 So, yesterday we did this psi 0 theta 1 psi 0. I want to make sure that everybody understands how to do it, not you, but let me see others. So, let us see how we do it. We are not going to use whole particle creation, normal creation analysis operator. So, how do you write this? You write this as ij psi 0, ai dagger aj psi 0 and some h ij. So, your theta 1 is sum over h of i. This is in first quantization. Again, note the difference. This is in first quantization, this is in second quantization, where the sum over ij are now no longer electron coordinates but above spin orbitals. So, this is the first thing. So, this is the second quantized notation. So, now how do you show that this is equal to sum over h a? That is what we have to show. So, what is the next thing that we do? So, what should be aj, first of all? So, j should be occupied. What should be i? Should i also be occupied? Does everybody understand it? If j is not occupied, then you cannot destroy. So, in our later language, this becomes a whole creation operator. It will become xj dagger in later language. Ai dagger also has to be a occupied orbital i because it is acting on the left side to psi 0. So, if you look at this part, it is the adjoint of ai psi 0. So, if only i is whole, this will survive. The adjoint will survive. Otherwise, the adjoint will also become 0. So, this can be now written as sum over ab psi 0, a a dagger ab psi 0, h a b. So, we have put j as b. Now, note that this a and b by our notation are occupied orbitals. Later, we can change it to whole operator, whole creation, whole annihilation. So, that is the standard notation but right now we are doing. Is it clear? Otherwise, there will be 0. If any other ij, it will become 0 anyway. Then, how do I manipulate this? Because ab can now act on psi 0. Psi 0 is not my vacuum now. Psi 0 is actually, my vacuum is a physical vacuum because this is a normal creation annihilation operator. So, ab can act on psi 0. If I would have made this vacuum, then I have to write this as a whole creation operator. So, of course, it can create. It is basically creating a whole. So, it is possible. So, what you do is actually to write a whole annihilation operator on this side. So, to do that, you do an interchange. So, do the commutation. That is the next thing. So, write a a dagger ab as delta ab minus ab a dagger psi 0. Is it okay? So, I am now using the anti-commutator that this plus this equal to delta ab. So, I am writing this as delta ab minus this. So, then the first term is very easy. Delta ab comes out with h ab. Psi 0, psi 0 is 1. Psi 0, psi 0 is 1. So, basically this is multiplied by psi 0, psi 0. Then, I have a second term which is minus psi 0, ab a dagger psi 0 h ab. That is the second term. Psi 0, ab a dagger psi 0 h ab. Now, you can see this is actually a whole annihilation operator. If this is a whole particle vacuum, it will become 0 or you can argue that I cannot create an occupied orbital in psi 0. Hence, this part is anyway 0. So, this goes off. So, you are left with only this quantity. This is already 1 because of delta ab. The result is sum over, sorry, there is a sum over h a. So, there is a sum. I forgot to write the sum. So, there is a sum everywhere. Sum over ab is there. So, sum over ab is there. So, this becomes sum over a h a, which is my first quantized result. See, any one particle operator can create one spin orbital and annihilate one. So, that is the idea of the one electron operator. So, that is the reason this has come. There has to be, it is a matrix. So, it is like a linear combination. My string of operators, so basically when I am writing theta 1, my string of operators are ai dagger aj with some combination coefficient like a matrix superposition. So, I am writing one operator as a product of a set of operators. The set of operators, the only idea is the set of operators cannot be just creation or annihilation, but one annihilation, one creation because it is eventually a one particle operator which acts on a wave function not to change its number of electrons. So, when theta 1 acts on a psi which is an electron, the result should be an n electron function. Operator acting on a, so my number of electron must be conserved. So, I have to have one pair of annihilation and creation. Two electron, I have two pairs, but always number conserving. So, the question that is asking is that why ai dagger aj? The rest is just a linear combination. So, there has to be coefficients. We are calling it Hij. So, it is a general expansion. Like a function we expand in a complete set, we are expanding operator in another set of annihilation and creation operator. And it is self-consistent because of the anti-commutation rules that everything gets clear. You can actually do the same with theta 2. I will not do it, but please understand that you will exactly have the same thing, ai dagger aj dagger al ak. And again you can argue that al ak must be occupied orbital, ij must also be occupied orbital. So, it will become a dagger, a b dagger, a d ac. All four are occupied orbital just as here. And then show by anti-commutation that you will have two sets of integral. One is a regular integral a b a b, another will be a b b a with a half factor because that theta 2 already has the half factor. So, I think