 So, now let me revisit this letter's room for matrix element and in particular of course for expectation value. So, we wrote down this where psi is chi 1 chi 2 2 chi n. Remember whenever I am writing this is a determinant with a 1 by square root n factorial. So, I must mention that this is in full form it is 1 by square root n factorial and then a determinant which is which looks like this chi 1 1, chi 1 2 to chi 1 n, chi 2 1, chi 2 2, chi 2 n and so on till chi n 1, chi n. So, whenever I am writing this average form this essentially maps into this. So, please remember this including the 1 by square root n factorial. So, this is basically normalized function I wrote down the Hamiltonian which was sum over h of i plus 1 by r. So, our matrix element was now you add this was the matrix A, A point 1 and A point 2, A point 1 plus A point 2. So, can you tell the first term sum over summation i chi i h chi i that is the first term. Note that these i's are indices for spin orbitals. These are not coordinates of electron, coordinate of electron is only 1 and that is a dummy variable. So, whether it is electron 1, electron 2, electron 3 does not matter. The n spin orbitals actually take care of any electrons. Then you have the next term which is tell in any one form whichever form you remember half of that is one form half of summation all i all j all i all j i less than j. Then half won't be there if it is half it must be all i all j correct. So, then this is 2 electron integrals chi i chi j 1 by r 1 2 chi i chi j minus chi i chi j 1 by r 1 2 chi j chi j. So, this is 2 electron integral 1 and 2 these are coordinates of the electron these are again indices of the spin orbital and these have 1 and 2 which should match with these coordinates of the electrons. If you write by chance 1 by r 2 3 somebody can write that then it is assumed that this is 2 this is 3 this is 2 this is 3 I hope it is clear this is a dummy variable it doesn't matter. At the end of these integral will not depend on the coordinate of the electron, but the integral will depend on which spin orbital it is because each spin orbital is a different function. So, that is as it is dependent on spin orbital and not on the coordinate of the electron I again repeat so that there is no confusion. This is called the Coulomb integral and this is the exchange integral. The same thing can also be written in a form which is computationally simpler because it does less job i less than j chi i chi j sorry chi i chi j 1 by r 1 2 chi i chi j minus chi i chi j 1 by r 1 2 chi j chi j and we noted why it is so because for i equal to j the entire term is 0 of course each of them is not 0, but the entire term is 0 and for i less than j and i greater than j they are identical ok. So, I can write only one term and knock off this half so half into 2 becomes 1 is it clear because of the symmetry if i and j are interchanged each of the integrals remains the same. So, some of the symmetries you must understand so the symmetry of the two electron integrals if i have chi i chi j 1 by r 1 2 chi i chi j that is identical to chi j chi i 1 by r 1 2 chi j chi j correct if i interchanged both left and the right side both here for example they are not same because i have interchanged only one pair in the right. So, it produces a different integral why are they same this has one this has one this has two this has two in this case this has become one and this has become two, but they are identical because this is not because i and j can be interchanged, but one and two can be interchanged. So, instead of chi i 2 I can always call it chi i 1, but then you will say that this also has to be interchanged which is correct I can make r 1 2 and that is equal to r 2 1. So, if I interchange this it is actually identical. So, that is the reason this symmetry is there not because i and j can be interchanged remember chi i and chi j are different spin orbitals, but they come because the coordinate interchange is actually assumed it is sometimes not obvious because I am not interchanging this, but this is actually implied that it is r 2 1 because r 1 2 is same as r 2 1 and that is why all the integrals for i less than j and i greater than j both Coulomb and exchange together are equal each of them is equal. So, I can write only one pair is it clear and at this stage please understand the symmetry very important computationally you would like to have this because the number of loops please remember in computer loops are very important how many operations you are doing number of operations that you are doing is n C 2 here the number of operations that you are doing is n square right each i going to 1 2 n j going to 1 2 n so n into n n square this is n C 2 so this is obviously less in the other point is that even here when the number of operations are less you are again multiplying by point 5 which is an additional operation for each of them. So, it is actually in all senses it is a bad formula to program so it is good to of course understand this is how it came further a technical term I am going to introduce kai i kai j now I am not going to write 1 by r 1 I will write a double bar kai i kai j I am introducing a term which will simplify the algebra later as kai i kai j 1 by r 1 2 kai i kai j which is the Coulomb integral minus kai i kai j 1 by r 1 2 kai j kai j this entire part Coulomb minus exchange I am giving a new nomenclature so that I do not have to write each time please understand this is only for simplification. This integral is also a 2 electron integral I am not deliberately writing 1 by r 1 2 but there is a double bar and assume that there is a 1 by r 1 2 but it is not like a regular integral so it has to be given a different form so it is Coulomb