 are calling Mila plus at second order and I want to note that we are divided h equal to h0 plus v where h0 was sum of the Fock operator, Hartree-Fock orbitals as the starting point of h0 Eigen function okay. So basically our psi 0 0 the psi Hartree-Fock and e 0 0 is sum of the orbital. So I am using A as an index for occupied orbitals. Then we had e 0 1 which was psi 0 0 v psi 0 0 which is psi Hartree-Fock d psi Hartree-Fock and your v was essentially since h0 is sum of Fock operator v was sum over 1 by Rij minus d v Hartree-Fock one particle Fock at the attendance seat okay. So if I put this and apply Slater rule this was can somebody remember this value what would be just e 0 1 applying Slater rules who can tell. So v is just 1 by Rij this would be added to the orbital energies eventually to get total Hartree-Fock energy. So what is this value in terms of spin orbitals minus half yes sum over all a b right a b anti-symmetrize a b where these are the spin orbitals right. So minus half this if you add this to the orbital energy you will get the total because e a was h a a again in terms of spin orbitals I am writing plus sum over b a b anti-symmetrize a b. So you had the b 2 so a is not summed up of course for orbital energy of a when I sum the orbital energy of a this summation will come plus this is all a b if I add the first order perturbation correction you will get half of a b a b okay. So that will give you Hartree-Fock energy. So I just want to make sure that you understand in terms of Slater rule of course in terms of e not 0 it is size 0 0 h not size 0 0 and e not 1 is size 0 0 v size 0 0 so addition gives you Hartree-Fock energy that is of course easy to see. So your e 0 0 plus e 0 1 was the Hartree-Fock energy. So by our definition of correlation energy the e correlation starts at second order okay and then we went through the second order perturbation equation and we wrote an expression for the correlation energy which is the correction at the second order. So you can also call by our old definition just e not 2 e not 2 is nothing but the correlation energy at the first correlation energy contribution which is the second order because at the e not 1 there may be a value but then that just adds up to Hartree-Fock so our first correction terms comes from e not 2 and we derived the expression that it is 1 by 4 a b and r s where r s are virtual orbitals and a bs are occupied orbitals okay. So I hope to this point all of you can derive this was actually derived okay initially we wrote all excited determinants of h not but we found that the single excited determinants do not contribute because of Brillouin's theorem. So you are only doubly excited and then we use against later rules from the doubly excited which was simply the two electrons a b anti symmetrized r s mod square we had a less than b r less than a initially then we said all a b r s and since a equal to b r equal to a is anyway 0 so it does not harp. So then we have a factor 1 by 2 for this pair 1 by 2 for this pair we got 1 by 4. So I hope all of you remember this derivation so the homework essentially starts from here. So please give it on Monday that you have to now write this anti symmetrized integrals in regular integrals and write an expression see which are equal use the symmetry to make sure that you just do not should not just expand and one line that expands in is one line I can also do a b r s minus a a b s r r s a b but try to simplify in terms of regular integrals which is still in terms of spin orbitals then do a spin integration. So that was essentially so eventually land up into a correlation energy for in terms of special orbitals both for the occupied and virtual orbitals you have a special orbital remember the entire thing is of course done for ground state a closed shell. So essentially a closed shell system where the Eigen function of H naught is a single determinant in fact all Eigen functions are single not only that is a single determinant which turns out to be a restricted essentially means it is a closed shell so that is that has been the requirement for deriving this particular formula however if it is a single determinant and not restricted I told you there are other Hartree-Fock like unrestricted Hartree-Fock restricted open shell a similar derivation can be done it is not really a problem but I am just telling that this particular derivation is follows the restricted determinant restricted Hartree-Fock return. So you essentially get this as the first correction to the Hartree-Fock energy and that requires the in two electron integrals between two sets of occupied orbitals and two sets of virtual orbitals at this point there are spin orbitals but we will discuss what happens when you integrate and get this special orbital. So let us now go ahead and try to understand these I was yeah MP1 correction for unrestricted Hartree-Fock okay that will also we have still the same you know same situation first order anyway it will get back the Hartree-Fock because that is for a very simple reason you just add up H0 and V whatever is your size 0 0 your E0 0 is size 0 0 H0 size 0 0 E0 1 is size 0 0 V size 0 0 so if you add it up so that has nothing to do with anything okay in fact as long as you have this structure you will always get first order giving you Hartree-Fock or reference energy I would not call Hartree-Fock in that case in the case of UHF it is UHF but you can have a more complicated reference even then what you will get is a reference energy which is which is the size 0 0 H size 0 0 so that is what I call the reference energy yes so basically what I mean to say whatever you do these two quantities will give you size 0 0 H size 0 0 and as long as