 I had given this particular problem that E0 in terms of f mu nu and h core mu nu or h core mu nu. So many people got confused, I just want to kind of discuss for a minute. So basically all you have to write and this is it was written explicitly that it is for closure, so that was also not a problem. So either you can write in spin orbital and do a integration or you can directly write in terms of space orbital which is also given to you which is basically 2 times sum over h a a equal to 1 to n by 2 which I had called h i i does not matter, i i is nothing but phi i h phi i okay, phi a h phi a plus a b then you had 2 j a b right minus k a b, so this all goes to n by 2. So each of them are in space orbital, what you are expected to do is to now expand this in terms of atomic orbitals and then prove what I had given, there are several expressions that one can give but that is an expression that is very useful because at the end of the Hartree fog, the reason it is useful, at the end of the Hartree fog we have f, we have h core mu nu and of course you have p right, so these are already available. So we do not want to write another expression for energy, any other expressions you want to write the energy of expression in terms of these quantities which are already available just multiply. Remember many of these things that have been done are also done with an understanding that the computation would eventually become simplified. I can write in various form but I have to make the computation simplified, so already certain things are done, so we already have p mu nu okay, you already have f mu nu and of course we have h core nu, no question about it okay, at the point where the SCF has converged, this of course has nothing to do with convergence, this is one time calculation, these two are converged quantities, using these three quantities you directly write the energy, that was the essential idea of that paper, of that particular question and you can start from here and do all your exercise and do a little manipulation to get that, it is not very difficult okay, little bit of manipulation was required. I may also mention before I go forward that the fog operator in spin orbital, so I call it f of let us say x1 is equal to h of r1, those who have difficulty in writing, sum over j or a equal to 1 to n by 2, now complete it, complete this, complete this in terms of spin orbital, I have used a, so please remember, chi A star 2 correct, then 1 by r1 2, 1 minus p1 2, chi A 2, d tau 2, I think some people had even difficulty in writing this, so I want to mention, we had used j, I am using a, it does not matter, see that is one thing that you have to be very comfortable, you should write the indexes, indices and please remember the integral in terms of coordinates and spin orbital, you have to very, very carefully, the coordinates are dummy, the spin orbital is not a dummy, because spin orbital is a, each spin orbital is a different function, so if I integrate different function results are different, so if I have f of x or let us say f1 x dx and f2 x dx, they are not same, remember they are all definite integral, let us say A to B, simple calculus, they are not same, if f1 is different from f2, however f1 of x dx, A to B is equal to f1 of y dy A to B, here the function form is same, coordinate has only changed, some people has confusion even of this, so when you are using the coordinates, the coordinate is not an issue, this spin orbital is an issue, so I have to sum over A, because for each spin orbital, the form of chi A is different, chi 1, chi 2, chi 3, they are all different in terms of spin orbital, this is 1 to N, so for each spin orbital, the form of chi A is different and hence I have to explicitly take care, this is no longer dummy, but whether this spin orbital has electron 3, electron 4, electron 1, electron 10, it does not matter which electron, because that is integrated, one of these integrations is being done and I repeat, there is only one electron which is integrated, you might wonder what about the electron 3, electron 4, it does not matter, because I am summing over this spin orbital and that is what turns out, that when I am summing over spin orbital, it is effectively summing over all electrons, so that is what it turns out, so instead of summing over electrons, I am summing over spin orbitals, electrons are just dummy, so you take one electron, take its average on this spin orbital chi A, another electron is actually in a different spin orbital, that is also being integrated, because there is a sum, it is just that I am calling that into electron also 2, because if I call it 3, it makes no difference, the result would be identical, so I am basically summing over all the electrons, so that is another confusion, people think that okay, there are N electrons, where am I averaging over all the N electrons, I am averaging over all the N electrons, it is just that I am not calling them electron 3, electron 4, electron 5, okay, I can call that, so then you will have a very simple picture, as if electron 1 is first spin orbital, electron 2 is in second spin orbital, electron 3 is in third spin orbital, if I write it