 you can follow along with this presentation using printed slides from the nano hub visit www.nano hub.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 all right so this is lecture 5 on energy bands and this is today is a very interesting lecture in the sense that today at least finally will be solving a real problem it's not real yet but it gives a lot of information about the real thing we eventually want to know about so we'll start by talking about how to solve Schrodinger equation in periodic potential UX now remember why we want to solve the Schrodinger equation to begin with because as I said not all the electrons participate in the conduction process equally so if you want to know the resistivity of a material then in that case we'd like to know which fraction of the total electrons are actually taking part in the conduction are available for conduction so in this case we want to know where the electrons sit to begin with and then subsequently we'll figure out what fraction of them moves and the periodic UX well because it's crystal and in a crystal we want to solve the Schrodinger equation to see where energy levels stay within a crystal and then we'll talk about a set of set of other things now this is again the problem that we had been talking about in the last class as well and this is a problem where the red electron is sort of swimming through a series of ions or the core of the nucleus and they are being periodically pushed and pulled by these respective positive charges and the electrons are of course negative so and what we did in the last class in solving the toy problem we assume the atoms are so far apart that we can just take one of them don't worry about the neighbors and just take one of them say where they are if they are very close to the bottom or way up in energy how they transmit and where they stay but of course atoms are not far apart they are only five angstrom apart or less than in many cases maybe two angstrom apart they are not far apart so in this case it is impossible to sort of neglect the effect of the neighbors and that is essentially the essence of this problem that we are going to solve now let's consider this full potential in in a in a simplified way remember again the the reason we write it as a square potential going up and down is because it's an idealization in practice it will be like a series of Coulomb potentials you remember this hyperbolic type relationships and they will one over art type relationships then they will sum up but this is idealized version just for simplicity but it will give us many information which will be correct also in real crystals okay now this is ux goes up and down in position now if you look at the period let's define them a is the bottom of the well so this sort of the extent of the atom let's say and the b is a barrier between them and the period then therefore is a plus b we'll call it p remember this is not momentum this is for this case just a distance you know maybe five angstrom from one atom to another now you see something funny here that I can of course solve for the Schrodinger equation here in one shot if I take an energy which is below ux in that shaded region as you can see then when the energy u is higher than k your energy then you know how to write the solution e to the power alpha x plus e to the power minus alpha x that we have seen in the last class what should we do when the energy is above the potential you know in the white regions in between well we'll write it as sine kx and cosine kx and remember I have to know which white region I am writing at so I put a little index n plus 1 just to say that this is next to n n region so I put an index so you know for every set of pair of this potential there'll be one two three four and so on so forth so I have a series of solutions now if you didn't know anything else then and simply applied what I told you last time that is take a boundary condition and plus infinity another at minus infinity two unknown constants gone every interface match the wave function match the derivative right then you can essentially solve this problem no problem but you can immediately realize that even in a centimeter cube of material even in one dimension they are probably hundred million atoms sitting now you have to solve hundred million multiplied by two equations now the boundary condition may give you two less but that's not really helping you so this is an impossible problem to solve and there's no way then in 1940s when these problems were actually solved you could possibly solve it there is no computer but it was of course all that time and this is how and and what helped the solution of this problem is this beautiful theorem called block theorem will not prove it or anything but just show you the content of the other theorem that how it solves it it's very simple you'll see now remember the four steps there are five steps of both solving for the energy and the wave function the first four were determining the determining the eigenvalues of the energies now we write the same equations but as I just explained in the previous slide that if I just use this rule blindly too many equations cannot solve it and therefore I cannot know where electrons sit what I can do in a state is in a state of using two and three we can replace it with something very simple using the fact that this is a periodic potential it's not any arbitrary potential things repeat and that information we should be able to use somehow to reduce this humongous problem to something that you can solve essentially in 15 minutes now remember where that comes from or see how it comes from so the main feature of this potential is that if you take any given location ux position x and potential corresponding to that ux assuming that x equals 0 is the origin on the left and that can be an arbitrary origin then you can see that when x is displaced by