[Music] you so having done most of the approximate methods in treating quantum mechanical problems we embarked on the last one that is scattering theory now scattering theory is known in some form in the context of classical mechanics where you have reduced two body problem into a one body problem in the center-of-mass frame and have defined an effective mass consisting of the the two masses and then have considered scattering in the CM frame or the center of mass frame and finally the cross section was computed in the center-of-mass frame and then it was converted back into the lab frame to be able to establish a connection with experiments here of course the approach will be quantum mechanical but very quickly let us see or do a recapitulation of the classical scattering theory that you have been familiar with in in the center-of-mass frame so it's about two particles just drawing one slightly bigger than the other so M 1 and M 2 they have velocities initial velocities u 1 and u 2 and they scatter and after scattering they go to some so there are you know I mean so if it's a three dimensional scattering problem then of course we have to consider that there are three dimensional scattering problems are slightly difficult but we still need a lot of input about the initial velocities in order to solve the problem because if in one dimension of course you have momentum conservation equations and the energy conservation equation in 1d you have one momentum conservation equation and one energy conservation equation and we are talking about elastic scattering at the moment however in three dimensions there are three momentum conservation equations and one energy conservation equation the number of unknowns are however six because the three initial components of or rather the initial velocities and of each of the particles in three dimensions are the unknowns given the initial velocities for the start of the problem that is before collisions say the velocities are given which are known as the initial velocities and after collision the velocities are known as the final velocities so our aim is to calculate the final velocities and their angle with respect to the incident directions and so on and this problem is it is somewhat complicated in the sense that a number of equations need to be solved and they should be consistent with each other now in quantum mechanics before we go to quantum mechanics of course the solution of the problem in the CM frame is somewhat easier and what happens in cm frame is that the velocities actually so in in a three dimensional collision problem so there's a particle that comes from here and there's a particle that comes from here and they go and they scatter and so this is say m1 and this is m2 and then the scatter in different directions it could be m1 here and m2 here or the vice-versa I mean this could be one or two and this could be two or one and you need to calculate all these final state velocities whereas in CM frame what happens is that one usually has a scattering like this so there is this is say v1 or rather u1 see let's and this becomes you to see and this is the so this is the v1c and this is a V to see and this angle is Theta so these are the in the CM frame the center-of-mass frame these are the initial velocities of the particles 1 & 2 correspond to the two particles like mass as M 1 and M 2 and C corresponds to the CM frame and in the CM frame the particles are basically the total momentum is 0 in the center-of-mass frame and the particles actually scatter like this with the initial with the final velocities V 1 C and V 2 C and it makes an angle with respect to the incident direction by an angle theta and the problem becomes reasonably simple and it can be solved in the CM frame so of course we are going to do quantum mechanical scattering theory so let us physically discuss that what happens in in a typical collision experiment so what happens is that there is an incident beam that comes and this incident beam so this actually corresponds to the particle one that is mass 1 if you want to say and then it goes and strikes a detector or rather a target okay so this target is actually the particle 2 and so this is called as a target and then it gets scattered at various angles and so this is the incident direction and it can get scattered at this theta 1 and theta 2 and so on and these are the positions of the detector okay so and let's let's just for convenience call this as I said direction as we will use this direction right here all right so this is a typical scattering experiment such as an alpha particle is being bombarded at the nucleus or or say an atom is being bombarded with some particle and so the atom becomes the target and the incident particle is the particle with which it is bombarded and then there are detectors which could be kept at discrete angles theta 1 and theta 2 and so on or there could be a detector which is placed at all solid angles and then it is able to detect the scattered particles within angle between an angle theta and theta plus D theta so the scattering cross-section is defined in this case as we'll call it a Sigma Theta Phi because in general they'll be our dependence on the theta and Phi angle as well so Sigma Theta Phi in a solid angle D Omega so which is given equal to the number of particles scattered in a solid angle / the incident intensity which usually is denoted by i0 so that's the definition of this the scattering cross section Sigma Theta Phi or it's called as a differential scattering cross section and a total cross section