 So, I think there are a couple of very important things that we did yesterday and I want to revisit them before we go to any new topic. One is to calculate these later rules of the type 2, where one of the determinants differs from the other by one occupancy. So, if one of the determinants is psi 0, which has chi 1, chi 2, chi A minus 1, chi A, chi A plus 1, chi N. So, there are N, N spin orbitals. I have specifically marked A, a particular in spin orbital which I am going to replace and hence I have put A minus 1 and A plus 1, otherwise there is no meaning that A can be 2, A can be 3, if A is 3, A is of course 3 then this is 2, if A is 1 then of course this does not exist. I mean I hope you understand just the meaning of this, I have just specifically marked this to make sure that you understand the A. Then, so this is one determinant which I call just a reference determinant with respect to this, the other determinant differs by one occupancy which I called A R and that was chi 1, chi 2, everything is same including chi A minus 1 here, but this chi A is replaced by chi R and the rest is chi R. I hope everybody understand this nomenclature, so it means exactly in that column of chi A I am replacing it by chi R. So, there is a question if I interchange what will happen? So, I could, I can put chi R in some other column and after that I interchange this, so then of course there will be negative sign, so that is a mute point, but at this point what we are saying that wherever the column of chi A is there, I am replacing it by chi R or the row of chi A is there, it depends on how you are doing it. We probably defined the spin orbitals as the row and the column was coordinates, so it does not matter, so either that row or the column depending on how you write the determinant would be replaced by chi R, the coordinates are still 1, 2, 3, 2, n, because that is a dummy coordinate, again I repeat, coordinates are always dummy, so it does not matter, important thing is there are n electrons, there are n spin orbitals, so here also we have n spin orbitals except that the chi A has been replaced by chi R, so when you have these two determinants and these are now arbitrary spin orbitals, they are not necessarily hard to recall when I am using slated rules, so they can be arbitrary spin orbitals, the slated rules now says that psi naught H psi A R is nothing but chi A H chi R plus sum over B equal to n, chi A chi B anti-symmetrize chi R chi B, is it clear? Obviously I have written B equal to 1 to n, but B cannot be equal to actually A, so I can say B naught equal to A and obviously B is not equal to R also anyway, they would become 0, so this is the type 2A, this is the 2B, this is for the one electron part, this is for the two electron part and clearly from the expectation value you saw that it was chi A H chi A sum over all A that is gone here chi A AB chi A AB sum over all B with a half that has changed, so now it is a single term for this and just one single summation for this, so this is the first important thing and this is a slater rule for any arbitrary determinants of this type, we are not suggesting that the spin orbitals are Hartree-Fock or anything, so this slater rule is now applied to show Brillouin's theorem, so Brillouin's theorem now specifically says that these are Hartree-Fock orbitals, so now we are saying that apply slater rule and the next condition is if psi 0 is Hartree-Fock determinant and of course chi R is orthogonal to all chi A, any chi R, I mean in principle we have all unoccupied orbitals which are orthogonal to occupied orbitals, so if chi R is equal to chi A then we are saying that the psi 0 H psi A R which still follows this rule of course, this is a general rule but this will now become 0, so this is the content of the Brillouin's theorem, so this is just a slater rule, however if psi 0 is Hartree-Fock and this spin orbital chi R is orthogonal to chi A and all occupied orbitals actually then it so turns out that this expression becomes equal to 0, so that is the, so you should first separate this, so if I give you any arbitrary determinant and ask you to calculate such, you should always use this, do not write 0, the Brillouin's theorem only applies if one of them is Hartree-Fock. So if I am only using Hartree-Fock orbitals then only Brillouin's theorem applies but this is a general rule for any slater rule just like you had given slater rule for expectation value, so the proof of this why so because in case this is Hartree-Fock then we showed that this expression which is of course a general expression is nothing but chi A F chi R, so this was actually equal to chi A F chi R in case psi 0 is Hartree-Fock written, that is easy to see because we have defined our Fock operator as H plus sum over partial integration over d tau 2 chi B star 2 1 by R 1 2 1 minus 2 chi B 2, so that is exactly this is, please note that this is the expansion, the expansion is just this for this double bar, so that is always there, so that is what we mean by anti-symmetrize, so 1 minus P 1 to maxed anti-symmetrize, so we do not write it but whenever I am writing double bar it has a 1 by R 1 2 into 1 minus P 1 2 and of course if we write for the spatial orbitals then we have to specifically write 2 times 2 minus this, so since you have a chi B star 2 chi B 2 with this stuck in the middle, quite clearly if I integrate this with chi A star on the left, chi R on the right, so it becomes chi A star 1 chi R 1 then the entire quantity is nothing but an integral chi A star F chi R, this is a number, this is also a number, it is a complete integration, whereas a fork operator is an incomplete integration remember fork operator is an operator, so it depends on coordinates 1, this is a number, so the further integration is done