a plus b the potential is the same it's one one period away and I'm instead of a plus b I'm writing yes p in order to save some save some time now if the potential is the same then although I don't know the solution yet I don't know the solution of the Schrodinger equation but I could say this that whatever be the solution at location x the solution at location x plus p must be the same remember the definition of this Bravais lattice and the periodic solid that every point you sit looks around you is exactly the same as every other point you sit now if every point you sit looks exactly the same the solution whatever it is we do not know whatever it is that must also be the same so if that is the same so therefore I can say the probability of finding an electron at a given location is exactly the same as it is one lattice site away what does it say about the wave solution itself the psi this is this is psi squared what about psi itself well psi you can see the wave function is almost the same except I'm allowed to have a phase factor e to the power ikp again p is not a momentum it's a distance why ikp why is the same well if k was 0 no problem right on both sides I have the same thing square it no problem but remember this is an absolute value squared that means it's a complex conjugate psi psi star and so when I have psi star then obviously on the right hand side I will pick up a minus ikp and when I multiply them to then of course this ikp e to the power ikp will go away so therefore the solution on the other proposition on the left hand side that the probability of finding an electron is the same one lattice site apart implies that the actual solution one light side apart must be given by this wave function multiplied by a phase factor you see so take this this is an important step and so you should try to understand that clearly okay and this is although also by the way this is k is not this h square k square divided by 2m as we saw before this is just a number k because k could be anything I could multiply with anything and when I multiply with complex conjugate anything I put it will always go away so k doesn't have to be any specific number it will be in a second but that's not the old k okay so now let's use this information see how it works I just told you that if I know the solution psi x at any point if I knew then at x plus p distance away the solution will be related to each other by the relationship shown in the bottom you can see at x plus p this simply one phase factor shifted for the original solution at x well if this is true then isn't this true as well if I wanted to go two step not one then the second step I could relate back to the first step and I could write that psi at twice the step is simply psi at the first step multiplied by the one phase factor right you see this and if the first one is already known from the left-hand side you can see and therefore this I could write do you say this that when I have gone two steps I could in sin in cascade the solution back and essentially pick up the phase so if I am to step away I will have a pick up a phase factor compared to the first one 0th one is by 2 kp so you can see that if I am n step away and n could be a billion if I'm n step away I don't know anything else but obviously I could say this n step away this statement should be obvious because what I'm doing here you can see I have picked up a phase factor which is kp of course but now multiplied by n so I have n phase factors and this you can see where it came about the n was connected to n minus 1 n minus connected to n minus 2 all the way folded back until you hit one and that's where you picked up all this phase factors well I haven't done any solution yet but simply I'm setting up the problem and in a next three or four slides we'll see how to solve use this information to solve the problem this is something replacing the boundary conditions you see continuity of the wave function and the derivative of the wave function that is something we'll use on top of this equation now let's look at this and then say the following statement now this part is not obvious and this is sort of one of the tricks and we'll see how it works but I'll explain it later on so assume you have taken this string of atoms now you pull it back now how do you pull it back in real crystal you couldn't possibly take the atoms and pull it back in a string right so in principle you couldn't do it but what you are telling here is that when the atoms are so far apart let's say one and a million unless the atoms are very close to the surface in between you know one side you have a million plus 20 another side you have hundred thousand or something it really doesn't matter what you have at the very end because it's so far away so therefore whatever you do pulling it back twisting it whatever you do it's not really going to change the solution inside of course we are not going to know that unless we begin to compare it experiment later on that whatever we did this is correct or not but for the time being we'll proceed and then we'll come back and see whether the solutions are real or not now if I pull it back then what will happen you see so I'll start from one two three go all the way but now the n minus one the nth one is again the first one and then two n one is again the first one you see so this because it's a periodic boundary condition that's what they say in that case what should happen well in that case you can see that if the nth one is exactly equal to the first one then psi x plus np should not be just a phase shift away it's the original thing because on the left hand side I should be able to replace this with psi x again because it's the first one and if the first side left side and right hand side has to be equal then you can see then e to the power i k p m that must be equal to one only when I pull them back together and if it is