 In our discussion of variation method, we have taken a few examples. Now we go back and see how we can nicely represent functions as linear combinations of other functions and how this gives us a better result for energy in case of our good old particle in a box. So, this is something we have covered already. We have said that it is possible to express trial functions as linear combinations of other functions. And in this case, what we said is that we are going to work with fixed functions and we are going to use the coefficients as the variational parameters, which means that we are not suppose we are using a lot of Gaussian functions, we are not going to change the full width half maximum or position of the Gaussian functions. For each Gaussian functions, those parameters will be fixed. What the only thing that will change is the coefficient Cn. So, this is our variational parameter. And what we said is that by changing this, we will try and get the minimum value of the functional epsilon 0. And that is going to provide an upper bound to the ground state energy. And we have demonstrated this with a simple example, where there are only two terms in the summation. So, phi equal to C1 f1 plus C2 f2. And also, right now, we have kept only real coefficients. But actually, the mathematics is not very vastly different if the coefficients are not real. Imaginary coefficients are a reality. I mean, it is a strange way of saying things perhaps, but that is actually correct. So, you do have imaginary coefficients, does not matter. You always end up taking their complex conjugate when you talk about contribution. So, what we have shown is that for this system, we have been able to construct the secular equations for the two variational parameters, the two coefficients. And we have written these equations in terms of h11, s11, h12, s12 and h22. So, these are essentially integrals. What integrals are these? I actually should have written here. Sorry for having forgotten, but I hope you have not forgotten. Hii, let me write in general term is equal to integral fi h fi. So, hij would simply mean, okay, I have to close this, this. And sij, I think ij is a more general way to write because I can always be equal to j. That is simply integral fi star fj over all space. So, we have formulated these secular equations in terms of these integrals. And see these integrals are numbers. And the good thing about working with these integrals is this. Let me underline this word here. These functions f in terms of which we have expressed our trial function, these are arbitrary, fine. Of course, they are, but they are known. Arbitrary in the sense that they may not be the correct functions. I mean, there is no particular reason like we had used a particular function, right, e to the power i phi. That was the eigenfunction of this Pz operator. So, when we tried to construct the phi dependent part of the wave function of hydrogen atom, we had taken a e to the power i phi, im phi. And that made perfect sense. Here, it may, well, it has to make some sense, but it may not make perfect sense to use some particular function f, right, that way they are arbitrary. But they are known. So, if I use a Gaussian function, I know where the maximum is, I know where the, what the full width of maximum is. If it is an exponential function, I know what is the constant associated with that exponential decay, so on and so forth. So, I know the functions. Also, I know Hamiltonian. So, if the function is an eigenfunction of the Hamiltonian, I can hope to actually work out the value of these integrals. And if I can do that, well, this is C1, a variable multiplied by a number plus C2, a variable multiplied by a number. So, it is just plain good old two linear, two set of two simultaneous linear equations, that is all. It is just that it might look a little scary for somebody who is new because we are writing enigmatic things like H11 minus ES11 and so on and so forth. Nothing to worry about that. Now, see when you take secular equations, you could do two things. First is you can try to find the coefficients. After all, coefficients are the parameters here, right? But do not forget, one thing we do not know yet is E. So, what we do is, right now we do not worry about coefficients that might be the story for another day. If you go through our lectures in symmetry, for example, there is this NPTEL course that we taught in symmetry. There we have actually found out these coefficients as well for pi molecular systems. But here we do not bother about coefficients. Our goal right now, the only goal is to determine the ground state energy. So, E is our target. And to target E, what we do is, we write it in matrix form like this. And if you study the theory of linear equations, for this to be, for this C1, C2 to be non-zero, that would give a trivial solution, this secular determinant of this matrix, that must be equal to 0. That is very well known from mathematics that existed, that has existed for a long, long time. So, this secular determinant equal to 0. That is what you have to solve. And now see, H11 I can figure out and we will for particle in a box. S11 