this is a reference function this is a reference energy so I am not calling it correlation energy and at least for a single determinant I do not call it correlation energy so whether it is RHF or UHF it does not matter okay so one of the things that we have seen is of course the importance of doubly excited determinant note that in this process of getting E0 2 we had first got size 0 1 which is the first order correction to the wave function remember size 0 0 is psi Hartree-Fock but we also got size 0 1 which is a linear combination of all eigenstates of H0 except Hartree-Fock correct so if when you when you did this we remember our first correction just like here was actually from the doubly excited determinant okay and so we wrote this as psi ABRS and the matrix elements or the linear sorry linear combination coefficients for this determinants was this matrix element divided by epsilon A plus epsilon B minus epsilon R minus or essentially the difference of these two energies okay in a reverse order so this minus this so this is size 0 1 and actually if you remember by E our E0 2 was psi Hartree-Fock V psi 0 1 from there we got an original expression and then we applied slatter rules here so we can apply this slatter rule of course here just like we did here but that is not what I want to show what I want to show that the first order correction to the wave function which is the first important correction to psi Hartree-Fock to from the psi Hartree-Fock is actually starts from the doubly excited determinants the singly excited determinants which you might have guessed should be the first correction it does not come because the corresponding matrix element becomes 0 because of Brillouin's theorem so Brillouin's theorem essentially dictates that if you start from Hartree-Fock you have a Brillouin's theorem and then the first important determinant that contribute to the lowest perturbation order is actually doubly excited determinant psi ABRS so that is why I want to highlight the importance of doubly excited determinant and part of it will be discussed again when you do CI and is also the content of Synonal Glue's review that I had given you to read which is essentially known today as a pair correlation theory. So this is actually much broader perspective essential idea later we will see is that this correlation essentially means that I have two electrons let us say they are AB spin orbitals AB they get excited so this is somewhere here let us say I am just taking a two level system so one orbital two spin orbitals one orbital two spin orbitals so they get excited together to some RS to form what is called the doubly excited determinant so in a very graphical manner the two pairs are getting excited together to form doubly excited determinant they get excited essentially correlation essentially means they are getting excited so this is why in general it is called a pair correlation theory later we will see for a many particle problems we can have another set of pair which can also get excited and these two excitations can take place at the same time so time is of course here just for understanding but if it takes place at the same time what you will see is actually a four fold excitations which will come which will be a quadruply excited determinant but the probability of this taking place is nothing but since they are taking place simultaneously and independently will be a product of this pair and this pair so this is the content of the synonyms later pair correlation theory but at this point let us let us say that the more important determinants come because of the two electron excitations now why excitation is related to correlation now is one to tell very physically what does the correlation mean okay physically let us try to understand so you have an Hamiltonian and we are trying to solve the Schrodinger equation note that I have been mentioning this and this is a very important part of the correlation let us again understand two particle problem if I have two particles r1 r2 okay and in the limit of r1 tending to r2 which means the coordinates of the two particle of course the particles cannot be traced the particles have only a probability being anywhere but whenever the particles are at r1 and r2 what will first happen to the Hamiltonian so Hamiltonian can be defined for a given r1 and r2 right if you look at this Hamiltonian you have a 1 by r1 to term right so at r1 tends to r2 this 1 by r1 2 will become infinity so this is what is called singularity so this becomes in the limit a singular Hamiltonian if you have a singular Hamiltonian essentially means it is an infinite because 1 by r1 2 is becoming infinity and then normally for an infinite operator the wave function whatever the wave function is energy will become infinity it is just like multiplying an infinity times a finite number you get infinity so even if psi is not in infinite function it is a finite function since it is a singular operator when it acts on this this will become infinity so which means energy will become infinity of course such energies are not allowed we are looking at finite energy system so if it is a finite energy system then what is the allowed wave function so the allowed wave function for such a case when Hamiltonian is singular is that the psi of r1 r2 must go to 0 because that is the only way this can survive because infinity of course acting on 0 can be finite so energy can still be finite but the wave function cannot be finite I hope you understand it is just like a multiplication it is a very loose manner I am saying like infinity acting on a 0 can be finite but infinity acting on a finite is infinity so just like that the only possibility is that