 You can follow along with this presentation using printed slides from the Nanohub. 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. This is lecture four for solid state devices and today we'll be talking about solution of Schrodinger equation. Now there is quite a few topics to cover today. So we'll get started by talking about time-independent Schrodinger equation and we'll see how to frame the solution and understand the algorithm of solving the problems. So I'll take an example and show you how to solve it. The idea is not necessarily that we just focus on that problem per se but understand the steps or algorithms of solving Schrodinger equation in general form. Now if you remember the original motivation of the problem, in lecture one and two we talked about the crystal structure, how to count atoms per centimeter cube that is density and we found this particular problem that if we just take the number of electrons and multiply it with the number of atoms, number of electrons per atom and multiply the density that doesn't really explain the resistivity of a material. So resistivity of copper versus silicon you wouldn't really be able to explain it and what we realize from that discussion that while all electrons are sort of created equal but they do not behave equally and that's understandable. You have heard about valence electrons which sit very close to the core of the potential and those of course don't go from hopping from one atom to another but there are electrons higher up which essentially can easily travel from one atom to another and so therefore you understand that they will not these electrons will not contribute equally to the conduction process and that is exactly the problem that we are trying to solve and here I am seeing again showing again the yellow circles representing the core of the atoms the nucleus let's say for silicon and then on the right hand side the square potential going bobbing up and down that's the idealized representation of the Coulomb potential because as you realize and when the electron is close to the atom the yellow one the red circle which is supposed to be the electron when it's close to the yellow one then there is a lot of attraction because positive core and the negative electron so the potential is low so you can see the square going down and when you are a little bit further out if you are not as close then the potential is somewhat higher and you can see I have drawn the three positions of the electron one is way down into the square this periodic potential you can see that that electron will essentially remain bound to that atom they are not going to move these are just once s state 2 s states remember this from your high school chemistry those are the states and I have again shown you one electron which is way up way up on the top that essentially doesn't care about what this potential is is so far above the ground that essentially that potential variation will not mean much and in between of course this will be a balance of the two so we'll solve these problems and see how to what energy is the electrons these respective electrons can sit here and that will be that will give us the information about how they contribute to the transport problem so let's start by thinking about the Schrodinger equation you remember in last class we derived this simple second-order differential equation on the right hand side I have the time dependent part first derivative of time and on the left hand side we have a second derivative in terms of position you realize that had it been a Maxwell's equation the time derivative on the second right hand side would have been the second order equation Schrodinger equation is slightly different from the equation of light and you will be doing a homework based on that now let's assume that we are just interested in time independent properties we just don't suddenly turn things on but let's say electrons are in steady state you've applied a voltage and the electrons are slowly flowing through from one contact to another you want to know the time independent part and one trick that people often use is to say that the entire wave function psi for the electron can be decomposed into two parts just the position dependent part psi of spec x which is a small psi and then the time dependent part is given by e to the power i e t divided by h bar this i omega t essentially that's what you have there so if I insert this equation on the insert this substitution on the left hand side then you should be able to see that this is the result I will have do you say this for example on the left hand side you can see when I take the second derivative with respect to position and I'm not taking a any derivative with respect to time so that e to the power i e t by h bar that the whole thing will come out on the left hand side and the same is true for u of x but look at the right hand side because I am taking a derivative with respect to time now so therefore I will pull out a i e divided by h bar in front and as a result you can see that this is a pretty simple substitution and very quickly you can see that this exponential factor is the same both on left and right hand side so I take it out and once I take it out it becomes this very simple equation because all the e to the power exponential term all of them are gone and I have very simple equation I can even simplify it a little bit more do you see how the simplifications have come about I have cross multiplied to m not divided by h squared throughout with a minus sign and therefore the above equation the equation on the top becomes the last equation on the bottom now one thing I will say before we start solving all the Schrodinger equation and other things that many time you know when you hear about quantum mechanics people hear about all sorts of