 Greetings, we began discussion on Fano resonances and we have already come a significant way in this discussion. We have discussed the Fano parameters related them to the general formula coming from collision physics. And we began to discuss the Helium atom spectrum as a prototype of a many electron system. So, this is a two electron system, two electron is the smallest many electron system. So, it is best to develop the formalism in terms of a two electron system. And I will spend some time at the beginning of today's class to discuss the Helium atom spectrum. So, here you have two electrons and these are the singly excited states of the Helium atom. So, you have got one electron in the 1s and then the second can go into the 2s or the 3s or the 4s or it can also go into the 2p, 3p, 4p or 3d, 4d, 4f and so on. So, as you go from left to right in this diagram you are going to higher orbital angle of momentum states. So, l equal to 0, l equal to 1, 2, 3 and so on. Now, this is part of the story because here you have in the two electrons one electron in the 1s and the second in the 2s. But then there is there are two possibilities because the two electrons can pair into a singlet or a triplet. So, this is the singlet panel and there is a likewise a triplet panel. So, you can have singlet as well as triplets. So, you can have these two combinations of any two electron system. So, I am going to look at some of these diagrams a little closely magnify some parts so that we can focus attention on details. So, here are the possible configurations of the two electron system. You can have the 1s 2 two electrons in the 1s then you can have one in the 1s the other in the 2s, one in the 1s the second in the 3s and then in the 4s and so on and then finally, into the continuum which is indicated by EC. But then it can go to l equal to 1 state. So, you have the 1s 2p, 1s 3p and so on these are the things that I showed in the previous diagram. And I will strongly recommend this book by Macy and Burhup volume 1 which is a very good reference for these details. So, these are the different configurations based on the 1s state, but then you can have you know these are the singly excited states. So, there is a bound to bound transition and over here there is a bound to continuum transition. So, these are the singly excited states one of the two electrons is excited to the other electrons, but then you can have excitations of both the electrons and then you have doubly excited states. So, these doubly excited states you can have both of them into 2s 2 then one in the 2s the other in the 3s, one in the 2s the other in the 4s and then one in the 2s the other in the continuum s, but then you can also have one in the 2s and the other in the 2 in the 3p this should be 3p and not 2p no it can be 2p and then you can have the 2s 3p that and then finally continuum p as well right and then you can have 3s 2 and so on. So, all of those I have not shown. So, there are these several singly excited state configurations and several doubly excited state configurations and here two quantum numbers are involved two principal quantum numbers are involved both are bound states. So, they have the principal quantum number and one is indicated by an uppercase n and the other by a lowercase n. So, here is another look at the same diagram and here you have got the 1s 2 and these are the singly excited states of the helium. So, this has got an ionization threshold of 24.6 electron volts then you have the doubly excited states and these are based on the 2s 2 2p 2 and so on possibilities right and then you have some more of these like the doubly excited states based on the 3s 2 and so on. So, they have different ionization threshold. So, this is 24.6 this one is 64.6 65.4 this is 79. So, these are the different thresholds for these limits then again as I showed in the previous diagram you can have the orbital angular quantum number moving from left to right. So, you have the s states over here the p states over here and then d over here and so on. So, as you move to the right you can go to higher angular momentum quantum states. The labeling as you notice is done in terms of independent electron picture. So, the quantum numbers are the hydrogenic quantum numbers although this is the many electron system and the hydrogenic quantum numbers are not really good quantum numbers. So, what gives you a good quantum number is a configuration interaction between these states. So, the real wave function will be a superposition of these different states and some of these could also include the continuum states and it is this configuration interaction which is of great interest to our discussion today. So, now in these diagrams which I showed until now today I have not shown the continuum but I will show it in the next slide which we had a glance at toward the end of the previous class and this is from Fano and Cooper's review of modern physics which is very awfully cited literature in this and here this is the same spectrum except that energy is moving from energy increases from left to right in the previous diagrams it was from bottom to the top. So, I have rotated this diagram through 90 degrees and here energy is going from left to right. So, this is the first ionization threshold which I showed you in the previous figure