 Let me write down what is the problem. So again to do this, there are certain rules. So these rules were derived by John C. Slater and they are usually known as Slater rules, Slater rules for matrix elements. In fact, these rules should be required even when I do C i because remember when I do C i, this I have to calculate matrix element of the Hamiltonian in terms of the d i d j which are like determinants. So I must know how to write the matrix element of the Hamiltonian between two determinants. So that is the essential problem in terms of the spin orbitals which form the determinants. So first rule A is the expectation value. In the matrix element, one important point is expectation value which means both sides are identical. Left and the right is identical. Again I am assuming that they are all orthonormal set. So all the rules are derived using orthonormal set. So psi tilde is chi 1 tilde, chi 2 tilde, chi n tilde while chi i tilde, chi j tilde equal to delta i. So that is the problem. So the rule is that if I can expand psi tilde in terms of chi, I can expand psi tilde in terms of chi on both left and the right. I can write h as sum over h of i plus 1 by sum over 1 by r i j and then do lots of integrations. Now it is very complicated because you have n factorial, you have one term here but this one term is actually determinant. So there are n factorial terms, chi 1, chi 2, chi n, then chi 1, 1, chi 2, 2 is interchange, etc., etc. n factorial term with a 1 by square root n factorial on each side. So 1 by n factorial. So if you see, it may look horrendous but actual analysis will tell that this is very easy and this is what John Slater did. What I would do is to simply write down the results. So let me write the result first. First let me write down h again because the symbols must be correct, sum over h of i plus 1 by r i j. So this goes over n electrons, this also goes over n electrons. h of i is a kinetic plus electron, whatever external potential, electron nuclear attraction or any other external potential. This is my total Hamiltonian. Remember total Hamiltonian has only a sum of one electron term and sum of two electron terms. There is no three electron, four electron term because this is a Coulomb interaction. So that is the hallmark of Coulomb interactions. I have two parts to this Hamiltonian, this and this. Let me call this a 0.1 within a. So a 0.1 is just the psi tilde sum over h of i psi tilde. So the first part I am doing. Remember this is sum over h of i. So there are n terms here. So how many terms are there which have to integrate? There are n terms here, n factorial here, n factorial here and of course everything will be divided by 1 by n factorial. So it looks like a horrendous number of terms. But remember when I am doing this, let us say I do two electron. So I have got chi 1 tilde 1, chi 2 tilde 2. Let us assume that everything is real. It does not matter. We can write star. So one of the terms, then chi 1 tilde 2, chi 2 tilde 1. That becomes your determinant and then you have a sum over h of i and again the same term chi 1 tilde 1, chi 2 tilde 2, minus chi 1 tilde 2, chi 2 tilde 1. So this is what it is. So please do this. You should try to do this at least two electron problem manually because there are not that many terms. You can expand each of the terms. So what is the first term? Let me expand. So chi 1 tilde 1, chi 2 tilde 2, let me write in Dirac notation. Actually if I write in Dirac notation, I do not need to write the stars. I can knock this off. Chi 2 tilde 2, h of 1, chi 1 tilde 1, chi 2 tilde 2. So that is the first term. How many terms are there? 2 into 2, 4 into 2, 8 because h 1 plus h 2. There are 8 terms actually. I am writing only one term right now. There has to be h of 2 here. I have to do the integration. Integration is over d tau 1, d tau 2. Now you can see it is a trivial integration because h of 1 can only connect one pair of spin orbital. The rest can be directly integrated. So these two can be directly integrated and that gives you 1. So what is the result of this? The result of this is just chi 1 tilde h chi 1 tilde. Is it clear? Just a simple one electron. Now it does not matter if there are n electrons here. n spin orbitals in here, they would all have given you 1. 1 into 1 into 1. Let us take another term where this is h 2. This is h 2. So let us take the same term where this is h 2. You should be able to do these integrations. Now in this case, you will get this as chi 2 tilde h chi 2 tilde. Instead of chi 1 tilde, because chi 1 tilde, chi 1 tilde will become 1. So you have a chi 2 tilde h. Now this was a coordinate 2. This was coordinate 1 which survived but that does not matter. I am going to integrate. So it is a dummy variable at the end of the day. The integral result will be identical. So this is a number which will be over chi 1 tilde. This is a number which will be over chi 2 tilde for each one of them. Now if I interchange, it is very easy to see. If I interchange only on one side, chi 2 tilde 1, chi 1 tilde 1, what will happen? This will become 0 because now h 1 will connect chi 1 here, chi 2 here. 2 which is not connected will be chi 2 here and chi 1 here and they will become orthogonal. In fact, all such interchanges will become 0. So the only thing that will survive is whatever you have written here from these four terms must be the same as here. So one more will survive. When the left is this, right is also this. So that will survive of course and h 1, h 2 will be there. For both of them, four terms will be there. So you will actually divide by 2. 