you know complicated things philosophically intrinsic things that if this electron is here where is the electron where when if one electron is at a given point at a certain time what happens to the fraction of the electron at some other time at this site of probability is an all sorts of popular literature you hear about we are not going to just here I'll discuss any of those things for us this is a simple second-order differential equation will learn how to solve it how to use it and be done with it of course when you want to use quantum mechanics for your research and other things you will have to learn more but those you learn much more in your other courses for this purpose we'll just simply consider it second-order differential equation very simple thing don't worry about the philosophical implication of quantum mechanics that's for another course okay so how do I solve this second-order differential equation very simple because if e the energy e and u is the potential remember that the square potential going up and down if e is greater than u then the second term on the right is a positive and I will call it k I'll call by definition that whole thing so that I don't have to write it over and over again I'll call it k and this equation becomes simply this I mean this equation we have solved million times before and the solution the solution is also well-known right the solution is that generally it is a sin kx plus b cosine kx or if you wanted to write it between terms of exponentials you can write it as a e to the power ikx plus e to the power minus ikx and this you know how this come about this Euler relationship you use them you can go from one to another that's it when so we'll remember this right anytime we see an electron above the potential we'll directly write this solution what about the other case you know electrons cannot always be yes go ahead no it doesn't need to be a constant if in this case particular case if u is a constant then we'll assume k to be a constant but in general if u is a variable changing with position in that case we'll have to use numerical solution of the Schrodinger equation and that we'll discuss towards the end of the class now on the other hand if the potential you and I'll show you example how to use the potential is larger than the energy in that case you immediately realize looking at the top equation in the green box that the second term will become negative we'll call that thing negative thing alpha just a name and the solution of this equation well again no rocket science necessary the solution is e to the power minus alpha x plus e to the power plus alpha x and you can easily insert it in the other equation and see that that solution is valid so we'll use them often as you'll see so I'm going to show you five steps of how to solve any time independent Schrodinger equation and those five steps are first thing is somebody has to give you the you the potential how it's going up and down well how will I get it for example if you think about silicon right remember the crystal structure let's say you are going in along 1 1 1 direction and you will encounter atoms at a certain gap right certain spacing so from that you will get the value of the spacing of you how will you get the depth of you well remember silicon has a certain number of protons some other material let's say copper will have some other number of protons so therefore depending on the material that will give you the depth of the potential so you will have to get it from someplace that's what I'm saying once you have gotten in you and the boundary condition I'll explain that in a second you'll solve the Schrodinger equation then once you solve the Schrodinger equation you will get the value of psi you'll interpret psi psi star as the probability of finding an electron at a given point and then once you have the small psi you will multiply it with e to the power i e t divided by h bar remember the time independent time dependent part and get the whole side the large side and from that you can see there are prescriptions for calculating momentum p or energy level e and all other things now why did I get all this formula well that is something you learn in quantum mechanics for this we'll just consider it as an algorithm so let's solve some problems let's first solve a problem in which the one in the middle in which the electron is such that it's not exactly on the bottom of the well neither it is far away on the top and that will give us some exercise of how to use the five rules that we just talked about now the first thing you will notice in solution of these problems that we will divide it this will become clear as we go on that will divide the potential often in a periodic potential into the two n regions and that will give us n regions and that will give us to an unknowns you'll explain understand in a second we'll use the boundary conditions at plus and minus infinity that will reduce the number by two and then we will use the continuity of the function and the derivative of the function to get two n minus two equations and then we'll take the determinant and finally we'll calculate the wave function so let's see how it works I mean it looks like very complicated thing but you will realize in a second how it works so let's think about this for a second assume for just for discussion that the atoms are far away from each other so far away that essentially they don't tunnel from one to the other they essentially stay within their well but at the same time the well is not infinitely high well is finite but the atoms are far away in that case I can just take one atom itself and solve the Schrodinger equation for that case for example I have shown on the top that if I take a box out of it and that is what I have shown in the bottom as well and I have