the wave function must become 0 so whenever these two particles come close together wave function becomes 0 what does it mean wave function is a sine of probability mod psi square is a probability so we can now make a statement that there is a vanishing probability of two particles coming together right so let us make this statement for the correlation energy that the probability that the two particles come together is 0 in any system because you imagine that they have they have only we can only talk in terms of probability so the vanishing probability probability or probability density of two particles coming together or you know in a very loose manner and this is how synonyl actually argues in a very loose physical manner that the entire all the n particles are actually moving they are colliding so whenever they come close together the probability is 0 what it means that they cannot come close together so in the Hartree fog let us see in the Hartree fog what happens if you look at Hartree fog let us say again two particle problem so so restricted Hartree fog a bar so that is my Hartree fog wave function right I hope you understand the terminology a is a alpha a bar is a beta so you have a restricted Hartree fog a is the 1s orbital or whatever so you have alpha spin you have a beta spin now obviously the two particles are in the same spatial orbital although their spins are different they are in the spatial orbitals so there is a possibility of the two electrons which are anti-parallel okay coming close to each other which means the then when I when I calculate the probability there is a probability of the two electrons coming close for anti-parallel spin however for parallel spins they are unlikely to come because their orbitals are different so they will avoid each other parallel spin so the real problem in the Hartree fog is that for anti-parallel spin this part of the exact nature is not being followed okay by the Hartree fog wave function so this is basically for the Hartree fog so in Hartree fog theory this there is a vanishing probability so how is that how is that done in the perturbation theory when you go beyond Hartree fog is that as soon as these two particles come close together they will anti-parallel spin they will try to avoid and the one way to avoid is to actually get excited so if you get excited at least there is a less probability of crowding please remember we are only talking in terms of probability so we can't say that there is there is that they will actually come or not come the probability is 0 so when you give a more space there is a possibility of avoiding each other if you have to send a crowded space if you look at n particles in a crowded space they are going to collide with each other so the virtual orbitals give you that space virtual orbitals essentially tell you that I can actually go away of course you may argue that if they are going away to two virtual or same virtual orbitals again opposite spin that is what the singlet does then again they will be together but the point is now the probability is being reduced there is a some probability of being here some probability of being here some probability of being here because there are so many virtual orbitals so it is only a question of probability it is whenever of course there is a singlet the two electrons will be in the same special orbitals with anti-parallel spin okay but if I have many options like I have only one option here in the Hartree fog A bar but now all RR bars which are virtual orbitals are opening up as an option so the probability gets reduced so essentially probability just gets reduced so that is how the argument of synonyl glue went again it is a very qualitative argument it is not a quantitative argument it is qualitative argument to understand what is correlation and why W excitedly important. The further argument that synonyl glue gives the single excitation are of course not important because of brilliance there so of course one can argue why not three particles and again synonyl glue had a very brilliant argument he said three particles in any case cannot come close together because they cannot be in the same orbital even in Hartree fog because of Pauli principle so three particles cannot come close together that is number one number two is that the V or the Hamiltonian which makes them correlate has only a maximum of two particle interactions so it has only one particle and two particle interactions because of cool off so it is very hard for the Hamiltonian to correlate three particles together so that is another argument that he gave that why three particle onwards is also going to be small why it is only the two particle which is important. So the entire correlation he said is dominated by what he called pair correlation theory and I think if you read this very nice you know synonyl glue actually told in physical terms what we will talk later about the couple cluster doubles and so on but after that he did not unfortunately as I told you he did not give a full rigor he gave a rigor to some extent but not the complete rigor he did not actually you know set out equations and so on and that was his weakness and then later other people came but the idea of the two particle correlation being the dominant correlation really came from the synonyl glue theory so that is why I just thought I will spend a few minutes though that is not really something that I would like to dwell those who are interested should read the synonyl glue paper advances in chemical physics it is a lot of other insights that you will get. So I think this is very important we will see this again in the context of configuration interaction that the doubles are