which is 24.6 electron volts and then you have the second one at 65.4 and so on right. So, these are the ionization thresholds which move from left to right as you go to higher ionization thresholds. Furthermore you have we have shown also in this these discrete bound to bound states and you will notice that they have an overlap with the ionization continuum of the singly excited states. So, there will be configuration interaction between the doubly excited states and the singly excited continuum. So, let us have some further look at this now there are some more details coming up in this diagram and that is the reason I am showing this diagram in pieces you know part by part because if you put everything in the same figure all at once there is so much information over there that we tend to miss out on some of the details. So, here we have in this diagram we have shown the 1s2 ground state of the helium atom. So, these are the singly excited states as we have seen. So, this is the singly excited state continuum this is the doubly excited n equal to 2 continuum right. So, the 2s2p states and so on are over here and then you have got the continuum for this and then likewise for the n equal to 3 states. So, these are the ionization thresholds and you can have the ionization threshold is of course given by the usual Rydberg-Bammer formula. So, you can just put it over here and you will get the corresponding ionization thresholds. Now, for helium of course z is equal to 2. So, you get 4 over here z square and that is that is what it is. So, here is the energy scale and this is not to so these scales are different. So, you have got 24.56 over here which comes here then you have the n equal to 2 limit which is here shown on the scale over here and then you have got the n equal to 3 and you have the doubly excited discrete bound states all of these are embedded in the continuum on the left side. So, all the doubly excited states on the right are embedded in the continuum of what is on the left and that is the configuration interaction in which you will have a bound to bound configuration having a configuration mixing with a bound to continuum. So, here is the magnification of this part alone. So, this is the singly excited state configuration right converging to the series limit at 24.58 and it is this part which is magnified over here, but now I show not only the 1 S 2 S and 1 S 2 P separately, but I also show the singlet and the triplet. So, there are actually if you notice over here there are 4 rid bulk series, but they are so tiny that one could miss out on that and that is the reason I have magnified this diagram. So, you have got the 1 S 2 S singlet and the triplet and then you have got the 1 S 2 P again singlet and triplet combinations as I mentioned earlier. So, you have a number of possibilities. So, all of these are built on the 1 S 2 bound state. So, this is the magnification of this part which is the singly excited configuration. Now, if you look at the doubly excited configuration. So, here is a magnification of this part of the doubly excited configuration based on 2 S 2 S 2 S 2 P etcetera, but then again the 2 S 2 S you have a singlet and a triplet and likewise for the 2 S 2 P also you will have a singlet and a triplet P. So, you will have a number of rid bulk series coming out of this. So, here is a so this is again a magnification. So, there are actually 4 rid bulk series in the double excitations. So, this is what leads to the configuration interaction between the bound to bound and bound to continuum because all of these discrete states doubly excited discrete states they are embedded in the continuum. So, here you will have a similar spectrum also for the negative hydrogen atom which is again a 2 electron system. So, I am not going to discuss this in great length, but just to point out that the hydrogen minus will also have similar very similar rid bulk series. So, when you work with photo detachment and so on you will be using a spectrum of this kind. So, now this is the main question of interest that you have these 2 electron configurations and a many electron system you can you can discuss important correlations in terms of these 2 electron configurations. So, when you have the energy of a doubly excited state in 2 possible configurations 1 of which is a bound bound and the other is bound continuum then you will have resonances and what is going to happen is that the electron may spend some time in the excited state configuration before escaping into the continuum. So, the electron can of course escape into the continuum because energetically that is possible. So, it would happen, but before it does so there would be a little bit of time that it spends in the doubly excited state and what this time delay will do is to cause an additional phase shift which will be a resonant phase shift because the double excitations belong to the discrete bound states and they are therefore, at very sharp not not really sharp they have their lifetimes, but they are at specific energies at the resonance energies. So, there will be a resonant phase shift coming from this and it is this that we are going to learn from Fano how to analyze this. So, we mentioned in some of the earlier classes that there are these 2 kinds of resonances the Feshbach resonance or the