 Let me now turn my attention to another type of determinant. Let me also first tell you that if I have A as an occupied orbital and it acts on let us say determinant which already includes that orbital like Hartree-Fock, then the result is of course 0. I mean that probably you understand. If A is an occupied orbital, normally you are always using A and B as occupied orbital. So, I hope that is a killer condition. In a way that A dagger psi Hartree-Fock must be 0, because I cannot create an orbital which is already there. Now let us try to understand how do I create a determinant which is psi A R from the Hartree-Fock because this is important. Now I have to come to the actual picture, singly excited W determinant. I am essentially creating a singly excited determinant. What are the things that we are doing? We are destroying A and in that same column we are creating R. The first thing that we do is to destroy A and then to create A R. Let us see how it works. I have a determinant which is again chi 1, chi 2, etc. and here it was chi A, here is chi n. I want A R dagger A to act on this. Before I do this, I use my trick. I interchange chi 1 and chi A, just one interchange. This becomes minus A R dagger A A and this becomes chi A. chi 2, chi 1, chi n. Please do very peacefully. The algebra has to be very correct. Now I can annihilate chi A and then I can create chi R A R dagger. I can very quickly write this as minus chi A will be replaced by chi R. I hope that is very clear because A will destroy immediately to the right. R dagger will create immediately to the left and remember this creation is allowed because R is a virtual orbital. R is a virtual orbital because I am writing chi A R. I hope it is understood that A is occupied, R is virtual. Obviously it is not present in Hartree-Fock. So I can create and it will create right here with the minus sign chi 2, chi 1, chi n. Then I do exactly opposite. I push this to the left and again chi 1 and chi R. Then I will get equal to plus. This minus will again now become plus. I will get back chi 1, chi 2. Then wherever A was there, wherever A was there, one has gone. Now wherever one is there, R will go. So R is replacing A and I get back this determinant. This is exactly my chi A R. The second quantized notation to write chi A R is a singly excited determinant. It is A R dagger A acting on chi Hartree-Fock. Is it clear? Now in the same manner, you can write a doubly excited determinant. So let us try to do that. I also want to tell you that it is very important first to annihilate then to create. You might actually wonder why not others because R is anyway absent here. I might as well have created then annihilate it. In fact, it may be a good idea to show you what will happen. Let us do the other way round. A A A R dagger acting on determinant chi 1, chi 2. Somewhere there is a chi A chi n. I can of course bring chi A here first. So minus A A A R dagger chi A chi 2 chi 1 and now I create. So I have minus A A chi R comes here chi R chi A chi 1 chi n. Now I have to annihilate. So I have to do one more switch. I cannot annihilate otherwise. So this will become plus A and now chi A will first come then chi R then chi 2 etc. chi 1 chi n and now I can annihilate chi A. So I get plus chi R chi 2 chi 1 chi n. Now you see the problem. I have to now push chi R here. I get an additional negative sign. This actually becomes minus of chi A R. You will get chi A R. What is the negative sign? Yes, correct. It is supposed to be like that because there is an anti-computation between them anyway delta A R minus this. So it is supposed to be like this because of this negative sign and I just show this. So if you try to annihilate, create first and then annihilate, you will actually get the same determinant what is the negative sign. So it is important to realize when I have excited determinant first annihilate then create even though both are allowed because this is occupied, this is virtual. So both are allowed but first annihilate then create the virtual orbiter. Not the other way round, you will get the same determinant but with the negative sign. So you have to then take care of the phase. So let us now do the doubly excited determinant. So now you should be able to tell me how to do that. chi A B R S. So how will I do this? Remember in place of A R has to come, in place of B S has to come. So now you have to be very careful. It is not just creation annihilation. So there will be two annihilation, two creation. So first I will annihilate. Let us say A A, A B then I will create. What is important is you create A S dagger immediately and then you are acting on the chi A. So I annihilated A, I annihilated B then you create S so that the S goes in the place of B and R goes in the place of B. It is just a question of plus and minus sign. So I think you can see this again that if you do it only in this manner you get A B R S. You can do that. So what is telling is let us do it the other way round A A A R dagger A B A S dagger that is also fine because then I ensure that the pair is done one at a time. So that is also fine. The reason I wrote in this manner because this will have some significant later when I write the Hamiltonian but you can do it like this also. There is no problem. I annihilate A replace R then I annihilate B replace S. So it goes in the right column. So there is no problem. So either way it is okay. So that becomes my chi A B R S which is the W X R E determinant. Is it okay? So you should be able to handle the wave functions using the second quantized notation. So the last item that is there is actually to discuss the operators because unless I write the operators in terms of the second quantization I cannot do the physics because it involves wave function and the operator. Wave function I now know how to write. I will come back to writing the wave function using somewhat different types of operators which are called holes and particles instead of just electrons. So I will come back to that. That will simplify the electron correlation problem but that is little later. Right now we know using the creation and annihilation operators of electrons I am able to write any wave function and of course remember