 Over the last few classes we have written a lot of mathematical expression, done some algebra, done some calculus and we have not done most of it. Most of it I have said that it can be done this problem in this book, that problem in that book you can do it yourself, to tedious and we have not done it. We have not done it for a reason. If you try to do it then it is impossible to stick to the limits of NPTEL and also it will be very boring to go through all that mathematics and then we lose the site of chemistry. As I said many times what we really need to understand is the philosophy and the logic that goes into this. Mathematical detail and we do not really need so much. We do but we do not have to work out everything at least for now. But today it is more or less a storytelling session. Well the story also involves a lot of mathematical expressions but we will just do some very very simple algebra by and large I will just tell you what happens when something is done. And to start with I show you a picture of an elephant and you might be wondering what is going on here. Do we have an elephant in the room? No actually look at this picture carefully. It is really a plot is not it? x axis is something independent variable y axis is a function of x and to know more about the functional form you can visit this website. But the point that we wanted to stress here is that by using proper combination of mathematical functions one can actually generate whatever shape one wants. What you need is sufficient number of terms. This if you go into this website you will see this is the picture of an elephant generated using a single parameter. What that is? What is the detail that we will not get into? I was actually looking for this picture of an elephant that is drawn using 30 exponential functions I did not find it on the net I found this. And the reason why we show this is that this is the way we are headed now. Until now we have been talking about exact solutions, approximate solutions, worrying about analytical form and trying to make sense of it out of it. Slowly we are entering a regime where we cannot do that any longer. It will all boil down to drawing the shape we require like an elephant using a suitable combination of functions. And while doing that the saving grace of course is variation theorem or upper limit theorem which tells you that you can never get an energy that is less than the actual energy. So, let us see how far we can get today. Last day we had talked about the perturbation theoretical treatment of atoms. Today we will talk about variational methods and we will need maybe one or two more lectures before we can complete our discussion of atoms. So, to recapitulate we now are sufficiently familiar with the Hamiltonian of many electron atoms. We know that it is a sum of n number of one electron Hamiltonians and there are these Nc2 combination of electron-electron repulsions. These you cannot ignore you have to account for them somehow and the way we are accounting for this so far is considering something called your shielding constant and considering that this the effective nuclear charge is what an electron sees and effective nuclear charge is the actual nuclear charge minus the repulsion that an electron experiences from the other electrons. And doing that what we have done is we have modified the wave function instead of z will write z effective and then we showed you some result remember this value minus 78.99 this is the energy of your helium atom. So, we are trying to get close to this 78.99 electron volt that is where we want to get in this simple shielding constant kind of calculation we are showing you minus 77.68 electron volt and today finally we will see how we can get to the shielding constant using variational method. And also this is something that we will not use today but in the next class we are going to get back to this later determinant determinant way of representing wave functions after today's lecture. And one thing that we talked about yesterday is that from now on to just make things a little simpler we are going to use atomic units where mass is the mass of an electron unit of mass is mass of an electron unit of charge is charge of an electron unit of angular momentum is our good old h cross unit of permittivity is 4 pi epsilon 0 electric potential has unit as potential of an electron in the first pore orbit and I have not corrected this m to n yet where it is of copy paste and Bohr magneton is the unit of magnetic moment energy unit is Hartree remember and one Hartree again I am stressing this because it is very easy to lose sight of this one Hartree is actually 27.21 electron volt substantial in atomic scales. So when you use all this Hamiltonian for helium atom that we have written earlier was down to something simpler in atomic units but while doing this please do not forget that this m h cross square 4 pi epsilon 0 they are all there if you want it in if you want to write it in the expanded form you should not forget which goes where of course when we calculate energy then we do not have to worry because we have this conversion 27.21 electron volt is equal to one Hartree all right and with that background I have kept this same as what we have