 You can follow along with this presentation using printed slides from the Nano Hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Okay, so let's get started. And today's lecture six, and this is a continuation of the energy band and its properties. And the idea is if you remember that we are in the process of calculating the total number of electrons that can respond to an electric field so that we can calculate current. The number of atoms multiplied by the number of electrons per atom didn't give us the right results in terms of conductivity. And we are trying to find what fraction of the electrons are actually moving in response to an electric field. And we are already making headway towards that goal because we have seen that by solving the Schrodinger equation in a periodic potential, we can find that the electrons do not sit everywhere. Some of them sit at a very low level and they have certain properties separated by band gaps. And then we have a series of these levels. And what we want to explore is that looked all look very complicated whether there is a summarized version of that information that we can use later on. The energy bands are essentially the solution of the Schrodinger equation that tells us at what energy and what wave vector levels the electrons can sit. So today we'll continue on the properties of the electronic bands. We'll also talk about this notion of an EK diagram and constant energy surfaces before concluding. So let's talk about this a little bit. Again on the left hand side you remember that this is the solution of the Schrodinger equation in a periodic potential. We have y axis is energy and the plus and minus k are the wave vectors where the electrons can sit. Do you remember that this came from the periodic boundary condition when we bring the string together, the first atom connected to the last atom. Only certain k's are available plus k moves one direction in the chain and the negative k goes the other way. Now let's think about this about when electrons sits in these bands how they behave. I have shown here something called an EF which is Fermi level. Now at this point you don't need to know what it means. But let's say it is some level above which there are no electrons and below which there are all the electrons sit there. So for example now in this case what would be the current for such a band. Now the first current associated with the purple band we are calling it J33 because this is sort of the third band from the bottom. And you can immediately see that all the electrons should have a minus q and I must sum up over all the velocity because charge multiplied by velocity gives me the current. But here I don't have any electron. So of course I do not have any current because there is nothing here to carry the current. So I have it zero. There is no problem. What is more surprising however that electrons on the bottom on band 2 and band 1 also carry no current. That's strange because there are lots of electrons you see. So in principle they could have carried current and the reason they don't carry current is because of this. For example if you focus on the first band I am again summing up of all the electrons carrying different amount of current because they are moving with different velocity. But when I do I quickly realize that I have equal number of positively going electrons with plus k and negatively going electrons. And so therefore although each one of them is carrying current the sum of them is zero. It is like going from Chicago to Lafayette. Let's say both lengths are have cars going in the opposite direction but one way there is two ways there could be no net flux going from one point to another. If there is no cars which is band 3 or if there are equal number of cars going in the opposite direction in that case it is again zero current. Two quick points now this argument requires the number of states that I have in the plus k lane and the number of states I have in the minus k lane they are exactly the same. Now for all Bravais lattices which has inversion symmetry that if you put a mirror and reflect it it remains exactly the same. In those cases this will always be the case that the same number of plus and minus k state this is not generally true. Second you could argue couldn't they simply when they move couldn't they just sort of pile up on top of each other you know two electrons occupying the same state that way you could have a current. But of course because the poly exclusion principle and these electrons are fermions therefore unless one vacates the space the other one cannot move in. So therefore in this case there would be no net current although there are lots of electrons. Now what about partially filled state let's say you have a band you make the temperature a little bit higher so that some of the electrons which are sitting in band 2 goes up in band 3. Now in this case even in the beginning it might look like that there is no net current you see the number on the red side on the positive. And that on the negative side there again exactly the same so in principle there shouldn't be any current and the same is true for blue. Except that when you apply an electric field what will happen let's focus on the red points because the electric field is in the positive direction. And the electron has a negative charge and a negative effective mass sorry positive effective mass in that case electrons will be moving opposite to the direction of the electric field. In this case you have more electrons going to the negative side