 Let us go back to the Lagrangian once again. So, I hope all of you can write the Lagrangian. It is first the energy and the constraints. So, energy was if you remember again we are going back to spin orbital after knowing how to do the spin integration we are going back to the spin orbital for a general Hartree-Fock equations. So, that is the reason we are doing. So, chi i H chi i this is the one particle one particle summation over all the n spin orbitals plus half of anti-symmetrized we have written it minus sum over. So, I hope all of you can write this of course, this expression can be written in different form and I have not written the exchange or coulomb explicitly I have used the anti-symmetrized. Of course, eventually we will expand. So, then we said that we vary each of the chi i to chi i plus delta chi i. So, this is an infinitesimal change and we make the first order change which I call delta L equal to 0. So, that is that is basically ensuring the stationary conditions. We discussed that last time that if I just make first order change in the variable the first order change in the functional represents the derivative first order change divided by delta x represents the derivative. So, of course, I can put this equal to 0 it means derivative is equal to 0 right. So, that is my condition that I am going to. So, then what we started to do is to write what will happen if I chi i changes to chi i plus delta chi i since these are the trial functional. So, we put for every chi i some tilde but again that is really a question of only symbol. So, I put everything as then we rewrite delta L. So, when I write delta L I ensure that I take only the first order change. So, that is the delta L this is the first order change. So, if you look at the first order change. So, I made chi i to chi i plus delta chi i. So, the first is delta chi i h chi i plus sum over i chi i h delta chi i I am not writing this summation index exactly 1 to n every time but I think it is obvious that these are 1 to n. So, I have two terms from this there is no change no first order change due to h change in h because h does not depend on chi i h is a fixed operator and then we had plus half of now to make sure that we understand what we are doing we want to write everything explicitly. So, we wrote this as delta chi i chi j 1 by r 1 2 chi i chi j. So, we had four terms if you remember and then you do the same thing there are same thing on the right hand side I am not complete this and then we had the exchange. So, the summation goes over then I have the exchange which is now delta chi i chi j 1 by r 1 2 chi j chi i plus. So, this is minus of course this is within the within a full bracket plus chi i delta chi j 1 by r 1 2 chi j chi i and then two more terms. So, this is what we did and the two more terms come basically from the differentiating the right hand side delta chi i chi j chi i delta chi i and so on. So, I am not writing this now we noticed on the on that day that this term is a conjugate of this term complex conjugate of this term because of the factor h is hermitian and these two terms which I have left a conjugate of these two terms and similarly these two terms which I have left on the right hand side are also conjugate of these two terms. So, we we decide to write this as delta l equal to delta chi i h chi i. So, I will not write this conjugate term plus half of sum over i j delta chi i chi j 1 by r 1 chi i chi j plus chi i delta chi j 1 by r 1 chi i chi j minus half sum over i j delta chi i chi j. So, it is just writing it do not worry a plus complex conjugate plus the whole thing complex conjugate I mean you have to be just careful in that way nothing great about it you just have to ensure that the indices are right. So, I I I am changing the first term and the second one these ones I am not changing lambda sorry here also the lambda is there similarly of course. So, I will have only one term from the lambda minus i j lambda i j delta chi i chi plus half sum ok. So, make sure that this algebra looks long but it is really nothing once you understand what you have to do it is just manipulating the indices correct and then seeing what is equal to what that is the most important. So, now we will what we will try to do is to look at these two terms this term I have a sum over i j ok and this term also has a sum over i j. So, if I sum these completely not for each i j but if I sum this completely and sum this completely they are actually identical by dummy variable interchange i and j ok or you can say both and one and two also coordinate interchange. So, both are both interchanges are ok if I sum over i j they are dummy variables i and j I can call i as j j as i similarly I can call coordinate one as two coordinate two as one. So, it is very easy to see that these two are actually identical and in the very same way in the exchange part this is identical to this. So, actually same way this is identical to this exchange part exactly I dummy variable interchange. So, only one of them can be written with the factor half removed in each case. So, what we will do we will write only one of them and the factor half will remember the way