we can, H12, S12, H12 everything. The only unknown now in this determinant is E. So, I will expand the determinant, equate it to 0. I will get in this case, I will most likely get a quadratic equation. And I will solve it to get the answers. So, this is what we do. We will get two roots of E, if it is a quadratic equation, they will be in terms of the integrals. If I can find values for the integrals, which we will for particle in a box, then we will get two roots as particular numbers or defined quantities. Then we can evaluate the integrals. As I said, out of these two roots, the one which has a lower energy corresponds to the ground state energy. That is all. And this is the general form. It is not necessary that small n will be restricted to 1 and 2. It can go all the way up to any number capital N. So, this is what the function will be at that time, capital if phi equal to c1, f1 plus c2, f2 so on and so forth plus c capital N, fI capital N. In that case, we will just get a bigger perhaps more intimidating secular determinant. But this is a secular determinant of nth order. So, if we expand it, of course, it is going to be tedious once again. But if we can do it, if we want to do it, we can. So, if we can expand it, we are going to get n roots for E. And once again, the lowest energy, I should not say lower energy here, lowest energy should be the ground state energy. Now, with that background, let us see how we can use this kind of wave function, linear combination of functions in particle in a box and does that improve the situation at all. You might remember in one of the previous modules, we had actually tried to do a variational calculation of ground state energy of particle in a box. And there we obtained an overestimation by 4% and that time we said, okay, 4% is good, but not good enough. So, with that background, can we improve upon that 4% if we use this kind of a linear combination of functions as a, as our trial function. To do that, what we do is, we use f1 equal to x into 1 minus x, f2 equal to x square into 1 minus x whole square. The choice of these wave functions is inspired by the discussion we had the last time we talked about trial functions for particle in a box. Remember, we have said that the functions have to be symmetric with respect to L by 2. And here, we have said L to be equal to 1, which means whatever we write is in terms of L. So, there is an L square in the denominator for energy that has been said to 1. If you want the absolute value, you just bring that L square back in the appropriate position, okay. So, we need a function, we need functions that are symmetric with respect to the midpoint and also they must vanish at the boundaries. x into 1 minus x as well as x square into 1 minus x whole square both satisfy these conditions. I request you to recall what we had obtained in the earlier treatment. Remember, which function, which trial function we had used earlier? We had used phi equal to x to the power alpha into 1 minus x square, I think something like this, okay. And that alpha had come out to be, I think, 0.862 or something like that, okay. Let us remember that result. Now, we are trying to improve upon that 4% overestimation that we had by using a linear sum of two functions, both of which are sort of inspired by the function that we had used in the previous go. And this here is our Hamiltonian, you might remember that in the last time we had used Hamiltonian, we had written Hamiltonian in terms of atomic units. Well, here we have not using atomic units, I am doing it intentionally because we should actually be equally conversant, equally comfortable with all kinds of units, okay. So, this is our Hamiltonian. Now, the next thing to do is to see whether we can make this secular determinant any simpler. And that can be done if we can work out the expressions for all these integrals or you can say the matrix elements. Remember, we talked about it, these are all matrix elements. Let us try to do that. What I will do is, I will show you the steps for one. Well, the easiest one, H11. And then I will leave it to you to work out the rest if you want, you do not have to work out all, but work out at least one or two, one of the Hs and one of the SS, that will be good, okay. So, Hamiltonian is minus h cross square by 2m d2 dx2 minus h cross square by 2m comes out. I am integrating now between 0 and 1 because I have set L2 equal to 1. So, what do I have inside there? Phi multiplied by d2 dx2 phi. So, this is the first term that I will get, okay. This will be H11, okay. Because this is F1. Remember, H11 essentially means integral of F1 star H F1 overall all space, all x in this case, okay. So, x into 1 minus x into d2 dx2 of x into 1 minus x simple. So, this x into 1 minus x essentially is x minus x square. So, first of all, we will find the first derivative of x into 1. It is very simple. Perhaps many of you would have worked it out by the time I do it. So, essentially I am trying to get d dx of x minus x square and without having to know rocket science, I think we can see that this is going to be 1 minus 2x. So, the second derivative is such that we have to