 Greetings. So, welcome to the unit 5 of this course. This is again on collisions and I will call it as part 2 of quantum theory of collisions because the first part we already have some 11 or 12 lectures in the unit 1 and then we took a detour and did some second quantization, random phase approximation, Feynman diagrams. So, we did that in unit 2, 3 and 4 and now we are resuming our discussion on quantum collisions. So, in this unit we will begin with the Lippmann-Schwinger equation for potential scattering and subsequently we will also be doing the Coulomb scattering and then the resonances. So, this unit will basically have 3 classes, first on Lippmann-Schwinger equation, second on Born approximations and third on Coulomb scattering and then in subsequent units we will have about 8 classes on resonances. So, there will be 4 classes in unit 6 and another 4 in unit 7. So, there will be a total of 8 classes on resonances. So, now this is the picture of scattering that we have in front of us that you have got a scattering target and incident beam which comes and interacts with it gets scattered and you can think of the scattering region as a certain spherical region of space and different parts of the region would be responsible in some sort of a cause-effect relationship that that would be the central cause for scattering scene at a given point. So, you can think of this phenomenon in terms of a cause-effect relationship and we will introduce the Green's functions or the propagator to describe this. We have a different source point in the scattering region and I will describe the field point with the position vector r, the source point with the position vector r prime with reference to the scattering center and the difference vector is r minus r prime. So, this is the direction in which scattering is taking place with reference to the picture you have in this diagram. So, the essential equation quantum mechanical equation that we are looking at we are doing non-relativistic quantum mechanics and we have the Schrodinger equation with which we work. We can write it in terms of the reduced potential just to get rid of some constants like 2m and h cross square. So, we rewrite this equation in terms of the potential, but essentially it has got the same information as is available in the Schrodinger equation itself and you have on the right hand side of this equation the inhomogeneous term and when you set it equal to 0 you get the corresponding homogeneous equation for a 0 potential which is the problem for a free particle. So, this operator the del square plus k square operator this is known as the Helmholtz operator and we can have the Green's function for the free particle which is defined by this relation. So, this is it defines the Green's function which is the solution for the Helmholtz operator and now we have to find out what would be the appropriate Green's function which will describe the scattering process. So, this is a differential equation, but we can also cast it as an integral equation and what we propose is that if we set up and a solution in this form that the solution way of the Schrodinger equation which is this psi that this can be expressed as a sum of the incident plane wave and an integral involving the Green's function and the potential and then there is also a solution psi over here and this is a little this is some kind of cheating because what appears in the integrand is this wave function which is what you really want to determine. So, you do not know it in the first place. So, it is in some sense not really quite fair to put on the right hand side what you have on the left hand side of the equation, but that is the formal structure of this relationship and I will discuss this the consequences about this. So, this is the Green's function and this we have to determine appropriately with due consideration to the boundary conditions because the boundary conditions are essential to analyze the solutions you when you have a differential equation means you can get very many different kinds of solutions which satisfy the differential equation, but the ones that we are interested in are to be subjected to boundary conditions which are appropriate to the physical problem that we are dealing with. So, this is a tricky situation you have got wave function on the right hand side in the integrand which is actually the one on the left hand side and this is what one would regard as a catch 22 situation that you are asking for the solution to be written in terms of what you really want to determine and that problem is it falls into a vicious circle and you really cannot solve it. So, we will figure out how this is to be solved, but the first thing to do is that the formal structure of this equation is appropriate that it does indeed describe the scattering process and that is the first thing that we will consider. So, let us begin with what we have over here. So, we have the Schrodinger equation the first thing we do is show that this is indeed a solution of the Schrodinger equation although it is not a very useful solution because it is describing in the integrand the wave function which is what you want to find. So, it is not particularly useful. So, whether or not it is useful is a different story, but that it is at least correct that formally this is a solution of the Schrodinger equation is something that we first test and that can be seen by operating on the right hand side of this by the Helmholtz operator. So, as if you do that you can very easily verify that in fact it is a solution. So, let us do that as the first step. So, you operate on this formal structure formal solution that has been proposed by the Helmholtz operator and now you know that this is the homogeneous equation for a free particle when the potential is 0. So, the first