important so we are first arguing that the wave function itself will have only pair correlated doubles and of course the wave function has doubles and energy will also have similarly E0 2 E0 2 can be obtained from this. So that is a triviality the point that we are trying to say that the size 0 1 contains mainly WXRA determinants and not triples why not triples of course by Slater rule you know why not triples but what synonyl glue did was actually to give a physical insight to this formula that is comes from the perturbation theory itself that synonyl glue need not do synonyl glue are good from a different point of view that there are two particles which can be in the same orbital in the Hartree form particularly the anti-parallel spins that is number one so they have to be correlated because correlation is essentially making sure that such things do not happen. So two particles should not come close together so then you have to give a space for them to go somewhere so they can go to any other virtual orbitals so that space allows them to avoid coming close together that is number one. Number two is that three particles onwards are unimportant because they cannot come close together by the same argument and the Hamiltonian is only two particles. So actually what synonyl glue is trying to do is to justify what you get from the perturbation theory by physical insight he is not using this because if you are using this anyway it is very clear I mean perturbation theory of course you will tell you first order triples cannot come because of Slater rules but that Slater rule itself is being justified in some other way you can say because he has not used this formula synonyl glue did not use perturbation theory so the arguments are much more qualitative and this actually are what it turns out to be couple cluster theory which is not really perturbative. So you will see so you do not use perturbation theory perturbation first order perturbation only justifies what synonyl glue had told that the pair correlation is important. So he is not using this so we will come back to this when we discuss the couple cluster again qualitatively please remember that we will we turn out to have classes we are not going to discuss very thoroughly many of these new methods but at least I thought the perturbation will discuss and try to give a justification of the synonyl glue theory. I had also mentioned in the previous class that if you do a restricted hard to reform let us say for hydrogen molecule for hydrogen molecule at a large distance so at r h h tending to infinity the e r h f does not become twice e hydrogen atom so this is something that I mentioned actually it becomes greater than the two hydrogen atom it goes to h plus h minus configuration and mainly that happens because of the spins remaining attached it is a singlet wave function so they remain attached to 1s here and 1s here. So let us try to understand this problem little bit better so remember my molecular orbitals are formed out of linear combination atomic orbitals which are my two bases so let us assume that those bases are 1s hydrogen A 1s hydrogen B again by the same nomenclature that original energies are identical when before the molecular start molecular form then you have the phi 1 and phi 2 are the two orbitals special orbitals that are formed these are the special orbitals molecular orbitals so I all of us know that one of them is anti-bonding one of them is anti-bonding one of them is bonding and the bonding orbital is let us say phi 1 so your Hartree-Fock then becomes or r h f then becomes phi 1 phi 1 bar correct so I hope the nomenclatures are clear so when I am calling phi 1 it is a MO for the interaction region so what happens for the psi r h f as r tends to infinity so let us as let us try to understand this as the r goes to infinity what happens to first phi 1 and phi 2 these are the special orbital both of them go to the 1s orbital that is clear right because they were started from 1s orbital and they had a bonding and anti-bonding orbitals if I do the reverse MO diagram then phi 1 will go to one of the 1s phi 2 will go to other one which one goes to where I do not know so the molecular orbitals become atomic in nature at r tending to infinity which is not surprising because there is no molecule at that point so when we say LCAO MO approach although AO's were used only as a basis since I am using the AO's there is a physical interpretation to this that these MO's will eventually become AO so that is where LCAO MO picture has a chemistry inside it is not just a basis so my 1s was not just a basis it has a physical inside that if I take this molecule apart these molecular orbitals will go over to these basis functions because they are atomic orbital basis I hope all of chemistry you know if I would have some use some other basis crazy basis then it would not have happened because eventually they go over to the atomic orbitals but because I have used the atomic orbitals and basis these phi 1 and phi 2 go over to the 1s hydrogen so then what happens to the wave function at that region so psi Hartree-Fock at r tends to infinity will then become 1s hydrogen A and let us say 1s hydrogen B bar correct because one of the phi 1s will become 1s hydrogen A phi 2 will also become 1s hydrogen A but now you have 1s hydrogen A 1s hydrogen B if they go separately or what is worse is that they mean they will more likely become this because both of them are phi 1 so if phi 1 becomes 1s hydrogen A then it will become 1s hydrogen A 1s hydrogen A bar and this is precisely the configuration which is H minus H plus because what is it show it shows both the electrons are in one of the 1s orbitals by chance