 Let me go back and analyze the fact that when I said that this phi i is a sum over the basis and the overlap between the basis functions or the original basis functions is s nu mu. I must mention here that we started with spin orbitals which are orthonormal which means my space orbitals for closed shells are also orthonormal. So I must mention that phi i, phi j is also delta ij. However the atomic orbitals of course have an overlap matrix. If this was also orthonormal then of course you know that this coefficient would have a very simple relation that the C-dagger C would have been an identity matrix which is what happens when you expand one orthonormal function in another orthonormal set. But this is no longer true here. In fact what is true is that you have C-dagger S C equal to identity matrix. Again you should be able to prove this as a practice problem. I have just given you all the pieces. So you should be able to see very quickly that the C-dagger C cannot be identity matrix but C-dagger S C is identity matrix. Please practice this. Only I will not do here. So I am just mentioning but these are things to be proved. If that is so then you look at the P matrix. P is 2 C C-dagger. Because the name is P you may think it is a projection operator but it is not actually because projection operators are supposed to be idempotent. If you quickly see this is not idempotent. So P is of course a Hermitian matrix that is very clear but P square is not equal to P. So if you do P square it becomes 4 C C-dagger C C-dagger and of course this is not 1 and even if it was 1 this 4 would have remained. So P square would not be equal to P because of this factor 2. So it is quite clearly it is not an idempotent matrix that is something that you should get out of your mind because many of you moment I write P in projection operator it is not really a projection operator. What does this actually there is an equation it satisfies again you should be able to prove this. From this expression I give you the expression try to prove it at home. Again practice problem. P S P what is P S P? 2 C C-dagger S C. So you have C-dagger S C which is identity actually yes. So P S P but then there is a 2 here is equal to 2 P again I will not do this but I may ask in the Midsum exam it is very easy I hope all of you will be able to prove this that P S P is equal to 2 P which actually tells a very interesting thing that in case I would have walked in an orthonormalized basis in and I would have defined my P matrix in using the orthonormalized basis which means my A mu prime which is an orthonormal basis I had a C prime remember the same molecular orbitals are expanded in terms of orthonormalized basis. So then this will become C prime this becomes A prime. I define my P with respect to that C prime new P let us say P prime. Then of course you can very easily show that this S is identity matrix so you can very easily show that P square is P is still not an idempotent matrix but can you now tell me what is an idempotent matrix? What function of P will be idempotent matrix? In an orthonormalized basis let me see how quickly who can come up with an answer. See for an orthonormalized basis let us say P prime you know everything is prime this is identity but this is still not idempotent but some very simple function of P is idempotent can somebody tell me? I hope the question is clear so I have a P prime so P prime S prime is of course identity is still equal to 2 P prime but this is now identity I will not write it I am not deliberately written the identity because it will become even more easy. So can you tell me a simple function of P prime which is idempotent? P prime is still not idempotent you are pretty close 1 by 2 P prime because you see here P prime square is 2 P prime divide by 4 P prime square by 4 is P prime by 2 P prime square by 4 is half P prime square is it clear? But note however that this half P is not idempotent matrix P defined in an orthonormalized basis so I am talking of half P prime this of course is not idempotent matrix that is very clear. So these are some of the interesting things that you can just play around the algebra of the projection operators and I hope it is clear everybody can see this why it is so simply divide by 4 on both sides so you will see that the left hand side is P prime by 2 square so a very simple function like half P prime in an orthonormalized basis would have been an idempotent matrix but this by itself is not an idempotent matrix that is clear because of the fact that the basis is not orthonormal orthogonal and anyway this factor is there okay good so let us move ahead with further interpretation of the P you are now very important thing that we can do from charge density bond order matrix is what is called the population analysis population analysis actually gives you the number of electrons and each atom or the charge of an atom in a molecule. Now many of many chemists are always interested that after the molecule is formed what is the charge let us say water molecule what is the charge of oxygen what is the charge of hydrogen now I will quite clearly show that that is really an arbitrary thing but the chemist are so much obsessed with this charge the charge is an arbitrary thing neither can it be measured nor can it be calculated exactly but unfortunately whole of chemistry depends on that particularly in extreme cases like ionic completely ionic bond you can of course the results are invariant but in all intermediate