is obtained from this differential cross section by integrating over all theta and Phi so that's equal to Sigma and so this is equal to a sigma theta Phi D Omega and this can be written as D theta sine theta from 0 to PI and D Phi 0 to 2 pi and a Sigma of Theta Phi now we will see that the Sigma or that scattering cross sections may not depend upon Phi for spherically symmetric potential so most of these potentials are spherically symmetric so in most of the cases Sigma is only a function of theta and so that can be written as so for a spherically symmetric potential Sigma equal to 2 pi 0 to PI sine theta d-theta and a sigma of theta ok so that's the total scattering cross-section and we need to calculate this ok so let us just start with the quantum theory of scattering so strictly speaking we need a time-dependent description of the problem because the collision event happens as a function of time so before collision there is a time slot so we can talk about before times before collision and after collision so the wave function or the wave packet which is a collection of waves that represents the incident particle as T going to minus infinity that behaves that it has passed through a target and the wave packet corresponding to the final products as T goes to plus infinity so we have T going to minus infinity when it's only the incident particle which is traveling as a free wave traveling as a free particle maybe and say at T equal to zero scattering occurs and at t equal to plus infinity that is at very large times the incident or the scattered particle rather on the incident particle or maybe a scattered particle which may include the incident particle one of the incident particles as well the scattered particle detected by the detector detective okay so there are a few conditions that the scattering would involve so these conditions of or rather these properties that we are going to be particularly interested in is that we are going to talk primarily about elastic scattering just to remind you the elastic scattering the where the energy is conserved it's called as an elastic scattering momentum is always conserved both in elastic and inelastic scattering in elastic scattering energy is not conserved and it gets dissipated in the form of maybe heat or light or something okay and second is that we shall not bother or rather take into account about the internal structure of the particle the scattering potential is has translational invariance so V r1 r2 that's a two-body potential actually depends only on the relative coordinates and that's about it so these are some of the properties or rather the constraints of scattering and then of course we can also talk about the effective mass which is the which is given by mu which is m1 m2 divided by m1 plus m2 now for for P going to minus infinity we can assume that the potential is equal to zero so the particle that incidents on the scatter is is free and it's sufficiently far away from the scatterer and the the state of the incident particle is represented by a plane wave so by a plane wave or a wave packet a plane wave packet and so we are thinking of that the sy incident which is at T going to minus infinity is like exponential I K dot R and since the incident direction was taken as a Z direction in the last schematic diagram that we have shown this can be written as exponential I K said where K is given by 2 mu e by H cross square mu being the effective mass of the two particles so after that of course for when the scattering occurs and I'm just thinking that that is roughly at the time in the middle of minus infinity to plus infinity so at the time when the scattering occurs then the particle reaches a region where there is a VR exists and this exponential I K Zed gets significantly modified okay and how it gets modified it's very difficult to figure out because then one has to exactly solve the Schrodinger equation corresponding to a given potential now the problem with solving that is that the VR could be essentially very complicated in which case and a closed solution of the Schrodinger equation may not exist but however that does not impede us to write down that for T going to plus infinity that is the particle has gotten scattered from the target and have reached the detector it consists of two things one is that the plane wave which of course it could still remain is a plane wave and not get scattered at all by the by the detector so and number two a scattered wave and of course the first one we have discussed the important thing is the second one and we need to be convinced of the form of the scattered wave because the rest of the analysis is going to crucially depend on what's the form of the scattered wave that reaches a direct detector as T goes to plus infinity so in a given direction Theta Phi the radial form is something like exponential I K R by R this called as a spherical wave so it's basically it's an outgoing wave with the same energy as the incoming wave you mind then think that what is this one over R in the denominator as it's shown here this R in the denominator where does it come from now it's important to see that in in special I mean three dimensions two other three special dimensions this quantity which is the Schrodinger equation plus K square exponential I K R is not equal to zero for del square to be of the form R square del Del R of 1 over R squared del Del R del Del R okay however this is equal to 0 which you should be able to see very trivially if you simply put here so we'll write del square F and this is f so if you put exponential