if you can write this as phi 1 phi 2 bar then of course it will become 1s hydrogen A 1s hydrogen B bar but that is still not right because you know for a singlet wave function if I have a degenerate you cannot have this and this you have to mix okay so this is of course a singlet wave function there is no problem with this wave function this is a proper singlet wave function starting from a singlet but this is a wrong asymptotic behavior okay so wrong asymptotic behavior means it does not separate to the correct result because you know that if I have gas phase hydrogen if I dissociate it should become hydrogen atom plus hydrogen atom and not H minus H plus I hope all of you know this I have told this gas phase any diatomic molecule becomes A plus B not A plus B minus or A minus B plus okay never happens because ionization potential of any atom is greater than the electron affinity of any atom so you can never have A plus B minus is the ground state A and B together some would be much lower than A plus B minus because ionization of any atom is very high I mean you look at hydrogen ionization hydrogen ionization is how much 13.6 electron volt which is 300 kilo calorie roughly okay 320 whatever so no 300 kilo calorie I think 23 13 into 22 20 is 260 and 40 so roughly 300 kilo calorie which is already 1200 1300 kilo joules almost so you can imagine greater than 1200 kilo joules which is so high just to make H to H plus so H minus H plus will be so high how much of energy you can gain by H minus very little 1200 1300 kilo joules remember is larger than nitrogen bond breaking if you do a homolytic nitrogen bond breaking it is about 900 kilo joules CH bond breaking or hydrogen bond breaking is 400 some kilo joules we have done that exercise oxygen bond breaking is also 700 to 800 kilo joules somewhere 800 let us say nitrogen is about 890 or 900 but hydrogen ionization itself is more than 1200 kilo joules you know many times you do not realize these these numbers so ionization is so difficult okay hydrogen is of course quite large for other atoms it may even be less that is a different matter like cesium is most easily ionized but even then that number is very very large so that is the reason a heterolithic bond cleavage in chemistry is much more difficult than a homolytic bond cleavage so if I have a methane molecule I do a homolytic cleavage CH3 radical and hydrogen atom right that is much easy to do than CH3 minus H plus which is deprotonation right so deprotonation is much more difficult because it requires very high energy simply because of the hydrogen ionization that kills the problem so I just want to tell you why the H minus H plus is of course a wrong asymptote because in gas phase this is a very high energy so the ground state should actually go to H plus H which I am not getting so the question is how do you get H plus H singlet so let us try to see the wave function that we should get in the asymptote so let us look at what we should actually get in the asymptote so for hydrogen molecule again so the asymptote asymptotic singlet state so you had this phi 1 here you had the phi 2 here and this was my 2 1 s functions right is the usual curve that we plot so eventually each of these states will go to one of these which one goes where it does not matter so I will at the asymptote I will have these two states only what will be a singlet state I will have 1 s h a 1 s h b bar right but that is not enough I have already told you for a 2 electron singlet in a degenerate you have to do the other so you have to also add or subtract 1 s h a bar 1 s h b and of course with some combination that is not important okay in this case it will be simply 1 by square root 2 that is unimportant the point is with an overlap integral that overlap will anyway go to 0 okay at the r tending to infinity so we already know this in this case so it will become simply 1 by square root 2 but the point is the asymptotic singlet wave function is actually not a single configuration that is the first thing to understand you have what we call a multi configuration or 2 configuration in the asymptote okay it is a linear combination of 2 different determinants so I am calling it 2 configuration so either if you use r h n or even if you use u h n these are all single determinants as soon as I break them apart none of these will actually go to the asymptote correct though I had already mentioned that the u h f energy goes correctly but it is not a singlet that is a problem I am talking of a singlet state so the u h f energy goes correctly to the exact energy but then it is not a singlet so if I want a singlet asymptote of course u h f cannot give r h f cannot give even if I would have put phi 1 phi 2 bar as my determinant there okay you would still not get because you will get only one of these you would not get a linear combination so right in the beginning you have to start with phi 1 phi 2 bar and phi 2 phi 1 bar your original r h f original reference wave function then only you will get these as the asymptote is it clear but this is against the spirit of Hartree-Fock this is not Hartree-Fock right this is not Hartree-Fock because Hartree-Fock has always one determinant so the point that I am trying to mention that if I start with one determinant Hartree-Fock and try to split or try to separate the molecule I will never get the asymptote correctly okay and particularly I am giving an example of a singlet and I cannot get the singlet either way whether I do r h f and u h f though energy in u h f is close to the exact energy because of certain variation method that the energy goes down it comes closer but the wave function you will never get so if you want to get a proper singlet