compared to that of the positive side you can easily see and as a result the current will no longer be 0. And so that's no longer 0 what precise value it is we'll calculate later on but the point is that in this case on a partially filled band the current will not be 0. Now let's focus on the blue one what happens in the blue one is again of course if you apply an electric field but remember these are electrons which is fine minus q charge but these have negative effective mass so they will move in the opposite direction. And as a result when they're like the electrons in the valence band they move in the opposite direction correspondingly there'll be an asymmetry between plus state. And the negative states minus state and together once again there'll be a net current but there's a very nice way to think about this and that's something called a hole and let me focus discuss that. How many states do I have per band you remember this is the total number of atoms that I have in a crystal right now that could be millions of atoms let's say or a billion of atom in a string. So therefore the number of blue points I have shown is on the order of a billion let's say in a one dimensional chain although I show here only five or six. Now if I have a billion summing them all up to get a current is very difficult so therefore I could do the following. What I could do is that instead of summing over them all I could partition them I could sum them over all the states they are they are regardless of whether they are full or not. And that's the field or not that's the first term see the other below the sum I say all it means all state regardless with their full. And then I have added something because of course the first term is negative I have over counted because some of the states there were not any electron I shouldn't count them. So I have taken them out in the second state but this time I have added them as empty. Now the first term you see which has the summer below the sum there's all in this case there are the same number of plus state and same number of minus states inversion symmetry we're in zone and all the Breville lattices and all that and as a result the first term will go to zero. See and therefore the second term remains and the second term says that when I want to calculate current in a field band which has a negative curvature or the negative effective mass then the best way to describe them is to focus on the ones that are empty. Because then you see there are millions of them so I don't sum them all there are few empty states I focus on them. I assign a plus q charge look at that that I have instead of a minus q over L I have a plus q over L left size assign a plus chart and let the electron move let the holes move in this case or empty spaces move in the same direction as the electron with plus velocity. And so this way I can again calculate the total current and this way of viewing things are called the holes this this particles fictitious particles that I am thinking about these are called holes or empty states. Now things are a little bit more subtle but let me give you an analogy at least for this course I think that should be sufficient assume and this is a example shock Lee originally gave that assume that you are on the top of a building. And you are seeing a few cars from the top of a building on the parking lot and the cars are black and the background parking lot is white. Now when there are few black cars moving in one direction another you're taking pictures from the top you can you will see a few black points moving around in the white background. However assume that the parking lot is essentially all filled all filled with cars. Now there are few spots left and those will look as if it's white there are white spots from the top. Now when the electrons are moving in a particular direction or the cars are moving in a particular direction it will be a bunch of white space a black space with a few white spots moving focusing on the white spots is like thinking about holes. And you can see the same direction electron goes is the same direction the white space goes because white space cannot move it's moving because the electrons are moving or because the black cars are moving. So that conceptual framework gives you this idea that why the charge is positive and why they move in the same direction as the electrons the empty states move in the same direction as electrons do. Now this is a very important concept and I will hope that you'll spend some time thinking about it clearly. Now one thing I want to point out that about this effective mass is effective mass is not equal to the free electron mass. Effective mass sort of encapsulates the information about the lattice and where the electron is in a given lattice at a given energy. Because the electrons had it been in the free space it is always had a net value of M0 9.1 10 to the power minus 31 let's say. Of course in each of these bands we have already seen the effective mass is variable sometimes is positive sometimes negative sometimes 0 all sorts of complicated things. It is because at each band the way the electron responds to the electric field is not the same when it is sitting deep in the potential. Of course you apply a field it cannot go anywhere so its mass is very heavy when you go up in energy it moves very easily from atom to atom so its mass is low. So some way it retains the information about the crystal within this effective mass. Now one thing is that if its varying all over the place it wouldn't have been a very