we are doing it is a following that I am trying to first if you want to really formally show that this is equal to this you first want to bring this here. So, write this as delta chi j chi i chi j chi i. So, that is a one two interchange then call j as i i as j. So, first do one two interchange then do i j interchange followed by so you exactly get this and what will happen that delta chi j which is coming here will now become delta chi i chi i will become chi j. So, you will have exactly in the same form is it clear to everybody. So, please understand how to manipulate the dummy variable a once again for a given i j it is not same because I am using this dummy variable i j. So, it must be sum over sum over i j this is equal to sum over i j this that is important to understand. So, first I my my overriding interest is to bring all delta chi i on the left. So, I want to bring the delta on the left to do that I am doing a one two interchange after I bring delta chi j on the left I want to look it make it look like delta chi i. So, I make i j interchange it is very simple. So, both that is also allowed because the whole thing I summed over i j again of course I repeat that within the sum for the given i j this is not true, but this quantity summed up this quantity summed up is same is it ok for anybody I mean these are very small algebra, but it should be very comfortable in doing it. So, if you do this then I can write I can have a further simplification and since these two terms are same now I will not write half. So, half will go off. So, of course the first term remains as it is delta chi i h chi i and the second term sum over i j just one term delta chi i chi j 1 by r 1 2 chi i chi j minus sum over i j delta chi i chi j 1 by r 1 2 chi j chi i minus sum over i j lambda i j delta chi i plus complex concept of all the terms when you note when you wrote the energy we could remove the half by saying i less than j here also half is removed, but it is not i less than j it is all i all j just just note this little different here half got removed in a different way because I have combined these two terms. So, what looked like a very monstrous expression you know 2 from here 4 from Coulomb 4 from exchange and 2 from the Lagrange multiplier so it almost 12 terms has now simply become 4 terms because I am writing this conjugate suppose there are 8 terms actually. So, but the 4 terms are relevant rest I can write as complex conjugate and I have managed to write it in a way that the left hand side everywhere is delta chi i that is a trick because now I am going to use the variation whatever I told. So, essentially we said that the first part for making delta i equal to 0 it is sufficient to make this 0. So, then we are writing the equation now. So, the equation now becomes delta sum over i delta chi i h chi i tilde now I will put tilde since I have a chance to write the equation all over again sum over i j delta chi i chi j 1 by r 1 2 chi i chi j minus sum over i j delta chi i chi j 1 by r 1 2 chi j chi i minus sum over lambda i j delta chi i. So, that becomes your Hartree-Fock equation finally for all delta chi i. So, this is a one way of writing the equation this is also Hartree-Fock equation. So, for all arbitrary delta chi i. So, there are 2 arbitrariness all i all delta I can repeat for each spin orbital all arbitrary delta. So, that is the meaning of derivative remember I said when x goes to x plus delta x delta x can be arbitrary change then delta y must be 0. Arbitrary means it can be any value. So, when one dimension you may have only one direction when these are multi dimensional problems. So, I can have any way I can change delta x infinitesimal otherwise it is difficult to interpret. But actually if I want first order it does not have to be I showed you I think I again repeat what I showed that if x goes to x plus delta x now tell me where is infinitesimal coming here. So, f of x delta f of x. So, let us say f of x is x square then what is delta f of x 2 x time delta x first order. So, that is equal to 0 which means 2 x is equal to 0. What did I use infinitesimal? Did I actually use explicitly? No. So, even if it was finite what is the problem I in fact said in the calculus they call it infinitesimal because they do not use first order change. They use if you write this as f of x plus delta x then you are in trouble then it is first order second order etc. Then they say minus f of x which is the total change divided by delta x then you have to use limit delta x density. I think I clarified that because here second order terms are surviving divide by delta x you still have a terms which are delta x delta x square delta x cube they have to be made 0 for the first order change delta x term does not survive just like here 2 x delta x divided by delta x will simply give you 2 x. So, the derivative first derivative can be expressed in two ways one is total change of the function this is total change of the function divided by the change in x as the change in x goes to 0 that is one way second way is that you simply change x to x plus delta x do not care how much is the change consider only the first order change divide