I K R by R then this is equal to 0 okay so which means that the Schrodinger equation is satisfied and this K square is of course the energy and we are talking about the free particle thing so this is the reason for having a a scattered wave packet to be of the spherical form which is exponential I K R PI now just to summarize what happened so far is that a particle of an incident particle which comes as exponential I K Z and so this is at far away from the detector I mean far away from the target so this happens at T going to minus infinity the scattering happens at say T equal to 0 and then the detector detects this this wave so if this detector detects both a free wave plus a spherical wave of the form this so it's basically a sinusoidal function or an oscillatory function with an envelope which goes as 1 over R so that happens at T goes to plus infinity ok so that is being detected by the detector so this is the target and this is the detector here let's write it here and this is the incident particle so we are not bringing in explicit time dependence of these wave functions or rather not we are not trying to evolve them with time but in our mind we are clear that far away from the target the incident particle is free and propagates like a free wave it is like exponential I K dot R because now the direction of propagation is taken as the z axis so that free wave is like exponential I K said then it strikes a target and we are unable to solve Schrodinger equation or rather find out any information because of the complexity of VR what happens at T equal to zero and we disregard that fact but however we know that when it's detected by the detector at t goes to plus infinity then the detector has it has two components or other two parts where one of them is like exponential I K Zed and the other is exponential I K R by R so these are the two now what happens is that if we simply take this two terms then we are not having any information about what scattering takes place there in the target or rather this as if the target is not there and we simply have a free particle propagating and then free particle plus a spherical wave being detected at the detector so there has to be some information of the scattering that had taken place in the target and at least that's what we are interested in so in order to bring that we take the wave function as R goes to infinity okay so R goes to infinity it's an exponential I K said plus a FK Theta Phi exponential I K R by r FK Theta Phi is called as the scattering amplitude and for spherically symmetric potential that is V of are simply depending upon V R theta in which case we have or rather it is V of it could be simply V of R we can simply write this psy k of R as R goes to infinity its exponential I K set which means this in there is an equality but as R goes to infinity so this is I K said plus f K theta and exponential I K R by R so that's the there's a wave that is detected by the detector and the information about the scattering that had taken place at the scatterer or the target is embedded in this FK of theta which is called as a scattering amplitude and now our main objective is to calculate f K of theta okay so in order to calculate so let's just write the objective so far now what is FK theta or how is it related to the the scattering cross-section basically the scattering cross-section is obtained from FK theta by taking a mod square which in general is a complex quantity so we can write that as well obtain Sigma of theta which is the differential scattering cross section is FK theta mod square that's that amplitude square and so we'll have to calculate that but before that we need to in order to calculate this we need to understand a few things and a study of greens function is very important in this regard so let's see how greens function comes in this we are going to give you a simplified description that is we have to solve an equation so this is a greens function we are just going to introduce it not speak too much about it just say that how is it relevant for studying scattering theory so we have to solve an equation which is like this del square plus V of R which is we can simply take that V of R to be equal to V of R u its even independent of theta and this sigh of R this is equal to e sy of R and so this is equal to del square plus K square minus U of R and this sy of R equal to 0 so where E is equal to H cross square K square over to MU and V of R is nothing but H cross square over 2 mu and U of R ok or U of R is 2 mu V R by H cross square in which case we can write down this equation as the same equation we are writing it in terms of this you so its laplacian plus k square phi of r that's equal to u of r sy of art and of course U of R is nothing but so this is this and this is nothing but H cross square by 2 mu u of R now let us define a greens function by this equation so as if greens function is a solution of this equation you can look it up in any mathematical physics book especially afgan gives a very good introduction to this the greens function so greens function is a solution of this equation which is laplacian plus k-square and acting on g RR prime which is the green's function and that yields a delta function so g r r prime is a green's function and delta R minus R prime is the Dirac Delta function okay so G R R prime is green's function of the operator this del square plus K square alright now we propose or beacon answers sigh of our equal to size 0 of our plus D cubed R prime G R minus R prime u R prime sigh of R prime is the solution of