useful concept except that you realize that although it varies let's say corresponding to the red band this is the effective mass or one over for the blue which is the whole band this goes the negative and similarly for the green and the other one third and a fourth one. Now first thing you realize that well the red one doesn't matter because this is full way below the Fermi level carries no current I don't care about that effective mass anyway. Second one is that way on the top I don't have any electron so I don't worry about that one either because it's not participating in the conduction process. The ones that I worry about are only a few only two or three or four. For example in this case we only worry about let's say the green and the blue one where the electrons can actually move but even in that case we worry about only towards the bottom a few little states towards the bottom. And you can see the corresponding region where I plot one over m star in that little part of K it is approximately a constant. So therefore in the top and in the bottom region this is approximately a constant and as a result you can say that those bands are characterized by that mass of course a given band doesn't have a complete one single mass. But for practical purposes for calculating the current when they are sitting on the bottom of a band I can represent each band with effective mass. So that's all I had to say about effective masses and bands. Now we will want to discuss a few other things again a preparatory information for for the next lecture where we'll talk about real crystals and how to think about it. But the first point I want to discuss is something called a Brillouin zone. By the way one quick point let me mention which I didn't do in the in the previous slide was that you remember that there was a place where the mass of an electron went to zero. You remember started from positive went to zero and then went to negative. Now what does it mean negative effective mass we just talked in the context of a whole that how to convert it to a positive mass and a positive charge we just did. Now what does it mean when you have a mass equals zero right or the current in or one over m star equals zero. All it simply means that when you apply a field you cannot accelerate it right that's what it means but that does not mean that the electron will sit there. In fact electron is moving with a certain velocity when it's like it has a positive mass let's say and then when it reaches that point its velocity will not change because it cannot be accelerated. That doesn't mean its velocity is gone has gone to zero it will cross that point in K and then continue on and respond to the field. So unless one purposely put an electron in that state itself which is a never possible then in that case the zero effective mass does not mean there is no electronic transport. All right so let's talk about EK diagram and constant energy surfaces. Now one thing you noticed when we solved all these problems that all the bands are finite in size here I have shown him in green green region towards the bottom the bands are very narrow high effective mass they don't electrons don't go from one side to another very easily. They are filled on both sides anyway so it doesn't really matter and as you go up the size of the bands changes fine. But the first thing you notice in the solution that the solution goes from plus K to minus K not on arbitrary values but only over pi over P P is the lattice spacing and minus pi over P that's where the solution is and what the solution is that's the next point. The first is that the solution cannot go beyond that point so that's called a blowing zone and you can see that it could be constructed this way. On the right hand side I have shown the position of the real atoms real space atoms with circles and they are spaced at P P is the spacing not momentum remember. I could do this following let's say just by by construction that because I have P I could take that P and make this minor 2 pi over P 4 pi over P just by construction I could calculate a set of numbers 2 pi over P 4 pi over P 6 pi over P and populate the plus side and remember that this is 1 over a real space dimension P so it has a dimension of a wave vector. And similarly I could create a series of points with minus 2 pi over P minus 4 pi over P and go along and so it's like a K space or wave vector space lattice. Now you see if I take halfway in between starting from 0 connected 2 pi over P and minus 2 pi over P sort of take a vertical cut that green region has the edges at plus pi over P and minus pi over P right. Now notice that the solution space I have on the left hand side is exactly the same so although I have not solved the Schrodinger equation at all right. I haven't done any equation but I already know just by this construction that where my solutions will be where I'll have to fish for the solutions. Now if this is the case in 1D then what happens in multiple dimension well it turns out that there is a very simple algorithm to find where the solution is the actual solution will come a little bit later. And these are the formulas by which you can use to calculate this now you will do it in homework so you will understand this in great detail when you do it you know for the homework. But the point is that there is this recipe that if you have a three dimensional unit cell with sides A, B and C then you can correspondingly create an inverse lattice or a K space lattice by using finding the formula for K X, K Y and K Z. Similarly