between 1 coordinate 1 which was lying unintegrated in the fork operator, so that gives you a number, so please try to understand this how to manipulate this, so this become this is a 2 electron integral, this is a 1 electron integral but when I write fork operator it is just H of 1 no integration and then only the partial second integration is done in the fork operator, so after I present this chi A star F chi R it is a complete integration and this is nothing but this, then we say that since these are Eigen functions of F or even if they are not Eigen functions F acts on chi A and transforms within the occupied set and the virtual orbital chi R is orthogonal to all the set, this quantity is equal to 0, so we say that even if it is non canonical F of chi A is nothing but sum over B lambda B A chi B, chi B which is actually 1 to n, so obviously your chi A and F is Hermitian operator is nothing but chi B lambda B A star sum over B and now we can quite clearly see if you close the integration with chi R then all there is a sum over B but these are only occupied so all chi R, chi R is orthogonal to chi B, this is a number which goes out so it becomes 0 either way and if it is canonical then of course it is trivial to show, so either in the case of a canonical or a non-canonical Hartree fork this quantity is 0, so these are the two essential things that we did yesterday, so I want to again write that very clearly the first part was a general Slater rule for any arbitrary determinant, second part was application of this Slater rule to the case where sinot is Hartree fork and the chi R's are of course orthogonal to this spin orbitals, in such a case you get the simple relation which can be derived from this general rule and that is what I did and this is called the Brillouin's theorem, so whenever now Brillouin's theorem says whenever I start from Hartree fork determinant, if I calculate the matrix element of the Hamiltonian between Hartree fork and another determinant which is generated by one change of spin orbitals and this is what we will call now singly excited determinant, the matrix element vanishes starting from Hartree fork, so that is the final statement, so then we will move forward. The Brillouin's theorem is a very important content which will be actually used when you go to the post Hartree fork theory, but remember the Brillouin's theorem is a consequence of Hartree fork, we have not done anything, we have not done any more new theory, it is a consequence of Hartree fork because this is something that is given I cannot help it and from here the definition of fork operator gives me this and automatically I get 0, so the Brillouin's theorem is not a new theorem, it is a consequence of Hartree fork and many times Hartree fork is actually stated as a Brillouin's theorem, in fact that is also interesting interpretation of Hartree fork, it is you say we said that Hartree fork is a determinant which minimizes the energy for a single determinant, you can say given a single determinant can you find a set of spin orbitals such that this is equal to 0, it is also a Hartree fork condition, so it is a way of putting it, so in a way this is also a Hartree fork condition, the Brillouin's theorem is just a consequence of Hartree fork theory, so that philosophy should understand that we are not doing anything new, it is a direct outcome of the Hartree fork theory, but you recognize this when you move forward, so that is what we will see later. At this point I have also introduced the term called the singly excited determinants, please note that they are different from singly excited state, many people get confused, we are still talking everything about ground state, exact ground state, of course exact ground state we are far from reaching there, Hartree fork is an approximation, so Hartree fork determinant is an approximation to ground state and these are singly excited determinant with respect to a reference, whatever is the reference, so our reference that we are usually bothered is the Hartree fork reference, so reference can be Hartree fork, but it can be any other reference, an arbitrary reference like any size 0, so with respect to that because whenever I say singly excited it must be with respect to something, so there is a given determinant and then only you are exciting, so this is a class of singly excited determinants and of course how many of them you can generate, we did that in the CI, how many of you can generate? N times M minus N, so let us say there are M number of spin orbitals, this is your total basis or whatever number of spin orbital that you generate, N is the electrons, then of course the number is N times M minus N, so N is the occupied, M minus N is the virtual spin orbitals, so that many number of, we are only talking of one of them, but just to say that any one of them will have the same property, any one of those N into M minus N determinants if I, with the Hamiltonian expectation value with respect to Hartree fork and the matrix element between Hartree fork and any of these singly excited determinants will be equal to 0, is it clear? So let us do a little bit of practice, so I gave some simple examples, has any of you verified this? Please do that, I told you just take a 2 electron system and please verify this, the rule itself, any arbitrary 2 electron determinant, so now we are going to do an application, so I said let us say chi 1, chi 2, H, chi 1, chi 3, so I have 2 sets of special orbitals from where I generate 4 spin orbitals, so let us say my special orbitals are like beryllium like which is 1s alpha, 1s beta, do not confuse the 1s with hydrogen 1s, it is just 1s, it can be any 1s, it does not matter phi 1, I could have said phi 1 and then the chi 2 may be 2s alpha, a chi 3 may be 2s