reproduced it the same one from the top and I have produced it in the bottom now let's think about it how to apply the solution methodology so what was the step one step one was that we'll have to solve the Schrodinger equation in various regions I have three regions here on the left my energy is below you do you see because my energy is you is let's say at that point a certain value my electron energy is below that value so my solution of that region will be what should it be remember anytime the energy is below you then I'm supposed to have e to the power alpha x and e to the power minus alpha x remember in five minutes ago we just talked about what about in the middle in the middle now the energy of the electron is higher than the local potential because local potential you can see is much lower right compared to the electrons as a result what would be the solution I could write it as a sine kx plus b cosine kx or I could write it as e to the power i kx plus e to the power minus i kx either form would be fine what about the right hand side the right hand side again the energy u is greater than the potential u is greater than energy e so I have the same form so three regions six unknowns so that was what I was saying in the last slide that anytime you have n regions you have two n unknowns that was step one what about step two we said that we'll have to use boundary condition at plus infinity and at minus infinity then that will take care of two unknowns let's see whether we can work it out here if you look on the right hand side for psi then on the red shown here in the red you realize that the wave function probability of finding an electron far from where the atom is is zero right and so the term with n e to the power plus alpha x the n has to be zero because if n is finite and if x is going to infinity that term will blow up and so that term cannot be cannot be a finite quantity so n is zero now what about the blue side on the left because we know psi at minus infinity is also zero right we cannot find an electron either too far to the right of the atom or too far to the left of that now in this case you have to be a little bit careful because you see we are going in the minus direction so in a state of the c being g zero you can see in the blue region the m will be now equal to zero do you see why because when I put minus x that quantity is what begins to blow up if I don't set m equals zero so these two are gone you see the boundary condition plus and minus infinity takes care of two unknowns I am still left with four a b c and d what do I do about those this is what I do about those that next condition would be that between any pair of regions any pair of regions you will have to use the continuity equation a continuity relationship that the wave function slightly to the left of the boundary and slightly to the right that one must be the same and so should be the derivative let's see whether we can use these two conditions to relate the four unknowns can we do that so let's first look at the top blue regions by the blue two blue equations and we'll talk about the red in a second look at x equals zero if the solution or the wave function has to be continuous then you can see that x will be equal to zero so the left right just immediately to the left of zero I have c because x equals zero what about under slightly to the right of x which term will be zero well you can realize that a has to be zero right because that for x equals zero that term is automatically zero a sin x and then b cosine x is this correct that's right and similarly what about the derivative conditions if you take the derivative first and then apply at x equals zero you can see you will pick up a alpha alpha c e to the power alpha x and you set x equals zero there and similarly you take the derivative on the right hand side and that gives you the equation the second blue equations from the top well I don't want to go through the next two equations but you can easily see how it works for example the first rate equation on the top this is evaluated at x equals a and at x equals a the solution of the equation wave function is a sin k a plus b cosine k a right these two terms and that must be equal to d e to the power alpha a because you're evaluating at a and similarly the derivative boundary condition okay this is no rocket science you can easily see match the wave function match the derivative be a little careful so that you put all the terms plus and minuses correctly that's it so how many unknowns I have I have four unknowns and I have four equations so I'm in a pretty good shape I can solve this equation right so we will solve this equation of course you can solve it in many ways but if you remember matrix algebra generally it's easy to solve it in this particular form so for example the c equals b you can see how I have written it in the matrix form in the matrix form I have one and minus one you can see the vector a b cd on the right hand side and if you cross multiply you will see that I have written the first equation on the top of the matrix is b minus c equals 0 that's what the first equation and you can read the other four equations from from that easily so you can see the term alpha and k how they come about with multiplied with a and c and that's that's it this is just a matrix form of the same equation now how do you solve it this equation when you have four unknowns and four equations you set the determinant of this thing to zero so determinant of the previous matrix equal to zero and that immediately gives you the solution of the equation in terms of tan of some quantity and on the right hand side you have another function now I have done two things oh by the way so let me before I continue you can see that only unknown in this equation only unknown is e because you can see the