 Greetings, we will have 4 classes in unit 6 and we will begin to discuss resonances in collision phenomena. Resonances are also identified as quasi stationary states and the principle reference for this unit will be Jochen's book chapter 4. Now, today's class we will go little slow and essentially what I will do is to remind you about some of the things we studied in an earlier unit in unit 1 of this course. Because we already talked about resonance states, we talked about the virtual bound states when we talked about collisions at low energies in particular the S wave scattering which we discussed in the unit 1 of this course. So, I will spend a few minutes revisiting that and so that will help us warm up to this topic. So, in particular have a look at lecture numbers 10 and 11 of unit 1 of this course and what is in the background in the slide is the Levenson theorem which we have discussed at considerable length in the previous unit. So, we will take our awareness of the Levenson theorem for granted for this unit. So, what we will do is to develop some further considerations which go beyond the immediate consequences of the Levenson's theorem and in particular we will discuss how the scattering phase shifts depends on energy at low energies. We know the low energy limit which is k tending to 0, but when it is not quite 0, but it is still low energy what is its energy dependence how does it change. So, we are now getting into some details beyond the Levenson's theorem and as I mentioned earlier this is the primary reference for this portion which is a very fine book I really love this. So, the phase shifts are measured with reference to a certain 0 of the angle and this 0 of the angle is the high energy limit of the phase shift and as k tends to 0 the phase shift would go to 0 for the lth partial wave. So, that is the reference level. What happens when k tends to 0? Now, that answer is provided by the Levenson's theorem. Levenson's theorem tells us what happens in the low energy limit which is k tending to 0. So, we know what is at the extreme k tending to 0 and then as k goes to infinity and then we will examine what happens in between and that is where law of interesting dynamics is. So, from the Levenson's theorem we know that for the lth partial wave if l is equal to 0 when we deal with S wave scattering and that is the most important contributing to contributor to the collision phenomena at low energies as we have discussed earlier. This phase shift goes to n times pi where this is the number of bound states. However, if there is a resonant bound state at that energy then it does not go to n 0 pi instead it goes to n 0 plus half pi where n 0 is the number of bound states which exist at that particular for that particular potential. This is the 0 energy resonance and we have discussed this in our previous unit. We have shown that the phase shift when you do have a resonant structure when the strength parameters the we discussed this in the context of a square wave potential and the strength of the attractive potential depends on the depth of the potential and the range of the potential the two together. In fact, it depends quadratically on the range and linearly on the depth. So, u 0 a square is the strength parameter or sometimes it is taken as a root of u 0 times a. So, this is the strength parameter and when this strength parameter becomes pi by 2 then you get the resonant possibility and the phase shift then goes to pi by 2. So, this is a virtual bound state. So, at any depth less than this you would not have a bound state, but then if you have the depth even a little more than this or at the range a little more than this then you will have a bound state. So, this is the 0 energy resonance and at this resonance the cross section blows up. It blows up as 1 over k square. So, you can see that as k tends to 0 it just shoots off. So, what happens if l is equal to 1 or greater than 1 then the phase shift is always an integral multiple of pi the integer being the number of bound states which exist for that particular potential depending on its depth and range. So, there is a peculiar behavior for the S waves as you have noticed which we have discussed in the previous unit and the 0 energy behavior is therefore, different for l equal to 0 and for l greater than or equal to 1. So, I will refer you to the lecture number 10 and 11 of the unit 1 of this course. We have seen in that unit that the tangent of the phase shift has for its lower energy limit with 2 l plus 1 power dependence on k. So, as k tends to 0 so that is just like a threshold effect and this goes into various you know these are results these results are very general and very powerful and they go into the framework of what are commonly known as the threshold laws Wigner made significant contribution to this particular area. So, these are the this is the threshold behavior and for l equal to 0 the tangent blows up when you have a 0 energy resonance if there is a bound state exactly at that limit. Now, we have discussed this that the cross section goes as 1 over k square and this phase shift then will be either pi by 2 for the resonant case or it could be 3 pi by 2 or 5 pi by 2 it is not an integral multiple of pi mind you in this case, but it goes in steps of pi by 2 3 pi by 2 5 pi by 2 and so on and how large it is depending on how many bound states the potential holds. So, just the fact that there is a there is an attractive potential does not mean that you will have a bound state because the potential will have to be sufficiently strong so that you can have a bound state. The potential being