by the change they are identical that is what that is what I did first. So, it is of course we are going to be infinitesimal change, but it is never really going to be used it is not required. So, when you said f of x x square I did not quite understand if I have a larger change also it will be divide by delta x it will be 2 x, but the point here is that these delta x can be arbitrary change that is the point that it can be any change any changes any direction and anywhere in you know it depends on what is x it can be a vector. So, in any direction and any change plus minus you understand so arbitrary means change can be positive negative. So, for any arbitrary change I am saying that this delta l must be equal to 0 that is what we mean and for all spin orbitals. So, that is why I have underlined this delta and underlined this i for all spin orbitals. Since it is for all spin orbitals I make a sufficiency condition to say that within the summation over i whatever exist I can make it equal to 0 then of course if I sum over i it is 0. So, one of the sufficiency condition is now this is very important now is that delta chi i now summation over i will vanish delta chi i h chi i plus summation over j everywhere summation of i I am just going to take out delta chi i sorry again tilde delta chi i tilde chi j tilde 1 by r 1 2 chi i tilde chi j tilde note that now the half factor has already gone out because I am writing only one of those two terms there are four terms two of them are any complex conjugate out of the two also I am writing only one term. So, there are lot of small small steps that I have written because of dummy variable interchange so please try to remember all the small steps. So, delta chi i tilde chi j minus sum over j delta chi i tilde chi j tilde 1 by r 1 2 chi j tilde chi i tilde because this is exchange minus sum over j lambda i j delta chi i tilde chi j tilde equal to 0 this is now a sufficiency condition because if this is equal to 0 now all you need to do is to sum every term over i that must also be equal to 0. So, this is 0 for again all i. So, I have now used this condition so if this is 0 this is now there is no summation over i so I must write this because this i is now a specific index because you may ask which i I am saying for all i it is 0 then of course if I sum over i each term that must trivially be 0 but this is not a necessary condition this is a sufficiency condition because of course necessary conditions are much more difficult to derive because there could have been a cancellation as I said sum over i so it could have still led to 0. So, I am only using a sufficiency condition to write this. Now, I am going to use the arbitrary of delta. So, there are two steps first I have used all i now for each i I am going to say that this delta chi i star is arbitrary yes sum everything is sufficiency everything is sufficiency that is all I said everything is sufficiency. Now, this delta chi i is not a sufficiency condition this is I am changing I do not know which one you are talking this one no this is not a sufficiency. What do you mean by that because this is what I want to do this is a necessary condition because that is what the variation theorem is that I change all delta chi i arbitrarily it should be still 0 that is the variation theorem summation over all i is 0 no no no the first term is summation over all i now so this is correct there is no sufficiency condition assumed here sufficiency condition comes here because of the fact this sufficiency condition there are two different sufficiency condition one is that this part is equal to 0 that is a sufficiency condition okay but then I am talking of something more so there are two sufficiency condition one is the fact that if this is 0 the conjugates so please try to understand very clearly if this is 0 the conjugate is automatically now I am saying that sufficiency 1 has been assumed so I have this equation now I am saying another sufficiency condition for each i if this is 0 then also it is 0 so I have a sufficiency 1 and sufficiency 2 sufficiency 1 is to merely leave out the complex conjugate sufficiency 2 is to say that for each i this is equal to 0 so it has also come under sufficiency condition but if you are saying that the sufficiency condition is that the conjugate have left out that is correct so there is a sum over delta so I have applied both the sufficiency condition one is that the real part is 0 or whatever this I am just calling it real part it may not be real part whatever so one of this part is 0 and then I am saying that it is sufficient to assume that under the summation for each i this is 0 so then I have actually arrived at the next set of equations which is now little bit simpler now what I am going to do I am going to say that within this equation now I have gone to this equation my delta is arbitrary that is all I am going to say and I want to write this equation in a form which is delta chi i star sum f of 1 chi i 1 and now his question will come we will see do not worry delta 1 I am now writing it in full form minus sum over j this second last term only I am writing it separately delta chi i star 1 chi j 1 