separately, so basically if I break this out, for one of them I write 2, another I write 3 and so on, you will actually see that all the integrals, all the electrons are integrated, so that is a very important part to realize and of course self-interaction has not been taken, one with one itself is not taken, because coulomb and exchange cancel and I repeated that part, so self-interaction is actually physically not taken, you might wonder what happened, why there are only, if I say 2, 3, 4, 5, what happened to the other electron, it actually is not there, so but on the other hand when I do spin orbital, I have A equal to 1, that will be taken care when F action is specific spin orbital chi B, it will take care of this spin orbital, that particular whenever chi A is equal to chi B, it will become 0, whenever that, so when I write F of F acting on a chi B to give epsilon B chi B, whenever this summation A will be equal to B, that will give 0, so there is no error being done, these are all some small, small problems that I want to plug holes, so this is a very important way to understand, how am I getting from a 2 particle coulomb operator and an N electron wave function, eventually a one electron operator by integration, and that N electron wave function is also a single determinant, otherwise you cannot get such a simple form, so N electron wave function is single determinant and that essentially leaves you with the Fock operator which in a canonical form is just this, so that again it does not matter whether F of 1, F of 2, F of 3 is an eigenvalue equation of an operator, so it is a one particle equation, so it does not matter whether that part coordinate is 1, 2, 3, again I just repeat, if I write this, this is identical to writing as F of 1 chi I 1 equal to epsilon I chi I 1 or F of 2 chi I 2 chi I 2 does not matter, it is an operator eigenvalue equation, it is a single particle operator, what is that coordinate of the electron does not matter, so many some of you I saw that were more worried about this coordinate of the electron, please understand that is the first part of quantum mechanics that the electrons are indistinguishable, so that really does not matter, it is an operator eigenvalue equation, what matters are the spin orbiters, they are different, so and please do not put epsilon I of 1, epsilon I is a number, the function is only chi I, so you can see that this is a number you do not write epsilon I of 1, epsilon I of 2, it has no meaning, because that is a constant it is a same, we will go forward assuming the core Hartree-Fock is done, there are a lot of issues of course, which you have to discuss before you leave Hartree-Fock and we will spend next 2, 3 classes on discussion on that, so one of the things that I first want to go forward is a technical issue is that note that we discuss this later rooms and we have used it in the Hartree-Fock, what are the later rules that we have used in Hartree-Fock that the Hamiltonian expectation value with respect to a single determinant is correct and this is something that we have done where psi naught is a single determinant chi 1, chi 2, 2 chi n, note again this means it is a determinant I have just written the main diagonal, again I have not written 1, 2, 3, n it is unimportant, I am again repeating, it is a function of n electron however and that is understood, so this is the main diagonal of the determinant that is how we write in a symbol but actually it is a determinant where one of the spin orbitals, each spin orbital is a row or coordinate is a column or vice versa it does not matter and you have a 1 by square root n factorial stuck in for normalization do not forget that, we do not write it in the shorthand abbreviation but when you write in the long form there is a 1 by square root n factorial so do not forget it as a determinant, so this basically is an average value of the Hamiltonian with respect to a particular psi naught, so what does it mean, average value means it is a matrix element of the Hamiltonian but the right and the left both are identical, so in a bigger picture of determinant space this is a diagonal element, you can say this is a diagonal element in a determinant space, if determinants are my basis which is what it is because you remember I discussed full C i where determinants are basis, so in that basis determinant this is only one element, how many determinants are there, if I m spin orbitals m c n that was a part of the first quiz, m c n determinants are there, out of m c n determinants if I write the matrix element of the Hamiltonian then what is the matrix size it will be m c n by m c n, out of which this is only one element but a very very important element, one element which is diagonal, why there m c n because when I do Hartree-Fogg or otherwise if I have a basis of m spin orbitals I can construct m c n determinant, even when I am doing Hartree-Fogg remember that the Eigen value equation gives me many more equations even in a Routhan basis, so I have more spin orbitals than I required, so I can construct more determinants out of the spin orbitals, what we want to do is that the Slater actually did a rules which almost all encompass