this is, I just wanted to tell you that you should be able to do this algebra. It is very simple algebra, but you should be able to do this. Later we said that because of this idea that we have to show that a dagger acting on size 0 is 0, which is little artificial because creation operator acting on something is 0 is artificial, we like to make it annihilation operator. So, that is the reason the whole and particle annihilation operators were created. And we called this as a vacuum instead of expanding this. Here I do not have to expand because physically I can understand what is there. But in a larger expression, it is convenient to work with the Hartree-Fock as a vacuum because all our correlation expressions are based on Hartree-Fock and using whole creation and particle creation or whole annihilation and particle annihilation. In which case anyway this should have become an annihilation operator x a and this should have become x b dagger. So, now if this is annihilation operator, if this is a vacuum, then automatically x a vacuum is 0. So, the idea would have come more mathematically. So, only for that reason we have switched to whole particle creation annihilation operator. But otherwise you can do exactly the same thing in normal creation annihilation operator, you have to be just little bit more careful. And in a longer algebra, it is much more difficult to do because the keeping track is very difficult. Keeping track becomes much more difficult. So, I think we will not go into much more details of second quantization, but we will only restrict to the form of the wave function, form of the operators in second quantization and all anti-commutation rules in the second quantization. That is very important. The rules of anti-commutation in second quantization, I think this may be enough as a background because we are not going to really implement second quantization in our correlation theory. That will take longer time to understand and do it, but eventually of course those who are interested have to do that. So, this is only a background that we will do. With that we said that we will come back to diagrammatic interpretation of perturbation theory. So, that is what we will do and that diagrammatic interpretation when you write the diagram has a throwback on the second quantization, that you write the operators in second quantization, expand everything in second quantization or MP2 energy expression, then convert this into diagram. Again there is a lot of rigor, lot of procedures to do that. What we can only do is to tell you the rules, how to do that. So, before I do, let me again recast your second order energy. Remember again, Psi 0 0 is my Hartree-Fock. I hope you remember that Psi 0 0 is Hartree-Fock. V Psi 0 1, this was the original expression of E naught 2 and then we wrote Psi 0 1 as a linear combination K not equal to 0. Remember it was intermediately normalized, so K not equal to 0, CK 1 Psi K 0. So, we find out CK 1 as overlap of Psi K 0 with Psi 0 1 or any specific L, I whatever you call it. This K is here a dummy index however. So, if I can calculate this integral, then I am in the business. I can calculate Psi 0 1, I can eventually calculate E naught 2. So, how did I calculate this integral? I went to the first order perturbed equation projected with Psi K 0. So, can somebody tell me this value, this we have already done. What is CK 0? So, it is Psi K 0 V Psi 0 0 divided by E 0 0 minus E K 0. That is a general form. So, that Psi K Psi 0 1 could be written as sum over K not equal to 0 Psi K 0 Psi K 0 V Psi 0 0 divided by E 0 0 minus E K 0. Note that for our MP perturbation, Mila plus MP perturbation, our V is sum over 1 by Rij. Again, I want to make sure that you understand sum over 1 by Rij minus the Hartree-Fock potential. That is your V. So, when I put this here, note that this quantity has to be only W excited determinant because of Brillouin's theorem. So, why is the Brillouin's theorem applied? Can you now tell me, because V has also one particle part. So, what I am saying is that when I calculate Psi K 0, V Psi 0 0. So, not K for an MP perturbation, we have Hartree-Fock, singly excited, doubly excited, all these perturbations are there. We are saying singly excited cannot come here. What is the argument? For 1 by Rij anyway, it is more than why it cannot come for 1 by Rij? It can come. Both of them will come, but there is a cancellation. There is a cancellation and that is a statement of the Brillouin's theorem that they will cancel because Brillouin's theorem essentially says that any Psi singly excited, H Psi 0 0 or Psi Hartree-Fock is 0. I can write H as H0 plus V. H0 is sum of the Fock operator. So, sum of the Fock operator, they are eigenfunctions of the sum of the Fock operator. So, this is 0. So, with V itself, it should be 0 and we have actually shown by elaborate calculation by a letter rule this and this that they cancel each other. Each of them survives, but they will cancel each other actually. So, that is the reason this is only doubly excited determinant. So, that is the first important thing to understand. I am going from Hartree-Fock to a doubly excited determinant and in my E0 2, which just comes as a cover for Psi 0 0 V Psi 0 1, I can now write this as sum over k not equal to 0 Psi 0 