 So, we stopped with writing down what you wrote the Lagrangian in the last class. So, the Lagrangian was in terms of the spin orbital. So, these are the one particle operator of the Hamiltonian. So, we take all these integral of this one particle operator for each of these spin orbitals and sum them. So, I is 1 to n. Again, note that given an energy expression, we also told you how to get a spin integrated form, but to derive Hartree form, we are first going into the spin orbital which is very general. So, the spin integration we can again do later. So, we are going back to the spin orbital plus half of ij, chi i chi j, anti-symmetrized chi i chi j, half chi i chi j, anti-symmetrized chi i chi j minus ij lambda ij chi i chi j minus delta ij. So, that is your Lagrangian. So, we have n square set of equations, sorry, just for correctness, I should put tilde everywhere. We have n square set of equations chi i tilde chi j tilde is delta ij and we have the Lagrange multiplier for each of them. So, Lagrange multiplier idea is that for every condition, you put one multiplier. So, if I have only one condition like I told f of x I want to optimize subject to g of x is 0, then your Lagrange L becomes simply f of x minus lambda times g of x because there is only one condition. Here actually it is not one condition because i and j are n, total i and j are n square in number. So, for each ij, I have one lambda ij. So, that is why it is a matrix. So, this is actually a matrix. I can write it as a matrix. It is n square in number. So, I have to now differentiate or I have to make this optimum or stationary with respect to chi i tilde as well as lambda ij. But of course, the derivative with respect to lambda ij is trivial because we all know that we will get back this equation and that is the reason the Lagrange multiplier is put like this. So, when I differentiate, so automatically if I differentiate del L, del lambda, any one kL let us say, the result will simply be chi k tilde chi j tilde minus delta kL equal to 0. So, that automatically gives me the orthonormality equation. So, let us say that I differentiate. I hope all of you are comfortable in this algebra. I am differentiating because I want to make this stationary and in particular of course, I want to make it minimum because of variation principle. So, the first condition is that make the first derivative equal to 0. So, if I do the first derivative with respect to a specific index of kL, this kL is a specific index. This was ij of the dummy index. So, I am going to do for all kL however, but one at a time, one set of indices at a time. So, let us say I take lambda kL. Then if you differentiate this, this part will not contribute because there is no lambda here. This is a partial derivative. So, only this part will contribute and I hope you know how to write this because this is a dummy index ij. So, only when i becomes equal to k and j becomes equal to l, this derivative will survive. Otherwise, it is again all the others are partial derivatives and in that case your i will become k, j will become l. So, you get this. So, for all kL, I can write down this equation. So, essentially I am not getting any new information. I am just ensuring that my conditions of orthonormality are satisfied. So, really this equation is of no consequence to us because that is something that I know in my mind that they will be orthonormal. What is important is of course the next set of equation which are basically the first derivative. So, then the next ones are any particular chi i, let us say chi k equal to 0 chi k at in day for all k. So, that will be my set of equations and we will actually like to see what we get out of this. So, this is the important equation. So, chi i chi k at in day whatever chi k at in day means. Now, before we start doing this, let me write the first derivative in a somewhat different manner which is easy. So, I hope all of you know this interpretation. Let us say I have f of x and I want to calculate df dx. So, this is my job of course in this case I do not write df because it is only one variable. So, how do you write this? There are many ways of looking at first derivative. Of course, if you have given analytic function you can directly differentiate. In this case such is not the case. So, one way to look at it is a following that as you change x to x plus delta x, some infinitesimal quantity, then f of x changes. So, the change of f of x to the first order I will define what is first order. First order is essentially the order of delta to the first order is divided by delta x is equal to dy. Change in first order. Now, this is very I think this is just a question for those who are mathematically well oriented they would anyway appreciate but for those who are not I will again repeat this. So, I make a change of the variable from the variable which I am optimizing to an infinitesimal delta x. Because of this there will be a change in f of x in general f of x will change to all orders of delta not just for the first order. I am only going to take the first order change and then divide by delta x I will get divided by delta x. Actually you can quite easily see let us say f of x in x square I am doing very elementary algebra now let us say f of x equal to x square. So, you know of course dy dx that is 2x I am going to find it out. So, how do I find it? I make x to x plus delta x. So, what is the total f of x? Nu f of x. So, nu f of x plus delta x is x plus delta x whole square you all agree. So, that is the new function. However, this function has first order x square this is of course no order 2x delta x plus delta x square. So, what is the change in this function? The function was originally x square the change is 2x delta x plus delta x. Out of this this is second order. So, I will not consider this. So, my first order change is only 2x delta x. Now divide by delta x you get 2x that is something that you already know divide x. For every derivative you can always find the derivative in this manner. I hope if you have done good calculus this is what they teach you and that is the reason we write y plus delta y. So, let me also relate that the derivative formula y of x plus delta x minus y of x divided by delta x limit delta x. Now, can somebody tell me how these and these are identical? Note that when I am writing this I am not taking it first order. I am taking it all. I am saying if you take first order and divide by delta x. Yes, this part ensures that all the second order onwards terms are gone because I am going to differentiate I am going to divide by delta x. So, the first order will not go. First order all of you know how to take the limit. It will be a 0 by 0 problem but actually divide by delta x. So, what remains is 2x. Second order onwards when I divide by delta x 1 order of delta x will remain all that will become 0. So, this is why I am trying to tell you. This is one way of looking at the derivative. This is what we have taught but delta x tends to 0 limit essentially means that we do not need not consider the change of f of x plus delta x from second order onwards. Because second order onwards I divide by only one order of delta x. So, one order will at least remain and they will all become 0. So, this is a different way of putting. This is how in the calculus in this book. So, I just wanted to show that this first order that I am taking is there is no error. I am taking only first order because eventually here I am not applying this delta x tends to 0 instead. Here I am taking a delta x and just taking up the first order. So, if you apply delta x tends to 0 then of course you do not have to care. So, you take all orders yx plus delta x automatically it will actually. So, both of these are identical. That is one thing that I wanted to show you and I exactly show you giving an example let us say x square that y this is identical that f of this delta x square will eventually get cancelled because it is of course not even there in my this thing. If you take it here then it would have become delta x and that would become 0. So, either way you get the same result. So, I am going to take this particular way of calculating the derivative. Now, since my derivative is going to become 0 essentially I do not need to bother divide by delta x. I hope you understand. I can only make the memory because I am eventually going to make