all types of matrix element, so what are the other types of matrix element, so first type this is a type 1 is when both left and right determinants are identical, so I want to put this now in a more general footing, they are identical determinants which means they are expect their average value with respect to that particular determinant, it so happens that this can be Hartree-Fogg but it need not be okay, in case that psi naught has this m spin orbitals then we know the result which we have done before is again I write sum over i chi i h chi i plus half i less than j chi i chi j anti-symmetrize chi i chi i, so that is the result whichever is the chi i, so I give lot of practice problems for this, for example if I have a 2 electron I can have 1 2 as the determinant, 1 2 is the spin orbital not coordinate, so if you just want to be very sure, let me write this as a chi 1 chi 2, this could be a determinant, if I have let us say 4 spin orbitals then I have other determinants chi 1 chi 3 chi 1 chi 4 right, there will be 6 determinants, so let us take a simple case chi 2 chi 3 chi 2 chi 4 chi 3 chi 4, so if I have 4 spin orbitals 2 electron then 4 c 2 is 6, so these are my 6 determinants, I can pick up any one determinant and calculate the Hamiltonian expectation value, it has nothing to do with Hartree-Fock, forget about Hartree-Fock, given the determinants, given the spin orbitals in the determinant I should be able to write it, so your ground state may be chi 1 chi 2 but if I give you chi 1 chi 3 you should be able to write the determinant chi 1 chi 1 chi 3 h chi 1 chi 3 you should be able to write this in terms of chi 1 chi 3, so your I will now not become 1 to 2 right, I will become 1 and 3 whichever so your 1 to n was just lexically ordered, if I order them because 1 2 3 4 n is just a number symbol for me but basically it means whatever spin orbitals are actually included they have to be summed up, so this will have chi 1 h chi 1 plus chi 3 h chi 3, similarly this will have chi 1 chi 3 and so on and this can be further spin integrated, I have already told you how to do that, you will get 2 times coulomb, 1 times exchange and so on and depending on whether it is a parallel spin, anti-parallel spin exchange will vanish or not vanish, so all those you should be able to do because I am not specified what is chi 1 chi 3, they are spin orbital, they are not space orbital, so for example your chi 1 can be phi 1 alpha and chi 3 can be phi 2 alpha, so then you will have a different form of space integration, you will have exchange surviving, coulomb of course will always be there but exchange will survive because they have a parallel spin. On the other hand, if your chi 1 is phi 1 alpha and chi 2 is phi 2 phi 1 beta, so that is for this determinant, then of course the exchange will not survive, you will only have a coulomb because we have already told you that all pairs of electrons have coulomb interaction only parallel have exchange but this parallel and anti-parallel concept comes only when I do integration of a spin, as far as writing in terms of spin orbitals I do not care, that is a generic expression, that is a big advantage in terms of spin orbitals, this will automatically come in as a result of integration, the fact that the parallel spins have exchange, so this is not a result that I am going to force it, remember this is an outcome of the spin integration, so that is why I always say the Hunz rule is an outcome, Hunz rule is not fundamental, Hunz rule is an outcome of the anti-symmetry, because I have an anti-symmetric determinant, I get this and then I do a spin integration, I get Hunz rule, so I am not going to impose that, I am just telling you the outcome that if you do this integration, you will have parallel spins having exchange, anti-parallel spins without exchange and that is the reason exchange stabilized, so the parallel spins gets stabilized compared to the anti-parallel spins in the same orbitals, so obviously it means that implies the Hunz rule, but that will automatically follow, what we want to do now is to have a slater rule where a type 2 slater rule where of course left and right are not identical, so that is something that you should be able to do, but here also requires practice, but you should be able to do, but if I have a determinant, let us say chi 1 chi 2, just give a same example here, h and another of the 6 determinants, so let us say chi 1 chi 3 or it could be chi 3 chi 4, so several possibilities are there, any 6 can come here, any 6 can come here, then only I will have a entire 6 by 6 matrix in terms of the determinant space, so slater of course found out all the rules, so slater rules essentially can tell you all possibilities, so type 2 is a possibility where one of the determinants either left or right does not matter, determinants differ from the other, differs from the other by one spin orbital, that is the occupation of the spin orbitals in two determinants are identical except one, I hope I clear, like here there is no difference between left and right because they are identical determinants, so I am now looking at a type 2 where one of the determinants either the left or the right