0 V and then push this Psi 0 1. So, which means this Psi k 0 come here and Psi k 0 V Psi 0 0. For the same reason, now k must contain only doubly excited determinant. So, you have to understand because later on these quantities will come everywhere. So, we then expanded the MP perturbation and we call this A B R S 1 by 4 A B anti-symmetrized R S, R S anti-symmetrized A B epsilon A plus. So, that was the final formula when we expanded this Psi k 0 in terms of this and we also said that if you construct this as a resolvent, then you can write it more nicely with including sum over k, then you can write this as a Psi 0 0 V resolvent V Psi 0 0, where resolvent is sum over all the states Psi k 0 Psi k 0 divide by the energy. Now, what we want to do is of course we can write everything in terms of second quantization and then follow this as a diagram. So, what we will do? We will directly write a diagrammatic form of the E naught 2. There are various diagrammatic conventions. So, I must first tell you there are many many diagrammatic conventions. I have already told some name like Feynman diagram, Huygen-Holtz diagram, Brando diagram, Goldstone diagram, several. What we are going to do is one of the simplest diagrammatic conventions and for this course we will just stick to that and that is called the Huygen-Holtz diagrams. So, I first want to show you that this expression, how it can be written in terms of diagrams. So, there has to be some rules so that from the diagram I can get back this expression and then we will see how to proceed further. So, the first rule is that for every one particle operator, in this case there is no one particle operator because you have only one by Rij. So, one particle operator like Hf whatever is there is represented by a vertex which I can just call some dot vertex with one incoming and one outgoing line. You will see that how the second quantization really comes in. So, for example, if I have to, I have a matrix element which is Hij let us say or Fij whatever depends on what is the vertex then this will be constructed as a vertex. The vertex will be either H or F depending on what I am going to do with one incoming and one outgoing line. So, what is the second quantized form of this Ai dagger Aj? Whenever I have Hij I know that the operator is Ai dagger Aj. So, that means J is annihilated, I is created. So, but J can be occupied, J can be virtual, I can be occupied, I can be virtual. So, the first second rule is that whatever is annihilated must be incoming. That means incoming is it goes into the vertex. The one which is created must be outgoing, it must go away from the vertex, away from the vertex. So, let us assume that J is occupied, I is also occupied. So, both of them are occupied. So, I and J are occupied, occupied. So, remember my one particle operator theta 1 is sum over Ij whatever Hij, Fij does not matter, Ai dagger Aj. Of course, for the Hartree-Fock perturbation it will become F of Ij. So, let me write this as F of Ij for the time being. Do not have to assume that this canonical, non-canonical nothing, it has everything possible. So, let us say my I and J are occupied orbital. So, this is now A and B. So, how do I represent this operator in terms of diagram? So, this operator wherever it is there will be a vertex and now this is actually Ai dagger A B, both of them are occupied. Let us just take an example. So, A B is annihilated. So, the line B must go into the vertex, the line A must come out of the vertex. Then there is a rule 3 before we do that, that all holes, now if this is a hole or occupied orbital lines are lines with arrow pointing downwards and the particles which are virtual orbitals should point upwards, arrow direction should be upwards. Now, you try to with the rule 3, you try to represent this. So, my A B is first an occupied orbital, but it is an annihilation operator. So, first thing is that it must be incoming, it must have an arrow pointing downwards. So, a line pointing downwards comes into the vertex. So, it must come only from the top and B like this. So, this becomes my B, it is occupied line. I am actually writing in terms of normal creation annihilation operator, because in terms of this it is actually whole creation operator. A line but it is being going down because it is a whole line, it is coming into the vertex. So, it should actually end in the vertex because this is the line which is annihilated, electron annihilates. So, this will be actually a whole creation. Then I have a A dagger, this A is also occupied. So, the line again should go down and it should go away from the vertex. So, then obviously it should be like this. And this should have a matrix element attached. So, every time I have an operator like this, I have a matrix element attached. Matrix element will connect these two lines, which is whatever is the value of whatever is the operator. So, if it is H, it is HAB, if it is Fock operator, Fock AB and so on, then of course there will be simplification for the canonical Fock operator that we will see later. For canonical operator of course it is very clear that A must be equal to B, otherwise it is 0, because both A and B are Eigen function of the Fock operator. So, FAB is delta AB times the orbital energy. So, that is a different matter but