attractive is a necessary condition it is not a sufficient condition for you to get a bound state solution. So, this theorem which is stated so simply that the phase shift is either n pi or n plus half pi depending on l being equal to 1 or greater than 1 or equal to 0. We obtained this theorem explicitly for the square well potential however, it holds also for many general potentials not just for the square well potential. So, which is why it is such a powerful theorem. Now, what we will examine is we know the limiting behavior as k tends to 0 what happens as k changes. So, this is k equal to 0, but if the k increases gradually and you are still in the low energy regime you are still in the low energy regime of the collision phenomena. If you plot the phase shift as a function of k how will this plot look like what is the low energy energy dependence of the phase shift is what we will discuss today. We will illustrate this by taking the example of the square well potential. So, you have got the depth of minus lambda 0 square and then you have a range of r equal to a. So, this is the square well spherical potential and we will re examine the bound state solutions. The bound state energy is intrinsically negative. So, h cross square k 0 square by 2 m is equal to minus of e 0 because e 0 is intrinsically negative and we will consider first the l equal to 0 solutions the S wave scattering and in this case the exterior solution of course, will die off exponentially. So, it will have an exponentially decaying solution outside and in the interior you will have the Bessel functions for the solutions and the Bessel functions the argument of the Bessel function will be kappa r where kappa will be determined by this k 0 which is the bound state energy and the potential depth. So, these are the two things which will determine the value of kappa. So, we have put a subscript b on kappa to remind us that this is the bound state solution. So, these are the two solutions and you can see that for us in for the geometry that we are dealing with lambda 0 square will be greater than k 0 square. So, that you will have a bound state there. Now, we have discussed the solutions at length in the previous unit. So, I will not revisit that in any great detail, but you will remember that the bound state solutions were obtained at the intersections of these two family of curves. So, there is one solution which is the relationship between xi and beta and then another set of solutions which is an equation to the circle and wherever these two curves intersect we get a solution and this being transcendental relations typically graphical or numerical solutions are obtained. And I will like to make an interesting observation over here by quoting some writers from books which you would have used David Bohm's book for example, he says in section 11 that this is a transcendental equation and must be in general solved numerically or graphically Now, this is David Bohm. Now, we have Branstad and Joshain and section 4.6 they say the energy levels are found by solving the transcendental equations either numerically or graphically which is pretty much the same thing. Then you have Griffith's introduction to quantum mechanics and what does he say that this is a transcendental equation it can be solved numerically using a computer or graphically as if there is no other way of doing that. And then I would like you to read this paper by Ken Roberts and Ram Valuri my good friend at the university of western Ontario and I have given the link over here I strongly recommend that you go through this work and they have shown using Lambert W functions how analytical solutions can be obtained for this problem. And this Lambert W function is a very powerful technique you can get analytical solutions in many situations they have a very wide range of applications not just in collision problems, but in a very wide variety of problems in fact even for the Coulomb scattering that we discussed yesterday you can have some applications in 4 organization cross sections you can use them. So, they make use of the poly logarithm functions and this is a very powerful tool which Ram has really mastered and I will strongly recommend that you are going to yourself with this technique. So, what was the result of the Levenson's theorem that as lambda 0 a is the strength parameter right this is lambda 0 is the square root of the depth a is the range and as this increases from 0 to pi by 2 then till it gets to pi by 2 you cannot have any bound state at all. So, the number of bound state is 0 and the phase shift will be 0 this is for the s waves. However, when this strength parameter exceeds pi by 2 then you have one bound state and then the phase shift the low energy phase shift as k tends to 0 will be equal to pi and what happens at pi by 2 that is something which I have indicated by this asterisk that is when you have a virtual bound state. So, that is a resonant bound state. So, it is in some sense neither here nor there or in another sense it is here and also there and it is like the Schrodinger's cat which is either dead or alive and the reason this happens is because of quantum mechanics because there is a super position between the bound state solution and the continuum solution. So, you have a resonant structure over there and these two possibilities and that is when you get the resonance. So, you have this you have the resonance at pi by 2 you have it again at 3 pi by 2 when the first bar state is already nicely bound and the next bound state is about to be bound it is not bound at the depth slightly below this and it is bound slightly