these are the unit cells lattice for the unit cells in the reciprocal space and I'll give you an example and then you can use the Wigner site algorithm but this time for the inverse space to show where the solutions would lie you see. So let me give you an example it will become very clear nothing complicated here. Do you remember this particular picture where we had in a real space a lattice with lattice constants A and B not necessarily perpendicular to each other. We started chose a central point connected all the neighbors and then took a perpendicular bisector for each one of the points and looked at the enclosed volume the smallest enclosed volume. And that was the unit cell you remember this right now the proposal in the previous slide said that in order to create a corresponding lattice space in the inverse reciprocal lattice what I should do is the following. This is a two dimensional space therefore I have in the third dimension I could say that I have a unit vector of size one right. So therefore for instead of C I have just put Z carat but other than that you can you can see the first one is K X 2 pi B cross Z carat and correspondingly A dot B and when you simplified you see immediately that the result would be essentially 2 pi over A. And similarly for K Y you see that it is essentially will be 2 pi over B right in this particular case. So if you have that you could correspondingly create a inverse lattice of points where you start with a central point and then create the points 2 pi over A 4 pi over A 6 pi over A keep going in that direction. Along the Y direction K Y direction you start 2 pi over B 4 pi over B. Now you have the two accesses and fill up the whole space by these points again this time use the Wigner site algorithm on this figure. And when you do same formula same procedure same mechanism when you do this would be called the Brillouin zone. And this is where within this space of K X and K Y all the solutions of the Schrodinger equation must lie. This is where the electron these are the only wave vectors the electrons can have when they propagate through the crystal right. So let me in that discussion let me quickly show the how the one dimensional one would look like. You see started with P I make the inverse lattice at 2 pi over P I do the Wigner site algorithm cut it in half and I have the corresponding E K diagram you can see pi over P and minus pi over P. Now one thing I want to point out here these are the solutions somehow I have calculated the solution of the Schrodinger equation. Now one thing I want to point out here very important which is I shouldn't really have plotted 0 to minus pi over P why? Because you see this is exactly the same 0 to plus pi over P whatever solution I have if I just mirrored it then I could have gotten all the solution. So I am really unnecessarily plotting the reverse side and anytime again you have inversion symmetry this will always be the case the solution and plus K will exactly be the same as a solution of minus K. So why waste space and trees in order to draw another half of the diagram right. So we shouldn't most of the time you'll see we'll just say 0 to pi over 2 and assume that my 0 to minus pi over 2 is exactly the same okay. Now let's think about 2D which is a little bit more complicated but this is going to get to the 3D what we are really interested. Now do you see in 2D our Brillouin zone would look something like this assume that I have a one of the five Breville vertices do you remember we had in 2D five and one of them was a not equal to b and the angle was 90 degree let's say I have taken one of those. And then my rule is that I should populate the inverse space as 2 pi over a 4 pi over a and so on so forth and then cut it in half to define the Brillouin zone and therefore I have pi over a to minus pi over a and pi over b on the y axis in a blue axis and minus pi over b. You see this is where the solution will be I don't know the solution but this is where the solution will be. Now the solutions of course would look a little bit more complicated and it might look something like this. Previously remember I had these four bands and I had just one axis plus kx or minus kx. Now in this case of course I will have a three dimensional thing it's like a energy is like a pole and these cups are hanging from the pole and if I had four bands how many caps do I have? I have four caps hanging from the pole. Now the point is that drawing such a thing is easy on a power point not even easy on power point but it's very difficult if you had to do it every day and draw it this way which is very difficult. It's difficult to read off information also so many times what people will do is the following. Instead of really plotting the whole thing they'll plot this three dimensional information on a two dimensional plane. It's like a weather temperature map in the newspaper you see let's say at different places the temperature is different. They put a counter map right they put a counter map that type of idea that representing three dimensional information in two dimensional plane how do I do that? So one is that let's say along the red axis I make a cut if I do a cut then the kx axis goes from 0 to pi over a do you remember? And I have the corresponding surface shown here is also with a red line corresponding to that cut in the cup right and I have only plotted 0 to pi over a because my 0 to minus pi over a is exactly the same. So I haven't plotted it one panel is enough. Now if I wanted it along ky direction what should I do? Well I