cases of covalent and ionic the results are not calculable exactly it is there are completely arbitrary so I will first show you one important population analysis which was developed by Mulliken I think many of you have heard this Mulliken population analysis it is very popular analysis used in Gaussian and many programs and many people actually give this population analysis to show what is the charge distribution of an atom in molecules so how is it done I will first show and then I will show why is it arbitrary. So let us go to the definition of density and this is something that we have been doing in density functional class but at this point I will merely say that given the single determinant if I integrate psi star psi over all coordinates except one all space spin coordinate except one and for that one also I will integrate over the spin then I will get electron density I will simply write in terms of the spatial molecular orbitals the result is simply i equal to 1 to n by 2 2 times phi i star r phi i it is very easy to show this actually again you use actually this later rules to integrate that from the determinant and you get chi i star chi i each chi i is alpha and beta so it is phi i star phi i 2 times. So it is very easy that take orbital density multiply by 2 simply because each orbital has two electrons and you get the total density for as hard to forget is very easy so this is for the closed shell and again I restrict the discussion for the closed shell but it is very easy to generalize but just by the occupation of the orbitals instead of 2 I will put actual occupation of the orbitals if I have more than n by 2 orbitals special orbital. So this is my electron density and that has a very nice interpretation that if I integrate the electron density over the volume element it is n. So when I do the when I calculate the electron density remember in my definition of integral chi star chi over all integration I multiply by factor capital M so that I get this normalization constant because all of you know that in a system the integrated electron density is total number of electrons. So that is the reason I have to ensure this and it is very clear that that is the case because when I do the integration you can see that each of the integration gives you 1, 1 multiplied by 2 is 2 and you have a sum over n by 2 so it gets n so it is very clear that this is following that this particular thing. Now I can expand start to expand everything in terms of I can integrate and I can write expressions of n in terms of charge density bond order matrix. So basically what I will do I will have to expand this phi i's in terms of the atomic orbitals I hope you should be again able to do you will get a coefficients you will get a mu a nu of r integrate because there is an integration you get an overlap matrix times the 2 coefficients. Multiply those coefficients make the projection operator practice problem can you show that n is equal to trace of p s or it is sum over mu p s mu over the entire basis this mu is a basis now so I will again not do it but it is a practice problem you should be ready before the exam. The strategy is very simple start from this integration this rho of r you substitute this so you have integral phi i star phi r whatever 2 sum over i then each of the phi i you expand in terms of the atomic orbitals so the integration will be over now atomic orbital basis which will give me s the rest of this c star c will give me projection operator you just show that this value this I will of course vanish because I will go into the projection operator sum over mu and sum over mu will come because of this expansion and you should be first able to show that this is equal to this and hence it is equal to trace mu mu is just diagonal element here mu mu not mu nu mu mu only diagonal element please you should be able to show this by substituting this phi is here and expanding phi is in atomic orbital then integrate of course it is an integration so it is a number n is a number so what we are seeing showing that if you calculate a new matrix called ps matrix I have a p matrix I have a overlap matrix calculate this ps matrix then take the diagonal element sum of the diagonal element that is your total number of electrons now this gives a very nice interpretation it says that now I can relate n to the basis remember my originally n was to the molecular orbital what was my how was it related all n electrons were distributed to per each molecular orbital for closed shells now I am saying I can distribute the n to the atomic orbitals saying that each atomic orbital has a contribution from its ps diagonal element and that sum is equal to n obviously it will not be 1 or 2 it will be much maybe much less because capital M is much greater than n its sum over all capital M is equal to capital M so I can for example have atomic orbital basis I have ps of 1 1 plus ps of 2 2 all these together will become now capital M so this gives a very nice interpretation and this is what molecule can did that after I construct the molecule I can now say that the atomic orbitals my atomic orbitals are such that if I calculate the diagonal element of ps on those atomic orbitals I will interpret that that is the population for the atomic orbital and I sum over all such atomic orbitals which of course are distributed in several atoms will give me total number of electrons and this is the atoms in molecule picture so for example in a water molecule I want to now get