delta 1 you agree if I can write it in this form then I am in the business because what I will say for all delta chi i this is equal to 0 then I will say that this minus this is equal to 0 remember this is energy so everything is integrated but if I have to write this in this form you can see that already f of 1 is h of 1 that is no problem the second term onwards integration over d tau 2 has to be performed explicitly and I say I said that last time if you explicitly perform the integration over d tau 2 this is actually a function of 1 so that is what I wrote so all the function of 1's I have collected the question is can I write this is chi i 1 I do not have to write initially I could have simply said some g of 1 whole thing that is that will satisfy you for the time being and I will see how to write that so I that is what I actually said that if something times g of 1 minus this is equal to 0 then I can say that g of 1 minus integral this chi j 1 is equal to 0 so if this is let us write this now what is g of 1 we will write it down little bit later so let us write delta chi i star 1 till day again g of 1 d tau 1 minus sum over j lambda i j delta chi i star 1 chi j 1 d tau 1 so we will see what is g of 1 g of 1 is basically all these quantities inter except for this quantity integrated by d tau 2 that is all that you need to know but if I can write this for all delta then I made a statement first to understand the statement that this quantity minus this quantity is equal to 0 because I could have punch this quantity inside this and could have said that the whole thing is equal to 0 because this is also a function of 1 so I did not do it only lambda i j has to be written explicitly so if this is equal to 0 then for all arbitrary delta chi i star note again I have always doing delta chi i star and that is same as doing a variation in delta chi i because they are complex ones you get so there what would be my equation g of 1 minus sum over j lambda i j chi j till day 1 equal to 0 you agree if I can write like this whatever is my g of 1 remember in the last class I had also said that when we did this delta chi i star 1 chi j 1 and the reverse we made a complex conjugate I had assumed that the lambda is a Hermitian matrix just remember that that was also a small point that I did not note today but I had done it last time otherwise you cannot write even there itself when you wrote that as a complex conjugate very in the beginning so that is sum over does you so then we get I can write an equation if this is true for all delta chi i star this minus this is equal to 0 I hope all of you agree to this this is a very important point essentially it means if I have a delta chi i star 1 sum quantity h of 1 delta 1 is 0 for all delta chi i star 1 it means h of 1 is 0 this is just like we say in the vector product if all integral remember vector product s star 1 b of 1 delta 1 is 0 for all a then the b itself is 0 we use this in vector product in fact very simply we use in this in the all scalar product of the vector a b equal to 0 for all a it means b equal to 0 so this is the same thing I am doing now actually we assume this because any arbitrary function if it is integrated to this a given h of 1 this is not arbitrary this is a given any arbitrary function if you multiply by this h of 1 and integrate and if it becomes 0 h cannot be 0 obviously so then what what has to sorry this cannot be 0 so delta chi i cannot be 0 so obviously h has to be 0 since it is arbitrary delta chi i that cannot be 0 0 is a definite only a given number so the only way this is possible if this is 0 and this is actually goes back to the scalar product of the two vectors of a and b it is 0 for all a it means b is equal to 0 it is very similar in linear algebra if operator acts on a function if operator acts on an arbitrary function to give you 0 then the operator is a null operator right remember it is very similar okay so the linear algebra because that is a very strong condition so that is what I am going to use provided I know how to write g that I will come to that that is not very difficult so I am saying that provided I know these then my final equation will be a single particle equation this is now a coordinate of 1 remember everything here was multiplied but when I writing the equation these are dependent on the coordinate of 1 okay so that something coordinate of 1 would be equal to 0 and that is what we will eventually call a single particle equation the rest the question that he asked is now involved in g so when we analyze the g so I have to find out extract this one particle part from each of this by carrying out the integration over d tau 2 carrying out the integration of a d tau 2 and there his question is there that how do I handle physically how do I handle the exchange part because there is something else that he knows that in the g of 1 I want to write chi i of 1 on the right right now that knowledge is not there but if you want to write then how do I do that was his question actually if I want to paraphrase the question so we will see that right now I do not care how I want to write because quite clearly this is a once I carry out