 is that many times this basis, A mu or A nu whatever is a basis of, is a set of real functions and this really helps. Although I want atomic orbitals, it will be useful to use real functions and this is where for example, the Cartesian p or d functions are really helpful. So you do not use for p z, you only use p z, p x, p y and do not use exponential i phi, exponential minus then it becomes complex. I will tell you why because then if it is real function, this star is itself and that introduces several simplifications. So one of them is the following. So for example, let me take one type of function, let us say mu nu lambda sigma, just does not matter what is the index. I am not writing 1 by r 1 2, but 1 by r 1 2 is there. So mu nu lambda sigma, let us say all are real. So this is 1, this is 2, this is 1, this is 2 in terms of dummy variables. Since all are real, I can do lots of interesting things. I can interchange between mu and lambda, I can interchange between nu and sigma. The result will be equal. So I can start writing very quickly certain symmetries and I hope you will follow me very quickly. Let me for again sake of, remember when I am writing mu nu lambda sigma, it does not matter. For the sake of brevity, I am not writing 1 by r 1 2, but assume that 1 by r 1 2 is there. So mu nu lambda sigma very clearly would become lambda nu mu sigma. Mu sigma lambda nu, please remember this is the lexical order from where I started, mu nu lambda sigma and this can be interchanged depending on how what these are, does not matter. So that is equal to lambda nu mu sigma because they are real, remember. And then I can interchange both this as well as this. So it will become lambda sigma mu nu, lambda sigma mu nu. Then I can do another interesting thing because of the fact that 1 and 2 are dummy indices, I can independently interchange 1 and 2 on both sides and that could be done even if they are complex, remember. These are only for real. But even if it is complex, I could write this as nu mu sigma lambda because that will only change dummy variable 1 and 2. I hope all of you are understanding. Once I can write this, I can write further prescription for the real functions using the same set. That means interchange now this sigma mu nu lambda, interchange this. So nu lambda sigma mu interchange both. So sigma lambda nu sigma nu mu, sorry sigma lambda nu mu. This is when mu nu lambda sigma, each of them is distinct that is mu not equal to nu not equal to lambda not equal to sigma. Then you have this what is called the 8 fold symmetry. 8 fold symmetry I have understand what you mean by 8 fold symmetry. Of course if mu is equal to nu, some of the symmetries will vanish because there is no point in interchanging mu and nu. For example this and this is identical in that way. So it will become 4 fold symmetry. Number of symmetry will reduce depending on the uniqueness of the indices. But as long as the 4 indices are unique different you have a maximum of 8 fold symmetry and that is the largest number that you have. So eventually your m to the power 4 is not m to the power 4. It reduces. Like if everything had 8 fold symmetry then it would become m to the power 4 by 8. But that is not true because some of them have 4 fold, some of them have 2 fold. If they are equal then there is only one number. There is no further reduction. So now can you tell me what is the actual number of integrals? Assuming the real function. So I add to your home assignment for the next Monday this one. Calculate the total. Don't worry about zeroes. Assuming there is you don't know about this per city that I am not asking. Assuming every integral is you have to do. But the symmetry is what you have to factor. If I have mu nu lambda sigma I will not like to have lambda nu mu sigma again because they are equal. I like to have only one of that number. So based on only the symmetry not the special symmetry based on the integral symmetry. So what is the number of total integrals? So the assignment is the following. So that is again for the home assignment. If this mu basis is a set of real functions calculate the number of unique integrals. So I hope you understand what is unique integrals. That means something should not be calculated twice if they are equal. Unique integrals of course when I say integrals it is two electron integrals. So let me clarify and unique two electron integrals in an M basis without taking into account any special symmetry. So you can assume that the molecule is such that it has no other symmetry. It is A1. Many of you have done group theory. It is A1 symmetry so no further symmetry is there. Point group is A1. So no further symmetry should be assumed. Of course if there is a further symmetry it will reduce even more. So what is the total number? Obviously from my discussion you can make out I can give you the hint that it should be something to do with M to the power 4 by 8 but it is not. As clearly I told you because all numbers are different because if all are uniquely different then those numbers have to be divided by 8 but that will not be M to the power 4 by 8. So what is the actual number? I