alpha, so this is chi 1 and chi 2, 1s alpha, 1s beta and now your chi 3 is 2s alpha, so basically what you are doing, you have an electron which you are exciting from one of the orbitals, one of the 1s electron, 2 or 2s in a simple system of 2 electron system of let us say helium, so then I have a determinant which is 1s alpha, 1s beta, H and let us say one of them I, you decide, let us say I have put up 1s alpha, 2s alpha, 1s beta, so I ask you to calculate this matrix element, this is the helium ground state, I mean in a very simple term, this is an excited determinant with respect to this, I will not call it excited state but many chemists will call it excited state, I mean no problem, it is an approximation to an excited state, to be honest it is an excited determinant just like this is also a ground state determinant, so now tell me what will be the result, applies later rule, let us do a practice, I have replaced 1s alpha by 2s alpha, so the first one itself I have replaced, see please remember this order really does not matter whether you write chi A, chi B or chi A but make sure that the order is same because if you interchange both sides it is identical, so do not be particular about writing, the point here is that the ones which are replaced must come in the same place, so if this is coordinate 1 this should be coordinate 1 and the rest should be all other of it, so it is very simple do not even look at the formula because after that it does not matter whether you write that as coordinate 1 or coordinate 2, so that is identical, do you agree with this, that is it, do you agree because this summation over B is only 2 but B cannot be equal to A, so there is only one summation, A is 1s alpha, so it has to be only 1s beta and this should be 1s alpha and 2s alpha which are replaced, so this B remains as 1s beta 1s beta, I mean I am not understanding what is the problem, so that is the reason I am little bit perturbed, I mean it is just a trivial example, what is the problem, no problem but just confidence, lack of confidence, that is not bad, if there is no problem and lack of confidence it is okay, eventually you will gain confidence but if there is a more serious problem I am worried of conceptual problem, everybody agrees with this, of course if it was 3 electron there would have been 1 motor because then summation over B you would have taken you further, I am just giving a simple thing, now somebody can do spin integration, now it is trivial just do the spin integration and write it in terms of spatial orbitals, 1 electron, there is a 2 electron, how there is only 1 electron, no, no, no, but oh my God you are doing separately, that is 2, if you are writing exchange it should be minus, if that becomes 0 then you do not have to worry about the other one, if one of them is 0 you do not have to worry, 0 multiplied by anything is 0, okay good, thank you, see it is very easy, there is a think of it, there is a coulomb and exchange term, exchange term will be just think mentally, when there is a coulomb term this is 1, this is 1, this is 2, this is 2, so just write the same thing erase alpha, alpha, beta, beta, that is doing nothing, so simply write 1s, 1s, 2s, 1s as it exists and when there is exchange you do not have to do it, moment this becomes beta 1 it is 0, so I am still not understanding why you are not able to do fast, so if I were you I would simply write this as 1s, 1s, 2s, 1s, I remove alpha, beta, alpha, beta and you have to be smart, street smart sometimes to write this, you do not have to do that exercise and this is exactly why we say that if there is a opposite spin there is no exchange and we have said this, so exchange cannot come and that is obvious, see try to mentally think that this is 1, 2, 1, 2, this is 2, 1, so 1, 2, 1, 2 everything survives because 1, 1 is alpha, 2, 2 is also beta, moment it becomes 2, 2, 1 none of this will survive, so the exchange is gone and when it survives alpha, alpha is 1, beta, beta is 1, so you do not bother about them, so simply remove this, you have the j term, you have the Coulomb term, unfortunately this term cannot be interpreted as a j, j is different, so you note that you have a 1s, 1, 2s, 1, so it is not a density, this is 1s density but this is not a density, it is a transition density, so this cannot be written as j minus k, that is very specific for expectation values, that is the only difference, so please do not try to interpret in j and k, it is a Coulomb term in some way that it is a regular term but it is really not Coulomb in the sense of expectation value, because average value of energy has a Coulomb term, this is not energy, why are you not getting it because this is not energy, this is a matrix element between one determinant to another determinant and that is the reason what you are getting is like a transition, 1s to 2s, which is what you expect, so do not interpret them as Coulomb in exchange but write in terms of this integral, so the result is this, so this result is finally 1s, so this result is finally 1s h 2s plus 1s 1s, 1 by R1 2, 2s 1s or 1s 2s, it does not matter how you write it, you can write it 1s 2s, it does not matter because both sides are identical, so that is the result of the spin integration, exchange vanishes for the same result, I could have done another integral, remember do not interpret these as whether the transition takes place or not, selection rules and all that, because these are not excited states, they are only determinants, we are only following rules to write later rules, so let us say my chi 3 is, so my let us say I am extending this from 1s alpha to 2s beta, I am excited, so then my integral is 1s alpha, 1s beta, h 2s beta, 1s beta, here after the spin integration, spin integration is very easy, just see