written earlier from next slide onwards you will see that we have written it in atomic units. So we discuss a perturbation theoretical treatment for ground state of helium atom where the sum of the one electron Hamiltonians is taken as unperturbed Hamiltonian and the term involving electron-electron repulsion is taken as the first order correction to Hamiltonian the uncorrected wave function or 0th order wave function is taken as the product of the 2 1 s orbitals it about helium so 2 1 s orbitals so again we are working within the ambit of orbital approximation and then we said that first of all e 0th is z square by 2 minus z square by 2 that is equal to minus z square in Hartree and we said that we just go through this problem 6 of chapter 8 of McQuarrie's book on quantum chemistry you will get to learn that the first order correction to energy is 5 8th of z in Hartree so energy in Hartree is minus 2.75 which translates to minus 74.83 electron volt remember what was the value experimentally minus 78.99 is it not 78.99 So that is where it is we are not really we are not really there yet and then we said that this is only first order correction you can work on second order correction higher order correction and 13th order correction more or less gets close to the value today now we are going to talk about variational method for treating this many electron atom problem we are going to talk about helium atom and we are going to arrive at what we had talked about earlier this effective nuclear charge here I have written the Hamiltonian in atomic units half del 1 square minus half del 2 square minus z by R 1 minus z by R 2 plus z by R 1 2 of course half del 1 square minus z by R 1 that is the Hamiltonian for electron number 1 half del 2 square minus z by R 2 is the Hamiltonian for electron number 2 and the z by R 1 2 is the term for which we have to do so much of discussion the electron electron repulsion term okay what is the trial wave function we start with the same trial wave function okay but of course we will play around with it as you will see same trial wave function we take 2 1s orbitals I hope you remember that 1s orbital is of this form root over z cube by by pi multiplied by e to the power minus z rj okay this is the 1s orbital written in terms of the jth electron if you do not remember the coefficient it is okay but one thing you should know is that this 1s orbital is essentially an exponential decay in r alright this is my trial wave function I start from there okay trial wave function again is the same wave function that we had in our orbital approximation 1 electron hydrogenic wave function orbital next thing to do in variation method as we know is to find out this functional epsilon of pi which is the similar expression as energy just instead of the exact wave function psi we are using the trial wave function phi so we will work with this functional epsilon of pi integral of phi star h hat phi over all function space in this case over all values of rj and phi star is equal to phi in this case we are talking about 1s orbital now what is the variational parameter what we will do is well even before that now when we talked about time dependent perturbation theory we had referred to problem 6 now we refer to problem 7 of the same chapter if you go through this if you work it out you will get this expression for epsilon of phi z square minus 27 z by 8 right so then looking at this expression and also looking at the expression of psi if you think what is the handle that we have what is the parameter that we can play around it what can be our variational parameter can we play around with rj makes no sense right rj in any case and we have to formulate the problem that way so the only thing that is left really is z okay nuclear charge and it makes sense to use z as the variational parameter because as we have discussed qualitatively already that this nuclear charge is not really the charge that is failed by an electron and electron feels an effective nuclear charge which is nuclear charge minus some shielding constant sigma and essentially we are trying to find out sigma so one way of finding out sigma is to keep changing the value of z vary the value of z and see for which value of z we get a minimum in the value of the functional epsilon we will call it E min as usual and that is going to set an upper limit to the energy of the system right so z will be the variational parameter also wherever we get the energy minimum that is going to be the effective nuclear charge so I am not really calculating sigma I am calculating effective nuclear charge and I will I can find sigma by subtracting the effective nuclear charge from the actual whole nuclear charge that is there in case of helium it is 2 we are talking about helium do not forget okay so z is my variational parameter this is the expression for the functional so what is the next thing to do I want to find the minimum of the epsilon of the functional epsilon so I have to take the first derivative with respect to z equate to 0 what is the first derivative 2z minus 27 by 8 right first derivative of epsilon with respect to z is 2z minus 27 by 8 very easy and that is going to be equal to 0 for us to get the minimum value of the functional