x. So that's how this term, the minus sign together that could be written in this form and in order to get the appropriate correspondence to this lattice what we need to do is take these linear terms in k and replace them with sine functions. You see, what we did before was we had quadratic terms in k and they got replaced by the cosine terms in the lattice model. Here you have linear things so they become the sine functions. So k y I could replace with sine k y a divided by a because for the small k sine x is just x, so you get k y a and then the a cancels out, you get k y. So, this would be the lattice version of that. Now you can take the sine functions and write them out in terms of exponentials, that is, sign of x is E to the power plus I x minus E to the power minus I x divided by 2 I and same here, with the sine k x a you write the difference between the exponentials and then a 2 I here. So now you see you have this part of the, I guess, dispersion relation written out in terms of exponentials and then the idea is you look at what goes with E to the power I k x a and that is what you want to put with beta x because if you start it from the beta x when we're calculating the dispersion relation, remember, what you do is write out that sum and you have alpha plus beta x times E to the power I k x a beta x dagger times E to the power minus I k x a, et cetera. That's how you'd calculate the dispersion relation. Now you're doing it backwards. You know the dispersion relation and you're trying to figure out what you should put there and so the idea is whatever multiplies E to the power I k x a, namely that quantity, that's what should go in there. And you can see that's exactly what I've written, eta over 2 a sigma y. Note that here the I was in the denominator, I put it in the numerator and that's why I lost the minus sign. Next, if you look at E to the power minus I k x a, that is the quantity that should go in there, which is exactly the same factor but with a minus sign, and then if I want the beta y and the beta Y dagger I should look at the coefficients of E to the power I k y a and E to the power I, minus I k y a and now it will involve sigma xs. So this is how I should choose the parameters beta in order to make the corresponding dispersion relation match what I expect from the continuum model. Okay, so that's kind of finishes what we're trying to do here. 

[Slide 6] What I want to do then next is just make a couple of comments about the nature of the eigen states and the eigenvectors for the spin-orbit Hamiltonian, you know, what we have discussed. That is, let's say you focus on zero magnetic field, so the Bs and as are gone so you have what we had before and then this new term, the spin-orbit term. And the question is what is the EK relation for an electron that obeys this Hamiltonian? Now, if the second term was not there then we know what it would look like. It would be just E equals E c plus H-Bar square k square over 2 m. Instead of p you put H-Bar k and you get the dispersion relation. And now we have an I so there are kind of two eigen states, an up spin and a down spin. Now, question is what effect does this have and that is the part where it's useful to know a very important identity or something that you could work out easily is if you consider this matrix, sigma dot V, where V is any vector, what you can show is that this will have two eigen values and the eigen values will be plus and minus magnitude of V. Doesn't matter what direction it points, the eigen values are the same. But the eigenvectors, of course, depend on the direction of V and the plus sign corresponds to an eigenvector whose spin direction is in the direction of V and the minus sign corresponds to one whose spin is in the direction of minus V. That means, supposing V is x, let's say, so in the x direction, so unit vector in the x direction. Then the eigen values would be plus 1 and minus 1 and the upper 1 will correspond to an, these would be the eigen values and the upper one would correspond to an eigenvector whose spin is parallel to the x direction and that you can show as 1 1. You know, in previous lectures we've talked about how given any direction you can write down these eigen spinors, the two component spinors. So for V in the X direction this would be like 1 1. Minus V would be like 1 minus 1, et cetera. So those would be the, this is the general result. So here then what we have is eta over H-Bar sigma dot this so what you'd expect then is that the eigen values for corresponding, because we have added this, will have plus minus eta times k where k is the magnitude of k. That is, you see, we want the magnitude of this quantity, this is the vector, and what is this vector? Well, it's the transverse component of p, the momentum in the xy plain. And so that would be H-Bar times K and you want the magnitude of that and the H-Bar cancels, this H-Bar that I have there. So these would be the 2 eigen values though, they have split. So instead of having one parabola you'd now actually have two things there, right? This part. And the upper one, the plus sign, would correspond to spin, which is parallel to minus z cross k because if the spin is parallel to minus z cross k then you see overall you get a plus sign because of that minus sign there. So that would correspond to this one and the minus sign would correspond to a spin, which is parallel to Z cross K. So, let's just explore that a little further. 