minus exchange the exchange that I am doing on the right side is implied here whenever I am writing this so it is actually combination of two terms so do not do not interpret this as just Coulomb. So, this is a very special symbol that I am introducing this symbol and I will explain why it is called so is called anti-symmetrized two electron integral of course these are two electron integral these are anti-symmetrized two electron integral and I will I will just now tell you why it is anti-symmetrized so this anti-symmetry essentially means that if I take this double bar integral which is Coulomb minus exchange and then interchange only one side of it right hand side this will actually become negative of itself. Remember for each of this integral if I interchange the right side it is not negative or anything it is different that is how I got Coulomb and exchange but for the anti-symmetrized integral when I exchange this is negative which can very easily seen because this is nothing but now the Coulomb of chi i chi j chi j chi i which is this minus the exchange so the sign is exactly opposite okay so I hope you can see that this is now negative of this which was not true for each of these integrals they were different because in this case this is this minus this in this case now chi j chi i will come first to the plus sign and then its exchange which is chi i chi j with the minus sign so they are negative of each other I hope everybody is convinced see if anybody has a problem in the algebra please tell me and that is the reason it is called anti-symmetrized which means if I interchange one of the pairs it is negative of itself okay just like your wave function when I interchange two coordinates of pair of electron it become negative so just like this but interchange here only on one side therefore interchange two sides it does not help it will get back the same integral you can quickly see here it does not help because each of them remain same so it does not matter of course I could interchange also the left pair similarly so this is negative of chi j chi i chi i chi j and it is same as interchanging both chi j chi i chi j chi i is it clear so each pair of interchange either on the left on the right produces a minus sign so if I interchange twice both left on the right it will come back to the same thing and please practice these integrals if you if you write this what will happen this will become chi j chi i chi j chi i which is same as this where I already told you that each of the integral I can interchange then minus the exchange which is again the same of this so this is actually exactly same as this but if I interchange one side it will give a negative sign because the exchange will become plus coulomb will come with a minus sign these are very simple mathematics very arithmetic actually just practice this so that you have no confusion so many times we like to write this only to simplify this and it also has a very specific in this form it has a very specific symmetry that anti-symmetry so that may be exploited at some point of time but the coulomb and the exchange are not explicitly seen here for example the coulomb and the exchange are explicitly seen so that part you have to assume but that is not very difficult okay so the same form one can write now as so I can write this psi i psi as in a more simpler form as sum over i chi i h chi i plus i less than j chi i chi j anti-symmetry that is it so in a more simple form so I take n c 2 pairs of i j spin orbitals remember these pairs are again not of the electron coordinates but whenever I am taking spin orbital pair physically actually I am generating electron pair it is one and the same because I have n electrons in n spin orbitals instead of looking at electron pairs I am looking at spin orbital pairs which tells me where the electrons are so all pairs anti-symmetry this anti-symmetry element now contains the coulomb and the exchange explicitly so I am not writing it each time is it clear these are again for some standard in a simple way to write the energy so this is the energy that you have for a wave function which has n spin orbitals okay is it clear because eventually of course you have to start doing a variation remember our objective of Hartree-Fock is to make this minimum subject to a condition that the chi i chi j should be equal to delta i j okay so that we will do so but so that the form should be very clear so we will take a pause here and look at this form little bit more critically for specific determinants so let me take some specific determinants and analyze them this energy whatever the chi of course the chi is will be obtained for Hartree-Fock the chi is will be obtained as I told you after variation okay the other class I discussed the variation theorem will actually give you chi I still do not know the basis okay but let us assume whatever is my chi i if I know chi i's this is the energy in terms of chi i's okay so we will try to analyze this form little bit more critically and actually there will be lot of good physics that we can see from there so let us look at a determinant which is closure okay there many times in this course we will actually look at closure so what is closure let me again define what is closure I have defined this in 4 to 5 so let me again define closure very properly so first of all number of electrons or number of spin orbitals is even that is very important you can talk in number of spin orbital because it is a determinants so that means number of electrons so that is even so n is even out of these n by 2 spin orbitals or alpha spin and the rest of the n by 2 spin orbitals are beta spin is it clear because each spin orbital has a spin part and a space part so the