Fano Feshbach resonance as I normally refer to it and the other is the shape resonance. So, the Feshbach resonances are because of configuration interaction between the bound to bound and bound to continuum configurations. So, the shape resonances are due to the form of the potential. So, this is when the electron can tunnel through a potential barrier and then it finds itself to be exposed to 2 different alternative states one which is trapped in the inner well and the other after tunneling through the barrier. So, these are the 2 possibilities and Feshbach resonances will therefore occur below the ionization thresholds where a shape resonances will be above the thresholds. So, these in both cases you do undergo a phase shift through pi as you go through the resonance. So, the Fano Feshbach resonances are relatively narrow and this is just a gross feature one has to get into the details because within the family of Fano Feshbach resonances there are very many different kinds of shapes and widths that we have already talked about in our previous class. So, what happens is that our interest now is in studying the electron correlations the configuration interactions which lead to these Fano Feshbach resonances and they lead to auto ionization these resonances that we are talking about are more specifically called as auto ionization resonances. And what happens is because the discrete bound state is embedded in the continuum at some energy the energy difference between the 2 states becomes exact and then it becomes impossible to use ordinary perturbation theory because in perturbation theory you often have the energy difference in the denominator and then things blow up. So, you need special techniques and that is what Fano introduced in this marvelous paper. So, let us refer to this configuration interaction between discrete and discrete and continuum. So, the discrete state I will indicate by phi d. So, get familiar with the notation I am using the continuum state by psi e the discrete states are a square integrable. So, I normalize them using the usual integral normalization integral and the dimension of the discrete states are of course, l to the minus 3 by 2 the continuum states however are not square integrable and these are then normalized on the delta function like a delta function scale. You can normalize them on a k over 2 pi scale in which this is how the normalization would go or we have discussed this normalization earlier as well in scattering theory and you can also normalize them on the energy scale and if you normalize them on the energy scale then the dimension of the ray of the functions would be e to the minus half l to the minus 3 by 2. So, mind you the dimensions of the discrete states and the dimension of the continuum states are different because of how we have chosen the normalization. So, these dimensions are different and that is something that you must keep in mind. So, this will also be similar case for the negative hydrogen ion and you can think of this 2 electron system whether it is a negative hydrogen ion or a helium atom as a 2 electron system and if you ignore the interaction between these 2 electrons then it is as if you are having a Hamiltonian for one hydrogen atom and another hydrogen atom. It is almost like that. So, the 2 energies would add up if you ignore the interaction between the 2 electrons. So, you have got the kinetic energy operator for one electron and the kinetic energy operator for the second electron and the potential energy of each of the electron in the field of the nucleus, but you have ignored in H 0 this is the unperturbed Hamiltonian the interaction between the 2 electrons. So, this is your unperturbed Schrodinger equation and it gives you an energy which is just the sum of the 2 energies. Now, you can write the 2 electron wave function as a product of these hydrogenic states and your product then becomes your 2 electron wave function becomes a product of these 2 and these are obtained as solutions of these 2 independent Schrodinger equations. Now, this is the case when both the electrons are in the discrete, but we considered various possibilities you can have 1 s, 2 s, 1 s, 3 s and so on, but both are in the discrete, but what if 1 is in the continuum and this is the 0 order energy for the 2 electron system when both are in the discrete, but then you can also have 1 in the continuum and the first order correction when you do include the interaction if both the electrons were in the discrete would then be given by the matrix element or the perturbation Hamiltonian in the unperturbed states which is phi d which is just the product of those 2 1 electron hydrogenic states. So, this would give you the first order correction. So, this goes into the energy matrix E phi. So, we are going to construct the energy matrix for various possibilities. So, you have now the perturbation H 1 and E phi will give you the energy correction due to the electron-electron interaction if both the electrons are in the discrete. So, this is the possibility we have considered over here. So, E phi is the correction and of this H 0 is nothing but the sum of the 2 energies that we got earlier and H 1 comes from this correction over here. So, this is what gives you the E phi which goes into the correction. So, this is when both the