does not matter, differ from the other by one occupation, so now tell me is this type 2 as everybody is in sync because you have a chi 1 chi 2, you have a chi 1 chi 3, so this is only one difference, is this type 2 the second one? No, because there are difference of 2, both are different, so this will handle later, so right now I am talking of only type 2 where there is a difference of only one spin orbital, so this is basically the type that we are discussing and very generically let me write down the type in the following manner and I think that is the notation that you should now understand, the same size 0, I am just using the same size 0, H on the left, on the right I am making an interchange of one spin orbital A replaced by one spin orbital R, now this will have a significance in the context of Hartree-Fogg, in the context of Hartree-Fogg whatever are occupied in sinor is occupied orbital, when I have solved the Hartree-Fogg problem, so let us assume that the sinor is Hartree-Fogg, in which case the A is an occupied orbital and R is a virtual orbital or unoccupied orbital, so that is what it means, so what is psi A R? Let me write down psi 0, so psi 0 is chi 1 chi 2 etcetera, chi A minus 1, chi A, chi A plus 1, blah blah blah up to A, so chi A is some number, let us say 4, so I have just written explicitly chi 3, chi 4, chi 5 and the rest I am not writing, around A I am writing deliberately, then what is psi A R? psi A R is exactly the same form except that A is replaced by R and the rest remains same, only one number is changed, one spin orbital is changed, so that is called I R, how can I change this? If this is already my Hartree-Fogg, then these are called occupied orbitals, so this occupied orbital can be changed by what? It cannot be changed by any another occupied orbital, it has to be changed by an unoccupied orbital, if I change by another occupied orbital determinant will become 0, correct, that cannot survive because two spin orbitals is bad analytical, I must change, so in the context of Hartree-Fogg, if psi 0 is Hartree-Fogg, these A and R will have a specific notation, they are occupied, unoccupied, however at this point I am not bothered, I am just saying that the psi A R is a determinant formed by changing one of the spin orbitals chi A in psi 0 by chi R, which was not present, so I would rather say at this point, more generally psi A R is a determinant formed from psi 0, whatever is my psi 0 by changing or replacing this spin orbital chi A by chi R, of course the chi R should not be present in psi 0, that is understood, otherwise it will become 0 and obviously A cannot be equal to R, because if A is equal to R I am not doing any change, that is also clear, so in the context of Hartree-Fogg however which is going to be the context that we are going to discuss, this will be always occupied, this will be always virtual, that is only because of our definition because context of Hartree-Fogg essentially means as long as this is Hartree-Fogg determinant, but the rule that this letter did is of course very general, so I am placing this as a very general thing that whatever is my determinant need not be Hartree-Fogg, need not, it has some orbitals, this has some orbitals, A R essentially means it is exactly same determinant, only A is replaced by R, that is the meaning of this symbol, then this letter gave the rule, as usual there is a 2A and 2B, what is A? A is the one electron part, B is the two electron part, so H also has a sum over H of i and plus 1 by R ij, so remember in this case this was my A, this was my B, so I am just going to write separately or together, it does not matter, the first part is very simple, this will become chi i chi A, sorry H chi R, that is it plus, so the one electron part is absolutely simple, unlike here where it is sum over N numbers, here it is only one number, that is a matrix element of the one electron operator H with the one that is replaced and the one that is that is replaced by, and that it is very easy to see which will come on the left, which will come on the right, because psi 0 contains chi A, this contains chi R, note that when I am writing psi A R, psi A R does not have chi A, psi A R has chi R, it is just that A comes here, but A comes here only to tell that from here I delete this, so A is missing actually, so whatever is coming on the left, on the bottom subscript is actually missing, whatever is coming here is actually present, so it is very clear how to write this, if I interchange this it will be chi A R, it will be chi A, so it is also very easy to see how do I write, so that will be my first and only one term, again I am not proving it, but the proof is very similar, if I do exactly the same multiplications, the expansion on both left and the right hand side, now you will see that every case this chi A and chi R will cause an orthogonality problem, because remember of course very important is that the chi R is orthogonal to chi A, I mean that I must ensure and which is obvious because my spin orbital basis, whatever basis I am replacing they are orthonormal, so they are orthonormal occupied orbital, virtual orbital, so orthogonality is assumed, so