epsilon okay as we have said that would be the upper limit of the energy calculated for the system and as we have also said correspondingly the z value for which the minimum is obtained that is going to be z effective okay so it is well formulated very easy algebra it is all done what is the value of z here this z is z effective as we said we will write z effective 2z minus 27 by 8 equal to 0 so a class 5 class 6 student can work out that z effective in this case is 27.16 what is 27.16 it is 1.6875 I hope I have got the division right please work it out yourself in case it is wrong I do not think it is wrong it is fine so the point is it is less than 2 what is the actual nuclear charge it is 2 what we see is we have calculated a value of z effective which is less than the actual nuclear charge and from here we can calculate the value of sigma shielding constant to be 2 minus 1.6875 that will come to how much 0.3125 isn't it yeah 3125 0.3125 that is the value of sigma that we get from this very simple variational treatment of the helium atom ground state problem okay so we know how to find the energy minimum which is the upper limit of energy actually do not get confused because I am saying energy minimum and upper limit energy minimum means minimum of epsilon the functional epsilon that sets the upper limit of the energy calculated by this method okay remember upper limit theorem so we have not only been able to find the energy that we are looking for we have also been able to determine the effective nuclear charge and that is great okay now what will we do to find the value of this epsilon we take this effective nuclear charge and plug it in there not to find the value of epsilon sorry to find the value of imine right so we know z effective we plug the value of z effective in the expression for epsilon to find imine let us do that imine turns out to be 1.6875 square minus 27 by 8 1.6875 75 that turns out to be minus 2.8477 which translates as 77.49 electron volt okay so we are getting close but perhaps not close enough that is the problem close but not that close also so to understand the problem let us have a look and let us compare with what we get the minus 2.8477 this value is perhaps wrong but anyway first order perturbation theory gives me a value of minus 2.75 second order perturbation theory gives me a value of minus 2.91 13th order perturbation theory no it is okay gives me a value of minus 2.90372433 and so on and so forth so this is where we should get and so what we see is by the simple variation method that we have used here we are doing better than first order perturbation theory but not as good as second order perturbation theory and definitely not good as good as 13th order perturbation theory so this is definitely not the end of the road this is definitely not the end of the story and also to make a point that whatever we have discussed first order or variation method they are not all that great that comes out nicely if you look at another molecular property and that is ionization energy okay we all know what ionization energy is in gas phase the energy difference between the ion helium plus and the uncharged atom helium that is ionization energy okay so now what we will do is we will work out ionization energy that is calculated using variation method as well as first order perturbation theory now energy of helium plus is very easy to calculate as helium plus means it is a one electron system right one electron system will have an energy of minus z square by 2 where z is equal to 2 okay there is no question of effective nuclear charge anymore there is no question of shielding there is only one electron so it feels the entire nuclear charge that is there okay so 2 square by 2 minus 2 square by 2 that gives you minus 2 atomic unit so energy of helium plus is minus 2 atomic unit ionization energy then would be minus 2 minus minus 2.8477 in case of your variation method 2.75 in case of perturbation theory and these are the values that you get and you appreciate the picture properly when we take it to electron volt remember electron volt is a larger number is not it one atomic unit is 27.2 electron volt so if I translate that we see it better that is all you might not think that 0.84 0.75 in our minds we think that okay it is little less than 1 what is the difference of course if I told you 7500 and 8477 you would have seen the difference but let us see the difference in electron volt multiply by this 2 27.2 the numbers you get are 20.4 for first order perturbation theory and 23.1 for variation method so you see a significant difference is there between the two calculations and then when you look at the experimental value 0.904 this is the correct value then you get an ionization energy of 24.6 so 20.4 electron volt is definitely very very far away from 24.6 electron volt ionization energy is 23.1 close enough well definitely better than 20.4 but see the difference between 23.1 and 24.6 the number calculated from variation method and the number obtained experimentally is 1.5 electron volt 1.5 electron volt is not a small number in this context because typically this 1.5 electron volt would be of the order of the bond strength strength of a chemical bond would be about 1.5 