term gives you nothing it gives you 0 and then you have the Helmholtz operator operating over here. So, now you know that you are going to pick up a delta function from this operation because that is how the Green s function is defined. So, del square plus k square operating on the Green s function gives you the delta function. So, this is the Dirac delta. So, you get a Dirac delta from this operation and then you are going to integrate this potential on this wave function multiplied by the Dirac delta. So, you will end up having a Dirac delta integration. So, here you have the Dirac delta inserted in place of del square plus k square operating on the Green s function and you find that when you carry out this delta function integration you recover the Schrodinger equation. So, we know that whatever we proposed as an integral equation is formally correct regardless of the fact that it is not particularly useful at this stage. But then we will figure out how to develop some sort of an approximation to it so that it will become useful. So, this is the Green s function, but we do not know what the Green s function is and this Green s function will have to be chosen according to appropriate boundary conditions because just like the differential equation requires appropriate boundary conditions to be solved the integral equation requires the Green s function to be defined with reference to appropriate boundary conditions and we know that in collision physics we make use of these outgoing wave boundary conditions. So, you remember that and we will put those boundary conditions to describe the Green s function in the collision process. So, first we have to determine the Green s function G 0 with reference to appropriate boundary conditions. So, what we will do is take the Fourier transform of the Green s function. So, let us define this Fourier transform and just to simplify the notation I will drop this K. So, instead of G 0 K comma r I will write this only as G 0 r. So, this is nothing but the Fourier representation of the Green s function this is the usual standard Fourier representation of the Green s function and we will also have the usual Fourier representation of the right hand side of this which is the Fourier representation of the delta function. So, which is a sum of these exponential terms. So, when you sum up all these exponential terms integrate over the whole volume in the K space you get the corresponding Fourier representation of the Dirac delta. So, instead of G 0 and delta on the on this side we use the corresponding Fourier representations. So, here they are. So, now you have got this integral of in the momentum space or the K space and then you have got these coefficients G K prime and now you have del square plus K square operating on e to the i K prime dot r which you can see immediately will give you this i K prime square from the del square term. This will come as a K square multiplier right hand side is pretty much the same the 2 pi 3 on both sides of the equations have cancelled each other. So, we are getting a pretty straightforward relationship now and now you can see that you have in the integrand both sides of the integrations are integrations in the K space and the integrand has got one factor which is common e to the i K prime dot r on both sides. So, if you look at this the this is the definite integral in the K space over the entire K space. So, the corresponding integrants will have to be the same and that tells us that this G K prime must be the inverse of K square minus K prime square. So, that the product will give you the unity which you have on the right hand side. So, this is what you get for the Fourier transform and here you can factor this into K plus K prime and K prime minus K and you recognize that the greens function is now given by this integral. So, this is mathematically completely equivalent to what we started out with using the Fourier transforms. Now, having got this notice that what you are integrating out there is a certain symmetry that you can exploit because in this symmetry you can you have the volume element in the K space which is given as K square d K sin theta d theta d phi. So, this is the spherical polar coordinates in the reciprocal space in the K space and if you exploit the azimuthal symmetry about one direction about which you can choose the polar axis then integration of the azimuthal angle about that will give you a factor of 2 pi. So, let us get that out of our way and the rest of the integration is now only over 2 degrees of freedom which is theta and phi. So, you have got the K square d K sin theta d theta right the integration variables I am using as a prime because that is what we started out with. So, these are the primed variables K prime square d K prime sin theta prime d theta prime this is the integrand 1 over K plus K prime K prime minus K e to the i K prime dot r. Now, look at the integrand over here. So, now we can separate out the integration over theta and K these are independent degrees of freedom. So, what you are integrating over theta actually theta prime. So, the range of theta prime is from 0 to pi range of phi prime was from 0 to 2 pi which is what gave us the factor of 2 pi. So, now you have separated the integration over theta prime and pi and it is convenient as we do in so many problems right from high school mathematics to change the integration variable to cosine theta prime and then write this integration appropriately you have to put the limits instead of the limits for theta prime you now put the limits for cosine theta prime, but mind you there is a minus sign here. So, the limits go from cos theta prime equal to minus 1 to plus 1. So, this is pretty straightforward as such. So, this is the integral over the polar angle theta prime. So, carry out this integration