above this at 3 pi by 2 it is a resonance state. So, this is the picture we get from the Levenson's theorem. So, let us our interest is in examining how the phase shift changes with k. So, if you plot delta 0 versus k not just look at the limiting behavior k tending to 0, but what how does it change with k as you change k in the low energy domain. So, that is our question here and we first consider a potential which is a very weak attractive potential it is attractive, but not sufficiently strong to have even a single bound state. So, all that attractive feature is not sufficient to bind a particle and in this case the phase shift will go to 0. So, if you now plot delta 0 versus k the behavior is given by this function which we have discussed earlier it will go to 0 and as you increase k the phase shift will increase because delta 0 k is given by this k a times tan kappa a by kappa a minus 1 we have derived this relation in the previous unit. So, the phase shift will increase, but it cannot keep increasing up to pi because for the phase shift to be pi you need the potential to have one bound state which it does not because the potential is attractive, but not enough attractive to have a bound state. So, the phase shift will not get to pi and what will happen to the phase shift is that it will keep increasing until it gets to pi by 2 and then start diminishing again because the high energy limit of course is 0. So, the phase shift will start increasing and then it will come down. So, you can already expect that the phase shift will have this profile that it will increase from here. So, this is the curve 1 it has got no bound state 0 bound state and the phase shift increases till it approaches pi by 2 then flips over and then falls to 0 eventually as k increases. So, that is how the phase shift changes with k what if the potential has either a little more depth or a little more range then it will pick up a bound state once lambda 0 a crosses pi by 2. So, that is the threshold value. So, that is the minimum strength that it should have. So, this is the picture when there is no bound state at all and when you do have one bound state then the phase shift as k tends to 0 will be equal to n pi n is equal to 1 now. So, the phase shift will be pi it will begin over here and then it will keep falling. So, the behavior of the phase shift changes in a dramatic manner in a very dramatic manner depending on the number of bound states. So, for 0 bound state it goes like this if you have one bound state it will fall like this. So, these graphs are straight out of Jochen's book this is in fact figure 4.6 from his book and now having considered the L equal to 0 behavior we will now consider the large L. So, let us consider the large L and ask ourselves how will the phase shift change with k the limiting value we already know that as k tends to 0 it will go as n pi that is 11 since 7. But if you plot delta as a function of k not just the limiting value k tending to 0 how will the graph of delta versus k look like. So, that is the question we examine now remember that as L increases the centrifugal barrier term will become more and more important and it can become so important that it becomes so much more important than the potential itself that the potential does not matter it is there, but it is coming the total potential is a sum of the centrifugal term plus the physical potential the two potentials add up in radial differential equation for the Schrodinger equation the two potentials add up one of which becomes ignorable compared to the other and the other takes over the dynamics completely and the physical potential is of no 11 it does not matter if it is a square potential or it has got some shape or no shape or you know whatever it is it just does not matter. So, what is going to happen is that if you increase the orbital angular momentum quantum number then the target potential becomes less and less significant and then we do know that for any given value of k you can always go to values of L large enough because in the partial wave analysis L will go all the way from 0 to infinity in the expansion of a plane wave you can always go to a value which is large enough and that value the potential as if it does not matter and the phase shift would go to 0 as L tends to infinity for any given value of k. So, that is the large L behavior of the phase shift. So, let us take some examples here and we have already discussed L equal to 0. So, now we are discussing L greater than 0. So, it will be at least one or more and we have deduced this relation in the previous unit in unit 1 of this course that the tangent of the phase shift is given by this relation we had introduced the symbols over here d plus and d minus. So, these are the double factorial which we have used earlier we had introduced this dimensionless quantity which is a ratio of k j prime by j these are the spherical Bessel functions and gamma was the logarithmic derivative of the function inside the potential region and outside if it has got a strict boundary at r equal to a. So, that was the logarithmic derivative. So, gamma is this and this is coming simply from the continuity of the wave function and its derivative. So, the large L behavior is given by this relation no matter what k is and this will also have a resonant structure which will depend on this denominator here because if this denominator vanishes then you will expect a resonant behavior. So, we will use the interior representation of the phase shift which we have obtained in the previous unit. This is given in terms of the spherical Bessel function the potential