one then you know that this one equals e to the power i 2 pi n do you know this formula this is so this is so I can set these two things equal because every time something is twice the multiple of pi 2 pi and when it's to the power i and in that case the result is always one you know because it's like 360 degrees coming back and coming back again and so I can write k equals 2 pi n divided by np is this true do you see that if you look at the exponent on both sides left and right and divide both sides by np then you can see that these solutions are correct it satisfies the solution what does it tell you what it tells you is that any arbitrary value of k is not allowed if the potential is periodic remember the phase factor I had e to the power k p now any arbitrary value of k will not work if it has a periodic boundary condition then it must have 2 pi n divided by np now little n it could be 1 2 it could be minus 1 all the way to n over 2 why could it not be n plus 1 over 2 you can put it over there and you can show that that's exactly equal to 1 1 again the value is 1 again another thing I want to point out that there is this solution of the k that has two sides positive sides 0 1 2 3 4 and the negative side 0 minus 1 minus 2 minus 3 so on so forth what does it mean the plus 1 means remember e to the power plus ikx wave going in one direction right positive direction e to the power minus ikx was what was it wave going in the other direction so this is like wave going on the ring in one direction that's e to the power ikx wave going in the other direction e to the power minus ikx so of course both traditional solutions are fine electrons could go in either way and that's what the positive and the negative strings mean okay so we are getting there not yet but getting there remember any arbitrary value of k is not allowed only certain values of k are allowed okay what is the maximum and a minimum value well we can see what is the maximum the maximum would be when little n is equal to capital n over 2 and if I put it over there in the equation I will see pi divided by p and on the other side minus pi over p that means all my solutions whatever they are cannot exceed this boundary I don't know what the solutions are but it cannot exceed this boundary and later on we'll call this brillouin zone because this is where all the solutions of the eventual equation this is where they live okay now finally we are able to solve the problem let's see how it works do you remember the solution that I have on the left hand side of zero zero to minus b sin beta x and cosine beta x now on the right hand side between zero to a few slides before what did I write e to the power alpha x plus e to the power beta x right minus alpha x remember I'm here instead writing in terms of sine and cosine is this right well if it is at if this is equal to a imaginary quantity then it's fine sine and cosine will be fine if we write want to write it e to the power alpha x in the with the exponent alpha and beta in terms of imaginary it could be real or imaginary in that case writing in this particular form should be fine you should check this out just by inserting the equation that you know you you see the definition of alpha and beta above you should check it out yourself that this is exactly the same as I said before in terms of e to the power plus alpha x and e to the power minus alpha x exactly the same now let's use the boundary condition first on the green line or the blue line I'm not colorblind yet but so the blue line and what do I have to solve here match the boundary condition I'll have to say wave function is continuous and its derivative is continuous right okay so I know the two solutions I should be able to do it in a second so at x equals 0 minus and 0 plus I can say the continuity of the wave equation and its derivative and I will not labor you with this statement that b sub a equals b capital b sub b because you can see if you insert x equals 0 in this particular equation and let the wave function be continuous that's the relationship you're going to get okay the other thing is well then I should have to go to the other side because now I have to match the boundary condition at x equals a then I will be done but you can immediately see there is a problem if I need need to go to the other side from 0 to a and need the information from the other side then in order to know that the information at a plus I will have to then solve the next way atom then I'll have to solve use that information in order to get that information I'll have to go to the neighbors and I'll have to keep going to the right now this is not good what I want to do instead is the information we just collected that the wave function at x equals a what is it equal to the wave function at x equals minus b right one lattice vector away right except that there's a phase factor remember this is a periodic potential so it's not exactly equal up to a phase factor it's equal so instead of looking at the right I'm looking back within the unit cell itself this is why you read unit cells spend some time Bravais lattices and everything this is why you do it because then we so once you know the solution within one cell you know the solution for everything for bulk of the material okay so I want a slightly right to the a and that's equal to exactly equal to something slightly right to minus b that wave function within my same cell and therefore see whether you agree with this statement what I wrote that psi at a at x equals a meaning a minus a little bit to the left is equal to psi b but see the phase factor e to the power ikp and why is that because I have folded it back one lattice psi I have that equation and of course I have the derivative equation the derivative equation again has the little phase factor sitting there I'm actually done with this because I know the wave functions I can put them in you see