electrons are in the discrete. Now, what if one of the electron is in the continuum? Now, if one of the electron is in the continuum then one electron state is represented by the bound state which is chi and the other will have an energy which is E minus epsilon 0 with a position coordinate R 1. So, the, so you still expect the 2 electron wave function as a product of the 2 wave functions, one for the discrete state and the other for the continuum. So, now you have this continuum and the discrete state is embedded in this continuum. So, this is now your energy matrix. So, you have got the energy contribution when both the electrons are in the discrete that is phi D which is what we indicated earlier. When you have one in the discrete and the other in the continuum right then you have V E and then here this is just the matrix element of the Hamiltonian in the continuum states. So, this will have a delta function popping up right because both of these energies are in the continuum. So, this is now the framework of the energy matrix. Now, the thing to remember as I mentioned earlier is that this is got the dimensions of L to the minus 3 by 2 and this will have a dimension of E to the minus half L to the minus 3 by 2. So, what is the dimension of V E? The dimension of V will be E to the half. So, this V has got the dimension of root energy and this is to be remembered because when we take its square you get the energy width which has got the dimensions of energy. So, that comes out as V square in our analysis. So, just remember these dimensions. Now, the configuration interaction wave function will be a superposition of the discrete state and the continuum state and the continuum you will have an integration over a range of energies right DE prime right. So, this is your configuration interaction wave function and now our task is to determine these coefficients A and B and go from there ok. So, here we are. So, this is our Schrodinger equation. Now, let us project this on a discrete state because this is the Schrodinger equation for the two electron system and it will have components on the two electron bound states or one in the bound state and the other in the continuum. So, in this case we project it on the discrete state possibility. So, let us take the projection of this Schrodinger equation on the state phi d. So, we have got two terms one coming from the discrete and the other coming from the continuum right. So, these are the two states. Now, we already know this what is it? This we have already determined because this is E phi right. This is our E phi which would go into the energy matrix. So, this is E phi. So, we put this over here and now you have got the matrix element of H in this continuum and phi d. So, this H over here is nothing but E psi right. So, I pull E common and now you have got a projection of psi on phi d which you get from here which will give you A E. So, now just I have written this Schrodinger equation. So, that on the left you have got E psi on the right you have got H operating on the superposition of the discrete and the bound states. So, this is the projection that we are considering right on phi d. So, here you get the E phi which is here. So, E phi comes here this coefficient is A E coming from here and then this factor is nothing but V which we have written earlier right. So, you get V over here and this is the projection of psi on phi d. So, the projection of psi on phi d is this coefficient A E which comes here right. So, you have got E A E equal to A E E phi plus this integral. Now, I will refer to this equation with this green star because I am going to need it in subsequent analysis. So, whenever we need to refer to this or I will pull it up with a reference to this green star it is like giving it an equation number. So, here you have got this configuration interaction. Now, we will this time project it on the continuum. Now, this continuum state is a mix of this is the configuration interaction state it is a mix of a discrete and a continuum right. So, it has got both the components and we now project this Schrodinger equation on such a state which is a mix of discrete and continuum. So, this is our projection now look at it term by term. So, this is the projection on the discrete part and here is a projection on the continuum part. So, now what do we get? We look at each term separately A E we can pull out as a we can as a factor we can factor it out and then you have got the matrix element of H n psi n phi d which is nothing but V. So, from this term we get A E V E and likewise from this term we will get a Dirac delta because these two states are in the continuum. So, you get a Dirac delta there. So, now you have this relation and now we have the left hand side which is E and these two will of course give you a Dirac delta you can see it coming and then on the right hand side you have this delta function integration over E prime. So, only the term in E prime equal to E double prime will survive. So, you get a coefficient B with E E double prime and this E becomes double prime that is because of the Dirac delta integration. So, here are the relations we have. Now, this is the result that we got on the previous slide and in this I now put this superposition on the right side. So, now I get two terms and from the first term