you have got e to the i K prime r cos theta prime over i K prime r and then you have to take the difference of cos theta over the upper limit less the lower limit. So, let us do that. So, this is the difference at the two limits you have got the denominator i K prime r right. So, now what do you have you have e to the i K prime r minus e to the minus i K prime r and you can write this as a sinusoidal function and then get rid of the common terms 2 i and so on. So, that is straightforward algebra and this is the expression that you get. Now, let us take this to the top of the next slide and if you look at the integrand the integrand is an even function of K prime. So, you can extend the integration from K prime equal to 0 to infinity to K prime equal to minus infinity to plus infinity and then take half of it. So, mathematically it is absolutely satisfactory physically it is meaningless because in the spherical polar coordinate system you have the space covered by the polar angle going from 0 to pi the azimuthal angle going from 0 to 2 pi and the third degree of freedom going from 0 to infinity in the reciprocal space it is the K or the K prime. So, physically K prime having negative values is of absolute no significance, but mathematically our interest over here is to determine what is the value of this integral and we can use complex analysis to evaluate this integral by exploiting the fact that you have got an integral in which you are having an even function. So, you extend the range of integration which is from K prime equal to 0 to infinity you extend it to negative values of K prime and let it go from minus infinity to plus infinity and then use contour integration to deal with the poles you see that there would be poles at K prime equal to plus K and K prime equal to minus K. So, then you can use contour integration to handle those poles. So, here you are so now because you have extend the range of integration is which was from 0 to infinity you now have the range of integration from minus infinity to plus infinity you have picked the factor of half. So, that mathematically you have exactly the same value for the net integration. So, this is the integral that you now have to determine which is from minus infinity to plus infinity and we once again express this sine function in how we had seen it earlier in terms of e to the i K prime r minus e to the i K prime r divided by 2 i we had put it in the sinusoidal form just to make the even nature of the integrand explicitly manifest. So, now we go back to this form and we now write it as two integrals one coming from this term and the other coming from this term. So, there are now two integrals to evaluate. So, the first one is coming from e to the i K prime r and the other is coming from this second term e to the minus i K prime r and then of course, you have this factor which is common to both. So, these are the two integrals which we now have to determine and these two integrals I have represented over here as i 1 and i 2. So, these are the two integrals that we have to determine what we have to keep in mind is that these are to be evaluated appropriately with reference to the boundary conditions which must be referred to without which the solution has got no meaning. So, here we are. So, we have got the two integrals to be determined i 1 and i 2 we have to be careful while evaluating these integrals because both of them have got poles and these poles are at K prime equal to plus K and K prime equal to minus K. So, we will make use of contour integration and use the Cauchy's residue theorem which you are all familiar with. So, this is just that if you have a Lorentz expansion of a function of a complex variable then this integral over here over a closed loop will be given by the residue which is just this factor 1 over z minus z 0 of in this Lorentz series expansion. So, that is the theorem of residues and in particular we will use a Jordan's lemma because we have got in the integration one factor which goes as e to the i a z we have already seen that. So, having seen that it would be obvious to you that appropriately we must use the Jordan's lemma which tells us that if the only singularities of the function are isolated poles then you can evaluate this integral according to this relation. So, it is basically just a straightforward extension of the Cauchy's residue theorem. So, we will make use of the Jordan's lemma and notice that the Jordan lemma requires this coefficient a to be a positive and we have to be careful about it because if it is not positive then the contour has to be closed in the lower half of the plane rather than the upper half of the plane. So, that is the only thing that we are going to have to keep track of. So, in this case we have two integrals. So, let us look at the first integral and in the first integral this k prime is to be integrated from minus infinity to plus infinity and the residue of f k prime at k prime equal to k is e to the i k r as you can see. So, this is the residue at k prime equal to minus k it will be minus i k r. So, depending on this we have to close the contour appropriately. So, let us have a look at what result we get from the Jordan's lemma and that will of course depend on how we choose the contour itself. So, we can choose the contour by hopping over these two poles and in this case you can immediately see that this integral will vanish. So, this will vanish if you choose this contour to hop over these two poles. You can of course choose other possible contours by including these two poles or including one at a time or both at a time and depending on what you do and how you choose this contour you will obviously get a different answer. So, here was our first choice of the contour when you hopped over both the poles and the solution for this integral