xi is e divided by u0 but u0 you know right remember the potential going up and down so the value of u0 is given to you already so in this equation do you know a you know a right the width of the potential you know the value of a so in fact in this equation you know everything except e so by solving this equation then you should be able to solve the for the value of energy e at which electron is allowed to sit now I did two things here which I didn't really explain and presumably you remember from your high school or not in university undergraduate days that why the determinant of a matrix is zero and why if when you set it in this particular form this gives you the solution I'll post a small note at the end of the lecture or with the lectures in the website that will explain this a little bit more in case you have forgotten but this is essentially an algorithm of how to solve it I'm not really explaining why how the mechanical or mathematical aspect of it now how do I solve this equation this looks pretty bad and how do I do it well there are two ways one is a modern way use call a MATLAB function and you will do it in your homework put that function in and it will send you back the energy e for which the solution exists pretty simple but many times in older days people use this graphical method and that's very interesting so let me explain how the graphical method work because we will be using it later on assume I want to solve this equation x square equals x plus 6 well you could solve it in 5 seconds by taking the root of that of the quadratic equation right you know how to solve this off however there's a very simple graphical way of solving it assume the left hand side I call it y1 y1 equals x square and the right hand side I call it y2 let me plot now y1 and y2 as a function of x so the blue you can see I have shown here the blue going up as x square y2 is x plus 5 so you can see that at x equals 0 the y intercept is 5 now the equation these two pieces have to be equal to each other left hand equals right hand side and therefore the point they intersect is the solution so it's a simple graphical way of quickly finding out where the solution is right yes certainly yes yes I should exactly right so the x the top equation should be x plus 5 now if you know that one you can understand that process you can easily see how the solution of this complicated equation would look like so I have on the left hand side tan of some quantity that's a function of energy e through xi I have just a function of energy e so I could plot y1 and I that is what I have plotted in red the right hand side I have plotted in blue and you can see this one when there is a particular value of xi do you see what value at which the whole thing will go to infinity plus and minus infinity and from the top side you can see when xi is equal to half in that case slightly from the left it will become minus infinity and slightly to the right it will become plus infinity because when 2 xi minus 1 on the blue one that when it becomes equal 0 so that's why this function has this strange form regardless you can see the solution is when the red line crosses the blue line that's the solution and that one I will rotate it a little bit because remember the solution I'm looking for is the energy e right that's the solution so my solution is really in the x-axis so if I just rotate this 90 degrees then I can easily read off that energy e and that's where the electron sits in the quantum work at energy e it will sit there why it cannot sit below or neither can it sit above right so that is where the energy will sit and you can easily calculate it graphically you can see that as soon as the potential is given and the value of a is given you can draw the blue and the red proportions and that immediately gives you the value of the boundary okay it's not nothing nothing complicated here so I know how to calculate the energy level where electron set now most of the time that should be sufficient but sometimes I would also want to know the wave function itself for most transport problem you will see that we will not need this additional information but if you needed it one thing you realize that this a b c and d if I somehow know that magnitude then I would know the value of psi completely but given that this matrix multiplied by the vector is equal to 0 of course these are not linearly independent and so there is no way I can get four values out of this what I can however do is get the value of b c and d any three in terms of the remaining one how would I do that well think about this sub matrix a minus 1 0 alpha and cosine kx let's take it down and remember that's the sub matrix that multiply b c and d right and anything that is left left out I take it to the right hand side do you see let's talk about the second the second equation for example 0 alpha 0 now in 0 alpha 0 when I write out the equation you will see the first term was k a on the left hand side do you see the second line k and a and you will see that k multiplied by a on the bottom equation I've taken to the right hand side with a minus sign you see that and similarly the other ones you can easily do the other ones now you don't have to take one sub matrix you could take any other sub matrices it will be fine I took here b c and d but you can see that after I solve it which is invert take invert this matrix then I will get b c and d in terms of a and also of course there is this alpha and k but remember the alpha and k depends on you and energy e but I just solved energy in the previous side right so in fact I know alpha I know k I know the whole thing except for a I know b c and d values exactly right now how do I get the value of a well that brings me to the last slide on this topic remember that I have said that you have to find the electron