because of orthogonality everything will become 0 except when the position of chi A here and the position of chi R here is identical, so the rest will integrate to 1, this part will only come out, but there will be enough number of terms because n factorial here, n factorial here, so for each term n factorial there will be one term like that, maybe one, maybe two, so n factorial such term will be there which again multiplies with 1 by n factorial, square root n factorial here, 1 by square root n factorial here giving you only one term, out of this n factorial term many will be integration over coordinate one, some will be coordinate two, some will but does not matter, I again repeated, I repeat that the coordinate integration is immaterial, they are dummy, so eventually all of them will have same value, there will be n factorial such term which will be multiplied by 1 by n factorial to give you only one term, so rest anytime this side expansion is different from this side expansion it will become 0 because chi A and chi R will have different coordinates which will not integrate with H and that will become 0, now that takes care of the fact that there is, this is actually sum over H of i, so H1, H2 it takes care of that fact that all those factors will now be taken care, so leaving you with only one term where chi A, H chi R, again you can do this for a two electron problem, I would again request you to do a practice problem, expand this, expand this and just see with all 1 by 2 everything, so that you have a and remember then this will have H1 plus H2, there will be two terms coming from the Hamiltonian, please do that practice, let me now just write the two electron part will be good practice, let me now write down the rest of the term, then you have sum over B, I will expand this term chi A, chi B, anti symmetries chi R, chi B, just like here but here each of i and j were summed up, here again one of the spin orbitals on left and the right will be fixed, so there is no summation remember A and R as specific indices, so there cannot be summation, the other one can be anything, however obviously B cannot be A, B cannot be R because if B is A or B is R this integral will become 0 because it is an anti symmetries integral, the coulomb and the exchange will cancel to be 0, not the integral will be 0, so I can write B as 1 to n formally because anyway R is not included and even if B is equal to A it is 0, so formally I can write this expression, I hope it is clear to everybody that if B is equal to A for example, then you will have chi A chi A anti symmetries chi R chi B, so it does not matter chi R chi A whatever it is there, so if A is equal to B then it becomes chi A chi B, B is equal to A and this is obviously 0 because you have chi A chi A chi R chi A minus chi A chi A chi A chi R which can be written again as a chi A, so it is actually 0, so I hope all of you are familiar in writing this, this will become chi A 1 chi A 2 1 by R 1 2 chi R 1 chi A 2 right, beta 1 beta 2 minus you can interchange this or you can interchange this chi A 1 chi A 2 1 by R 1 2 chi R 2 chi A 1, now clearly you can see this is same as this by another dummy variable interchange of 1 and 2, so you should be able to do all this practice but the point is that for all anti symmetries matrix elements either the left or the right pair cannot be same for anti symmetries matrix, but each of the integral will survive but their result will be identical, so hence eventually I need not write this and I can actually write an expression as B equal to 1 to N as if it includes chi A but no harm done because that is anyway 0 and I have not included B equal to chi R no harm done because that is also 0, I have included something which is 0, I have not included something which is 0, so that is a good thing about 0, include do not include nothing matters but remember when you want to program of course you should not program B equal to 1 to all N, you should program excepting chi A, so there you must write B not equal to A if at all you sum all excepting I mean to be more precise because that is meaningless this is 0, so instead of double summation you now have a single summation, only on one of this pin orbitals because other two are fixed and again this comes out again please check this with a very simple example like this and here there is only one term, so you see a very nice progression in the A type of Hamiltonian which is sum of one electron Hamiltonian I had a sum over N terms which is now replaced by only one term no sum for the two electron part I had a double summation which is now replaced by a single summation, so you see the nice progression of reduction of sums for this and note also that there is no half factor here that is important to realize this type 2 is very nice and just here I will spend a time before I go to type 3, type 3 will be obviously where two one of the determinant differ from the other by two spin orbitals and type 4 will be more than two and I will argue that the type 4 onwards everything is 0, so I need to only discuss type 3 after this but before I do that let me just pause here and reflect on this particular quantity that we have