the number of electrons on oxygen atom that is trivial because now what I will do I will not sum over all atomic orbitals but I will sum over atomic orbitals which are centered only on oxygen okay so that will now that sum will give you that number of electrons on oxygen and subtract that from the charge of the oxygen you get the formal charge okay so let us say so I can define a charge of an atom A as N A 1 minute here let me write this as some Q A as Z A which is the atomic number minus sum over ps mu mu but now mu is not summed from 1 to M mu is summed over all the atomic orbitals which are centered on A remember my M atomic bases are distributed on different atoms so I will only pick up that particular atom sum over all ps mu mu that becomes my total number of electrons and of course I divide from the charge Z A that gives me the atomic charge Z A that gives me nuclear charge that gives me the charge of the atom so if it is more than 8 for example this is more than 8 that means it is a negative charge you have more number of electrons than 8 on the oxygen atom which is what is going to happen if you do a calculation of water and that is what you report Mulligan population analysis alright very nice except that now I will tell you some stories that this is not the only way to do it and I can keep telling you any number you have to believe that number and that is why population analysis arbitrary because this is there is actually there is no experiment to show this that this is the charge of oxygen in water molecule that is the problem p is for the whole molecule s is also for the p is actually coefficient yeah for the its atomic orbit it is also an atomic orbit s is also an atomic orbital so ps is also over the whole molecule but of the atomic orbital basis now so that is the important thing once I can recover my number of electrons based on atomic orbitals I can now recover number of electrons for an atom by simply summing over those basis function which has centered an atom total sum will be n because this is very sum this is very simple yeah because because I am summing only over the atom a so I will so these are my each atomic orbital so the contribution of all oxygen atomic orbitals is my total number of electrons and oxygen that is how it is interpreted it is a population of that orbital so I am summing over only oxygen ps of this is the total sum the trace is the total sum I am taking diagonal elements of those atomic orbitals which are only on a so I will say that is my total number of electrons and a no I mean that is the okay I mean this is if you say that this is my interpretation yes I agree that when I have got this result I am now interpreting that each ps me me is the number of electrons on that atomic orbital that is an interpretation yeah I agree but that was the molecular interpretation because then the sum comes out very nicely okay so that that seems to be quite logical interpretation actually I have no problem with that interpretation but I will say even with that interpretation I have arbitrary definition that is what I am going to show now okay so note that however if this p was formed on an orthonormalized basis again I repeat that means if it was p prime then of course if you just take trace of p prime that would have been total number of electrons okay in an orthonormalized basis basis would have changed so if I would have formed the projection operator on that basis so I call it p prime okay then p prime would have been sufficient because s is not there but that is a triviality what we don't we don't I am talking of this one or are you talking of this one that's what I am asking this is integer this is what I have asked you to show so when you show this integer but this need not be an integer I am not said that the charge of an oxygen atom is always an integer they say minus 0.2 that's the reason this however is integer and they have to show this I am not shown this so this is obviously integer but when I am taking a sum over mu on centered on a particular atom it need not be an integer but sum of a sum of b sum of c whatever number of atoms a total like would be again integer and that's number of n that we have shown this. Mullik can use this for his definition of charge the charge of this is like this is the charge this is the atomic number z a and this is the p s what I was mentioning is that if this p would have been calculated in an orthonormalized basis then of course s is not required it will be only p or p prime I would call it p prime whatever so that is much better because this p s is more difficult to construct then p prime has a direct significance here I don't have so although I am calling it charge density bond order metric the in terms of charge density p s has a direct significance not p but that is because of only this problem is coming only because of orthogonalization of basis in an orthonormalized basis p directly would this but now what I want to tell you that this is even with this interpretation there is a complete arbitrariness because you know that the trace is invariant under commutation so for example trace of p s first I can write this as trace of p s to the power half let's say s to the power half you agree with s to the power half s to the power half in associative law multiplication and the trace is invariant under commutation so now what I can do I can push this here so it is very easy to show that this is also equal to trace s to the power half p s to the power half all