on the right hand side the first equation on the top capital a sub a sine alpha a plus capital b sub a cosine alpha a this is simply by inserting x equals a in the solution of the above equation in the well region and so on so forth you can you can just match it up please take this time when you go home to take this out I'm going very fast because this is simply algebra it's not the conceptual point that you need to know but check it out whether it's correct I have how many equations do I have here I have four equations right two blue two red how many unknowns four unknowns you can see b with sub a and b a with sub a and b I'm done actually so if I have that the step four was put them in a matrix if I put them in a matrix then how do I find the solution of the eigenvalues I make that determinant of this matrix equal to zero and this is a simple four by four matrix how difficult can it be people could do it in 1930s and we are more sophisticated right so we should be able to do it in a few minutes and when you do you get a complex looking equation like this but just like when we solve for one quantum well problem it's no more complex than that because you can see in this equation the only unknown is xi and xi is essentially e divided by u0 u0 is already given from the material that we have and there are of course lattice spacing involved there is this a involved in various places that is something we also know so the only unknown in this equation is energy level e and once I solve it then I know where the electron sits in a crystal and how can I solve this again two ways you can solve it matlab but if you are not in matlab age you can solve it graphically as they do in the book and let me show you how to do it in graphically and that's where the band structure will come in so band structure is the solution of the Schrodinger equation in periodic potential so that we know where electrons can sit not necessarily they will sit there but we will see where the electrons can sit there's a distinction there's a distinction between a chair we are calculating how many chairs we have sort of in a room we are not still talking about what fraction of them are occupied you know so that will come later and so right now we are just calculating the solution of the Schrodinger equation all right let's talk about the band structure this we should be able to do and if you please pay attention to this one because this is where people generally miss it so let's first plot the right hand side remember we'll plot the right hand side and the left hand side look at the intersection let's plot the right hand side first remember that p is given that is spacing and k has discrete values 0 1 2 3 and corresponding to that 2 pi n divided by np so when k equals 0 what is the right hand side cosine kp is 1 and that 1 does it depend on energy 1 is 1 and so with k equals 0 I should draw a straight line independent of energy because it doesn't depend on energy what about k little n equals 1 well I should then have 2 pi divided by np right n is the number of atoms in this string and then I should have a solution that's again independent of energy but goes straight all the way through now you realize first of all that this one would have also been the solution had it been n equals minus 1 because cosine theta equals cosine minus theta so wherever the value is a little less than 1 but that's straight independent of energy right I'm just plotting the right hand side you can see the right hand side is actually how many equations this is n number of equations so right hand side is n number of rate curves actually it's not just 2 n number of rate curves okay I'm let's say I'm done I have spent a lot of time of course you have to draw how many lines maybe a billion lines but don't try at home this is one thing you few lines would be fine what about the left hand side well left hand side is a little complicated but you put it in a any graphing graphical routine you see that it will go up and down as it as it moves around in energy but this time of course the function of xi and xi is a function of energy so therefore you can see there is oscillation in energy where are the solutions solutions are where the rate cuts the blue because that's when the right hand side is equal to the left hand side so I'm done actually I'm done solving it I'll just have to read off the numbers in a particular way what I have done simply is rotated it 90 degrees just like we did in the last class remember rotated in 90 degrees and let's look at the solution then so this is k equals zero right this blurred this red line the solid red line is k equals zero and so I'm drawing something with energy in the y-axis and the x-axis is k so those are the first two points that cut not really the first the ones that I show in the dotted line as the second and a third you can see there is also a point that cut is below which is the first point here very close to energy equals zero okay so this is my solution at k equals zero for all those electrons that has k equals zero actually has these energies fine now what about the next solution remember I have the next line also and from cutting from that next line I will pick up a bunch of solutions there also so I correspondingly shift it to the right and again blue blue cuts the red dotted line now this time what is the value of k the k value is 2 pi over np so on the right hand side right hand figure instead of plotting it as a function of k equals zero I should shift it a little bit and pick up those points that is 2 pi divided by np do you say this that that will also give me a bunch of solutions now remember k equals plus plus k and a minus k gives me the same solution is that right so therefore if I go to the left side on the right hand plot I will pick up a symmetric point on the other side just like the first point pick up the symmetric point and this I can continue