got. So, let us assume now psi naught is Hartree Fock we have done the Hartree Fock, so now there are Hartree Fock spin orbitals and note my Fock operator I had written down again, so my Fock operator was H of one plus sum over b equal to 1 to n chi b star 2 1 by r 1 to 1 minus b 1 to chi b 2 d tau, note the Fock operator this put it in the side line can you see that this integral psi naught H when this is Hartree Fock please remember this is Hartree Fock, so this is the unoccupied orbital of the Hartree Fock this is occupied orbital is nothing but chi A F chi r I hope you can show this it is very easy to see look at the Fock operator the first term is obvious because Fock has H I have chi A H chi r that is obvious the second term I have a summation chi b star 2 chi b 2 and if I do chi A F chi r or that means what do I get chi A 1 chi b 2 1 by r 1 to 1 minus 2 chi r 1 chi b 2 which is nothing but this please note this very carefully, so if I do so let me write the chi A the right hand side chi A F chi r, so that is an integral in a full form chi A star 1 F of 1 chi r of 1 d tau 1 correct, so the first term is chi A star 1 H of 1 chi r 1 d tau 1 that is clear because F is H the second term has a summation, summation over b and now I have a further integration over d tau 2, so I have two integration d tau 1 d tau 2 but first I have to write chi A star 1 correct and then from the Fock operator I write chi b star 2 then I write the 1 by r 1 2 1 minus p 1 2 and here I write coordinate 1 first chi r 1 and then again chi b 2 I have an integration over d tau 2 and d tau 1 correct just in the long form I have written quite clearly these two together is nothing but this I hope all of you can see this that this term sum over b chi A chi b anti-symmetrize chi r chi, so that is the meaning of anti-symmetrize 1 by r 1 2 is not written you have a 1 minus p 1 2 so anti-symmetrize makes it exchange here on the right side, so this is in the long form nothing but this expression this entire expression is it clear to everybody, so what I am arguing is that this quantity which is this expression is nothing but this quantity chi A F chi r please note very carefully I am trying to go as slow as possible because I know that everybody is doing this first time this leads to an extremely interesting theorem what is chi A F chi r chi A F chi r is nothing but 0 not because of this integral but chi A F can somebody tell me why note note that chi r is an unoccupied orbitals which is orthogonal to the occupied, so you remember we have got an eigenvalue equation F chi A epsilon A chi A where A is 1 to N and in fact all the r's are also eigenvalue equation right unoccupied orbitals are also eigenvalue equation, so obviously chi A F is nothing but epsilon A chi A correct, so if I put this here this chi A F chi r is nothing but epsilon A chi A chi r is it clear and this is 0, so provided the orbitals chi A set chi A are orthonormal to the set r any replacement from this orthonormal set actually gives you 0 because of the fact that these also satisfy an equation eigenvalue equation like this. Now what important thing is that I had a non-canonical equation remember let us say I have a non-canonical I have not solved the Hartree-Fogge eigenvalue equation but I have a non-canonical equation so B lambda B A epsilon chi B this is now note that when they did the non-canonical equation they were still only within the occupied orbitals they were number of electrons are N later on I canonicalized, so even in that case this will become 0 because F chi A F when acting on chi A brings them only within the occupied spade, so each of the occupied orbitals are still orthogonal to chi R and hence it will become 0, so let us do this exercise. So let us write these up as a non-canonical Hartree-Fogge equation, so what will happen to chi A F can you now write this chi A F will be epsilon lambda B A which is Hermitian matrix so lambda B A star or whatever if they are real number you do not have to write chi B sum over B, sum over B you can put it star it does not matter then what happens to chi A F chi R it will become sum over B lambda B A chi B chi R, so instead of only one term it will have a sum over B, but all chi B chi R's are 0 because chi R is an unoccupied orbital which are orthogonal to all the occupied orbitals, so as long as I have constructed unoccupied orbitals which are orthogonal to occupied orbitals I do not care whether it is canonical or non-canonical the only issue is that if it is canonical then it is automatically ensured because unoccupied orbitals are also coming from the same operator Eigen value equation since the operator is Hermitian all entire orbitals are orthonormal. In this case I get only occupied orbitals then I have to construct an unoccupied orbital basis which is orthogonal to each other, but as long as that is done it will give you 0. So it actually does not matter whether it is canonical or non-canonical what is important is the fact that the unoccupied orbital should be orthogonal. So let me state this that a chi 0 H chi A R is 0 provided chi 0 is Hartree-Fock provided chi 0 is Hartree-Fock I do not care whether it is canonical or non-canonical