E and A E are scalars they will come out and you will get a projection of the continuum state on the discrete, but they are orthogonal. So, that will vanish and then from here you will get the scalar product the scalar product of these two which will give you a Dirac delta. So, let us put both of these over here and now you get from the first term you get 0 from the second you get a delta function you carry out the integration you get only the term in E prime equal to E double prime. So, you get this equation here and this is the one which I will refer to as a red star. So, now we have got two equations that I am going to be referring to one with the green star and the other with the red star. So, these are the two relations one with the green one with the red. So, these are the two relations that we are now going to analyze. Now, V is real and here we have used two primes. So, there is no need to use two primes we will use only one prime. So, I will rewrite them without the double prime and writing them as real numbers. So, what do we get from this relation red star? From here we get V equal to A E V E divided by E minus E prime which is fine as long as E is not equal to E prime. So, all this will work as long as E is not exactly equal to E prime, but that is just the case we really want to handle that is precisely what we want to handle. So, to be able to include the E equal to E prime Fano used a very beautiful technique which was devised originally by Dirac and this is the Fano Dirac technique. So, they make use of principal value integration to address this situation. So, I will show you how it is done. So, the purpose is to address this particular case of E equal to E prime and this is the Dirac Fano technique and what this technique does is following Dirac and this is really beautiful device to handle such situations that instead of writing B like this you write it as 1 over E minus E prime. So, the A E V E prime is here which is here. So, this is the A E V E prime that factor is here and then over here you add a function z multiplied by the Dirac delta because you are going to use it in some integrant. So, that will leave a value only when E prime is equal to E which is just what you want. So, that is the trick. So, this is to be used in principal value integration. So, this is the Dirac Fano trick and you will see that it really does wonders for us and the function z is so chosen that it meets the appropriate boundary conditions because the boundary conditions are known to us. So, by putting appropriate boundary conditions you can determine the function z and now whenever you have BE in the integrant then you can use this Dirac Fano substitute. So, this is the brilliant mathematical device which Fano used in this context. So, this is to be done using the principal value integration whenever BE appears in the integrant and the principal value integral which I am sure all of you would have worked with in your mathematical physics courses or some other courses and collision theory and so on. So, whenever you evaluate an integral from E 1 to E 2 and then you have a problem over here when E prime goes to E then you go close enough and then hop over that point and then do the rest of the integration. So, that is the principal value integral that is how it is defined. So, this is our Dirac Fano proposal and what is understood is that the principal that whenever you have this type of an integral to be determined then you should carry out a principal value of this integration and use z appropriate to boundary condition. So, these are the two things which go into this technique. So, it is the idea is very simple but extremely beautiful and very powerful. So, let us see what it does for us in this particular case so here we have the principal value integral defined and our interest is in this relation here which we have identified as a green star and over here this integral will be replaced by the principal value integral. So, this integral this BE is to be replaced by this factor here coming from the principal value integration trick of Dirac Fano and we will work with this relation instead of this. So, here we are so now this is the relation that we have agreed to work with this in this we have already employed the Dirac Fano trick now first we cancel out AE which is common in all the terms. So, AE is cancelled out we need not carry it any further the rest of the equation is here after cancelling AE now in this we have got three terms one is E phi which is here then you have this term which is here and then you have got the third term which is integration over E prime Z V E square and the Dirac delta. So, there are these three terms in this expression are we all together. So, now here what do we get we make use of the principal value integration from this term you have got a Dirac delta integral. So, this is the Dirac delta integration and this will give you a value which is appropriate only for E prime equal to E because the integration is over E prime. So, you get a value which is appropriate only for E prime equal to E. So, you get Z V E square from this term and here you will now carry out a principal value integration because the value for E equal to E prime is already taken care of over here that is already taken care of. So, your problem is done actually