should cut the cup vertically with a knife along the y direction. Now the y direction is goes from 0 to pi over b and 0 to minus pi over b again symmetric so I'll just put half of them. I have put them in half of them and in order to look nice I have made the 0 to pi over b looking the other side of course you could do it in any way you want these are just two panels of the cut. Now of course these are not only two panels if you really wanted to represent the whole cup you should have taken a cut at 5 degree angle maybe and that way you'd have maybe you have 360 degrees right you have a 72 cuts and you could plot out not 72 half of them are exactly the same so maybe 36 cuts or no 18 cuts 18 cuts and you could have 18 panels and from that 18 panels you could always reconstructed the cup you see. So these will be called the EK diagram because it tells you the energy versus wave vector along any particular cut you see. Now it's 2D isn't that bad but 3D will be a little bit more difficult you will see but if you understand how the procedure of doing things it will be a piece of cake you see. Now this is not the only way we could represent this information we could also represent the information in a slightly different way so that's the EK diagram. Now we could represent it this way also instead of taking a vertical cut taking a knife and cutting through the cups vertically I could also have the information and cut horizontally. If I take a horizontal cut at any energy E and then look at the ring at which it has been cut. I could have shown that you can see on the bottom right hand side the solution space is again KX and KY pi over A to pi over B. But this ring with a constant energy E1 could be represented by this magenta circle. Now it need not be a circle it could be anything whatever it is at that energy I could cut it. Now in order to represent this one one cut is not sufficient right. I will start with K equals E equals 0 energy equals 0 and I will have to take horizontal slices as many as I want all the way to the top. But remember what I said a few minutes ago that only a few bands take part in conduction. So if you are an electrical engineer thinking about just carrier transport you will just worry about cuts in the conduction band and in the valence band. And the ones that are too much below they don't carry current on the very top they don't carry current so we will not worry about those constant energy surfaces. But of course spectroscopist and others they might want the information for those ones also. Remember the photoelectric experiment where the pump electrons from deep in the valence band all the way out. In that case you need that information for us for the time being we don't need that. You can by the way you could also construct this EK diagram for this constant energy diagram also from the EK diagram also. And the way you will do it you will take a particular energy and see at what points in the red and the blue they cut. And whatever point they cut you will in the corresponding axis to the right you will pick that point. In this case you have just two panels so therefore on the constant energy surface you can just have two points. But if you had the 18 panel if somebody laid down the 18 panel by having a constant energy cut you could instead construct from this two-dimensional plot this particular constant energy surfaces. And this we will often do. We will not really picture this three-dimensional complicated thing at all. We will always go back from EK diagram to constant energy surfaces back and forth without ever bringing in the complicated three-dimensional one. So you should be very comfortable about how to draw it because I will ask you to draw this repeatedly on various occasions. So that's it for as we start our discussion in the next class for actual 3D I think we have all the preparatory information for that. And very quickly the so what we saw that the electrons only sits in a given states and with a certain value of K. And what are those values of K? 2 pi divided by NP. N is the number of atoms so that is how far they are spaced by and they sit in a given state. The second thing is at a given band right in the given band the electrons the number of states is equal to the number of atoms because that is how many people came to the party bringing in their own food. And so therefore in each band they must have the same number. Actually the number of states is actually even because whatever the number let's say 11 atoms came together to form a solid. Now of course each state there is two spin up and spin down and that we will discuss later. So therefore this always number of state is 2n and so the number of states in a given band is always even if you count for the count for the spins. This is something important we will talk about the implication maybe in a few class down. Effective mass said that it's not a fundamental concept and there may be systems where the effective mass cannot be defined properly. Do you remember that the graphene relationship you should also convince yourself that in that case how to think about carrier transport in those cases and how to think about the negative effective mass or zero effective mass in those cases. The third point was only a few bands contribute conduction and balance band partially empty or partially full and then finally the most of the practical problems actually can only be solved numerically. That's something important because we did things analytically just to get started but numerical solutions is what we will need.