this equation which we call it as equation 1 let's call this as equation 1 and let's call this one as equation 2 this one is equation 2 and this one as equation 3 also make this additional requirement that del square plus K square acting on sy 0 of our should be equal to 0 and one can actually see that the this 3 that is equation number 3 which is here which is here this is a solution of 1 that is the the Schrodinger equation that we have written now alright so after that we'll write down the del square plus K square Phi of R that is we need to check that this is equal to Del square plus K square so we operate del square plus K square on both sides by this size 0 R now we're going to show that this is 3 is indeed a solution of 1 and plus del square plus K square G of R so this can be you know I mean instead of either you can write it as R minus R prime or RR prime either is okay so this is as we wrote our our crime so this is and then there is a U of R Prime and then sigh of R Prime and D cube R Prime so that's a volume integral over R prime now this is equal to zero by the condition above that is equation 4 and this is the definition of greens function which gives R minus R Prime so that tells you that this Plus this sigh of our it's equal to a delta of R minus R prime u R Prime sigh R prime D cube R Prime now that of course is nothing but u r sy R which is the which is equation 1 so that means 3 is indeed a solution for the Schrodinger equation ok so that way we are convinced that this is that this is the solution for that now let us also try to understand what the greens function for this particular case is so since the equation defining greens function is given by Del square plus K square G of R R prime it's equal to Delta R minus R prime we a priori write down this is a purely from experience that one can write down the G of R is something like exponential I K R by R okay and remember this relation that you might have learnt in your electrodynamics course is that laplacian of 1 over r is 0 unless it encloses the origin in which case it blows up so it's minus 4 pi Delta R ok so this can easily be seen that a del square plus a K square exponential plus minus I K R by R so that's equal to a minus 4 PI Delta R this is to show because then we'll be able to show that so del square of exponential plus minus we the plus and minus both our spherical waves it's just that we decide to take a plus because it's a forward moving wave exponential minus IKr by r would be a backward moving wave this since there is nothing to reflect back after the target after the incident particle has crossed the target we drop this exponential minus I care but mathematically this would be there so this has to be computed or to calculate this and this is nothing but equal to this is like exponential so this is like this so this and F G there are two scalar functions so they're two scalar functions exponential I care and 1 over R so this is equal to our divergence of gradient F into G Plus F gradient of G and so on so this is equal to aji laplacian of F plus F laplacian of g plus 2 gradient F into gradient G so we'll apply this to this two functions exponential I K R and R so a del square exponential plus minus I K R by R equal to 1 over R del square exponential I K R plus exponential I K R R del square 1 over R and a plus 2 1 over R and this one exponential I K R or plus minus I K R and so on okay so this is these are the terms so if you simplify then it becomes equal to so del square plus K square exponential plus minus I K R by are equal to minus 4 pi Delta of R that tells us that we can write down a psyche now we'll take the plus sign in the in the spherical wave so we'll write it just like a psyche of R which is a exponential I K said we could have written it a vector R but now it depends on scalar R plus Rd cube R Prime and G Plus R minus R prime or RR prime doesn't matter it's this and then it's a sigh R prime and so on so that's the that's the answers see it's a little tricky here because well I need to put a plus as well here so this corresponds to the plus sign in this spherical wave see we are trying to calculate side k plus R which is the wave that reaches the detector as R goes to infinity which means at T equal to infinity that is there in the left-hand side as well as that it is there in the right-hand side as well so one is automatically heading towards an iterative solution a solution that is self-consistent and it has to be iterated upon now in order to get that we are left with no choice but to make an approximation and this is called as a born approximation so before we do born approximation let us simplify this picture even more and try to understand the asymptotic behavior of the scattering function of the scattered wave function so so there is a point oh and there's a point say P and there's a point M where the detector is so P is a point within the you know where the potential exists or within the influence of the potential so let's call this as R Prime and this is our okay and M is where the detector is so let's just make this things clear M is the detector and P is a point in the zone of influence of the potential okay and so of course this is equal to R minus R Prime and just interested in writing the modulus of that and let's talk about this direction being u cap and this angle be theta so M is as you understand and oh is of course the origin M is very far away because that that's the position of the detector such that this angle M or P is very small and because of which we could say that this MP that is the