1 by 2 is already there. So you will actually get only two terms. Your final result will be just this. The final result is chi 1 h chi 1 plus chi 2 h chi 2. That is all. You have already got one chi 1, sorry chi 1 tilde h chi 1. You have already got this two already got this, but with a factor 1 by 2. Do not forget. You will do it once more. You will get the same thing. Negative and negative sign will give you positive. You will exactly get this twice. Cancel with 1 by 2. You will get them twice. You will get them just once. So this will be the final result actually. So for two electron, you can see the rest of the terms are all vanishing because of the orthonormality. The orthonormality is a very, very important part. This condition is very, very important. Otherwise, of course, the rule should be very, very complicated. See if I do this, I can now write the general rule. So I just showed this for a two electron problem. So now I can write a general n electron problem. So a point 1. So the general problem is chi tilde sum over h of i, chi tilde is equal to sum over i chi i tilde h chi i tilde. It is very easy to generalize. i equal to 1 to n. Just as you have got this for two electron. So it is chi i, h chi i, i equal to 1, i equal to 2. So I write exactly in the same manner. For n electron, you will get n matrix elements. Now you can see the entire problem at least for the first one when this is just this has become very, very simplified. It just has sum over n 1 electron matrix which is very easy to calculate and now it is directly in terms of chi. Now this is of course not surprising because this was non-interacting and in non-interacting, we know that the energy is just sum of the one electron. So that is what you have actually got. So it is more interesting to look at what will be the second one, a point 2 which is chi tilde 1 by r i j chi tilde. You can actually try doing this for two electron. In fact, I will give you this as an exercise to try this for two electron. I will write down the final result, but please check that you get this for a two electron problem in the same way that I have done. This is half of sum over i j chi i tilde chi j tilde 1 by r 1 2 chi i tilde chi j tilde. These tildes are just cosmetic. Do not worry about it because I am using trial functions or I am using tilde. It is not important. Minus chi i tilde chi j tilde 1 by r 1 2 chi i j tilde chi i tilde. That is an exchange term. Note here, this exchange term remains because this operator is 1 by r 1 2. So within coordinate 1 and coordinate 2, I can interchange they will not become 0 because the orthonormality does not work here. I cannot integrate these two separately, but all others get integrated either to 1 or to 0. So if the others are there, they will not interchange. So they will become all 1. If they interchange, they will become 0 and this number comes out correctly because of again this 1 by n factorial. That you can check. You have n c 2 term here. You have n factorial here, n factorial here. You can check this for two electron problem 1 by r 1 2. There is only one term here. So here, this is a very important result which I am going to use and the result says there are two terms. This term is called the Coulomb term where one of the electron is in spin orbital chi i. Another electron is in the spin orbital chi j. Remember again, I am always using the Dirac notation. I think I have first said that 1 2 1 2 notation. So it is star 1 star 2 1 2. So I have a density of chi i spin orbital interacting with the density of the chi j spin orbital through 1 by r 1 2. That is a classical Coulomb interaction of two particles. And of course, if it is n particle problem, so all the others are coming through this sum over i j. The dummy variables remain 1 and 2, but the spin orbitals keep changing because spin orbitals are important. This one I cannot interpret in the same way, but this is quite obviously it is a Coulomb exchange interaction which is basically interchanging the left and the right side. So this two electron part boils down to Coulomb and the exchange part. Note that this sum over i and j spans entire 1 to n, i equal to 1 to n, j equal to 1 to n. It is quite easy to see, however, that whenever i is equal to j, note that this Coulomb term and the exchange term become identical because i is equal to j. So the interchange does not produce anything. So for all i equal to j, each of the terms is not 0, but the negative, the entire thing is 0. So actually in this summation, so let me refine the a 0.2. In this summation, I need not write the i equal to j part. So let me write this. So a 0.2 then becomes sum over i not equal to j half chi i chi j 1 by r 1 to chi i chi j minus chi i chi j 1 by r 1 to