is 0. Now, if you now include both the poles then the solution will be e to the i k r and e to the minus i k r and what are these functions? Now, this is only the solution of the space part of the Schrodinger wave function there is also a time dependence right. For the stationary states the time dependence is given by e to the minus i omega t and when you multiply this by e to the minus i omega t this e to the solution will become e to the i k r minus omega t for this term and e to the minus k r plus omega t. So, this will become a spherical outgoing wave and this will become a spherical ingoing wave right. So, you will get a solution in this case which is a superposition of an outgoing wave plus an ingoing wave. What kind of a solution are we looking for? In collisions we are looking for outgoing wave solutions. So, the first contour did not give us what we wanted it gave us 0. The second contour also does not give us a solution that we wanted it does give us an outgoing wave, but it does not give us only an outgoing wave it gives us an outgoing wave and also an ingoing wave. So, that is also not the one that we are looking for. How about this? This contour if you include the pole at minus k, but exclude the pole at plus k then you get e to the minus i k prime r which is again an ingoing wave. So, that is not what we want. So, now if you exclude the pole at minus k and include the one at plus k then you get an outgoing wave solution and that is the one that we are looking for as a solution to our scattering problem. So, it is just a matter of choosing the appropriate contour. So, what the Jordan's lemma tells you is how to evaluate the integral in this particular case, but then it is up to you to use the appropriate boundary condition and that is always the case whether you are solving a differential equation for scattering or you are solving an integral equation for scattering because in both you know these are only mathematical inverses of each other, but both require the boundary conditions to be referred to one way or another. So, this is now the solution for the first integral. In the second integral you do not have e to the i k prime r, but you have got e to the minus i k prime r. So, you have to be a little careful in evaluating the Jordan's lemma and what you will have to do in this case is to close the contour in the lower half of the complex plane. So, let us do that. So, now we close on the lower half of the complex plane to evaluate the integral i 2 and now I will go through this little quickly because we know how these contours are being evaluated. So, if you take this contour C 1 in which you include both the poles, but then close the contour on the lower half then you get a sum of outgoing wave plus an ingoing wave when you consider the time dependence. Mind you that unless you plug in the time dependence the terms outgoing and ingoing have got absolutely no significance because the time parameter has to be there. That is the one which is telling you whether a surface of constant phase is a spherical surface which is moving out of the centre or it is converging onto the centre. So, that has to be done because it is a travelling wave and you have to refer to it with reference to the time parameter. So, make sure that you insert the time dependence e to the minus i omega t at least in your mind. So, that you know what you are talking about and why this is an outgoing wave and why it is an ingoing wave. So, you get a combination of outgoing and ingoing wave on this contour C 1. Contour C 2 hops over both the poles and you get 0 because it hops over both the poles. So, the contour C 2 value of I 2 is 0 and then you have 2 other possibilities. One is to hop over this, but include this and the other is to include this, but hop over this. So, you have got 2 other contour C 3 and C 4 and you find that it is the contour C 3 which gives you the outgoing wave e to the I k r. So, that will be the right contour to be chosen and the greens function will be defined with reference to this particular contour. So, you have these 2 integrals that we wanted to determine and we now know how to evaluate those with reference to the outgoing wave boundary conditions. Both of these integrals I 1 and I 2 are to be evaluated with reference to that and since we are focusing attention on the outgoing wave boundary condition, I put a superscript plus to remind me that this is the greens function appropriate for the outgoing wave boundary condition. So, this is indicated by the superscript plus on the greens function. So, the contour appropriate for I 1 was the contour which I had called as C 4 and the contour appropriate for I 2 which gave us the outgoing solution is what we had called as C 3. So, that is just a matter of simple notation in our context and these are the solutions you have to sum up these 2 terms. So, let us go ahead and carry out this summation and the common terms you have got a factor of 2 pi here you have got 16 pi square here you have got 1 over r in both the terms. So, you put all the terms together simplify and essentially it adds up to minus e to the i k r by r divided by 4 pi. So, that is the greens function with outgoing wave boundary condition properly incorporated. So, now we have got the greens function with the outgoing wave boundary condition this goes into the Lippman-Schwinger equation. So, we have got the formal solution now. So, everything is what is how we wanted it except for the 2 things which we still have to address 1 as to what we are going to do about the fact that the integrand has got what is there on the left hand side and we do not know it. So, you are giving a solution in terms of the problem. So, that is cheating. So, we are going to have to address that we will and the other thing we have to remember is that the reason