somewhere you know either is sitting in the middle or to the right or to the left but between 0 plus and minus infinity it will sit somewhere and the probability of finding an electron is psi squared and so what I have done is between integrated between infinity to minus infinity and set it equal to 1 now you can see I have four unknowns of course c a b and d four unknown that's good but remember I have just solved b c and d using a in terms of a so once I'm done with all this put this all in what is the one the unknown is a is the one the unknown and if it is the one the unknown I can set it equal to 1 and I'm done and as soon as I know a I know BC and D and I have done solve the whole problem you see so this is the process five steps this is the process if you remember you cannot go wrong you'll solve all the problems exactly in the same way now sometimes there are remember this as the electron I said is somewhere in between it's not too deep in the well not too far out but sometimes the electron is if it's really in the core like one s atoms right one s electrons very close to the core in that case a further simplification is possible so for example for the infinite quantum well you can again solve for the middle region as a sin kx plus b cosine kx that's fine but the left and right side remember where we had e to the power alpha x and e to the power minus alpha x the potential is so big that e to the power alpha x that will decay so fast that essentially I can think the wave function is zero on the left and zero on the right right it's so big that e to the power alpha is very big and so it decays so fast that I might as well consider the wave function going to zero so that simplifies my life because I can say that psi at x equals zero is zero because I cannot find an electron to the left of the well so so high potential is so high and similarly at x equals a which is the other side of the potential I can also say well electron cannot possibly go to the neighbor in between the potential and once I solve these two equations I can find the energy levels for this bound states and the energy levels is given by h squared n squared pi squared 2m naught a squared I hope you will check it out this is a simple solution check this out I will not go through this but this one is interesting do you remember how did the energy level in hydrogen go as a function of n you remember that when I said emm equals constant bracket of 1 divided by n square remember minus 1 divided by m square remember it goes as 1 over n squared as you are going higher and higher in energy they get closer and closer together look at this one so that was a three-dimensional hydrogen atom look at this one this one is a one-dimensional square potential and here the separations are going more and more because it is going as n squared so the first level is at 0.1 ev second one will be at 0.4 ev next n equals to the next one will be at 0.9 ev because n is 3 n square is 9 so 9 times the first level so you can see the difference between the two so these five steps that I talked about if you follow them you cannot possibly go wrong but if you don't if you try to take shortcut then what will happen that you will solve some problems fine but often you might get stuck if I give you a slightly complicated problem then you might get stuck follow the rules you will be fine and this problem you might think that this is not a very recent problem after all all this have been discussed about a hundred years ago actually not this is a paper I took from science it's a very prestigious journal from 2006 here people are talking about if you have silicon and then you put the silver on top of it then you can use the same formulation I talked about a few minutes ago to calculate where the electrons sit now I have not explained where the potential diagram comes from by the end of the course you will be able to solve all this problem but for now what I want you to focus on is this look at the rate potential you can see there is a region equal to a square potential sort of and also a triangular region you can see the various levels at 1 2 3 4 and 5 and you can see the energy levels are not getting separated as n squared because this is not a rectangular quantum well the triangular region is widening and so it doesn't follow the previous rule of course but this is a problem one can solve analytically it's a very easy problem to solve and presumably I'm not sure I may have assumed may have assigned a homework on this so you can take a look but another problem which anybody who are using this computer today I mean all of your computer essentially is suffers or not suffers is based on this principle on the on the right-hand figure I told you about silicon dioxide remember that amorphous region and silicon the well-ordered region and I had three regions remember silicon silicon dioxide and polycrystalline do you remember from before and in that case also what we'll see that between silicon and silicon dioxide to the right-hand side you have a potential well shown here almost like a parabolic structure and again you have this bound levels and you will solve it exactly the same way I told you about that you will get this potential and then follow the rules and you will get this this bound levels and in fact this is very important that tells you how many electrons there are available for conduction and as a result you can calculate whether your Pentium is going to have a certain speed or not right so these problems are not theoretical problems very practical day-to-day problems okay so we'll talk briefly about bound versus tunneling states so this is been the last I have solved two cases already right when the electron is inside this potential going up and down when it's very on the in the below in the bottom case intermediate case what