psi Hartree-Fock itself is A1 dagger A2 dagger blah blah blah up to vacuum. Don't forget this that this psi Hartree-Fock itself has a big expansion which I have written A1 dagger A2 dagger An dagger vacuum. So if you really write in terms of the vacuum it has these then A is dagger AB, A are dagger A where A and B belong to this subspace, R and S do not belong to the subspace. Remember they are occupied the virtual. So it is a long expansion but you can write it and that is what I will simplify later when I introduce holes and particles. I will just close today's class by writing the operator. So let us say that I have a 1 electron operator. Remember I have been always talking of a 1 electron and 2 electron operator. So theta 1 which is sum over 1 electron operator H of i, i equal to 1 to n. These i's are now coordinates of the electron. I am again repeating in first quantization. They are not the orbitals. I hope that is clear and this has been told many, many times. And what we are suggesting is that the expression of theta i in terms of the second quantization is now written in terms of the orbitals i. These are now orbitals i and j as iHj A i dagger Aj. Again first annihilate then create. Note that these ij's and I repeat these i are the coordinates of the electrons. So there are n electron problem. These are coordinates. However these are spin orbitals and this span the complete basis, whatever my basis is, not just the occupied orbital. This is very important to understand. So I am now writing an expression for theta 1 again without proof is the matrix element of this 1 electron operator in this spin orbitals that I know how to calculate now. Coordinate is now dummy. So that is the reason I am saying iHj. What is iHj? iHj is integral chi i star 1 H of 1 chi j of 1, right, theta 1. So you can clearly see these i and j are now spin orbitals. Coordinate is just electron coordinate 1 that is integrated. It does not matter. So these are now spin orbital but what is important is this spans a complete basis. Let us say capital M. Then I can write theta 1 in terms of this basis as this matrix element iHj times A i dagger Aj sum over all ij. So it is not occupied virtual nothing all ij. So what I have done I have transformed this theta 1 which was in first quantization dependent on the coordinates of the electrons to a second quantized form where this is now dependent only on orbital basis. I will explain again more in the next class but let me complete the writing down for the theta 2. The second one is theta 2 which is the 2 electron operator. So what is theta 2? It is 1 by r ij sum over 1 by r ij and again exactly the same thing. So theta 2 in first quantization is i less than j 1 by r ij. Again I am using atomic units. So these are now coordinates. This is in first quantization. I now want to write the form of theta 2 in second quantization and without derivation let me write it as half of sum over ijkl and now these ijkls are spin orbitals. So they are basis and complete basis. Whenever I am saying basis it is spin orbital. These are not coordinates now for the theta 2. You have a half here and then an integral ijkl which is a regular integral in Dirac notation a i dagger a j dagger a l a k. A string of operators. So let me just spend some time in explaining this. What is ijkl? ijkl is an integral which you have already done. It is chi i star 1, chi j star 2, 1 by r 1 2, chi k 1, chi l 2, theta 1, theta 1. So it is very clear that 1 and 2 are coordinates ijkl are actually spin orbitals. In that sense spin orbital basis but this is not an anti-symmetrize integral. This is a regular integral so there is no double bar. Please note this in the definition. So I sum over all such bases and then write the expression of theta 2 in terms of an element multiplied by a string of operators. I have again transformed the coordinate dependence of the orbitals or the electron dependence of the orbitals in the first quantization to the basis dependent in second quantization. So this is my second quantization. This is the first quantization and this is second quantization. So one of the important features of the second quantization is that the operators now do not depend on number of electrons and this is a very significant feature which will be used. You may actually argue that instead you are using so many basis functions but basis functions will anyway come in your calculations. You cannot avoid that. Better bring them in the operator so that the form of the operator remains invariant for n electron, n plus 1 electron, n minus 1 electron, whatever you are doing. In fact it is actually amazing to see this that this was originally coordinates of the nuclear electrons and now there is no electrons and it does not matter what is, capital N is not there. They are all capital N basis. So number of electrons is completely silent and this is an operator. This is an operator. The string of product of creation and annihilation operator is an operator. So dimensionally it is correct. These operator strings is just multiplied by number which is the matrix element ij k l or ij. So we will come back and do some practice with this theta 1 and theta 2. One of the things that we will like to see is that our slater rules using theta 1 and theta 2 in the first quantization is reproduced by the second quantization operator. So now what we will do? We will take a determinant in second quantization on both sides. We will take the operator in the second quantization, expand and we like to see the result that we get is the same result that we got by slater rules in the first quantization. So we will do one or two such practice problem like for example, psi theta 1 psi which is easy. Some psi 0 theta 1 psi. We know that this is nothing but sum over ij and i equal to 1 to n chi i h chi r. The coordinate is a dummy only spin orbital but only n spin orbitals. So I have to first justify because here are m spin orbitals. How does m become n? So we will do that. So some practice problem we will do to justify and also similarly the two electron integral. That is only to give you confidence that what I have written is right and it also gives you a practice. I think I will stop here.