chi j chi i. You can further see by symmetry, if I interchange i j that is i to j j to i, each of the terms remains the same because it all means just coordinates are integrated in change. This is a dummy variable. So both of them would be identical which means within i not equal to j, i less than j and i greater than j are same. So the entire thing can now be written as just one of them i less than j chi i chi j no not equal i less than j and the same term. And now because I am taking only i less than j, I do not need to take the factor half because I have divided this as i less than j plus i greater than j. Both of them are identical. So without if you do not write the factor half, just write this and this will give you the physics that I am now the two electrons are interacting and pairs. i less than j is basically two pairs of spin orbitals which are different. Of course they cannot be same because of Pauli principle. So they have to be different and those two pairs of spin orbitals only interact. So if I have a chi 1 and chi 2 interaction is only once between chi 1 and chi 2. This was giving actually a somewhat superfluous result that I am interacting chi 1 and chi 2 twice. Even if i equal to j is ruled out, when i equal to 1, j equal to 2, i equal to 2, j equal to 1 each time I am counting and totally unnecessary because the factor is half they are identical. So that now this gives you the chemistry that this is a pair wise interaction coulomb and exchange. So pair wise is actually defined. So these notations of symbols are very important. So when you write these please remember your factor will depend on how you write. All i all j, i less than j both are identical i equal to j is a 0. So you do not have to worry. So coulomb and exchange. So this is what is called that the self interaction cancels. In fact this is later will be used in DFT and several other theories. This is self interaction that i and j they do not interact with each other. The coulomb and the exchange term cancels due to this anti-symmetry. So i will leave you here. I have to of course come back to the then i am now going to put together a 1 and a 2. So i will get the full expectation value and then of course i will also look at the matrix elements where these and these are different which will be required later. So probably i will not do it now when it comes down i will come back to those rules. So they can be different also they really become matrix elements. What i am looking at here is a special class of matrix elements which are called expectation value. Expectation value is essentially both left and right are identical. So that is what i am looking at right now. And that is sufficient for the Hartree form. So my final answer is a 1 plus a 2. I have given you a 1 and a 2. I am going to put them together. These are the coordinates of the electron is 1 and 2. This is the coordinate of the electron. Chi i and Chi j are spin orbitals. The sum is not over the coordinates. I again repeat. The sum is over the spin orbitals. The coordinates are always 1 and 2. You can choose 2 and 3. You can choose 1 and 4. It does not matter because these are dummy variables. So you may wonder where are the n electrons coming? Why only 2 electrons? n electrons are coming because of this. Either you sum over electrons or you sum over spin orbitals identical. So they are indeed 1 and 2. So I want to make sure. That is a good question. So when I am writing this as Rij, this is meaning because this is a total Hamiltonian. So I have 1 by R12, 1 by R23, 1 by R13. Everything is there. But eventually when I come to matrix element, you can write everything in terms of a specific 2 electrons but sum over all spin orbitals. In fact, this is very important that you said this question because many people get confused with the indices and this sum is actually making sure that you are taking over all pairs of coordinates. That is why I said there is a factor 1 by a n. All that will cancel out just like it happened for. For example, if you look at my 1 electron problem, there are also I did not write this is 1, this is 2. You understand? Both are 1. Both are 2. It does not matter what it is. 1 can be 1 and n can be 3. That electron coordinate does not matter at the end of it because that is what is being integrated. Please understand. All integrations are definite integral. I again repeat. So all of you know in calculus, if I have definite integral, the coordinate of integration just vanishes. So it does not matter whether what is important is the level of the spin orbitals. They are different chi 1 tilde and chi 2 tilde. Is it clear? So what I am saying is the same question you could have asked here. This is not 1 and this is not 2. I mean this can be 2. I have no problem but both can have coordinate 1. Fortunately, this is 2 electrons. I wrote all the things. Here also I can write similarly for 2 electrons. You do this as your own practice so that you are able to derive this for 2 electrons where this 2 electron part is just 1 by r1 and chi tilde is just chi 1 tilde, chi 2 tilde. So it is very simple. You write down like this, expand, put this factor and derive.