wave function you can see that the Hartree-Fock itself is not good so the question then we can ask is if I do a m p first order correction to the wave function remember when I say m p 2 that is for energy for the wave function what is the first order correction first correction at the first order correct so remember this do not say m p 2 for wave function also okay only for energy first order correction gives back with the zeroth order Hartree-Fock but as far as the wave function is concerned Psi 0 0 itself is Hartree-Fock and Psi 0 1 is different okay and that Psi 0 1 gives you e naught 2 because of that formula so Psi 0 1 has to be something different from Hartree-Fock okay so if you apply this we would like to know are we improving the system so if I just use a r h f I hope you understand the problem we will discuss that next class if I just use r h f 5 1 5 1 bar and then do a first order perturbation correction to the wave function am I improving the result in the asymptote so we want to discuss the behavior of Psi 0 1 as r tends to infinity okay along with the Hartree-Fock so it is important of course that my final wave function at the first order will be Psi 0 0 plus Psi 0 1 so the question that I am going to ask does it behave even qualitatively in r h h the region which tends to infinity again we will take only hydrogen molecule as a prototype case and we can even simplify the problem by taking a two basis so only we will take a two basis just like we did here 5 1 and 5 2 two basis so we will take the simplest LCMO we have one virtual orbital where you can excite both the electrons right so this r s can be generated because you have two special orbitals you have to force spin orbitals actually so we can very easily study this problem you can yourself try to do you know at your end what do you get finally what I am interested in this wave function when I add to the Hartree-Fock and then add to the Hartree-Fock wave function this first order correction okay so of course your a b r s will change because there are spin orbitals so let us say I have a and b are orbital this will become a a bar this will become b b bar I hope it is clear if a and b are special orbitals then this will actually become a a bar this will become b b bar in terms of special orbitals so I mean this is just nomenclature you can use some other nomenclature for special orbitals alpha beta whatever so the point is that you can easily study this model problem that am I separating hydrogen correctly because this is a very very important problem because if I am not able to separate hydrogen correctly there is something to worry how do I separate I already told you one way to separate would have been to start from this which means have a have whatever you call it reference or Hartree-Fock but it is not a single determinant Hartree-Fock it is a two configuration Hartree-Fock and this is a very important to understand all multi reference problems in quantum chemistry why multi reference problems comes as a reference so you can see my reference function itself should have been two reference function that is one way to definitely solve but if my reference is just 5151 bar then do perturbation like here does it help that is the question that we will ask if I of course start with this I know my reference itself is good enough at least qualitatively to separate but if I do not do that and I start with 5151 bar just as we are doing here for the Hartree-Fock and then do a first order perturbation shall I get the asymptote correctly I hope the question is understood because this is a very important question if I would have started with this anyway I would have got the asymptote correctly because this would become 1sa this will become 1sb vice versa rather matter both the determinants are there so first thing to understand that in the asymptotic limit the singlet wave function is of this form that is very important to understand I hope that is very clear because we have already given you a theorem that each of this determinant is not spin adapted please do not forget that theorem that if I have n alpha and n beta in this case n alpha n beta equal to 1 so one number of orbitals must be doubly occupied which is not true in either of these cases so these are now like molecular orbitals the same there is no difference between MO and AO at r tend to infinity so do not get confused so this determinant is not spin adapted this determinant is not spin adapted however a spin adapted wave function would be a combination of these two determinants and this is very simple to understand that I am saying that neither of this is correct but a combination of this is correct where this can be alpha or this can be beta this can be beta this can be alpha for a degenerate space this is correct and that is also physically right because we cannot pinpoint what is electron 1 and what is electron 2 electrons are indistinguishable so you cannot write like this you have to add these so obviously any one of these determinants is wrong but a correction term is right the question is if I start with phi 1 phi 1 bar can I generate that okay by doing a perturbation correction you can actually find the answer in fact you will get the answer itself yes psi a b r s means phi 2 phi 2 bar in phi 2 phi 2 bar in this case so what is asking is the psi a what is psi a b r s a and b are phi 1 phi 1 bar you have two state problem so there is only one virtual state so this is nothing but determinant 2 2 bar or phi 2 phi 2 bar so you can easily see whether the result will come or not actually the answer is no you can actually see this because I need phi 1 phi 2 bar plus phi 2 phi 1 bar I do not need phi 1 phi 1 bar plus phi 2 