find the derivative, we have to differentiate the this thing that we got the first derivative 1 minus 2x, okay. And we have got minus h cross square by 2m outside the integral. What is d dx of 1 minus 2x? I will not even bother writing. It is simply minus 2. So, that minus sign comes out and the 2 cancels with the 2 in the denominator. So, I am left with h cross square by m multiplied by integral from 0 to 1 x minus x square. Is that right? Yes. Again, this is also very simple. I will not even go through the steps. I will just show you the answer. This is the answer. I am showing this step only so that you can check in case you go wrong somewhere. The answer that we get is h cross by 6m, okay. Beautiful. Let me raise all that bad handwriting. So, h 11 as promised has been worked out and it turns out to be h cross square divided by 6m. So, it is up to you to prove that h 21 equal to h 12 is equal to h cross by 30m, h 22 is h cross by 105m, s 11 equal to 1 by 30, s 12 equal to s 21 remember, s ij equal to s ji because it is just integral of a direct product. And h 12 equal to h 21 by using turnover rule, remember h is a Hermitian function, Hermitian operator, okay. This is s 22. What do I do next? I take all these expressions and I plug them into this secular determinant. While doing that, it helps if I work in terms of e cross e dash which is em by h cross square and you can see why h 11 what is h 11? h cross by 6m minus e into s 11. So, minus e by 30. So, naturally this m will go up and so the expression just becomes a little simpler if I write e dash equal to em by h cross square. And this is the determinant I get that is equated to 0. I am going fast intentionally because there is no point in me reading out every step or what you do is you just multiply diagonally then subtract the product of the other diagonal in the terms of the other diagonal and you get this quadratic equation 3 e dash square minus 168 e dash plus 756 is equal to 0. We all know how to solve quadratic equations, minus b plus minus root over b square minus 4ac by 2a use that and we get e dash equal to 168 plus minus root over 19152 by 6 which translates as 51.065 and 4.93487. Do not forget we have said l to be equal to 1. So, remembering that when we go back to emin the minimum value of epsilon 0 turns out to be 0.125002 h cross by m. Now that might not mean much to you but it will when I show you the exact result in the same kind of expression. So, the exact result that we had got for your particle in a box written in the same way for l equal to 1 turns out to be 0.125000 h cross by m. See this number is 0.125002 and this is 0.125000. So, if you are wondering why we are writing so many decimal points does it make sense? Yes it does. So, remember earlier when we use that phi equal to x to the power alpha 1 minus x square kind of function the agreement that time seemed to be good now it seems to be horrible. We had 4 percent over estimation and let us say only 4 percent over estimation. Now look at this you stop at the third decimal place or fourth decimal place or fifth decimal place the match is perfect only in the sixth decimal place the exact value is 0 and the calculated value is 2. So, this is a good match in fact it can get better. So, what we see is that if we use more terms in the trial function then we can get a better match to energy and that opens up a tremendous possibilities. First of all you do not care well you do a little bit but you do not care about being absolutely right on what kind of functions to use. See we are talking about wave functions here we know that wave functions have to satisfy bond interpretation right bond conditions boundary conditions. So, they have to become 0 eventually they cannot go to infinity anywhere they cannot be 0 in all places they have to be continuous. So, we are only going to use that kind of wave functions that kind of functions f we will choose that kind of f only but the good thing is as long as that happens you do not really have to care whether you should use a Gaussian or a log normal or an exponential which function you use whatever function gives you a good match. That is the beauty of our variation method that the nitty-gritty is not even required you do not need to know the wave function all that well. Now think what we had said at the beginning we said the wave function had all the information about a system you make the Hamiltonian operate on it that is a question you ask the wave function gives you the answer here we are going beyond that we do not even know the wave function we are making up some kind of a wave function and we are just changing it until we get close to the right answer. Here we are doing test cases exact solutions are known so that is what gives us the confidence that yes it works we are getting close to the value that we had got earlier using the exact solution. Next we are going to talk about systems for which you cannot find exact solution right and that is where the promise will be fulfilled. So it is a I do not know if you can call it enigmatic I say this is a an amazing situation right in Schrodinger equation the only thing you know is