these trace would be equal to number of electrons and no question but now I can use for example and this is what left indeed use this as the population analysis so now what left indeed was to calculate charge on z a minus s to the power half p s to the power half mu mu centered on a the problem is the following while this quantity is invariant this quantity is not invariant because this is not an entire trace this is some over only selected one and then the results will be different charge will now be different so what is a charge of course some of the total charges will be constant that has been defined by the molecule if it is a neutral water molecule some of the charges will be 0 but charge on each atom is arbitrary this particular one is called the left in population analysis this is very popular as well with Moolikan just like left in orthogonalization you look at left in left in did s to the minus half left in this s to the half p s to the half left in likes everything symmetric you'll get about that man this is symmetric this is unsymmetric you have only left in said why I should go only on one side of course trace is invariant should could have been sp that is also equally bad is a distributed beautiful alabdian was a great mathematician incident they are equally agree but this quantity is not equal why because the entire trace is invariant this is not the trace is only sum over selected number so the theorem says that sum over all is invariant the theorem itself is sum over all here you don't have sum over all so there would be a cancellation okay so that is why you get different charge so this is Moolikan this is left in and of course you can have your own charges no problem you can somebody can say no no I will put left little less I can put s to the 0.1 p s to the 0.9 no problem the overall it has to be s but the s can be redistributed anyway you like and I can keep getting population analysis just to show you interpretation is same I still get different results nothing called right because charge is not experimentally determined so anyway there is nothing called right charge and I dare say that is charge is an imagination by a theoretician theoretician does not believe itself but the experimental is believe that because experimental is can't calculate charge but entire interpretation of why a reaction has taken place is based on charges so they use Moolikan and Loughdin whatever it suits their need yeah but I must say these are the two most popular Moolikan and Loughdin there are many others yes there are Herschfeld charges I don't want to go into that I myself have written several papers based on the reactivity on Herschfeld charges and I will show that if you calculate reactivity which is Fukui function neither Moolikan or Loughdin is better according to me Herschfeld is the best but not for charges only for reactivity and I wrote three papers in JCP 99 2000 2001 you can check on three long papers on why Herschfeld is good yeah but dipole moment is an overall thing some of everything you can still fit it to you can still fit it to different charges I can still model in a different way that's not a direct measurement really speaking okay so I will come to dipole moment later I think tomorrow I'm going to also now tell you how to calculate dipole moment through the P it's also very nice all right so I think I will stop here so just to tell you that the charge density you can already see why charge density I will now tell you why bond order tomorrow and how this can be also used for dipole moment calculation if you just have P but P is always in atomic orbit so whenever I do P something it is all sum over atomic orbit no longer molecular okay so that is very important and this is also the charge density bond order matrix is very important because there is a whole class of theories which are called atoms in molecules AIM these atoms in molecules essentially mean that once a molecule is formed you want to get back atomic character and your P your P is the vehicle P gives you that so P is a very very important matrix everything has to be done through P because P is what it is the coefficient of an atomic orbital for a molecular order so that is why is sum overall so P gives you if you look at P expression of P you have P mu nu is two times sum over i equal to n by 2 C mu i C nu i star right so if you look at each of the P diagonal element let's say diagonal element it is a contribution to A mu is equal to nu contribution to an atomic orbital from all molecular orbitals so I am now doing the reverse I initially said a molecular orbital from all atomic orbitals now an atomic orbital from all molecular orbitals so P allows me to do the reverse so in all AIM picture P is a key quantity okay so just remember this charge density bond order matrix and so when you say bond order it's also atoms in molecule picture after molecule is formed now we are saying how many electrons are there between two atoms okay for bonding so all those pictures the P is very very important otherwise you could never get it but unfortunately as I showed you for charge there is an arbitrariness because of this mathematics this mathematics and and Loftin cleverly exploited this to find another charge and he claimed this is better and many cases this is better because his claim to this is basically because it is symmetric he said I have redistributed this one at point five and point five okay so that's why I think it is good okay so I will stop here today