hope the problem is clear. Based only on these two electron symmetry no further symmetry should be assumed. That is what I mean. So calculate the total number of unique integrals and that will really help. Then all it depends is of course putting together this code. Basically multiplying this with these and so on and so forth. That is very easy. After that we will go to the iterative part and then of course how to solve the equation. FC equal to SC. How to solve the Routhan equation that I wrote. That is something that we will do. So let me now write the Routhan equation once more. F nu mu C mu i. I hope all of you can write it. Sum over mu epsilon i sum over mu S nu mu C mu i. If you note this I have to write for every nu and every i because in this equation nu and i are specific integrals this is. Mu is a dummy index. One of them is a dummy index. So given a molecular orbital i given a specific nu atomic orbital I have written down this equation. And I noted also if this was a Kronecker delta this would have a structure of eigenvalue equation. So what is the structure of this equation? I already have written to you a matrix SC. I will now introduce another nu matrix called the epsilon matrix or I can still use epsilon or maybe some eigenvalue matrix which is simply a diagonal matrix epsilon just as I do for epsilon m 0 0. Remember although I am interested only in n electron problem when I solve this equation this is a m by m problem. So I will actually get m orbital energy just as I get in eigenvalue equation. So it is actually a m dimensional matrix. So I construct a nu matrix E matrix do not confuse this with identity matrix. It is a diagonal matrix containing these eigenvalues or these values epsilon 1 epsilon 2 and the rest is 0. So it is a diagonal matrix then this entire matrix can be written for all nu i can be written in a matrix form as SCE. So it is a matrix multiplication. Note the same thing we do for eigenvalue equation. I can repeat this if it was delta mu nu the eigenvalue equation note that the same equation if S nu mu were delta mu nu what would have been written? This would have been F nu mu C mu i for a specific number would have been epsilon i C nu i again for every nu and i. How would you write this equation eigenvalue equation? You write exactly the same manner F CE equal to CE. I think in many parts we have discussed this although when I write it here the orbital energy can be written anywhere because this is a number. When I write in terms of matrix these matrix must come right of C matrix otherwise you will never get this equation. Because this matrix does not necessarily commute to the C matrix although it is a diagonal many people have a confusion the diagonal matrix always commutes to the another square matrix that is not true. A diagonal matrix commutes to the square matrix provided the diagonal values are all equal. So a degenerate diagonal matrix only commutes to the square matrix so you have to be very careful and normally you write always F CE equal to CE AC equal to CE whatever, whatever matrix this is the form of an eigenvalue equation. So precisely if you understand this you should be able to write this. The same equation you should be able to write except that the S which was an identity matrix has just come in the left of it and this is the format in which you have to write. Again E goes in the bottom and again S and C do not commute so you have to be very careful in writing this. Again at some point of time you may come back and ask you to derive this from here to here. How can you write it? These are all good assignment problem but please note this that we can write this F CE equal to CE in a matrix form and in matrix form when I write I do not have to write these for all new y. Remember here I have to write for all new y that is now all contained because when I multiply row times column all multiplications are contained. Here I am specifically talking about a new eth row, ith column and so on that is all contained here except that it is not an eigenvalue equation. I know how to solve this, I do not know how to solve this because of this guy S. That is another issue that we will come to that but before we come to that we will let me recognize that just like if S were delta I could write F CE equal to CE I should be able to write this as F CE equal to SE and if you know how to do this you should be able to do this but if you do not know we will come back please practice at home but can you write this in this form, write a simple 2 by 2 matrix, expand this you will see that you will get these equations. So you have actually four equations here new equal to 1 to 2 let us say everything is 2 take a 2 by 2 so M is capital M is 2 so you will get everything. Very clearly when I have a structure of M now the fact that I was a molecular orbital and I was interested only up to N by 2 molecular orbital that information is gone because when I solve this equation I am going to get everything now capital M the entire dimension would be capital M so whatever I do eventually I will get capital M sets of C me y capital M sets of C me y M square C me y M by