about when the electron is far up in that case how will I solve the problem well I can solve the problem in a particular way you remember that in this case e is always greater than the potential you right always and moreover this is so far up that I do not even see the potential variation too much it is like when you are in the aeroplane and trying to look down on the building whether the building is going down by 500 feet up and down when you are in 35,000 feet or 45,000 feet you don't really care you can assume that the land is flat and then solve for the problem so here I can I'll assume you to be essentially equal to 0 because I'm so far up and if I assume you to be 0 for a particular value of energy e I can easily solve this problem at a sine kx plus cosine kx or the other form which is a little bit easier to do e to the power i kx or minus e to the power e to the power minus i kx you know this these two solutions now the positive going wave if you are just thinking about electron moving to the right I can easily assign it to the e to the power i kx remember that when you solve for the wave equations in 6 or 4 in the electromagnetics the plane wave moving the positive direction is e to the power i kx and correspondingly anything that is going to the left hand side is e to the power minus i kx okay so I have these two electrons and I'm essentially done with my solution that's it this my solution but now the problem is how do I get the value of a plus and a minus and whether there are additional issues associated with it so I will get the value of the Schrodinger equation in a similar way and if I wanted to calculate momentum you can see that if I insert a e to the power i kx use this relationship psi psi star how do I get psi star so if I have a function e to the power i kx what will be the psi star e to the power minus i kx is it simply the complex conjugate if I insert it in here you can see that the momentum will show that if I e to the power plus i kx moving to the right if it's e to the power minus i kx moving to the left from this expression now let's come down from the lofty height to somewhere a little closer to the ground still above the potential but not way up so I can see the variation in this case of course I can see the variation so in these three regions that I have remember that anytime I have n region I'll have to solve individual functions in this three regions in this n region so I have these three regions so I write the solution in in this three respective three regions but the distinction is that now all three regions are above the potential so therefore my solutions are all e to the power i kx and e to the power minus i kx form right so I have those three solutions now let's say the electron is coming from the left electron is coming from the left what will happen that in the first interface at x equals zero a fraction of them will get reflected a fraction of them will get transmitted just like light when you have light coming from one side on let's say incident on glass a fraction of gets reflected and fraction get transmitted similarly on the next surface similarly a fraction will get reflected a fraction will get transmitted but what about the right-hand side once it gets transmitted right then it's not going to come back it's going to keep going because there's nothing reflecting it back right to the right-hand side if it is coming from the left if it is never comes back then what should be the value of n because it is not coming back so the value of n has to be zero because it is supposed to be the component which comes back you see have you start ahead you started from the right-hand side then in that case the value of c should have been zero because once it gets starts going to the left it is not going to come back right so depending on the situation you will take one of them out and correspondingly you will match the boundary conditions right at x equals 0 and x equals a but now you have a problem because if you match the boundary conditions at each interface you have two interfaces you will get four equations how many unknowns do you have here you have five unknowns here remember in the previous case where I have psi equals 0 at infinity and psi equals 0 at minus infinity took care of two of the boundary conditions here I have been able to take out just one and therefore I am left I cannot solve this problem exactly right I cannot solve the have a eigenvalue problem and so there is no bound levels no eigenvalues because I cannot set the determinant equal to 0 with five unknowns and what I can do the maximum I can do is to express everything in terms of one unknown you know just I keep one and then I express everything else in terms of that unknown and that is the maximum I can do and again this is a problem and this you'll do in homework on it you know I all this seems a little abstract unless you sit down and solve a particular problem you will do problems in math lab and also in nano hub so this all will become clear at that point but the point is that again this is a huge problem in terms of understanding how semiconductor devices work and we'll see that later on these days the laptop gets very hot right you put it in the lab and it's really you have to even in Walmart you can buy the schoolers to so that doesn't get as hot the reason it gets so hot is because there is a lot of current tunneling out from every individual transistor so it is as if it is leaking every transistor is leaking a little that leaking process is essentially exactly this tunneling process so if you have this potential but on the right let's say on the right hand side there's nothing reflecting the electrons back so if you have certain number of electrons on one side in the silicon side that electron can good get out exactly using the formulation I just