 So we were discussing the population analysis and said that if you have the PS, if you construct the PS and take all the basis functions over particular atom, sum them and then subtract this from the charge of the atom, this becomes the atomic charge, then it becomes the total effective charge of the atom A in the molecule. So this becomes the total electrons and this is the atomic charge, so that subtraction is the charge of the atom in the molecule. However we said that we can, since stress PS is equal to N and the interpretation was that that we have distributed these diagonal elements over different atoms, we could do some other population analysis like S to the power half PS to the power half, which is also equal to N. So we could write, so this is Moolikan, we could write the Lafdin as charges as the same ZA minus sum over mu in A S to the half PS to the half and indeed it is possible to find many such population analysis because of the fact that I can construct trace of many such matrices which are invariant by splitting S, P, whatever many, many ways we can do that. So that is essentially the idea, however this number which is summed over particular atom is not invariant, so we will get different charge density in different kind of partitioning scheme, so it is not really a fixed partition, it is arbitrary and that is the reason we always say Moolikan population, Lafdin population, Herschfeld populations, the different population analysis, we will restrict only to Moolikan or Lafdin, of course there are several other Bader population analysis, so several other population analysis is there but we will currently restrict only to Moolikan and Lafdin for this course. So let us go ahead and see how the density matrix, charge density bond order matrix can be used for certain other properties, so one of the properties that we are interested is the dipole moment, so dipole moment is a useful property, so I hope all of you know what is dipole moment, it is actually an operator which has two parts, one is a nuclear contribution, one is electronic contribution, so if you write the electronic part, the nuclear part is actually fixed constant, so I can always add this later, so it is the electronic part is what we are really interested and this is basically minus e sum over i r i, where r i is the vector, so that is my electronic part, so if I take the expectation value of a Hartree-Fock wave function with respect to this operator, I will get what is called the electronic dipole moment and then I have to add to it the nuclear contribution to get the total dipole, which is just like the Born-Oppenheimer, I add the total energy, I will add the nuclear, so what is important is to eventually calculate with psi Hartree-Fock, mu psi Hartree-Fock, so let us keep mu as one operator, of course essentially mu is nothing but r i, so it is a sum of one electron operator, i equal to 1 to n for all the electrons, so x i, y i, z i it is a vector, mu is also a vector operator, so we have to find out the average value of this operator, so how do you find out? We now use our rule, if you remember the average value rule for the one electron operator, just as we did for the Hamiltonian sum over h of i, so we will exactly use that and that result will be sum over, somebody should tell me now, what is the result? Note that this is a sum of one particular operator, so you can, if you want you can write this as sum over i, some mu i, just to make it comfortable for you or you can say minus e and r i, so what will be the operator here? What will be the expression here? Yeah, sum over i, chi i, mu i, mu i is nothing but r i basically, nothing but r i multiplied by the electronic charge with the negative sign, so this is what the dipole moment operator is, this is a one electron operator, so once again I sum this one expectation value of this one electron operator over every spin orbital and sum, so calculate over every spin orbital and sum, just like I did for the one electron part of the Hamiltonian, so it is the same rule, the rule is exactly same, so this then for the closed shell system I can do spin integration, so when chi i has phi i alpha and phi i beta, I can do spin integration, so what is the result? i equal to 1 to n by 2, 2 times yes, phi i, mu i, phi i, right, no, it is an average value, just like I did Hamiltonian, Hamiltonian or sum over h of i, so how did I do chi i, h chi i, so it is exactly same, I mean I am just writing mu i, you can say just mu, but you have to understand what is mu, I have written mu as sum over mu i, so that is the reason I wrote mu i, nothing else, but basically if you construct mu as just E of r, so r i is basically not