for a general operator vertex, this should be the operator. Now, let us take i and j something else so that you learn how to draw the diagrams. So, let us say now j is particle, i continue to be occupied. So, now what I am doing? I am annihilating a particle orbital. So, this is basically now AA dagger AR. So, I am annihilating a particle orbital into the vertex, but what is particle orbital? They are upwards, but it must go inside into the vertex. How will you do this? So, it will be like this. All particle lines will be like this, all occupied lines will be like this, but it must come in. So, it must come from the down and go into the vertex. On the other hand, AA dagger is also an occupied orbital. So, its arrow must point downwards and it is now actually going away from the vertex. So, then what is the other option? It must go like this. See, every operator has one incoming and one outgoing line, one particle operator. So, one of them is and this is why second quantization comes. Every operator has one creation, one annihilation, because I told you any theta 1 is AA dagger aj. So, let us take that i as occupied, j as virtual. I had taken here both occupied. So, now I am changing j as occupied, j as unoccupied, i is occupied. So, then AR has to be annihilated. A particle orbital has to be annihilated. So, it will go like this. An occupied orbital is being created. So, actually hole is being destroyed. So, this must go like this. So, for one particle theta 2, we have given i dagger aj dagger kkl. So, in terms of normal quantization, they will be all identical, not in terms of whole particle, because then it is all creation. Yeah, so that is a different matter, because there are four subsets. If you average over the all subsets, again it is same. So, this is a occupied orbital. So, a sense of the line must be downwards and it must go out. So, the only way to make is like this. And AR, this is the only way to make it. Now, I can take the other one, AR dagger AA, just opposite, that is particle creation, whole orbital destroyed. So, what will be the line? This will be exactly opposite. So, this will be now AA, this will be now R, R is being created upwards, A is being annihilated going in and line should be downwards. And all of you should be able to write this very easily, because remember, this is a Hermitian conjugate of this. AR dagger AR, AR dagger AR, Hermitian conjugates. So, diagram wise also, it is a conjugate, you just flip it. Diagram wise, it is just flip it. So, diagrammatically you can see it very easily, that it is a conjugate. This is, this you can already see, it is self conjugate. This is Hermitian, nothing else to do, because both of the lines are going this. This can be flipped. The final one is AP dagger AQ, both are particle. So, very clearly this line will be both up going. This is P, this is Q, correct. And note that again, this and this are very similar, except that the lines are just reversed. Both are unoccupied here. So, once you get into that mood, you can very easily do this. Both are occupied, both are occupied, it will go down. Both are unoccupied, it will go like this. One of them will be whole particle, another will be creation, another will be annihilation, whole particle annihilation. So, these are the two operators. Remember, I am calling this whole particle creation, because A is being destroyed is actually a whole creation. So, in my nomenclature, this will be actually whole particle creation. This will be whole particle destruction. Actually, there is a one pair of creation and annihilation operators in terms of A. But in terms of whole and particle, this becomes both whole particle creation, this becomes whole annihilation as well as particle annihilation. In fact later on you will realize for any operator, one or two particle operator, if I have all lines up, if I have all lines up, it is a whole particle creation operator. It is very easy to understand. Because if I have all lines up, it must have two lines coming from the top incoming and they must be occupied orbitals. And the ones which are going up are also occupied orbitals. So, that is basically a whole particle creation operator. It is very easy to see. If you have turned it around, it must be whole particle destruction operators. So, from the figure, we should be able to say whether it is whole creation, particle creation, whole destruction, particle destruction, from the figure itself. And you will get used to it once you start drawing. This is for the one particle operator. Of course, you have to go to the two particle operator, which is the essential point here because mp2 has two particle operators. So, let me write further rules for two particle operators just as here, they will be represented by a vertex with two incoming and two outgoing lines. Now, they can be occupied, virtual, anything, but the two incoming and outgoing lines. And you attach a matrix element. Just like you attach a matrix element here for every operator, here also I should write for every operator that I draw, there is a matrix element. So, this is let us say rule 4. So, for every operator, there is a matrix element attached. So, for example, for one particle operator, the matrix element is Hij, where the creation annihilation operators are in this