spin part half of them is alpha spin half of them is beta spin now when I saying when I am discussing spin orbitals it is very clear what how the electrons are oriented if spin orbital is alpha spin that means electron in that orbital is up spin so that is very clear from the discussion then each of this spin orbital has a space part so that is the that is the number 1 this is number 2 then the number 3 is that the space part of space parts of n by 2 alpha spin orbitals are identical to the n by 2 beta spin so they are actually identical so the space part of the n by 2 alpha spin orbitals completely overlap with the space part of the beta spin orbitals is it clear these three together is called closed shell I mean you have a you have an idea of a closed shell doubly occupied now you will see that this is exactly what we are seeing I am just giving a more sophisticated language without bringing in electrons everything in terms of spin orbitals because spin orbitals define my determinant remember if I have n spin orbitals for a single determinant it automatically means it is n electron I do not have to talk about n electrons and how this n spin orbitals are chosen that defines your closed shell and this is your common definition of closed shell nothing new but you can check up alright so many many times in this this course will actually be defining closed shell so we will write Hartree-Fock also initially for closed shells the closed shells are many most of the many assistant that you deal with a closed shell of course there are many systems which are not closed shells which are open shell radical these excited states triplet lots of things are there so what is open shell open shell definition is very simple whatever is not closed shell so now you do not have to define so whatever is not closed shell is an open shell which means any of these three are violated it is an open shell remember closed shell is not either of these three this plus this plus this so if one of them is violated it is already an open shell for example if n is odd it can never be closed do not even discuss the second and third okay now after that what kind of open shell that will depend on 2 and 3 but at least it is not a closed shell if n is odd so right away discussion ends and many times we know they are doubly everything else then 2 and 3 will define what kind of open shell it is if n is odd so there you will see there is a restricted open unrestricted will come to those point when you discuss open shell okay depending on what kind of space parts but at this point for my discussion it is sufficient to say that anything that is not closed shell is open shell I will not define any more open shell okay there is only 2 things either closed or so obviously if anything that is not closed is open so let us now take up a closed shell determinant and let us see what happens to the spin orbital so when I say I know the spin orbitals what does that mean now that means I know the space parts because what is spin orbital in this case it is just space part into alpha and beta so I know the space parts so let us say that I know this space part so chi 1 is phi 1 alpha chi 2 is phi 1 beta and so on chi 3 is phi 2 alpha and so on right chi 4 is phi 2 beta so I know phi 1 phi 2 up to phi n by 2 is it clear because I require n by 2 space orbitals to form the n spin orbitals for closed shell because all I need to do is to attach alpha or beta to each of them so I will double the number of spin orbitals is it clear to everybody so that means this is known to us in fact when we will do heart refog for closed shell this is exactly what we will find out because I know that these are alpha beta so the real variables will not be chi the real variables will be phi but at this point I am avoiding this question how we will get phi let us assume that I know phi's hence I know the chi's correct and now I applying my energy formulas later rules okay so let us write this later rules energy again a sum over i first I have to write over spin orbitals chi i h chi i plus you can do half let us see half ij chi i chi j I will write it specifically in full form for the time being minus chi i chi j 1 by r 1 to chi j what I want to do now is to transform this energy into the space orbitals to do this what do I have to do I have to integrate over the spin coordinate only remember these coordinates of the spin orbitals are space and spin part although the coordinates of the Hamiltonian are only r the spin orbitals have space and spin part so the spin part I am going to integrate and you will see that this integration is trivial so you look at this integration this chi can be phi 1 alpha phi i 1 of the phi is alpha or beta both left and the right are same chi i see this is alpha this is also alpha this is beta this is also beta which gives you always 1 so I will have in terms of space orbital 2 times i phi i h phi i is it clear where I now goes over n by 2 that is important so each of the spin orbital is derived from the two space orbitals each case the integral that survive is only in space coordinate phi i h phi i the spin part is 1 so the first part is quite trivial now we look at the second part plus half now I do the same exercise here so chi i is 1 this chi i is 1 whatever is the spin orbital here the same spin orbital occurs here so if it is alpha this is also alpha this is beta this is beta similar thing happens for j this can be alpha this can be beta so if you write it in terms of the space orbital the same expression ij n by 2 now then how many combinations will survive 4 this can be alpha independently this can be beta this can be alpha this can be beta so 4 combinations will survive each of them can be alpha and beta