talked about and as a result you can have tunneling current that's what makes the laptop hot and so the things that you are learning here in terms of being able to solve that problem electron coming from one side going away and not coming back it's a very important practical problem and that's something we will look at now these again by the third part of the course we'll cover all this and how to calculate the magnitude but the homework in between ask you to solve the tunneling coefficient and the tunneling current for a structure like this and you will do that in so I will conclude here what we have discussed is the solution of Schrodinger equation now these are analytical solution and as he mentioned that in this one the way we have solved this particular problem it assumes that the potential is piecewise constant so that in a given region I can write the solution either in terms of sine x cosine x or eta in terms of e to the power alpha x plus e to the power minus alpha x if the energy is below and in this piecewise region constant region then I stitch things up stitch the solutions up at every boundary and then use the boundary condition plus infinity and minus infinity to pull them everything together remember psi square equals one that gives me helps me to also pin down the wave function so eigenvalues where the electron state and the wave functions all those come out now if it is more complicated it's varying up and down all the time then of course you cannot use this analytical method computers are much better than us in solving that type of problem I will show you in the beginning of next class that are continuation of this lecture is how to solve the problem numerically and you will also solve the problem numerically in your homework but given those two numerical and analytical you are in good shape any potential you are given any potential you can always calculate where the electrons will sit where that they are moving at a certain velocity or not and as a result how they contribute to the conduction process you see so that's the goal I we are going for this is all intermediate steps and what we will do in the next class or and the class after is to solve the real problem which is you see here I took one atom and try to see where the electrons sit of course it's not one atom these are series of atoms 10 to the power let's say eight atoms all in a line I'll have to solve that problem and that is something we'll see probably one or two class from now now what I mentioned so far is that most of the solutions we do are all analytical solutions because it's easy to do in the class and this is how historically things have been done because in old days I'm talking about 1940s they didn't have a big computer like you do so in that case doing analytical work is the only right way of approaching the problem but these days of course we have computers and therefore we can solve many more complicated problems and I want to show you how to solve the Schrodinger equation not analytically remember the five steps we had about putting boundary conditions and matching is not not that but numerically how to solve it on your computer so that you can do solve complicated problems quickly consider I have a complicated band the ux is not piecewise constant becoming a rectangle up and down after all that was an idealization now the question is how would you solve the Schrodinger equation in this particular case you can see that this is could be complicated in principle analytically at least because you can see if you look at the k values sometimes the energy level e is above u so k is greater than 0 sometimes it is below 0 so the k is that there's a decaying solution and in generally complicated mess how do I solve for this thing this is how to do it and you will do it in the second homework yourself as well assume first that you have a ux the potential ux somebody has given it to you it can come from the atoms themselves or the molecules that you are thinking about and in some way this reflects all the other electrons that you have except the electron that you are thinking about all the other electrons the protons and everybody the core net potential is given in this blue curve ux let's divide the space of this in x space let's say one dimension into n plus one points so let's say the whole dimension distance is 20 angstrom and let's say I divide it in 20 pieces so each piece is about angstrom I can do that and you can immediately see that therefore I can specify that at that point at that node what the potential is let's say it's 3.1 eb and the next one let's say 2.8 eb and so on so forth I can specify that now let me assume that although I do not know the solution of this equation but whatever the solution is I just sketch it out in the rate curve and whatever it is of course I can write that value to be in the third node as psi 3 or psi 1 psi 2 psi 3 these are still unknowns to me but of course in principle I could say that I know the values at this point if I do then you can see this is very easy because it's a continuous function the rate curve it's very easy to relate psi 3 to two neighbors psi 2 and psi 4 how by simply Taylor series expansion and this is how that let's say x0 is the third node point x0 plus a is the fourth node point and I could easily expand the solution because it's a continuous curve in terms of the fourth node point in terms of the third node point and you know the Taylor series expansion the first derivative multiplied by a the second derivative multiplied by a square over 2 a is the lattice spacing or I have sometimes used p so it's the same thing I could relate the third one to the solution of the second one I do not know them values but I could always write it now if I do a sum you see it's something very nice will happen because when I do