the integration over 2 this is a function of 1 exchange of coulomb both are function of 1 okay I am not writing this as something times chi i of 1 I have abundant that I want to write it in this manner but right now I have abundant I just writing g of 1 so that you have no question okay so let me write down this g of 1 I want to keep this so I am not erasing this part of the board so I am going to erase this I am going to erase this part of the board and then what I will do is to see that the first part of g of 1 is just this okay so I can write down the g of 1 now the first part of g of 1 if you if you just make an analysis from here that I wanted to write remember equation generically looks like this g of 1 d tau 1 minus sum over j lambda ij delta chi i star sorry delta chi i star 1 chi j 1 equal to 0 so that is what I want so this part I am not going to touch so now I am going to write the g of 1 by comparison so what is the first term first term is simply h of 1 chi i of 1 right so when first term will map is it okay the first term is simply that the second term of course is 2 electron integral so I have to carry out the integration over d tau 2 then only I will get a function of 1 keep this here so what will be now g of 1 sum over j right sum over j also has to be carried out and that is already there actually I can see from here I need not go there so again I write the chi i tilde so now we can see sum over j I will write it in full form chi j star 2 1 by r 1 2 chi i 1 chi j 2 d tau 2 note that sorry d tau 2 note that I cannot forget this chi i tilde 1 because that is a part of g of 1 g of 1 should be everything other than delta chi i star 1 so of course only this part I can remove or this part I can remove so chi j star 2 1 by r 1 2 chi i 1 chi j 2 d tau 2 is a part of g of 1 is it all agreed so that is my g of 1 minus sum over j now let us look at the exchange chi j star 2 1 by r 1 2 chi i 2 chi j 1 d tau 2 agreed I have just interchange the right so that becomes my g of 1 the three terms so I have just interchange this part so again writing the tilde is chi i tilde 2 chi j tilde 1 this is chi j tilde 1 so up to this point there is no problem here any problem in this there is no problem so if I simply write it d like this and then push this g here that is my Hartree-Fock equation because once I put g that is my Hartree-Fock equation I already have the Hartree-Fock equation all I needed was to write g of 1 so I have given you the Hartree-Fock equation here is a 1 by r 1 2 is a space orbital if they are if they are opposite spin that is 0 if they are same spin spin will not integrate there will be space orbitals of these space orbital these which has to be integrated along with 1 by r 1 2 because space or if I have a phi i star r r 1 ok 1 by r 1 2 phi i r 1 d tau d tau 1 is it 0 let us say let us say they are orthonormal let us say it is i j I do not care even if it is there orthonormal is it 0 and that is a very fundamental point in fact we do two electron integral this is only coordinate of 1 we did two electron integral where this is phi j star r 2 whatever is there not 0 because the space part depends on r that is what you are repeatedly missing only the spin part can be taken out because your Hamiltonian is spin independent even that cannot be taken out if there is a spin orbital interaction in Hamiltonian but this 1 by r 1 2 does not have a spin so only the spin part can be taken out and that is why repeatedly we are saying that the exchange comes only from the parallel spin because then the space part does not matter spin part will make sure that it is 0 is it understood to everybody yes yeah what is the last question that is completely different that is a completely different what is this algebra that is only interesting I have not made anything 0 so I do not understand what is algebraic I just gave an example that this is exchange integral i j j i if they are real orbital i star 1 j star 2 j of 1 i so they are all 1 1 2 there is no star here so I can simply interchange 1 and 2 so this is 1 this is 2 this is 1 this is 2 this 2 I can bring here so this is equal to i i i i i i i j j but 1 by r 1 2 is there this j star 2 this j 2 and i 2 I can flip it is commutative but they are commutative I think you are making a mistake of commutation and this they are commutative with 1 by r 1 2 so this can come but I am not saying they are 0 I think it is commutative with that they are all function of x so every operator function of x I always commutes with an operator which depends on x so this is a commutative operator so I am only commuting it is just that I have not written 1 by r 1 2 explicitly but it is there there is no problem so it has nothing to do with the octagonality and then I said then these are 1 1 2 2 so I can further do a dummy variable interchange so this is because of dummy variable interchange this is simply because they are real if they are not real you could not have done that so the question is they are real orbitals yes so I hope you have given the right logic I only want the right logic but it is nothing to do with octagonality then it should be 0 I think you are getting confused in