electron volt or little more little less so since you are talking about atomic and molecular systems this difference is not really small this is sufficiently large so we have miles to go before we sleep we need to do better than this simple variation method that we have adopted. And to just convey how long it has taken to do better we will show you some timelines shortly. So now what we do is we start thinking of more general trial wave functions remember our previous discussion of variation method we had said that you can actually use any function to construct the wave function that is there remember the elephant that I showed in the opening slide okay to get the elephant I can use whatever function whatever polynomial whatever exponential I want as long as it finally gives me the shape of the elephant. So there is no need for us to stick to hydrogen atom wave function okay so we introduce the first wave function which is sort of a deviation from hydrogen atom wave function okay sort of but not a very far deviation when you do any theory you do not do something in white today and tomorrow you do something in black that it can take you very very far away from reality. So we always take baby steps and the baby step we are talking about is that of Slater orbital let us have a look at this form of Slater orbital here we are writing S for Slater S it is characterized by NLM three quantum numbers very much like your hydrogen atom wave functions orbitals atomical orbitals and it is a function of r theta phi if you work in spherical polar coordinates the expression for Slater orbital is the normalization constant which is a function of N and L multiplied by r to the power n minus 1 very similar to what we had for atomic orbitals hydrogen atom wave functions one electron wave functions multiplied by well I will read the last factor first multiplied by y l m of theta and phi this is familiar to us the angular part of the wave function form might be different that we will see and the second factor is e to the power minus zeta r what would that function have been in case of atomic orbital it would have been e to the power minus z r is not it so z is your nuclear charge right atomic number so instead of z we are using a variational parameter zeta okay it replaces z in the expression for an for a hydrogen atom wave function and again to think of it we have already done this what did we do in our previous treatment of variation method for helium atom we use z as a variational parameter didn't we so essentially we use z as zeta we just didn't call it zeta we still called it z here Slater called it zeta that is one but is that all as we saw just by using this z or zeta as variational parameter we do not get very close to the real real value we do not get very close to the reality so what Slater did was that he introduced one more variational parameter he used n as variational parameter as well now this might sound counterintuitive we know what n is for hydrogen atom n is the principal quantum number it has to be 1 2 3 4 so on and so forth how can I use it as a variational parameter and get something like 0.9 0.8 1.2 what sense does it make well the sense it makes is this we are not dealing with one electron systems any longer we are trying to stick to one electron wave functions as long as we can because they are familiar to us we know how to handle them and also many electron wave function is sort of a it builds upon one electron wave functions right you can think that you start with one electron wave function then add some complication you get many electron wave function many electron atom starting from one electron atom but it is not completely uncorrelated is it right there has to be some kind of a correlation there has to be some kind of a gradual evolution from a one electron system to a many electron system so we want to go gradually as well right the problem is we cannot write the exact form so the right thing to have do have done perhaps is to have added a term we do not know which term to add so what we are doing is we are essentially scaling the wave function by a variational parameter we are multiplying by a factor to arrive at a wave function that gives us the correct energy right now we only bother about energy okay we do not worry about how exactly we get the wave function it is a and ends justify the means sort of approach you if you want to call it that right and again we have said this earlier I will just remind you again it is sort of using sort of like using the activity coefficient right and pretending as if not all of the concentration of a reactor of a molecule is available for doing reactions whereas the reality is that all of the concentration is available it is just that they run slower okay it is similar kind of a scaling problem so nothing very new this has been used for a long long time okay so since we cannot work with one parameter we are trying to compensate by playing around with another okay that is why so do not we do not feel it is a sacrilege that we are going to get non integral values of n and the problem is I think I wanted to show you a value of n that we get after optimization but I have forgotten to write it well I remember I will show it the next day otherwise