we are doing mathematics or physics or quantum mechanics is because we are you know we are doing an experiment. Means as physicists we want the mathematical theory to simulate the experiment appropriately. So, that it describes the experiment it gets you it helps you get the solutions and when you carry out your measurements because measurement is fundamental to physics and when you carry out measurements you keep a detector somewhere and then in this detector from the observations in this detector how do you interpret the results and get information about the target. So, that is the whole reason why you are doing quantum mechanics it is not just because you want to solve a differential equation or an integral equation, but because you want to do some physics and what is the physics over here that there is a detector which is where is the detector it is very far it is in the asymptotic region it is in the region for all practical purposes as r tends to infinity. So, we are really not interested in the solution as it is, but we are interested in the asymptotic solution. So, that is the second thing we have to put in one is to figure out how to address this function second to get the asymptotic form and just to remind us that we are using the outgoing wave boundary conditions I will always keep a superscript plus on the greens function and also on the wave function. You know this is very important because when you do photo ionization you have to make use of the in going wave boundary conditions, but in this case we make use of the outgoing wave boundary conditions. You can also carry out the integration by displacing the poles slightly and this is just a tiny mathematical detour which may be of interest to you and depending on what book you are referring to you will find solutions which look very similar, but what they do is they they displace these poles by an infinitesimal amount and then carry out the integration entirely along the real axis rather than hopping over the poles. So, mathematically it is a completely equivalent process and then you can take the limit if you have displaced it through this epsilon prime. So, you add this I epsilon prime or subtract it then you have to take the limit as epsilon prime goes to 0 and then carry out the integration. So, essentially you get the same result. So, it will not give you any fundamentally new thing. So, now we are interested in the evaluation of this integral and this integral is what I have represented by the letter J. This we already know this integral this overall integral is J the green's function with outgoing wave boundary condition is minus of e to the i k r by 4 pi r. So, the 4 pi is here and the r is here and we will now seek the limit r tending to infinity. Now, you remember what this capital R was in the picture of scattering which I showed you I had a field point which was r and a source point which was r prime capital R is the difference r minus r prime. So, it will tell you how things take place at the field point because there is a cause for it there is a reason for it and what is the reason that there is a scatterer and where is the scatterer it is at the source point which is spread out it is not centered focused at a given point, but it is in the entire scattering region. So, it is located at r prime. So, this is capital R which is the distance between r and r prime the difference between the field point and the source point which is just the square root of the inner product of r minus r prime with itself and you can expand this factor to the power one half you can carry out an expansion and then develop an approximation as r tends to infinity because that is a region of interest. So, this expansion so, this is the picture that we have. So, this is r minus r prime this is the same picture that I showed you earlier and I will not discuss these steps in between you know how to do that. You would have done that a number of times by when you dealt with multiple expansions for example or so on. So, there are large number of problems in mathematical physics, electrodynamics and quantum mechanics where you make exactly the same kind of power series expansion and develop an approximation. So, I will not discuss this in detail I do have most of the intermediate steps on these slides. So, you can always refer to the PDF and then take it from there if you need to but you would not have to I am pretty sure about it. So, these blue arrows on the side they remind me that I can skip this part of the discussion and they will also remind you that I am going to skip the next slide for exactly the same reason. So, it takes a little while to work it out especially because it is always easy to make careless mistakes somewhere and get a wrong power somewhere and then you mess it up and spend another 5 minutes figuring out where it was or 2 minutes trying to do it all over again and we save all that time by skipping over this. So, all this is homework for you you will work it out it is already here but then these I will follow these blue arrows skip it and here again because there are all these unless you combine the terms and take factor out the common terms and so on basically that is high school mathematics. So, I do not want to spend any time working it out but I am sure that all of you can see at a glance what the logic is and during the class I want you to focus on the logic and work out these intermediate steps on your own instead of watching a movie for example. So, here we now have an expansion and then look at how these terms pop up. So, you have got terms in 1 over r and then 1 over r cube and then 1 over r square and so on. So, now you can see how an approximation can be developed and you have much to do to rearrange these terms I mean there are 3 or 4 or 5 slides which I am skipping as you can see