Hamiltonian you do not know the wave function you are making it up you are changing it you are doing all sort of things you do not know energy by doing all sort of things to the wave function the method takes you close to the value of the actual energy and in the process if you work out the coefficients as well for that value of energy you will also have a particular form of wave function okay and that is something that you can plot and you can see what it looks like and using it you can draw these maximum probability contours and what not even though you cannot solve the equation. So you might remember that this this cliche people pull the legs of quantum chemist saying that these people have only one equation and they do not even know how to solve it now let me say that that is that sounds very tongue-in-cheek smart thing to say but it is absolutely unfounded right because here we have a situation where we do not have to solve directly we can get the solutions by using beautiful approximation methods and in doing that we actually get important parameters that we are going to use later also this h i j h ji s i j s ji everything is going to come back and everything is going to be useful in many different applications of quantum chemistry quantum mechanics right all in good time but for now the take home message is that if you increase the number of terms in trial function you get a closer match to energy. So before we can close this discussion and go over to actual multi electron atom system we need to discuss two more things one of them I have sort of summarized in this slide first of all see the functions that we had used earlier remember were arbitrary even though they were known now I can choose a set of functions that are orthonormal aren't they can't I what happens when I choose an orthonormal function especially when I am faced with this daunting task of working out this kind of a humongous determinant does it help? Well as you can see what do we work with we work with these right and I have not written s similarly I can write s i i and s i j right. So here if I use these first of all if I use eigen functions of Hamiltonian and if I use if I choose such eigen functions that form an orthonormal set basis means this is what I am working with this is the set of functions on which I am doing all my calculations based on which this basis set. So if I choose an orthonormal basis where all these f functions are orthogonal to each other and are they themselves normalized then what happens then I know very well what h f i will be I am choosing eigen functions right I will write the others as well. So let us work with this h i j let us say let us say Hamiltonian operating on f i gives me what will I write epsilon i this epsilon that epsilon are not the same set different epsilon i f i. So this turns out to be so Hamiltonian operating on f j will give me epsilon j f j good old eigen value equations. So what will h i j be this will be integral f i star I know what h f j is it is epsilon j f j and epsilon j is a constant so it can come out f j remains inside the integral and this is what I get now see if I have chosen an orthonormal basis do not you know that this is going to be delta i j this integral is going to be equal to 1 if i equal to j and equal to 0 if i is not equal to j so what I am saying is that many of these h i j integrals well all h i j integrals where i is not equal to j they will become 0 and h i i integrals will boil down to just epsilon i so this entire determinant is going to become so simple they are only going to have this some epsilon term and you are going to have zeros right everything will be written in terms of those so that is what we are going to study in little more detail in the next module. So let us conclude today by saying that if we use orthonormal basis we actually get simpler expressions and the secular determinant becomes a little easier to handle there is another advantage but let us talk about that after we have discussed what happens when we after we discuss how orthonormal basis simplifies the problem further for us because remember the whole idea here is that we do not want to tedious a calculation we want it to be as simple as it can get next day we will see how beautifully this problem plays out if we use an orthonormal basis another advantage let me say this at least think of an orthonormal basis that we know think of say orbitals hydrogen atom that is an orthonormal basis suppose I write a multi electron system wave function as a linear combination of hydrogen atom orbitals then what happens we know exactly how to go about this yeah and we can actually hope to think we already know so much about orbitals the moment we bring in those orbitals then we know exactly how to proceed and we sort of know what to expect eventually right that is a very important paradigm in talking about not only multi electron atoms but also molecules. So remember the reason why we are allowed to use this linear combination is that variation method allows us to add as many terms as you want to the trial wave function and vary no matter what we do we can never do better than the best.