how many maybe a billion times I will not do that but you can see that if I kept doing this then I will have a bunch of solutions and these are all these red dots now one thing you will see that I have connected the red dots with the black line why did I do that because black line means a continuous that for all k values as if there are solutions but actually not right for red points are discrete these are not a continuous line the reason I can do it is when you have 2 pi over a billion n n is a large number and when you have 2 pi over billion plus 1 then essentially these are so closely spaced that even with a microscope you cannot probably see they were part so they this red lines essentially look like a continuous solution and therefore many times you will see that they are joined together people don't put this little dot points they just simply join together as a continuous point now what is the right hand side maximum right hand side of this k remember it was pi over p maximum k max and k mean so all the solutions stay live in this space and that's why the in the x-axis will call that region the k region is a brilliant zone associated with the solution now electrons can sit in many places I see but there are also places where electrons cannot sit you can see okay now how many solutions now I'm going to describe to you we have them done we are done with the solutions we could go home but rather is let's stay with it and try to understand a few features of the solution first thing is how many dots do you have in a given band what is a band well a bunch of solution which are together you can see how many banks we have we have four bands of solutions in the bottom second third and fourth and you can see each is separated by a certain gap we'll call that band gap it's a band of solution separate by a gap so therefore it's called a band gap and you can see how many solutions do I have let's try that so remember number of states per band how many solutions I have so I'll have to take the maximum region k max and the left hand side minimum region came in k max minus k mean is you can see 2 pi over p do you see on the bottom side for the graph the maximum side is pi over p shown here in the red and the minimum side is minus pi over p shown in the blue so the difference is 2 pi over p and 2 pi over p but the delta case where the solutions are available is 2 pi divided by np and therefore if you divide it up every band has n points you see so how why did it come about well remember that every atom there's let's say once you remember in the square potential I had first bound level second bound level third bound level when I bring a lot of atoms together these bound levels essentially all mixed together if I have n atoms then I have contribution from nn of them and so per band per band therefore it gets n fold degenerate right there are n solutions per band it is like potlack you know you can have let's say you do your own cooking at your home but if five people get to the potlack and then naturally what we'll have is five five different type of dishes right and if everybody prepared three and then you bring them up all together and mix it up then you can see for each type of food let's say the entrée and the desert and everything each type of food you will have five fold if you had 20 friends then you will have 20 of them if somebody if everybody's really cooking and it's the same thing now if you had to solve this Schrodinger equation every time you would be in trouble because that will be a lot of work and what we'll see that down the road it's actually not we don't really need all this information different people need different pieces of information for example remember the photoelectric experiment light coming in kicking an electron up and the electron coming up they shine x ray on it let's say so in that case i need to know the bottom level all the way to the vacuum level how far they apart all the gaps and everything i need to know in that case i really need to know the whole band structure but many cases in transport you will see when the electrons are really on the bottom side they don't really participate in conduction so for us many times we don't really need to know all of them just the ones that are relevant so we want to capture the information that is in over there into simpler things so that we don't have to carry over this all this information so that's what we're going to do now and that's the properties of the energy band from an electrical engineer's perspective not from a spectroscopy's perspective so let's take the question we want to ask now in the next four or four slides or so is if our electron was sitting in a band and i applied the field or i let the electron move what velocity does it approximately move with that's what i wanted how do i do that well i know the solution at every point and its neighbor so i write it as a e to the power you can see as an exponent i k x and minus i omega t what is that i omega t remember there's a plane wave with a given k that wave is moving if i put at one level that's the plane wave i will have at that state let's combine two of them so one is in red another is in blue on the top the first one is at k and the second one is slightly higher at delta k apart now when i put them two together what will happen since their k is not the same their energy is not the same they will interfere each other it's like two lights with two slightly different wavelengths so they will interfere with each other and when they do let's say the red one looks like this there's a plane wave right at a given time that's the snapshot and the blue one looks like that why because the wavelength is slightly different right the k is slightly different wavelength is slightly different so therefore even if you started them off both at the same point they will not stay in sync and therefore if i combine them it