sequence, a i dagger aj. So, if i is created, j is annihilated, the matrix element is just Hij, which is basically iHj in the first quantized notation, just to make sure that what is Hij. Similarly, for two particle operators, remember the whole and particle convention remains the same that arrow down is whole, arrow up is particle. That is a general rule. For two particle operator in Hugen-Hull's convention, remember now we are talking of Hugen-Hull's convention and here the convention to convention it differs. The attached matrix element is an anti-symmetrized two electron integral, ijkl for a sequence of operators which is ai dagger aj dagger al ak. So, basically k is going to i, j is going to l, l is going to j and the attached sequence is ij anti-symmetrized k. I will write the two particle operator right now. So, let me write a vortex of a two particle operator by taking a specific indices ijkl because otherwise I cannot write because the arrow line direction must depend on occupied or virtual orbit. So, let me say that it is a p dagger, a q dagger and a a or a b a a. So, I am trying to write this vortex. Of course, the attached matrix element is very clear. It will be pq anti-symmetrized. So, that will be the attached matrix element for this and I am writing this two particle operator now. p and q are created. So, p and q will go away from the vortex and they are up going line. They will go away from the vortex. There will be two pairs of up going line. Now, the two particle operators with two incoming and two outgoing lines. So, two outgoing lines are already defined. My two ingoing lines are a and b. So, let me write this as p and q. This is a and b. So, a and b now must be arrow direction going downwards and incoming. So, they must be like this a b. Both of them are coming. So, if I write little bit nicely, all four lines are upwards. Two incoming, two outgoing. So, this is a, this is b, this is b, this is q. All four lines are upwards to the top of the vortex. So, if you look at the table, there is nothing below but two of them are incoming, two of them are outgoing. So, that defines this operator a p dagger a q dagger a b a. Actually, it is very easy to read from the diagram. From diagram, it is very easy. Just see the arrow. I hope you can read this. p is going out. You can easily see it. Start to think in terms of figures. A p dagger a q dagger a b a. Now, whether it is a b a or a a b, that is the only question. That will resolve later. The anti-symmetric matrix element will only change sign depending on what is the sign. There will be sign convention which will take care of that. So, do not have to worry right now. But the sequence of operators is very easy to write that there will be a p dagger a q dagger. There is a a a b or a b a a. Which sequence they are destroying annihilating? That is the only issue now. Actually, diagrams are much easy to understand. It is very easy to understand the diagram. Let me do the other diagram. So, let us say a a dagger a b dagger a q a p. So, two lines are now annihilated which are particle. So, what will be the diagram? So, this will become p and q. They are up going lines. Up going lines annihilated. a and b are down going lines created. So, they will also go like this, a and b. So, exactly opposite to this. This is again not a surprise. If you look at the string of operators, this is exactly conjugate to this. I have two particle creation, two whole annihilation. Here I have a whole annihilation particle creation. Exactly opposite. So, diagrammatically you just flip it. It is very easy. Diagrammatically thinking is very easy. Of course, there are many other possibilities. All four can be particles. All four can be whole. There are so many possibilities. You can have so many lines. For example, this this can be another way of writing. So, this can be like this. So, all four are occupied. I can see from the beginning all four are occupied because lines are all going down. Two occupied are created, two occupied are destroyed on the top. You can have a situation like this, even this asymmetry. Two incoming, one outgoing, one outgoing, two outgoing. I can make incoming outgoing any way I want. Now, you can see these two are occupied. This is particle line which is created. This is another occupied line which is created. Actually, this is a vertex which has three of them occupied, one of them particle. You can call it three whole one particle kind of line. Where i, j, k, l are such that three of them are occupied, one of them are virtual. I am not drawing all diagrams because you can clearly see there are many, many possibilities. All holes. So, i can be whole particle, j can be whole. There are sixteen possibilities basically. So, all sixteen possibilities will transfer into sixteen shapes of diagrams. Then, I start writing the MP2. I have some more rules. Order, I am not going to suggest right now. The order will be taken care of by the sign convention because it does not matter. If it is PQAB, it is just a negative sign. The sign convention, when I write the entire diagram, there will be sign convention which will take care of the order, which will fix this order. I have to sum. This is only a vertex. I mean, there will be this plus this plus this. It is an operator, but then only one will be used. Finally, only one like an MP2 will show. Only one of them will be used. If all of them are used, then another will be used. Another diagram will be there. Let me write the actual energy expression now. Let me go back to MP2 diagram and write again the E correlation at second order. 