the sum then you can see that the sum on the left hand side is obvious and but the first derivative because they have opposite sign they will disappear right but the second derivative which has the same sign they will come down very nicely which is on the right hand side a square in the second derivative of psi at x at the node 3 so if I want to evaluate the second derivative of a function at a point I'm just taking the two neighbors and essentially constructing the second derivative from there okay now in general therefore for any node I if I wanted to evaluate the second derivative I could just use the information from the two neighbors divide by the lattice spacing square and I'm done so now you see this I could write I've written the same Schrodinger equation except there is this term t naught to capture the h square over 2 m naught a square that's the initial part remember these are flay electron mass we're not talking about effective mass here and the second derivative well I could write it in that form just I just derived and so when I insert that second derivative in the first equation you can easily see for the ith node it will connect the solution from the two neighbors and it will have the corresponding form now remember how many points I have I had the whole thing greeted up in n plus 1 points and so therefore in principle I should have n equations everyone connecting the two left person the left and person on the right and so you connect the whole thing up and then you have the solution and so this is just one so you will have a series of equation now this is of course if you try to do it on the pencil and paper this would be horrible but the computers are not as smart or do not get as bored so they can solve the problem easily I have the equations n unknowns of course the psi values are the unknowns but remember I have to put the boundary condition right on minus infinity and plus infinity now of course in computer there is no minus infinity and plus infinity you should take some value so on the left and right and then move it a little bit to see your original solution does not change if it does not change as far as you are concerned that is plus infinity that is minus infinity and I will set this to equal to 0 and once I set them to equal to 0 I can write this equation so that is the equation for any node that is the equation for the first node but fortunately I already know that sin not is 0 so that is gone that is good so I just need to know about the neighbor on the one side and similarly on the other side that was gone and the whole equation set of equations I can write it as a matrix n by n matrix that is the h they call it Hamiltonian and then you can solve this eigenvalue problem and when you solve the eigenvalue problem you have all the energies right now how many eigenvalue duo will I have a quick thing equal to the number of node points because you know it's an n by n matrix so I should have n eigenvalues which is good a lot that's one and second is remember how this is connected with what you did before remember setting the determinant to 0 right what I did over there is took the e on the other side so in our case it was h minus e multiplied by psi equals 0 and we said the h minus e the determinant of that equal to 0 the setting the determinant of that equal to 0 is exactly the same as finding the eigenvalue of a problem so therefore these two procedures are exactly equivalent I'm just following that five steps remember just exactly the five steps in a numerical form here so I have let's say 20 levels I'm very happy and I bring it to my professor by solving matlab or you know you have the nano half tools and all those very impressed but it turns out the professor says that you know how to program but you don't know how to the physics and this is why remember the equation of a solution of a Schrodinger equation the first level it was like a half sinusoidal curve right in the second one it was two humps one going up another going down so each of the solutions actually had a certain number of nodes the first one zero node the first second one one zero crossing and so on so forth now think about the 20th one the 20th one will have 19 zero crossing let's say going up and down let's say now of course that will have a lot of energy it's really going up and down very fast and that's way up there now when the energy is way up there on the top remember my boundary condition my wave function was supposed to go to zero at minus infinity and plus infinity think about the wave function on the top that you have just calculated of course for that one that one it is not going to zero in computer it is going to zero because you put by hand a hard boundary where you said wave function has to be zero on that first node so what has happened in that process in the computer what you just did you didn't not only solve this potential but on x equals zero you put a infinite potential up forcing the wave function to be zero therefore all these spurious solutions above you came about because you put an artificial boundary condition what will happen that if you change the boundary a little bit the ones in the bottom will not change right because they for them that left boundary was really infinity but on the top because the square potential as soon as you move it a little bit all those will keep changing those are not physical boundary solutions so you have to get rid of them all so being able to solve take a big supercomputer and solving a big problem doesn't mean that you always get the right result you have to think about the physics first and only then you are going to get the right results okay so let me end with this this was the extra material for I guess lecture four and we will continue then on the energy band and other issues in the next lecture thank you