magnitude of this R minus R prime which is R minus R prime that is equal to projection of almost equal to projection of MP on mm okay so R minus R prime that's equal to R minus you are so that's R and minus this thing that that is that projection which is u dot R so the G Plus corresponding to the plus sign of the spherical wave is R minus R prime that is equal to minus 1 over 4 pi exponential I K R minus R Prime and divided by R minus R prime this so for R going to infinity G Plus R minus R Prime it's equal to 1 by 4 pi with a minus sign exponential I K R by R and now I am writing this R minus R prime as R - you kept dot R so that is R - you cap dot R u dot R so the answers for the for the wave that reaches the detector so this is for R going to infinity is exponential I K said which was already there and minus 1 over 4 pi exponential I K R by R now there is an integral over this and this I K you cap dot R prime u R Prime and psy k plus R Prime and so on so we have been able to plug in the greens function that we have obtained into this expression if you compare this with this expression with this expression where there is a forward moving free wave and then there is a exponential I K R by R and then there is a F K theta if you look at that and compare with the one that we have just obtained so with this then this FK Theta Phi simply becomes equal to this entire integral with this minus 1 by 4 pi factor and this is Mike minus 1 by 4 PI and D cube R Prime exponential minus I K U dot R prime U of R Prime Saiki plus R Prime and so on just a little more simplification in the in the given present context if you look at the incident wave vector is in the Z direction this is what was assumed right at the beginning and there is a scattered wave vector which is in this direction where the direction in which the detector is placed and this is the KT which is the transferred wave vector and this is of course that angle theta then ki s is actually K you cap because we have shown that the u cap is in the direction of the detector so it's K u cap and of course your K I the magnitude of that is same as K s and which is equal to K because of the elastic scattering so KT here is the transferred wave vector and that is equal to K s minus K I equal to K okay so that is the the K that we are talking about here in this F k theta that's the transferred wave vector and now we are almost all set and going to talk about that approximation that we need to undertake in order to solve this problem that is the problem the problem of finding the wave function or rather finding the scattering amplitude and hence the scattering cross section so the born approximation [Music] so a sy k plus R equal to exponential I K I dot R plus D cubed R prime G R minus R prime M once again writing this R prime psy K dagger not not dagger DS plus I'm sorry so this plus and this so so this was the original Unser's that we had so a simple change in notation change our to our prime and our prime to our double Prime in which case sy k + R Prime it's equal to AI K I dot R prime plus D cube R double prime G R prime minus R double prime u r double prime sy k + R double Prime and so on now you see I have got the psyche dag as I cap plus R prime which I can put it in the first so this one can be actually replaced here or rather can be plugged in here and we get an equation which is a sy k plus R equal to exponential now I can I can if you have seen that I have started writing the ki dot R instead of KZ but that means the same thing so it's i ki dot R so that is the the part of the scatter the incident wave itself and this is Prime and this is a G plus R minus R prime u R prime exponential I K I dot R I prime plus a DQ r prime D Q R double Prime and G Plus R minus R dub R prime u R prime G Plus R prime minus R double prime you are double Prime and psy K R double Prime and so on so you see that if you compared between the first line on this slide that is the last term that is here that is here and the the two terms are two rather three terms that we have written here this one in the first approximation has been replaced as the plane wave and there is a second term and each term comes with the with an additional factor of you see this is a function of or rather depends only on one power of U this depends on u and u prime or uh rather two powers of U U and u are prime nu R double Prime and a sec at third approximation or third iteration will get in another U and so on so let us consider that you to be small and drop all terms that are quadratic in U and onwards so in which case we only keep the first two terms in this expression and that is called as born approximation so just one power of that and so using born approximation implies that we stop at the first order and in which case we can extract out the scattering amplitude from this second term and then FK theta and in the born approximation which so we write a B there in the superscript and this and an exponential minus K U dot R prime u R prime exponential I K I dot R Prime and so on so that now a little bit of adjustments in notation it's a d cube R Prime and because this is K s this term is K s so then we can write it as minus I K s minus K I dot R prime u R Prime and so on so so this is equal to so this is equal to minus 1 over 4 PI D cube R prime exponential minus K T dot R prime U of R Prime so KT is the transferred wave vector we have already told so Sigma the scattering cross-section is as we have told it is the Mod Squad so this is that be theta mod square so so what comes is that it is simply the scattering