1 by 4 sum over ABRS, AB anti-symmetrize RS, RS anti-symmetrize AB divided by epsilon A plus epsilon B minus epsilon R minus epsilon S. Now, what I am doing with these basic rules, there will be many more rules which will come up. I am translating this into the correlation energy diagram. First is to write this vertex. This vertex is ABRS. You can clearly see this is like this, ABQP. RS is unoccupied, AB is occupied. This vertex will translate into this. Is it clear? All four lines to the bottom just like here because it is A dagger, AB dagger, AS, AR. It is exactly like that. Two of them will be AB, two of them will be RS. Of course, the lines will be like this. This will be outgoing, these two will be incoming. You can see that they are being annihilated, R and S and the arrow direction is pointing upwards. These are created, A dagger, AB dagger, arrow direction is pointing downwards. These are occupied, these are virtual. Is it clear? I have also got the basic rule done, two incoming, two outgoing. You draw this vertex, RS AB. The rule is that for every vertex that I have, when I multiply, order them on the board one after another. How will this look like? I have already drawn this, exactly conjugate of this because that is a conjugate of this. Again two of them will be here. You can see that this is my RS, this is my AB. It is AR dagger, AS dagger. These two are going out. These two are coming in. They are occupied, they are virtual. Is it okay? Now I have a sum. I must need to know how to sum in the rule. The sum is performed in a very simple manner by what is called the contraction. This is something that I will just loosely define. The contraction comes because of the second quantization. The contraction is now the product of the second quantized operators. I have a second quantized operator here. I have another second quantized operator here. Only the matrix elements are what is seen in the actual expression. The summation is done by summing over lines, joining lines which are equal in sense. You can see this is occupied, this is occupied. I can join this line. This is occupied, this is occupied. I can join this line. I can join this line. Let me write this little bit better. This is a joining of two what is very simple. These two are AB, these two are RS. You can see automatically one represents this, one represents this. With the same sense you can join. This means I have summed over all AB, RS and I have an anti-symmetrized matrix element attached here. I have an anti-symmetrized matrix element attached here. Everything else would now vanish. The actual sense of creation and annihilation operators will no longer be there. What will be there is only the matrix elements when I write the energy expression. A further rule between the two what is says draw a line. Draw a line between the two what is says. The line will intersect some holes in particle's line. What is it intersected here? This, this, this, this. These four lines. This is just a line between the two what is says which intersects these four lines. Then attach a denominator where you put sum of the whole lines minus sum of the particle lines. This is now A and B. These are whole lines. So we have epsilon A plus epsilon B minus epsilon R minus epsilon S. Attach a denominator. That is a rule. Then multiply the matrix elements. If I give this rule you can automatically see that I am actually getting to this MP2 result and sum over all lines where whatever is joint sum that automatically comes. Of course I have to still explain one by four and all that. I will come to that. There are some more rules are there. So every pair of vertices, so this is very important. Between two vertices means every pair of vertices. So this is important when I go to MP3. There will be three V terms. So there will be another pair of vertices. So I have to draw another line. So later on it will be useful. So now let us go to the other rules. The next rule is that count the number of holes. Please write down in the rule numbers and all that I have lost it but count the number of holes and number of particles which are summed up. So these are called internal lines. Internal lines are lines which are summed up completely. So how many holes are there? After contracted holes, this is one line now. So you can count the number of contracted holes and contracted particles basically. So do not think of four, two here and two here. They are now contracted. So here are actually two hole lines, two particle lines. Then attach a sign which is minus 1 to the power h which is the number of, so h is the number of hole lines, number of internal holes. This is called internal holes. Particle does not come. It is only minus 1 to the power h. The next is count the number of loops, L, number of loops L. So let me show how to count. First of all I must write here when you are drawing this energy diagram. These energy diagrams must be closed and that is understood here. There are no open lines. Closed means all