important, what is important is the coordinate, so if this is 1, this is 1 then this should be 1, so in a simple way to understand, so r 1, so that is my, that is a dummy variable, so your mu i, mu i is just r 1, but it has x, y and j component, so basically it will become 2 times sum over phi i, so you have x, so you have minus E, let us say 2 times minus E into 2 times, phi i x phi i for the x component, so in a vector sense I can say that mu x equal to, mu y equal to similarly minus 2 times E sum over i, phi i y phi i and so on, you can write mu z and so on, so these are vector quantities, so each x, y, z you can write in this manner, so unlike your h, that is only different, this is a vector quantity, so I have to write for x, y and z, but each x, y, z after I can write a separate functions, your dipole moment is a vector quantity, so I have mu x, mu y, mu z, so whichever direction you want to calculate, you can calculate and then you can take the overall dipole moment as an average mu x square, mu y square plus mu z square square root, so that is an average dipole moment, so I can calculate this as an overall, so the point is that calculation of dipole moment is fairly straightforward, but I also need to write this out in terms of my atomic orbital, because my entire expression I want to do for Hartree-Fock without molecular orbitals, so I can do the same thing here, so let us start with 2 times sum over i, phi i mu phi i, so mu is now a one particle operator, mu is just one particle operator, so if this is 1, this is 1, this is 1, so it is just ERI, so mu is not the total operator, but just the one particle operator, so how do I expand this now, so I do the same expansion, for phi i star I bring in C mu i, so mu C mu i star C mu i, then mu i, then mu i, then mu i in A mu mu A mu, yes sum over i, chi i is not summation, psi is sum over i, psi is a product, is an anti-symmetric product, no, no, the orbital operator, that is why this sum has come, I think you are going back to this later rule, if you remember this later rule we said Hamiltonian is sum over h of i plus sum 1 by Rij, so when I took this part, let us say I call it H0, okay, what was the result, it was sum over i, chi i, h chi i, it is just that, I mean I am not doing anything new, it is just that it is a vector quantity, so these are all vector quantity, that is the only thing, these are all vector quantity, so I have to have x, y and z, other than that I am just applying this, okay, so I am again repeating that there is a sum over electron that need not come, sum over spin orbitals, takes care of that, rest of them by orthogonality all become 0, you may say that n times this should come because here there are n factorial but all that is taken care because there is a 1 by n square root n factorial in psi, please understand, when psi, I expand psi I have n factorial here, n factorial here, you have n terms here, I have not proved it, so when you do this there is a n factorial terms here, there is a n factorial terms here, there is a n terms here, however a term here with an off diagonal term here would become 0, so only there are actually not n factorial square, there are only actually n factorial terms, okay, and there are n terms here, so how many terms are there, if you have number counting there is a problem, n factorial into n, here I have only n terms, that n factorial is being cancelled because of the 1 by square root n factorial on either side, that is all, the number count is very easy, okay, so your total number of surviving terms are n factorial into n, which then eventually cancels 1 by square root n factorial gives you this term, and the terms are grouped such that these are the n groups, so out of these n factorial n terms, one set of n factorial is same, another set of n factorial is same, another set of n factorial is same, so when I cancel this 1 by n factorial, these n non-equivalent terms come, they are non-equivalent because the spin orbitals are different, yeah because the spin orbitals are different, so there is a simple logic, I have not proved it, but it can be proved very easily, so these are of course non-equivalent, kias are changing, so only those sets of n terms are there, but total number count is matching, so there will be no further summation, okay, so when I do this, now you can realize that I have 2 times sum over i, so this becomes my charge density bond order matrix, so I can write this as sum over mu nu P nu mu times A mu dipole operator, one electron dipole operator, okay, so many times just to avoid this symbol, we can write this as a D as a dipole operator, just to avoid this recurrence of mu and nu, but does not matter, so you can