will sort of in one place it will look like it will interfere constructively and in the rest of the places it's going up and down all the way and so it will gradually die off so i can say by combining the two electrons at two states i can say as if i have an electron at that location right otherwise i don't know where the electron is the plane wave probability of finding an electron is same everywhere in the other case so by interfering them i can say where the electron is now what will happen a little bit later a little bit later you see what will happen is because the delta because the e and delta e are slightly apart or this delta e apart so what will happen that a little bit later the red one will move to the right but the blue one will move at a slightly different velocity to the right because it has a slightly different energy slightly different omega and so it moves slightly to the right now the point where they are interfering is not constructively is no longer at x equals zero you can see they are interfering at a slightly downstream constant so as if the original electron which is sitting at x equals zero that as if has moved down to a certain distance and if i could somehow capture this information then i would know what velocity an electron moves if i put it in a certain band at a certain point now in order to stay on the peak what i need to make sure that the right hand side the blue side of this equation that phase does not change with time so that delta kx minus i delta e h bar multiplied by t that that remains a constant because then that will have a constant shape that i can follow i'm sure this step you probably didn't follow this you have to think a little bit that why on the peak and if you plot out some of the way functions very easily you can see it numerically but if this has to be a constant then i can say that the velocity v is delta x over delta t right this is by definition that at a certain time how far it moves delta x over delta t is velocity but you can see from the top from the blue exponent that delta e over delta t is essentially can be given by delta e over h bar divided by delta k do you see this from the blue just do this one step and then you'll see that this relationship emerges and from here you can you can see that the value of a the acceleration the acceleration is simply delta v over delta t which is the Newton's law simply Newton's law take one more derivative and when you take one this derivative and follow this equation through you will see that this actually looks like a Newton's law you can make it look like a Newton's law the this is sort of difficult to go through the equation but it's something i i guess you have to just write it in a page but the bottom line is that you could rewrite it such that that derivative of h h bar k what is h bar k that's the momentum derivative of h bar k with dt so change in the momentum as a function of time what is that that's force so anything that is left anything that's left if i call it en masse if i call it en masse then i can make it look like a look like a Newton's law as if in a like you know that is i have an atom i have an electron and then i apply a force and it keeps moving with a certain effective mass not the original mass you can see on the left hand side there's no definition of mass at all there's no or no notion of a mass at all there's no m not sitting here right so free electron mass but only thing that is sitting here is delta e over delta k that's how the solution changes in each of the band right so let me show you how it works it's better to show it with an example so let's take the any two bands let's say the third and fourth band i'm not showing the first and the second band let's take the third and fourth band let's focus first on the red band now velocity i have told you velocity is given by delta e over delta k divided by h bar that's the velocity if i look at the red curve if i change the k a little bit to the right hand side the energy goes up so therefore my velocity should be going up do you see do you agree with this statement that if you have if you are sitting on the bottom then the solution is symmetric on the left and right it goes through a minima so when you take a derivative the value is zero as you move away from plus in the plus k direction the velocity generally goes up and then eventually it will come down and similarly on the left hand side the solution is symmetric so the velocity would be going negative and then going back to zero does it make sense it should be in some way right because remember with plus k electrons are supposed to go to the right and the velocity should be positive because it's going to the right you can see indeed the velocity is positive because on the right hand side you can see velocity is indeed positive with minus k velocity should be the electron should be going the other way and the velocity should be negative and it is indeed negative you can see right because below the zero axis so that makes sense what about the effective mass the effective mass is a second derivative of energy with respect to k and you can see here if you take the second derivative of that first equation then you will get something like this and you can see that the mass this particular mass actually is not a constant it changes with position because the curvature of the solution changes and therefore this one changes as well but the main point is very close to k equals zero you can see it's approximately a constant and we'll use that information later but throughout the zone or throughout this solution space of course it's not a constant because the solution of the equation is changing throughout and similarly you can figure out that for the holes this should be the opposite the effective mass should be negative in one case and the electrons should be going in the other way so the for the blue bands and implication