lines are contracted. As you can see here, nothing is dangling. All are contracted to each other. That means basically there are four lines here. Four lines here, they have a handshake. So all completely handshaken. So all the MP2 diagrams or MP3 diagrams must be a closed diagram. That is very important. With the closed diagram now you have to count the number of loops. What is a loop? Loop is the following. You start from here A, go to R and come back. If you start from here and come back to the same point like A to R, then it is one loop. Then again you can come back with another line. B, come back here. B to S. So that is another loop. How many such loops are present? Two. You go from here and come back here. So that is one. That is over. I had another chalk. I could see the color. Then you go from here and come back here. I have traversed everything. I have traversed all the four lines. There is no other loop. You cannot start from here and come back again. It is the same loop. You cannot traverse the same line twice. If A R is there, B S is there, then A S cannot be there. So it is a question of how you write your element. So you can see my element is A to R, R to A, S to B, B to S. So you follow the lexical order here and that is your loop. Number of electrons? Yes. It depends on how I write. That is also important. I could have one loop also and I will tell you how. That will answer your question, the previous question. I can go from A to R, A to R but R comes back to B. B goes back to S. The loop is completed. When I go from A to R and then R comes back to A. So A goes to R but R may come back to B. Then it is a different loop. B goes to S. So that would essentially mean it is A B anti-symmetrized R S, R S anti-symmetrized B A. Then you see the loop. A just follow this. A goes to R, R goes to B, then B goes to S, A goes to A. How many loops are there? One. I have started from A and coming back to A, I have traversed all the lines. So this actually answers your question. How do I write the second quantized rule? It does not matter how you write. Had you written this R S anti-symmetrized B A, my loop will become one less and then the sign is minus 1 to the power L. So there will be a negative sign because your loop is changing depending on how you are writing this. So at the end you need not even see. You write the anti-symmetrized matrix elements any way you want. Keep the outgoing here, keep the incoming here, any way you want. The actual sign of loops will take care of the value. So that is how it is quite sophisticated rule. So let us assume that I am writing it in the manner I wrote. In the manner I wrote, you do not even have to see this. You just look at the algebra. Start from A, go to R, R back to A. I have finished one rule. Then you start from B, go to S or S, go to B, whatever. B go to S and then S comes back to B. So another loop. So two loops. See remember, it is a direct notation. A1, R1. So that is how you should see. So A is going to R, R is going to A. If on the other hand you flip it, then A goes to R, R goes to B, B goes to S, S goes to A. So that actually does the change and that brings the anti, the minus, because you have a minus 1 to the power L, the sign will be automatically taken care. So if I have internal holes H and internal loops L, my actual sign becomes minus 1 to the power H plus L. I mean I can do it together instead of saying minus 1 to the power H and minus 1 to the power L. So that takes care of the sign. In this case, I have two internal holes. I have two loops. So there is no further sign plus. Minus 1 to the power 4 is plus. So it remains plus here. The way I have written this. This is very important. The way I wrote this, because if I write it opposite, then loops will change. That will automatically take care of the anti-symmetry that we are talking of. So I don't have to be specifically careful when I write the two electron integral. What is RSAB or RSBA? I don't have to bother. The loops will automatically take care. Is it clear? Now the final one, 1 by 4. You still have 1 by 4. 1 by 4 comes for the following reason. How many pairs of equivalent lines? So that is the last rule. Count the number of equivalent lines. Count the pair of equivalent lines. What is an equivalent lines? Equivalent lines must be either holes, both of them must be holes or particles. They are outgoing and incoming on the same vertices. Now tell me how many equivalent lines are there? If you look at A and B, that is an equivalent pair of lines. Two lines which are both holes which are starting from this vertex, ending at this vertex. So that is one pair. There is a second pair which is R and S. Both of them are particles starting from the same vertex, ending the same vertex. So in this case, I have two pairs of lines. So let us say, count the pairs of equivalent line and I call these pairs that is a P, then attach a factor which is 1 by 2 to the power P. So in this case, of course, it is very clear that there will be 1 by 4, 1 by 2 whole square and sum is over all lines. So I don't have to specifically bother about A less than B or less than S, etc. Is it clear? So you can see you have so many facts, so many rules to of course eventually write this. But once you understand, the writing becomes very easy.