but it is very straightforward rearrangement of terms. So, I will leave it for you to work it out I think it is more than 3 or 4 but that is alright I can always say it was only 2 slides are you all comfortable good. So, here we are. So, we have the end result what is the end result that you have terms in 1 over r and 1 over r square and so on. And if you now retain only the leading term only the leading term because as r tends to infinity you can always ignore 1 over r cube compared to 1 over r square and you can always ignore 1 over r square with compare to 1 over r. So, keep the leading terms that is what the asymptotic region is about and this is where the field point is where the detector is that is where you are carrying out the measurements that is the place where you want to know what the solutions will turn out to be like. And in this limit as r tends to infinity you have an i e to the i k r by r and yes we have met this term in our first course in quantum mechanics and in our early discussions in scattering theory. So, we are beginning to see a form that we know how to exploit and then we have some remaining terms. So, you have got this e to the i k r by r and e to the minus i k f dot r prime. So, this will go where e to the i k r by r goes. So, this is the e to the i k r by r. So, the minus sign is here. So, that comes here the 1 over 4 pi is there that comes over here 1 over 4 pi. Then you have this e to the i k r by r which is good and this e to the i k r by r we know we can factor it outside the integration because the integration is over the variable r prime which is changing for different source points in the scattering region. But they are all causing an effect at only one field point which is r and that does not change as you integrate over r prime. The integration is over the source points for a given field point. So, e to the i k r by r can also be factored out. So, we will do that very quickly now. So, now we have got this e to the i k r by r outside and then we have this 4 pi to be properly accommodated. So, that has been done rather significantly. So, that you can get a plane wave over here with a useful normalization because, but it does not matter what the normalization is in the end the normalization can be factored out. So, now you have this e to the i k i dot r this is the incident plane wave and we know that a scattering solution we are in our very first what I called as a phenomenological solution. The phenomenological solution of the scattering problem is what you expect that you it will be a superposition of the incident plane wave plus a scattered wave which is a spherically outgoing wave and the conservation of flux tells us that since the area of the sphere increases as r square as 4 pi r square the amplitude will diminish as 1 over r. So, you factor out this 1 over r. So, that is over here then you have got the spherical wave which is outgoing why is it outgoing because there is an e to the minus i omega t time dependence. So, you have factored that out and what is the rest of it that is an angle dependent scattering amplitude. So, that is the scattering amplitude f that we have used in the phenomenological solution of the scattering problem. So, now we have got the solution to the scattering problem in a form that we have been using all along and we have obtained it using the Lippman-Schwinger equation we know that the form is correct we know that it gives us what we want except that we do not really know how to evaluate this integral because the integrant has got the psi plus which is there on the left hand side. So, we are going to have to worry about that, but at least we have addressed one of the two issues which I mentioned that we have to seek the asymptotic form because that is where the detector is and the other factor is how to address this issue. So, we will do that. So, this is the scattering amplitude and this is the solution I have used Jochen's quantum theory of collisions for this discussion. So, it is a very nice book that you will find very useful and this term is nothing but the plane wave. So, this scattering amplitude I write as an integral of which the first factor is the plane wave corresponding to the wave vector kf because the wave vector here is kf mind you over here it is ki that is the incident direction the kf exit direction is different. So, you must remember that. So, this is the kf here this is the ki with the incident wave vector, but this is the complete solution to the target scattering with the superscript plus which we do not know as yet, but we know that formally it is correct it is acceptable. So, this I have written in the direct notation and this is in terms of the reduced potential we can write it in terms of the physical potential v which is why the 2 pi and the h cross and the m pop up one more time which we had eliminated. So, now in terms of the real physical potential it is this and this is where I will conclude today's class and we will take it from here tomorrow to determine how we are going to deal with this problem because we have got a solution alright, but it only gives us the solution in terms of the problem that is not very good. So, that we will discuss tomorrow or in the next class and this matrix element is often referred to as the transition matrix element. So, this scattering amplitude is proportional to the transition matrix element which is t and the differential cross section is just the modular square of the scattering amplitude as we have discussed earlier. So, you get the modular square of this. So, I will be happy to take a few questions otherwise will discuss what to do about this cheating if you might want to call it which is to give a solution in terms of the problem, but then there are ways of handling it which is what we will discuss in the next class.