independently so alpha alpha alpha alpha beta alpha beta alpha beta beta beta beta is it clear all will survive because integrations are taking place like this again remember it is all Dirac integrations in terms of coordinates of electron it is 1 2 1 okay so coordinate of electron is 1 here 2 here 1 here 2 and these are complex conjugate normally but let us assume the everything is real orbital then the symmetries become easy but does not matter they can remain complex they can remain conjugate complex so 4 times this will survive and this will then become phi i phi j 1 by r 1 to phi i is it clear this is again a Coulomb integral but now in terms of space orbitals alone that is the only difference from this Coulomb integral interpretation is exactly same but in terms of space orbital this comes with 4 times but of course there is a factor half so let us not forget that we will bring that factor half later after analyzing the exchange part in the exchange part however if this is alpha this is this need not be alpha but remember this is 1 this is 1 so if this is alpha and this is beta it becomes 0 so out of those 4 combinations 2 will not survive so only thing that will survive is alpha alpha beta beta beta if this is alpha this is beta then it becomes alpha beta beta alpha and that is 0 because alpha beta will be integrated right or beta alpha will be integrated so only twice this exchange term will survive and giving you phi i phi j 1 by r 1 to phi j phi i so in terms of spatial orbitals this will be my final expression it is very very easy to do this this procedure of deriving this expression from this expression is what is called spin integration I hope it is clear what is spin integration that I have integrated only the spin variables and once I have integrated the spin variables the spin orbital becomes an expression in terms of spatial orbitals is it clear so this part for the closed shell I can now write the final expression and give them a name so energy becomes sum over 2 phi i h phi i this is very easy to see there are n by 2 spatial orbitals so there are n by 2 1 electron integrals since there are 2 electrons in each of them you are multiplying by 2 that is how this 2 has come alpha and beta right so that is the interpretation plus sum over i j remember I have picked up this first form half of all i also so remember that 2 times coulomb integrals phi i phi j phi i phi j this is ith orbital interacting with jth orbital but space now not spin so remember these are space orbital I call these 2j ij so this is a coulomb integral involving phi i and phi j moment I say it is a coulomb integral involving phi i phi j you should be able to write the actual integral since it is coulomb it must be phi i star phi i phi j star phi j so you should be able to write ij ij in this lexical form Dirac notation right ij ij essentially mean i star and i here are 1 1 j star and j here are 2 I read it from the left to right as ij ij is it clear and that would be simply written as j subscript ij this is another common notation so I am just introducing it half into 4 becomes 2 then I have half into 2 minus and I call this exchange integral k ij this is again just a notation see if I write k ij this means ij j i lexical order you read ij j i hope all of you are not understanding it is just simple way to write if I write k ij you should be able to write the integral if I write k12 you should be able to write the integral so that becomes a very simple form for a closed shell in terms of spatial orbitals of course this is n by 2 the summation index change now the number of spatial orbitals are n by 2 which is which actually came from the n spin orbitals here these are n and the process of spin integration is it clear is it clear to everybody okay good now you see for i equal to j just like in the spin orbital I noticed that kai kai j the spin orbital was identical so it was 0 but here it is not exactly 0 because j ii is equal to k ii indeed j ii is ii ii ii k ii that is equal but unfortunately in the energy i equal to j survives as one term so this becomes now 2 times sum over i n by 2 phi i h phi i for i equal to j there is one term which survived j ii because one j ii cancels one k ii one Coulomb term survived which is actually not surprising spin orbital nothing would surprise because they are space orbital so there can be two electrons in the sitting in the same space orbitals they will have a Coulomb interaction so this is not at all surprising because we are rewriting this expression in terms of space orbital in terms of spin orbitals this could never have survived because no two electrons can be sitting on kai ii because of Pauli principle but in terms of space orbital there can be two electrons of course I will beat one of them would be alpha one of them would be beta that is clear because now my information of the spin is gone I am just saying j ii but it is clear that they are opposite spins for the self that they are opposite spins then i equal to j is gone after that I can write the rest of them which is i not equal to j 2 j ij minus k ij note again by the similar argument j ij is equal to j ji k ij is equal to k ji that is if I interchange i and j the j and k remain the same that is the same argument that is basically the because one and two electron coordinates are dummy variable so that part is still correct so I need not write I can only write i less than j and multiply this by 2 I can say 4 j ij minus 2 k ij I less than j there okay so there are several expressions that you can write again because of symmetry but remember that this comes out separately or you can continue to write this and then simplify as and when required it does not matter I am just kind of showing you.