it is really brilliant idea. So, now it also tells us what Z should be because Z is now a solution of this. So, this principal value integral you represent for compactness by a function E this is a function some function of E there is no difficulty in evaluating it there is no singularity in that right. So, you evaluate this integral and you represent it by some function of E and that tells us what Z should be it is given by this. So, now we have almost all the pieces that we need and now it is just a matter of rearrangement of these terms to extract our final state configuration interaction wave function. So, here the Z is given by the Dirac Fano trick the configuration interaction is this as we know. So, what we are going to see is that this particular state which is in the continuum. This will look very similar to a continuum state solution that we have worked with in scattering theory all along you will remember that this goes a sin k r plus delta right. So, you have got the sin k r k depends on energy. So, it is sin k r plus delta which is a background phase shift, but then there is an additional phase shift which is coming from the configuration interaction this is the resonant part. So, there is this additional phase shift which is coming because of the time delay that the electron may get bound into the doubly excited bound state for some time before it escapes into the continuum. So, you have got a background phase shift and then a resonant phase shift due to the configuration interaction. The resonant phase shift it will turn out and you will see this coming in the next you know as we go further in this class that this resonant part of the phase shift will turn out to be given by negative trend inverse pi over z that is what it will turn out to be, but that is a result which you will see coming. So, now we have to determine this coefficient a according to normalization and this is our normalization here. Your configuration interaction state is you have a combination of the discrete state and the continuum state. The discrete state is weighted by this coefficient a. This is what we want to determine. These states are of course Dirac delta normalized. So, if you take this Dirac delta on the left, then on the right side you have got a projection of this on itself. So, you will get A e star A e from the first term phi and psi are orthogonal and then you will get the B e star B e times this. So, we have used these orthogonalities earlier as well. So, this is the left hand side is nothing but the Dirac delta and then the right hand side you have got the A e bar A e. There are two different energies I am considering mind you both are in the continuum e and e bar. So, you have got two coefficients A e bar and A e and over here because you carry out this delta function integration over e double prime. So, that the only term that survives is the one for which e double prime is equal to e single prime. So, you have got the e bar e single prime and then the B e and the subscript here is again e single prime. That is the Dirac delta integration. Now, here is the relation which we have obtained earlier which is from the previous slide which I indicate by this diamond here. The only thing I have done in moving this to the next slide is I take this A e bar on one side. So, I have delta e bar minus e over here. So, this is the relation I get from the previous slide. This is the Dirac-Fano trick. So, we have putting all our pieces together. Now, in this relation which is indicated by this diamond relation here, this over here is I use the Dirac-Fano inclusion of e equal to for equal to e prime because both are here you have got e prime over here. So, you have one for e bar and the other one for e. So, you have two substitutions over here one coming from this relation and the other coming from the corresponding relation for e bar. So, this is for e and this is for e bar. So, these two go into these two pieces. So, I substitute this and now I have got an equation which looks slightly large, but we can handle it on the screen. So, remember that you will get quadratic terms when you multiply you will get a z over here and another z over here you have a v over here and another v over here. So, you will get quadratic terms and z and v. So, let us write all of these terms. So, you have the a e v e and a e v e prime over here. So, there is an a e bar and a e. So, a e bar comes here the a e comes here the v e bar the v e prime is inside the v e prime is inside because it is under integration. So, it does not come out and then you have two terms over here. So, the v e prime remains under the integrand. So, you will get v e prime over e bar minus e prime here. Then you have got the second function there is a Dirac delta here and then you have got a v e prime sitting on top of this e minus e prime and then this function z and then this Dirac delta which is here. So, now you have got 1 2 3 4 4 terms. So, 2 terms multiplying 2 terms over here. So, you will get 4 terms. So, they are here 1 2 3 and 4 you have just multiplied them out and I think this is when I was preparing the slides I realized that it is really nice to be able to use in this technology enhanced learning program that we have the provision to flash these slides at the click of a mouse. So, that we can concentrate