and the radial solution I have put v not equal to 0 as a superscript because you deal with radial functions of two kinds one is the solution for the free particle which is the absence of the potential that is the reference phase and all the phase shifts are measured with respect to the free particle incident beam phase shift phase value. So, you have the interior representation this is an integration over the entire space 0 through infinity and when you have a large value of L then we have noticed that the potential does not matter because of the centrifugal barrier effect and if the potential does not matter then this solution over here which is the actual radial solution for the Schrodinger equation with the potential with the scattering potential, but then if L is large it does not matter. So, the solution will be pretty much the same as you would have if v is equal to 0. So, the radial solution for v not equal to 0 is pretty much the same as the radial solution when v is equal to 0 which of course is this spherical Bessel function. So, you can put over here for this case when L is large when L is large you can use this radial function to be given by the Bessel function. So, you will have one over here and another Bessel function over here. So, this is like the Born approximation in Bessel and you can see that this works at high energy this works as we discussed in the context of the Born approximation it is a high energy approximation, but the same expression appears even if you are dealing with large L. So, the large L behavior is similar to the Born approximation at high energy. So, these give you essentially the same kind of relationships. So, here when L is large you can use the same result and now let us plug in the explicit form of the Bessel function. So, what is the spherical Bessel function for large L? So, for large L what will it be? So, here this is the power series expansion of the Bessel function in terms of its argument z in our case the z will be k a or k r and so on right. And here notice that for any given value of k if you look at the terms inside this bracket they are all multiplied by another factor here, but then inside the bracket you have L in the denominator and all of these terms are going to be compared with the number 1. So, you have got 1 minus something plus something minus something and so on, but each term has got the L in the denominator and the value of L is pretty large it keeps getting larger as you go to higher terms, but even the very first term has got 2 L plus 3 and L is large. So, no matter what k is if L goes large enough you can throw the rest of the terms. So, what it means is that you can approximate for large L the Bessel function to be given by just this other factor which is outside this bracket multiplying unity over here and this is nearly equal to z to the power L divided by this 2 L plus 1 double factorial. So, this approximation would be applicable not just when z is equal to 0, but also for some given value of k when L is large. So, when L is large you get the same expression that you can use, but for a different reason the reasons are different for k going to 0 the reasons are that these higher powers of z become increasingly small. So, that is the reason for the low energy behavior. So, now our interest is in the large L behavior and we have justified that this expression for the Bessel function can be used in this context. So, we put this Bessel function the explicit form of this Bessel function over here you need the square of it. So, you put z is k R you put the square of it which is here k R to the power L upon 2 L plus 1 double factorial to the power 2 and now you take out whatever does not depend on R because the integration is with respect to R. So, whatever is not dependent on R integral you pull out of the integration. Now, this is the integral you have to solve and in this integral if you put the square spherical square well which is the potential of finite range then you do not have to carry out the integration from R going from 0 through infinity because the potential has got a finite range only up to a and up to a the potential is minus u 0 it is not a function of R for the spherical square well. So, you take it outside the integration. So, u 0 k to the power 2 L plus 1 comes here and you have a simple integral of R to the power 2 L plus 2 to be evaluated from 0 to a. So, you get a to the power 2 L plus 3 divided by 2 L plus 3. So, that is your tangent of the phase shift in this case. So, this is a nice result notice that the potential strength parameter appears explicitly and you have got this famous 2 L plus 1 power of k which appears in the phase shift. Now, we have to be careful because we have used this result for finite potential if the potential does not have a finite range and it has got a sizeable tail and you will have those cases like in a coulomb potential or something then you cannot use this. Let us consider two partial waves, two partial waves with adjacent quantum numbers for the orbital angular momentum. So, one has got an orbital angular momentum L then next one is L plus 1. So, we will write this expression for both tan delta L which we just obtained and we will also write the corresponding expression if the angular momentum was L plus 1. So, wherever you have L over here you will write L plus 1. So, you had L over here. So, now we have written L plus 1 over here you had 2 L over here. So, you have twice L plus 1 over here and we do the same thing over here. So, this is the corresponding expression for L plus 1 and if you can determine the value explicitly. So, you have 2 