[Slide 7] So the idea is that supposing that a given energy we ask which k is our eigen states corresponding to a given energy. Now, if you didn't have this part you might have said, okay, for a given energy the allowed allowed ks lay on the circle, right? All those which have the same H k square, okay. Now, what would happen is if the spin is parallel to z cross k then you have this minus sign here. So what that means is for the same energy you have a negative sine, something subtracting, assuming that eta is a positive number and so for the same E you'll now need a little bigger k. So if we draw all the states that contribute to a given energy you'd find that the circle got a little bigger. You'd have to use a slightly bigger magnitude to get the same energy. On the other hand, if you looked at the other one, the plus, then there's a plus so for the same energy you kind of need a smaller k. So the outer ones have spins that have shown the way I get this direction of the spin is I look at z cross k because if I'm here the k is in this direction, in the x direction, the z is coming out of it so when I take z cross k I get that vector. So that's the direction of the spin. So an electron whose momentum is in this direction, the spin would be in that direction and that would correspond to this state. If the spin were opposite that's also an eigen state but that would be, would have a lower, would be somewhere here for the same energy, and so on. So as you go around, of course, the direction of the spin is always perpendicular to the k. Now, what that means is that the ones with the spins that are in the direction that I've pointed out, those will have a bigger magnitude of k and you can see it's a bigger circle so there's bigger, more number of states and more number of modes as well. And so spins that are in the direction I've shown will have a larger density, number of modes, density of states, et cetera, compared to the inner circle, which corresponds to exactly the opposite spin. So this one corresponds to a spin pointing downwards and that one will have a lower number of modes, lower density of states. And this is the point that I'd made in the introduction briefly that because of spin-orbit coupling, what happens is usually whether you're going to the right or to the left you have the same number of modes but now, I guess, the spin and the direction of travel are coupled so if you are thinking of states that are moving to the right they will tend to have their spins pointing upwards, more than pointing downwards. On the other hand, if you look at the states that are pointing, that are traveling to the left, the minus k x 1s, they will have their spins pointing downwards rather than upwards. And this situation is very different from what you'd have if you had a term like this. You see, here, as you know, this term came from a term that looked like sigma dot z cross k whereas here it is like sigma dot say a magnetic field. Now, in that case you still have kind of two circles but one of them corresponds to spins that point along B and the other corresponds to spins that point along minus B. And it's nothing to do with k. So magnetic materials you could think of as if there's an internal, huge magnetic field, which causes splitting of that type and in that case what happens is all the up spins have some more modes than down spins. And up meaning those that are along magnetic field, down could be those that are opposite to the magnetic field. So it's the magnetic field that determines things but in spin-orbit coupling it's like every state fills a magnetic field that depends on it k vector. So if you're moving in this direction you'll feel a magnetic, effective magnetic field that's like z cross k. So let's say it's in this direction, whereas if you're moving in the opposite direction you feel effective magnetic field in the opposite direction. So every state feels a different magnetic field, which is why, you know, the eigen states, the spins of the eigen states keeps changing. So this is the, I guess the physical implication of the mat that we discussed and this is often useful when you're trying to understand or interpret experiments. 

[Slide 8] So with that then, I guess what we have done so far is we have talked about magnetic contacts, talked about this nature of spins, these vectors and spinors and we have talked about the spin-orbit coupling and how you include it into the Hamiltonian. So given all that, I guess, you could use the NEGF equations to calculate any quantity you want and what we'll talk about in the next lecture is how you interpret those calculations. Like, if you calculated the gn, which, as you know, is like the electron density, how do you extract information about spin density, for example, from g n? So that's what we'll do in the next lecture. Thank you.