focus of a discussion on the physics and not get lost in carrying out mere substitutions that itself takes such a long time on a blackboard if you were to write it using a chalk. So, this comes at the click of a mouse and we can actually concentrate on the physics and on how the analysis is done. So, here we have these 4 terms right and you can of course, look at the pdf at leisure and make sure that everything is done right. So, here you have 4 terms 1 2 3 and 4 out of these 4 terms these 3 have got Dirac delta and you are carrying out integration over E prime. So, they will give you terms in which only E prime equal to E will survive right and then there will be a 4th integral which will be left coming from this quadratic term in V right. So, those are the 4 terms. So, here is the result of those 4 terms. So, here you have got the first one you have got this integral over E prime this is quadratic in V. So, here it is and from the remaining 3 terms you carry out the Dirac delta integration and you are left with only those terms corresponding to which E prime is equal to E bar in this time over here E prime must be equal to E and over here E prime must be equal to E bar. So, those are the 3 terms which you are left with, but remember that in the last integral you have carried out integration over E prime. So, E prime must be equal to E bar. So, when you write this Dirac delta E prime has been written as E bar. So, that is something that you have to keep track of. So, now we have got all the terms and I have now I have got these 4 terms, but I rewrite them. So, this first term which is indicated by this 4 point green star which comes here. So, it is just a rearrangement of terms, but the color code will show you which term is where then I have this term coming over here with the difference that here I had E minus E bar, but now I have E bar minus E. So, I have plugged in a minus sign to compensate for that and then I have this term which comes here as it is and then I have this term with the Dirac delta E bar minus E coming here, which is a quadratic term in Z and V, good. So, now we have a double singularity here. So, what do we do? We use identity which is well known in mathematics. So, when you have a double singularity of this kind you use this identity that which lets you take advantage of the Dirac delta integrations. So, I will not prove this identity I will use it. So, this is the identity that we will use and this was used by Fano in this work. So, instead of this 1 over E bar minus E prime multiplied by E minus E prime you use the right hand side of this and this will give you the Dirac delta or product of Dirac delta in the numerator and then the rest of the terms. So, we will use this identity over here and using this identity this integral now takes this form. So, instead of this integral you have the D E prime V E prime square coming here the D E prime V E prime square and here this 1 over double singularity is now written in this beautiful bracket which is this entire right hand side of this identity. So, this is the relation that we now have to work with. So, I brought it to the top of this slide now. Now, here again if you play with the Dirac deltas you can show that the product of these two Dirac deltas is completely equivalent to this. So, we will use that and using this combination instead of this we simplify this further and let us now analyze this relation. So, this is the relation that we are going to analyze and I will I have used most of the discussion is from Fano's paper there is some from Kavans paper and some from some other sources. So, I will cite all the references and it is good to use them in combination. So, in addition to Fano's paper I recommend Kavans book theory of atomic structure and spectra and the notation is slightly different which is why I am alerting you to that that the principle value function this integral has the same notation in both Fano and Kavans, but the difference is here that what Fano calls as z instead of that Kavans uses eta, but his function differs from Fano's by a factor of pi. So, that is the only difference and if you are using a combination of these two sources then you have to be careful with the literature. So, here what we will do is we have this relation here in which you have an integral of 0. Integral over E prime of a quadratic term in V over E bar minus E and this is where you can use the principle value integral which is represented by the function F. So, we will use F of E instead of this term over here because the E equal to E prime or E equal to E bar we have already taken care of using the Fano Dirac trick. So, now by using this F instead of this we now get all of these terms are now over here. So, first I write these two terms which come here. So, these two terms come here then you have this term these terms in which you have got V square over E bar minus E. So, these are replaced by the principle value integration F E minus F E bar. These are the two which are coming here and then you have this term which comes here and then what are you left with? You left with one term over here which has got the pi square V square. So, the V square is here A E bar A E is here and the pi square is here and then you have got the E prime equal to E. So, instead of E bar minus E prime you have got E