 If you look at this function, this can be written in a form what Slater which I actually introduced in the last class, Slater determinant and I had also mentioned that in the 425, but again because of the fact that there is a little bit of heterogeneity, so I am doing it all over again and this is important for my course. So I write these two functions this minus this in the following form chi 1 of 1, chi 1 of 2, I will explain chi 2 of 1, chi 2 of 2 and take a determinant of this. So if you take a determinant of this, you can see that the diagonal is chi 1 1 into chi 2 2 and then you take minus chi 1 2 chi 2 1. So this function automatically becomes a determinant, if I write it in this form the way trick to write this determinant is the following that once I know the two spin orbitals, one row should be characterized by one of the spin orbitals. So let us say chi 1, the second row should then be characterized by chi 2, a column may be characterized by the coordinate of electron. So in this column only coordinate 1 comes, in this column only coordinate 2 comes. Again when I say coordinate, they are the coordinates of the spin orbitals, so they are basically the space spin coordinates R and omega, both are there. So when I interchange, I am interchanging both. So this is the way to write determinant and this way I can actually generalize very easily for n particle problem. Again this is not anti-symmetrized, this is sorry this is not normalized but we can be trivially normalized. If each of the spin orbitals is normalized then I can simply multiply this by 1 by square root 2 factorial and this whole thing will be normalized. So this will then become a normalized by function psi of 1 2. So it is very easy to normalize it, so that is something that we will not worry right now. Assume that the spin orbitals are orthonormal and we can actually normalize it very easily. So this is the form of a slater determinant and in this form lot of things are now obvious. For example chi1 could never be equal to chi2, you know something that I analyzed and found that does not have it becomes 0 because moment I do this it is of course two rows are identical so the determinant would become 0. So that was the reason the wave function would have become 0. Because of the fact that it is determinant the interchange of any two coordinates will always become negative sign. So it is the anti-symmetry is automatically guaranteed and then we find that to guarantee the anti-symmetry the two rows cannot be equal, so two spin orbitals cannot be equal. So that is the statement of Pauli principle. So the determinant automatically guarantees lot of things, so you do not have to analyze the wave function and keep writing what would generate a anti-symmetric combination right. Here for two particles I could do that but for n particle it will be very very difficult because when I go for n particle I require that the wave function must be anti-symmetric with respect to interchange of any pair of coordinates any pair any two particle. So that will be very very difficult by a manual analysis to write down the combinations. So what we would like to do is to use this idea of the determinant for n particle. So now from two particle I am trying to generalize this to n particle. So for n particle how many spin orbitals will be required? n now from the discussion it is very clear like two particle at least n are required otherwise it will become 0 and determinant also shows that any two rows become equal they are 0. So minimum n particles are required. So let us assume that for n particle problem note that we are still talking of non-interacting problems where it is exact for interacting of course this will not be exact because the theorem will no longer be true the product for only non-interacting case the determinant is an exact remember. So we are still discussing non-interacting we will come to the interacting part in the next class what happens for the interacting part and that is where approximations will come. So let us now say you have n electron non-interacting then I am still talking of an exact function now and I require n spin orbitals chi 1, chi 2, chi 3 etc which are generated from the eigenvalue equation of H. Remember I have an eigenvalue equation for the one particle operator if you note I have written down H of 1 chi i of 1 epsilon i chi i of 1 there are many, many eigenfunctions like every one particle operator has many eigenfunctions not one one of them is ground state then excited states and so on. So I can generate n number of eigenfunctions, n number of spin orbitals and then I can construct my wave function for a non-interacting n particle problem or n electron problem as an anti-symmetric product. So let us say I take chi 1 to chi n then I can write down this determinant again this is a normalization constant you do not have to bother too much but just like that it is 1 by square root n factorial and just like that we can generalize it first row is chi 1 everywhere and the coordinates will keep changing till chi 1 of n, second row will be chi 2 1, chi 2 of 2 so on till chi 2 of n and then like that you can go up to chi n of 1, chi n of n I hope all of you can write this very easily now. So this has the hallmark that each row is characterized by one spin orbital each column is characterized by coordinates note again while I said this you can trivially interchange the rows and the columns. So for example you can say that one column will have chi 1 one row will have one coordinate it does not matter many text books may write it like that it does not matter if follow one particular convention okay so we will follow this convention so one of the row has one particular spin orbital then automatically all products will come. How many products you have like for the two particle problem we had only two terms this minus this for this determinant how many product will have how many set up n product will have like one of these chi 1 1, chi 2 2 etc to chi n n how many n that you have n factorial terms in fact that is the reason this 1 by square root n factorial comes to normalize the whole thing. So every determinant if I expand the determinant I will have n factorial terms each of them will be a product of n functions okay so first term will be a diagonal product chi 1 1, chi 2 2, chi 3 as if 1 is in chi 1 2 is in chi 2 3 is in chi 3 and the rest of them will keep interchanging because electrons are indistinguishable so that fact is recognized by the various other terms of the slight determinant okay. So this becomes now exact solution for the non-interacting problem however but for the interacting problem I cannot still get an exact solution but this will then become now become a basis for the interacting problem to understand the interacting problem and that is where we will start in the next class.