phi 2 bar so you know all since the question is already put it is very trivial to see that you will still not get it so that is a very disturbing thing that my r h f is not good for dissociation I do a first order perturb correction it is still not good okay and I will do it in detail but you can clearly see from the discussion it is still not good so what is good is actually to start from this there is no other way that means I do not do this I do not start from this r h f at all I do not start from any single determinant the original determinant original function itself is this and then I try to improve in fact very loosely I would say these are the basis of all multi reference problems like you have multi reference CI you have multi reference perturbation theory and so on okay so I think this is something that is qualitatively very important to understand because we cannot really spend a lot of time on multi reference because we are going to go back to single reference CI after this but I just thought I would mention why multi reference is important at least in one context of dissociation and these are these are very important problems where a molecule is closed shell but is dissociated into two open cell fragments I had given you this theorem that is when r h f fails if it was a closed shell molecule I do want to dissociate into two closed shell molecule then there is no problem so for example if I would dissociate h 2 to h plus h minus then it would have been no problem I want I do not want h plus h I want an excited state no problem because both of them are closed shells so the problem is whenever a closed shell molecule fragments into two open cell parts there is a problem and the problem cannot be washed away by simply doing m p 2 or first order correction to the wave function second order it cannot be washed away so we will see how do you handle such problems in fact we will talk not much about the multi reference so that is why I am just bringing the relevance of the multi reference why people still have to do multi reference most of the course we are going to do single reference starting reference functions are single determinant as we have done for the perturbation okay so what we will do now after discussion continuing the discussion closing it we will go to actually CI but I think what I like to do also is to describe this perturbation energy that we wrote E naught 2 of course in terms of special orbital by the time assignments will be due but more importantly how to represent them in a very convenient language and that is something that you should learn and that is called the diagrammatics there is a diagrammatic perturbation theory which means the same algebra that you are deriving in a very laborious manner and that is important at least from E naught 3 onwards the things will become more and more difficult how do I conceptualize these by drawing a simple diagrams and I have been telling these the diagrams are I told you certain diagram like scattering goes from 0 to excited state come back a more I would say a rigorous way of writing at least E naught 2 and E naught 3 in terms of diagrams is something that I will introduce introduction would be slightly ad hoc because to rigorously learn diagrams how they have come you should also learn what is called second quantization which also we will do later but I think next class I will I will in ad hoc manner I will introduce the diagrams and diagrams are very pictorial so all of you will be able to understand very easily there is no algebra actually so the maths there is no maths the maths is actually written in terms of diagrams so whatever we wrote E naught 2 that formula is written in terms of diagram and after that only we will go to CI so 2 configuration alone is not enough that is what I am trying to say if I write just 1s a 1s a bar 1s b 1s b bar that is also not enough then I then what I am doing I am getting only h minus h plus in 2 different ways once putting these 2 electrons here once putting but that is not enough for singlet I have to have in this manner if this is 1s a this is 1s b this manner which is not coming by the RHA which is not coming but even mp1 wave function first order corrected wave function and I think I will do that I will complete it but since the discussion already took place you can see the answer okay but I am just trying to tell you the context why people do in many many cases this is at least one context there are other reasons people do there are other cases where multi references also comes in but this is at least one and this is the simplest that I can tell you hydrogen nothing can be simpler than that hydrogen in 2 basis that is what you learn all your life but you do not realize what are the problems okay that it cannot sigma g sigma g bar is all that you have read it cannot dissociate if you add sigma u sigma u bar it still cannot dissociate that is what you are doing right by mp1 and that is what I will tell so because you need sigma g sigma u bar sigma g bar sigma u right so 1 alpha 1 beta 1 beta 1 alpha and that is not coming because each of your functions is just a singlet function to start with so if you are determinant itself is been adopted by adding you will not get this because each of them is not spin adapted here the combination is so this is a little bit of a physics here so this cannot be obtained by simply adding two spin adapted determinants no I hope you understand what I mean the combination is spin adapted although each of them is not so that is a part of the problem here so this physics is slightly complex we will have a discussion again in next class and then probably introduce diagrams before you go to CI okay all right.