just read McQuarrie's book please it is given there when you get n something like 0.995 for helium okay so using this and this is where you get n equal to 0.995 so you use a product of slater orbitals instead of wave functions and you vary n as well as xi zeta sorry when you do that you get energy of minus 2.8617 atomic unit ionization energy of 23.4 electron volt which is a little amount of improvement from 21.23.1 electron volt or something that we got earlier okay so this is sort of a limit which is called Hartree-Fock limit and for the first time in this course we take the name of Hartree and Fock and in the next one or two modules we are actually going to discuss in as much detail as we can the Hartree-Fock approach to this multi electron problem many electron problem okay for now let us just say that this is the Hartree-Fock limit that one can reach using this kind of an approach okay and again what kind of a problem is it? It is a minimization or an optimization problem I have two parameters zeta is one n is the other I need to minimize z sorry minimize energy with respect to zeta as well as n okay so you know how to do it take derivative in this case partial derivative equate to 0 in many cases it will not be able we will not be able to do it like that we have to do it numerically as we have briefly mentioned earlier perhaps we will come back to it in more detail when we discuss Hartree-Fock in the next modules all right. So this is what is called Slater orbital and using it you can get a little bit of improvement but not really what you want so you want to try out other forms of orbitals and now that we have opened the flood gate we have said that we do not have to stick to hydrogen atom wave functions then there is actually no need to even retain the form here the rough framework of hydrogen atom wave function is retained with the introduction of the variational parameter. So in the subsequent kinds of orbitals and we will discuss maybe a couple of more in the next class but today let me show you at least one more Hylerus for example use this kind of a wave function is a function of r1, r2 and r12 the inter nuclear separation and the functional form that he used was product of two exponential functions one in r1, one in r2 multiplied by 1 plus a function g a polynomial g in terms of these three independent variables g is a polynomial now see you can play around with the number of terms in polynomial each term will be associated with the coefficient each coefficient will be a variational parameter and of course you can play around with z and arrive at what effective nuclear charge you get the only issue with that is that some of this correction for effective nuclear charge might go in the coefficients of g. So it becomes a numerical problem and you need computation you need to do you need good computers to handle systems like this because you try to do it by hand you will grow very old before you can solve even the helium problem. So as you see as computational power grew from 1929 to 1957 to 1959 this was a timeline I was talking about earlier Hylerus used 10 terms he could use only 10 I do not know what kind of computer Hylerus had if at all perhaps he did everything by hand he got an energy of minus 2.90363 and ionization energy of 0.904 which is a remarkable improvement over what we had earlier remember then Ginoshita in 1957 published this result with 39 parameters and you get minus 2.93 7 225 for energy atomic units ionization energy remains 0.904 Pecker is used 1078 parameters that means 1078 terms and published this result in 1959 you see at this huge increase in number of terms and the improvement is there right minus 2.903 I think I missed a 0 here I am sorry about that it is not minus 2.9372525 it is minus 2.9037225 so you get something like this. So what we see is that you keep on increasing the number of terms at one point of time it is going to saturate it is not going to get any better it will get as good as it gets and you cannot it makes no sense to increase the number of terms beyond that I mean you can no harm done because upper limit theorem tells us that in any case you will stop at the actual value of energy right you will stop a little more than that you will never go beyond so the chance of overshooting is not there but if you increase the number of terms the question is how good a computer you have what is your computational power so that is always the tradeoff more number of terms gets us closer to the realistic value but it becomes computationally more expensive okay if you have better computers be my guest use as many terms as you want does not matter okay so with that I do not know why I have animated this with that the field is set for us to now try to see how we can draw this elephant well elephant is allegorical of course how we can generate whatever wave function we need using the Hartree-Fock method and since we are running out of time I will not elaborate now next day we will also learn that Hartree-Fock method uses something called a self consistent field model very exciting challenging and opens up the field for addressing more complicated problems like molecules at later time that is what we are going to discuss in the next couple of classes