of this we'll discuss a little bit later now up to this is what is in the book and you are supposed to know all this but the professors are very fond of setting the following problem for the qualifying exam what if you do not have second derivative what do you do generally at that point you see the question and you say okay today the exam is not going to go well for example this particular case on the left hand side I have an energy by the way the x axis is k the wave vector right remember the solution space plus pi over p minus pi over p that's the two boundaries of the solution now instead of a straight line instead of this nicely curved solution I somebody gives me a straight line solution there is a material called graphene that people are very interested in that has some of these features at least in part of the zone it has this some of the solution what do you do now well first thing is that if you try to we'll look at the equation in a second so don't worry about it but first let's try to see velocity first derivative of the energy versus k well that's fine no problem because it is a straight line I can take a derivative and if I take a derivative then on the right hand side plus k I will have a constant velocity on the left hand side with minus k I will have a negative velocity no problem now what are you going to do when you are want to take the effective mass if you try to take a derivative of this what will be the derivative it will be 0 what will be the effective mass infinity right 1 over effective mass is the second derivative second derivative 0 effective mass infinite does it mean you pump and you put an electric field and electrons and want to move actually that's not the case so that very definition of effective mass doesn't hold anymore because we cannot write the Newton's law as we did in that previous case when you have a continuously carving back but the solution is still available because remember what we want to know that as a function of time how the energy how the position and the momentum of the electrons are changing that we already know from the first two equations for example you can see that if I if the force is the derivative of k with respect to time you know change of momentum with respect to time if I knew the force at a given point the electric field let's say coming from the electric field I could just integrate for that time for a certain time delta t and I will know the new k so it's starting with the momentum I know the force so after a certain time delta t I will know what the new momentum is similarly as soon as I know the velocity I could integrate the velocity for a little bit of time and I will know where the electron originally was and where the new electron is right so in fact I do not need effective mass I can solve the whole problem not even needing an effective mass and everything will work or just fine so variation of this problem are often said just to test whether you understand that effective mass is not fundamental it is helpful many time it works it's a historical baggage many modern materials don't have effective mass but not having an effective mass does not mean that electrons do not move easily through the castle in this case in fact electrons moves very easily through graphing you see this point okay so it's a long lecture but let me conclude so we first said the solution of the Schrodinger equation is easy if we have well defined periodicity 80 percent of the material of interest is not periodic but it's easy to solve we'll solve it here and then for many other material amorphous polycrystalline and we'll say well I cannot solve it over there but it's almost like a periodic crystal because most materials are now we said that electrons can sit in narrow energy bands and not necessarily narrow in various energy bands and we showed how to calculate that and these bands are separated by a gap so electrons cannot sit everywhere remember we said all electrons are not equal so therefore different electrons have different band curvature in which they sit in the towards the bottom electrons don't want to move remember they were relatively flat so when you take a series second derivative and when you have the effective mass inversely proportional the electrons in the bottom band they don't want to move these are valence electrons they don't want to go from one one atom to the next as you go up and up and up the curvature becomes larger or curvature becomes larger and the effective mass becomes smaller electrons can move through the crystal easily see and the final point is that the effective mass is not a fundamental concept I just emphasize that and we shall see how to solve k dot p model in real crystals because this was a fake crystal one-dimensional well what what is the last time you saw a one-dimensional crystal in a such a simple form but many of the features of the actual crystal are actually reflective of this variation everybody has bands everybody has this brilliant zone everybody has effective mass if it can be defined so the real crystals in many ways contain many features which are very similar to this k dot p model okay k p model or chronic penny model the people who first solve this problem okay take a look I skipped a few things but take a look in your book and understand this clearly it looks very simple and I am sure some of you have read this information before in undergraduate but to really deeply understand this is very important because most of these concepts we are learning now towards the end of the course you will see and also in other courses that this is a basis most of them don't apply anymore but if you don't know what are the approximations built into the classical theory you will not know when they are breaking and therefore when new theories have to be applied so understand every assumption with the details not just glossing over it right okay thanks