amplitude is simply the Fourier transform of the potential so once we know the potential we can do the Fourier transform that should give us immediately the the scattering amplitude and from the scattering amplitude one can easily calculate the the scattering cross section of the differential scattering cross section and given that our U of R equal to two mu by H cross square V of our Sigma B that's a scattering cross section so let me just write that just to remind you once more differential scattering cross section so this is equal to Sigma B theta and so this is equal to 2 mu square by 4 pi H cross square whole square and there is a I mean there's a mod of D cubed R prime exponential minus KT R prime V R prime mod square and so on so this is equal to so it's basically the scattering or so the scattering cross section is a Fourier transform of the potential so the potential that exists act the target that is influencing the the incident particle so the physical meaning of the greens function is that that it represents the amplitude at a point R of a wave that is radiated by a point source situated at R Prime so pictorially if you want to understand what born approximation mean is that so there is an incident so here and so there's an incident particle it gets into the target and there is a it interacts with the point in the target and here is the detector so it goes and reaches the detector carrying the information about the scattering and there is also another free wave that reaches the detector so this is that detector so this is born approximation and what we are neglecting is that there are multiple scattering events that could take place in the so there is one level more than the born approximation is that there is this and then this reaches the detector and there is another one reaches the detector okay so this is so so there are two scattering events and there are multiple scattering events which are beyond born approximation so this is beyond born approximation and this is quadratic in the potential okay so born approximation finally gives a very important and a very simple result that the scattering cross-section is simply equal to the mod square of the Fourier transform of the potential so if the potential is supplied suppose the potential is Coulomb like and I mean you have to consider a screen coulomb for the reason that you would understand when you try to do that problem then it is it is simply a term which is you have to calculate the Fourier transform of the potential and and that's that's about it multiply this by the to MU square mu whole square and 4 PI H cross square and all that and that is that is most of the things that are needed for scattering Theory the quantum scattering theory in the born approximation of course one can go beyond the born approximation however if the potential is weak then those terms which are square in potential and cubic in potential etcetera etcetera they will start contributing lesser and lesser okay so that pretty much completes the approximate methods in quantum mechanics just a very brief recap we have done the perturbation Theory the time-independent perturbation theory where we have talked about the degenerate and non-degenerate perturbation theory and then we have talked about the variational principle where a variational state is chosen which when minimized the energy being minimized with respect to that tunable variational parameter one gets an upper bound to the ground state energy and then we have also done wkb approximation which is for a slowly varying potential we have learned how to compute the the connection formula of of the you know the solution connecting the two sides of course these solutions are only valid asymptotically there is far away from the turning points and from the connection formula one could actually calculate what's called as the Bohr Sommerfeld quantization condition and then we have also done the time-dependent perturbation theory where we have calculated transition between different our different states the transition probabilities and in case the final state falls into a continuum of states one needs to invoke the density of states for the final states and then use the Fermi's golden rule in order to calculate the transition probability and here we have learned how to deal with scattering problem in 3d where the the an incident particle comes from minus infinity where it's far away from the range of the potential behaves like just like a free wave and then it reaches the detector also again at far away locations from the position of the the scatter or the target and there one gets actually a spherical wave but in addition to that one also gets a factor which is called as a scattering amplitude associated with the spherical wave such that the spherical wave has the knowledge of the scattering that it has the incident particle has gone through in the scatter and this there is a way that we have shown to calculate this scattering amplitude and hence the scattering cross-section within an approximation called as born approximation in this context we have also included the definition of greens function and for this particular problem shown what the greens function can be it's actually nothing but the spherical wave that we have been talking about so using that and an iterative solution I were terminating the iteration and the first order one can get the expression for the scattering cross-section which is nothing but the this Fourier transform of the or rather it's a mod square of the Fourier transform the potential [Music] you