M so I will actually get capital M molecular orbital is it clear eventually when I solve I will get capital M molecular orbitals because everything is now capital M whereas I required only N by 2 which is fine because all the rest of them will become unoccupied orbitals and I have already given interpretation for unoccupied orbitals more they come better they are and they will have a great use later although for Hartree-Fock I would only require N by 2 of them the result of the solution will give me capital M. Capital M molecular orbital space which means 2 times capital M spin orbitals and if I now do full CI then I will have 2 MCN determinants right when I have started with this atomic orbitals A they are not normalized once you normalize of course I am going to do that that is the next thing I will do how to solve but you have to remember after normalizing you have to get back the chemist picture otherwise the normalized orbitals are do not belong to any atoms so we will do that that is exactly what we will do but then we have to know how to get back the chemistry otherwise the chemist will not be interested so that is why that is why we started with an orthonormal basis otherwise a non-orthonormal basis otherwise I could have started with orthonormal basis okay in fact if you do methane calculation with the basis of only carbon atoms this is very fascinating a physicist would love that then the entire atomic orbitals are orthonormal and a physicist would argue that if I take a complete set of only carbon atoms I should be able to describe methane molecular orbital to chemist it will be atrocious that you are constructing a methyl molecular orbital CH4 without even having any hydrogen atomic orbitals but tell me what is wrong there is nothing wrong as far as mathematics is concerned I could use only carbon atomic orbitals and construct methane molecular orbitals and they would be orthonormal except that number capital M would have been very very large to get a meaningful information you require you would have required a much larger basis because eventually have to approximate so the approximation works well when I mix now this is where I always say that the chemist have intuition okay so chemist have intuition they will do it but you know their intuition has brought in some problems again but mathematicians can solve it so you have to help take help of mathematics people to solve your problem messiness that you have created because of intuition and there will be solution to this but yeah you are right I mean that lot of physicists say that you know take only one atom and expand the entire molecule in but you see you lose the entire picture and there is no lcao there is no understanding of the contribution of atoms in a molecular orbitals everything that the chemist love to do would vanish you will only get a energy but the interpretation is very very important remember so our interpretation is what we would like love to do so that interpretation would not be there very easy unless you do that okay so I hope you please try this at home I am not giving this as a formal home assignment the number of integrals I have already discussed so please do this once you do then you are in a good business then we can start solving this equation so two there are two parts how do you solve it and of course whichever way you solve it the iterative this remains because your f matrix depends on the coefficients don't forget this so we have to do an iterative solution of the Haltry-Fogh-Ruthan equation and then we have actually solved the molecular problem completely almost completely except that I still have to worry about what are these basis functions you know right now I have just told you these are the basis functions which have written mu nu lambda please remember also that in in short hand notation when I am writing mu lambda nu sigma they are actually a nu a lambda I hope that is very clear that this these these are basically a nu a lambda you know and so on 1 by r1 to a mu a sigma in a in a long notation this integral okay but but I am using just nu lambda mu c because that's really unimportant now because they are in the basis yes how do you have non-iterative because you can't have because that's a physics yeah that's a physics the the Haltry we have we have an iterative solution because your original Haltry-Fogh equation of the fog operator came has the field from the other orbitals if it was not then of course it would not be variational optimum so that is the first thing that I learned that to have a variational minimum energy I require the field of one electron operator of an on an electron must derive from the field of the other orbitals other other electrons so that's the first thing I have learned now that cannot vanish okay these are all then versions of Haltry-Fogh for atoms numerical for molecules expansion of bases okay but the basic spirit will not go so somewhere that iterative thing will remain in fact if you do that vanish then there is something wrong physics how did it vanish so the dependence on the other orbitals are coming now through the coefficients okay yeah alright