and the unoccupied orbitals chi R set chi R are orthonormal or orthogonal actually to chi A that is all that matters okay. This is automatically satisfied provided I have a canonical equation because chi R comes as a result of the operator Eigen value equation and you know that if it is a Hermitian operator all Eigen functions are orthonormal automatically chi R become orthonormal otherwise I have to impose this, but as long as that is true there is a very nice result that when I do CI remember when this determinant will come when I do CI MCN MCN. So one of these will be Hartree-Fock with singly excited this is called the singly excited determinant I will give a name for this now and that is very clear why it is singly excited from the Hartree-Fock you are exciting one electron from chi A to chi R so all singly not singly excited state please remember singly excited determinant state is a different meaning because state should be Eigen function of the Hamiltonian these are not Eigen function of the Hamiltonian none of them because even size 0 is not an Eigen function of the Hamiltonian it is a Hartree-Fock approximation. So singly excited determinant and the Hartree-Fock determinant cannot be coupled through the Hamiltonian so this matrix element is now 0 so that is a very important theorem provided this is Hartree-Fock and I have again repeated the theorem so that is the statement of the theorem and this is very often called Brilloise theorem. So this theorem is called the Brilloise theorem and we have already done the Koopman's this is called Brilloise theorem and in some sense Brilloise theorem is a consequence of the Hartree-Fock it is very clear that this is a part of the Hartree-Fock so as long as size 0 is a Hartree-Fock I construct a set of spin orbitals which are orthonormal and then this determinant is 0 so that is the Brilloise theorem and Brilloise theorem will have a serious consequence when you do the later part of the correlation theory so Brilloise theorem essentially this statement of this is just the Brilloise theorem. So I have proved it by first applying Slater rule and then showing that that expression is nothing but chi A f chi R and if it is chi A f chi R because chi R are orthonormal to chi A it is 0. I think the mathematics is only really to write this Slater rule here and identify that that is nothing but chi A f chi R and please practice this Slater rule. So for example take chi 1 chi 2 h chi 1 chi 3 note that you can again do spin integration I am not done that again you can do spin integration so for example here the first term will be chi 2 h chi 3 correct that will be the first term then the second term will be sum over b equal to 1 to 2 of course here it is only 2 so it is very easy to write and then you have the one that is I just fix this here chi 2 chi A anti-symmetrize chi 3 chi A where chi A will be 1 or 2 but of course it cannot be 2 in this case it is trivial it obviously cannot be 3 anyway so it has to be only 1 so there is only one term which survives. So for this case I can straight away write it so this b equal to 1 to 2 goes off so I only want to write correct and then you can do spin integration depending on what is chi 2 what is chi 1 what is chi 3 up spin down spin you can do the spin integration so again same thing will happen what will vanish what will not vanish but I hope this is an application of the Brillouin's theorem I hope you understand because I had a sum I had a sum over b but b has to be either 1 or 2 it cannot be 2 so only 1 survives in fact sum over b not equal to A that is the easier way to write so again note this expression psi 0 H psi A R I write it again it is chi A H chi R plus all orbitals which are in psi 0 b but not equal to A 1 to n chi A chi B anti-symmetrize chi R chi that is it now you may write chi B first chi A later chi R chi B first chi R later does not matter that is identical it is an anti-symmetrize so of course you have the exchange term when you write it in full so I have so this is the application to a simple two electron term okay yeah any problem you are not able to read anything just let me know so sum over b b is not equal to A so for example here it should have been 1 or 2 but it cannot be equal to 2 so only 1 survives and whatever will come here will come here and these are the 2 which are different so here chi 2 is changed to chi 3 so that comes here and you can of course write it the other way round in both case that is identical result so you have a cooldown minus exchange integral so this is how you should please practice this because these are new things that you are doing unless you do many practices you will not be able to write you can now take 3 electron so you can take 3 electron for example chi 1 chi 2 chi 3 one determinant another you can take chi 1 chi 2 chi 4 so the change is only chi 4 so this term is very easy chi 3 h chi 4 this term you have to write properly now because b will be not 3 but b can be 1 and 2 so there will be 2 terms so there will be just number of terms will increase otherwise is very easy so this will become chi 3 chi 1 chi 4 chi 1 chi 3 chi 2 chi 4 chi 2 so there are 2 terms will come okay alright.