here, see here, even here you can make out, you do not have to know that, from the 2 determinants you can make out, 1s alpha, 1s beta, 2s beta, 1s beta, good, agreed or all agreed, what is 0 now, 1 beta, why that is 0, oh that is alpha and beta, good, very good, everything is 0, fantastic, no exchange also, good, so I hope you learn something, that what happens to such an integral, if I expand 1s alpha to 2s beta, the matrix element becomes completely 0, just that happens because of spin integration, okay, all right, good, so I hope you can do many practices, of course you can excite, you can excite both 1s alpha, 1s beta to something, but that will be type 3, which you will do later, yeah, because of spin integration, but those are all rules that will come, but actually when spectroscopy come, they are dipole selection rule, that is completely different, in the place of H, so those are different rules, but here it just turns out because of spin integrations, for the closed shell, yeah, because spins will, that is no other reason here, physically it means that when you have a closed shell singlet, you are trying to make a state which is triplet, so basically singlet and triplet do not mix, so in a way, in a way this is something that we expect, okay, so that is a pure triplet, this is a pure singlet, so they do not mix, with Hamiltonian, Hamiltonian is spin commutes to the square operator, all the physical Hamiltonians, so that is the reason it happens, another closed shell triplet state, then they will not mix, that is the meaning of this, matrix element, yeah, I mean I can, I can always have a matrix element of any operator between two basis functions, right, this is a N electron basis, this is another N electron basis, because all of these together will be full CI, when I do actual ground shed web function, it is not Hartree-Fock, Hartree-Fock plus singlet excited plus W excited will be also there, I told you MCN, so that time when I construct the total energy, I will have to construct the entire Hamiltonian, which will also involve such off diagonal terms, these are off diagonal terms of Hamiltonian, so they will be shifting my energy, so physically they have an important role, if these were very large, that means ground state will be shifted by these determinants in a very large way, significant way, so what it shows is that if I start with Hartree-Fock, then the singlet excited have very little role, because they cannot shift it significantly, but you will see that there is a role, it is not absolutely that there is no role, they have a role because they are matrix element with W excited determinant, like this is W excited, then that will not be 0, so they will shift into the higher order, they will not directly shift the ground state, but they will shift in a different way, so that I mean we will discuss when it comes to CI, I do not want to worry about it, but it is important at this point I want to say that this is important to evaluate the off diagonal matrix elements, and one consequence is for the Hartree-Fock that we get is a Bravais theorem, which just shows that if I start with the Hartree-Fock, these off diagonal matrix elements are completely 0, and this has one significance that if I do CI only with Hartree-Fock and singlet excited determinant, the ground state energy will not be changed, because all the singlet excited determinant block will not couple with the Hartree-Fock, so that entire block will become 0, so this is called a block diagonal structure, so if I diagonalize one of them will be energy, so basically if I do a CI I told you that CI is a Hamiltonian matrix element in that space, so one of them is Hartree-Fock, and the rest of them are all singlet excited states, so they are all 0, these are all 0, here of course there will be values, these are Hamiltonian between one singlet excited and another singlet excited, all values will be there, but then you can see that this quantity is already decoupled from the rest of the matrix, because this is 0, this is 0, so this is what is called block diagonal structure of matrix, and one of the eigenvalues will remain Hartree-Fock, so if I do what is called CI s, the ground shed energy remains Hartree-Fock, you do not change anything, what you will get is the other states excited states, when you diagonalize the rest of them, so we will come to that when you CI, so its significance will be more understood when you come to CI, at this point you need not worry about it, but since you ask the question I thought I would just answer, but the significance will be discussed in greater detail when you come to CI, so please practice more such thing with 3 electrons and so on, the basic result will be matrix element of the Hamiltonian and eigenvalues of that, so there it will have a block diagonal and matrix element, matrix will become block diagonal, so one of the eigenvalues will be unshifted, the Hartree-Fock part will never couple with anything, if you do CI s, so that is the reason for ground shed energy we never do CI s, if you remember, everywhere you do CI sd or DCI, CI d, double CI, you have never heard of single CI, single CI is done only for excited states, so those who have run Gaussian and so on you realize that for ground shed energy you never do CI s, because it does not change the result and this is the reason is the Hartree-Fock, of course all CI s are done from the Hartree-Fock as a reference, that is one thing I want to tell, so Brillou's theorem holds good, if of course CI 0 is not Hartree-Fock then it will be different result, but this particular thing that happened was because of spin, that is nothing to do with Brillou's theorem as per say, so because we are trying to make singlet and triplet and it should become 0, so that is fine.