bar minus E because E prime has to be equal to E. So, that is what you get here. So, now I think we are getting it in a mathematical form which is becoming compact and we can address it in a tractable form. So, notice that these terms both have got this factor common this is A E bar A E over E bar minus E this is A E bar A E over E bar minus E. So, you can combine these two terms you can factor this out and write it write the remaining terms separately. So, here the principle value how are we including E equal to E prime because the delta function is equal to E prime is here it goes into the Z that is taken care of already. So, when you take the principle value integration you are hopping over the singularity you go as far as the singularity and then pick up the integration from the point next to it and the point of singularity is taken care of by the function Z. So, here you are and now you have got this term as common in this factor. So, we can factor it out and I have factor it out and written the remaining four terms. So, they are now in terms of f and f E bar. So, here I have here I have f and f E bar and here I have Z V square and here I have Z E bar V E bar square. So, these are the four terms which will be in the remaining factor. So, those are the four terms in the remaining four in the second factor this has been factored out and this is the residual part which we had written here which had the Z square plus pi square and the Dirac delta. So, here it is. So, you have the A E bar A E V E bar V E Z square plus pi square the Dirac delta. So, we have got all the terms Z we know we already have seen that Z is given by this relation and we are now able to anticipate a very important very nice result. So, this is the value of the function Z at E, but we have two energy indices one is without the bar and the other is with the bar. So, this is Z E bar and you put both of them. So, you have got two expressions because you have got Z V E square and then a Z E bar V E bar square. So, you have to use the one with appropriate notation and the difference between these four is then nothing but E minus E bar because of this cancellation because when you cancel this the E phi drops out and then you have only the E minus E bar. So, now we have got a simpler relation. So, this is the left hand side which comes here and the right hand side you have got A E bar A E or E bar minus E which is here what is under this red under lining is now just the difference between E and E bar which is here and then you have got these two terms over here which is Z square plus pi square direct delta multiplied by A E bar A E and V E bar V E. Now, I rewrite this by moving this term to the left. So, I have got on the left hand side this term minus this, but I have it with the plus sign because I have E minus E bar in the numerator and it is negative in the denominator. So, I have got A E bar A E with the plus sign on the left. So, this is the relation that I will bring to the top of the next slide. So, here we are. So, we have got these two terms of the left. Now, this is the relation that we have referred to earlier from slide 68 the which I had indexed by a diamond and now if you look at these two relations it is obvious that this factor over here must be equal to unity the left hand sides of both the equations are the same the right hand sides are very nearly the same except for this factor which must be equal to 1 and that tells you what the coefficient A E must be this is what we were looking for. We wanted to know what is the weight factor in the configuration interaction mixing. So, this is the weight factor and this now comes out as a square root of 1 over V square Z square plus pi square. So, here you can write Z because we know what Z is Z is E minus E phi minus F over V E square. So, you can rewrite in terms of this function of E instead of Z. So, this is your weight factor A. So, we now have this and now we have taken care of the E equal to E prime and now we get the configuration interaction we have all the terms we have the B this mixing we have got this coefficient. So, we now get the complete wave function and in this we make use of the principal value integration as we have done. So, you have the complete configuration interaction wave function in which you have the A E phi D which is here and then you have this piece over here. But because of this Dirac delta integration you have got only the term corresponding to E prime equal to E which is why you get this Z E V E psi. So, this is the configuration interaction wave function that we have now reduced and we will use this in our next class when I will discuss the resonance phase shift. Because I mentioned that what happens as a result of the configuration interaction that the electron which has the possibility of escaping into the continuum because one of the two configurations belongs to the continuum. It has the possibility of spending some time in the doubly excited bound state. So, this is the bound to bound and bound to continuum configuration interaction and because it spends some time over here that time delay manifests as an additional phase shift. That is the resonant phase shift which I had anticipated little bit toward the beginning of this class. But we will see it more explicitly in the next class. Any question?