now write this as a trace of, you can quite easily see, you have a sum over mu which is summed up, okay, so let us write this as a dipole operator, so let me call this set of terms A mu, sorry mu A nu, this is one electron operator, again I remember this is one electron term, okay, let me call this D mu nu as a matrix element of dipole operator, so then this can be written as sum over mu nu P nu mu D mu nu, which is nothing but trace of PD, when I multiply PD one summation is done and this other summation of a nu ensures that it is a trace, okay, so it is trace of PD, so all you need to do is to calculate the dipole matrix element, one electron operator again, X i, again this is a vector quantity, so X i, Y i, Z i between the basis sets, call this some dipole matrix, so I will have three sets of dipole matrix, one for X, one for Y, one for Z, okay, and then take this trace of the three sets of quantities, I will get mu, overall mu for X, Y and Z depending on what you use, DX, DY, DZ, so just a vectorial quantity, yes, same, same, because I have used the same, I have used the same definition, 2 sum over I, C mu I star, C mu I have called P mu, so it is same, I mean I could have still called it P mu nu and as long as I multiply right quantities, so then I have to change my multiplication rules, that is all, so this is to be consistent with the matrix multiplication rules, all right, so all you need to do is to calculate this X i, Y i, Z i, X, Y, basically not X i, X, Y, Z, what is X i, there is nothing called X i, it is X, Y, Z, because i is dummy, electron coordinate is dummy, so all you need is to calculate the integral of X, Y, Z, integrals of X, Y, Z over the basis for dipole moment and that those integrals are stored in D, D matrix, D is a vector quantity, vector matrix, so you have DX, three matrices DX, DY, DZ, okay, so you calculate the integrals of X, Y, Z over the basis and store them DX, DY, DZ, you use this as a trace of PDX, PDY, PDZ to get mu X, mu Y, mu Z, okay, so that is your electronic contribution and then of course you have to add to it the nuclear contribution, so nuclear contribution is also a vector quantity, so you have RA, capital RA, so you have to do the ZA into capital RA, so again RA has XA, YA has ZA, so add the nuclear contribution and XYZ, so you get the total dipole moment in XYZ, is it clear? So the strategy is very simple, so everything that I have I can, I am just reducing spin integration and then in terms of the orbital, space of these, the basis orbitals, yes, yeah, so if I have, this is a vector quantity, right, so your mu is a vector operator, it has XYZ, mu is nothing but XYZ, okay, RI vector is nothing but XYZ, so it is a one electron operator of XYZ, so I compute X integral over MUA nu, why not, no, you do not have to do plus, you can do separately and then later on also you can sum, it does not matter, you can write it as a vector quantity but basically you have to do XYZ integrals, okay, so that is your problem, how will you write it? So once I am MUA, I can always write MU equal to MUXI, okay, plus MU, that is your, that is a separate point, but what is required is basically to calculate the integrals of XYZ, that is the new thing that is required, which I have not done in Hartree Fock, remember when I did Hartree Fock, I had integrals over half Laplacian, I had integral over 1 by R for atoms, 1 by RI sum over RI, R minus RI for molecules, whatever, so H of I and 2 electron integrals, 1 by R12, but this is a new integral that you have to calculate, R which is XYZ, so that is what I am saying, that calculate the integrals over XYZ, call them dx, dy, dz matrix and simply take a trace of this, so that becomes your very simple idea, again you do not need molecular orbitals, I can write everything in terms of atomic orbitals, yes, any question? No, we are not, we are not worried about nuclear spin here, no, no, no, no, there is no, in fact this, nuclear is just a positive charge, fixed positive charge, okay, we are not, in fact we are not even taking, there is a coupling between the spin and nuclear, those things are not taken, so those are all relativistic regime, in fact that coupling is also concerned with relativistic regime, here it is very simple, they are point charges, fixed point charges, okay, so just as, this is a classical dipole moment, so the nuclear part is just classically calculated and added to the electronic dipole moment, electronic part has to be quantum, so that is important, because electrons are not fixed particles and that is the first thing that we have reached, right, the electrons are particles stationed at one point, okay, so that is all we are discussing, alright.