commutation of course they are commutative all these fundamentals is absolutely clear I mean we do that all the time that even operator is a function of x it commutes with another function of x that is what we do problem is only the momentum the momentum guy does not come with the function of x because it is a delta x so whole problem of quantum mechanics come because of the px if Hamiltonian has only v of x nothing would have happened then all functions would have been function of Schrodinger equation I can function but unfortunately no system can exist without the kinetic energy and that is where the px guy will come and create lot of havoc because it does not come into the x so I think that those parts of quantum mechanics should be very very clear now it is good that you are asking those questions because initially I did not understand but it is good so if I have been able to write g of 1 all that I have to do is to plug in here so let me plug it in here and then I will explain what was this question actually little later but let me plug it in here there is still no confusion but he has some something in his mind that the chi I 1 must be always on the right which is not getting for the exchange so that is something that I will explain later so now what I am going to do just write this g of 1 and get write down the Hartree-Fock equation so it is H of 1 chi I 1 plus sum over j integral I deliberately write it in full form because of the reason and then what I do is that anyway 1 and 2 commute so I push this guy on the right because then it looks nice then I am integrating over chi j 2 d tau 2 I will also finish and then write this as a chi I 1 I hope you understand that this is same as this it is just that I am first completing the d tau 2 integration then writing the chi I 1 because chi I 1 cannot be integrated when I am doing the d tau 2 integration chi I tilde 1 is a constant so it can come out of the integration sign that is what I have done again to be consistent let us put it in this then the next part which is sum over j now the problem will come what is trying to say chi j star 2 1 by r 1 2 now the integral over 2 explicitly consists of chi I 2 I cannot write chi I on this side so it has to be actually chi I tilde and neither can I write j outside because the sum sum sum sum is over j so everything has to be done together chi j 1 d tau 2 of course it is d tau 2 it is still d tau equal to sum over j lambda i j chi j some of the text books from the beginning they use this as a lambda j I but do not worry about it it can be used right from the beginning lambda j I or lambda i that does not matter you continue to be consistent that is all right so again putting the tilde this actually is my Hartree-Cock equation okay this is actually an Hartree-Cock equation which is a somewhat long standard Hartree-Cock equation then what you know in the sense that this is not an eigen value equation of a one particle operator because I am not able to extract that one particle operator yet first of all this right hand side does not have a eigen value structure the left hand side also these two terms are able to extract but here I am not able to extract chi I tilde 1 so I cannot write something acting on chi I tilde 1 assume that this was the diagonal assume this was diagonal I am still not able to write something acting on chi I tilde 1 equal to some number times chi I tilde 1 because of this exchange term and that was his question am I right okay so we will we will we will see but this is actually a Hartree-Cock equation if somebody says this is Hartree-Cock equation right absolutely right it is actually a very general form of Hartree-Cock equation and this is what is called in general Hartree-Cock equation sometimes it is called non-canonical Hartree-Cock equation we have also may have heard canonical and non-canonical this is actually non-canonical Hartree-Cock though what is practiced in all the textbooks all the program is a canonical Hartree-Cock equation so we will actually later see how to convert this into a canonical Hartree-Cock equation but this is correct the result energy that you will get out of this chi I tilde 1 after solving will be same as the canonical Hartree-Cock equation energy there will be in varying so this is correct there is no problem with this equation let us only try to understand the equation clearly let us assume this is diagonal for the time being so that the understanding would be very easy and let us assume this is Coulomb and exchange terms are not there then what would happen in your equation a very simple H of 1 chi I 1 equal to number times chi I 1 which is basically the non-interacting picture that I told you forget about 1 by R I j your Hamiltonian is sum over H of i then your spin orbital should have been Eigen function of H of i that will come if you neglect these two terms and make this diagonal that is important so it is you can see that that structure is already there but because of the Coulomb and exchange we ask this question is it the best one now we have got the answer this is not the best this is the best because we have applied we have obtained this by variation method remember