L plus 5 in the denominator and 2 L plus 3 power of k a. So, if you take the ratio of these two angles then notice that it falls pretty rapidly as L increases because the ratio goes as 1 over L square and this is what one would expect because it is completely in line with our understanding of the fact that as the centrifugal barrier becomes strong then the higher partial waves can be ignored. So, all that is fine the only thing is that if the potential has got some strength even as you go as far as L over k then you may have some significant contributions to scattering. So, otherwise you can use this in those situations where the potential has got a finite range. So, any potential which has got a tail like the Yuccava potential or the Coulomb potential you will not use this particular approximation. So, we will study the behavior of the phase shifts as a function of k for large L and at lower energies. So, the scattering solutions we consider and the scattering solutions we can again rewrite the Schrodinger equation for the reduced potential u instead of v. So, that we get rid of the m and h cross and so on. We set up the differential equation for y instead of r, but the information in the differential equation for y is essentially the same as that in the Schrodinger equation. So, this is the effective effectively the Schrodinger equation or what comes straight out of the Schrodinger equation and we have to examine its solutions. So, you have the solution for r less than a will be given by this because the potential in this case is minus lambda 0 and for r greater than 0 the potential is 0. So, there is no term corresponding to the potential and instead of this kappa square kappa is determined by the depth of the potential and the energy both, but outside this region the depth of the potential does not matter because the potential itself does not matter there is no potential outside r equal to a. So, these are the two equations for the two regions of space and we can find the solution. So, for the inner region you have got r times j as the solution and outside the region the potential will generate a fish shaft and the solution will be written as a superposition of two linearly independent functions you can as usual use the Bessel function and the Neumann function you can also do it in some other basis if you like, but essentially the two pieces of which will be which will be superposed will be superposed by a coefficient which will be determined by the scattering phase shift because that is what is generating the potential the scattering phase shift itself. So, this is the solution for r greater than a and so there is a this relationship between kappa and lambda 0. So, these are the two solutions for the two different regions of space the solution however will have to be continuous at the boundary at r equal to a. So, the usual way of setting this up is to get the logarithmic derivative and then explore the continuity of the logarithmic derivative itself and whatever be the value of this derivative from inside it will have to be the same as you get from the outside and you can for the outside you have got the solution explicitly. So, you can see that this is kappa j prime kappa r by j kappa r kappa is coming from the depth and the energy itself. So, this is the logarithmic derivative and we had expressed this logarithmic derivative we had introduced this dimensionless parameter q because it offers a good deal of mathematical convenience it is. In fact, the logarithmic derivative itself if the solutions were just the free particle waves. So, that is in the numerator you just have to remember that the derivative is with respect to r and therefore, there will be an extra factor of k when you write the take the derivative with respect to r because argument is k r and not r. So, when the potential is 0 the factor q will be equal to 1 because these two will become equal that is how we have defined it. So, this is what we have got so far and you can rearrange the terms and put it in a somewhat simple form. So, you now have l into this l which is in the denominator here I have multiplied this 1 by l. So, I get l plus 1 plus l over here and this l is now in the numerator. So, this is the little bit of rearrangement we have made use of the explicit form of the Bessel functions we have taken the derivatives appropriately and by putting this these the ratio of the derivative and the function itself and the Bessel function you find that this ratio becomes l over a. So, now what does it give us for q? q is this. So, q you can get this numerator now goes as l over a and you will now get this q the dimensionless quantity which we defined to be l over a gamma and gamma we will take the limiting value of gamma is k tends to 0 plus some other terms which we have ignored and those will be of the order of k a square those are the terms that we have ignored, but then we take the limiting value of gamma which is represented by this gamma carrot or gamma hat. So, this is the result we have. So, this is the limiting behavior of the dimensionless parameter q what is ignored in the tangent of the phase shift. So, far are these terms of the order of k a square other than that all the terms are taken care of. So, I will like you to recognize that there will be a certain resonance condition and at the resonance if the terms that we have ignored will they be of any importance and we had some discussion on this in the previous units. So, I am going to stop here at this point so that when we come for the next class we have brushed up some of these considerations, but essentially we know the terms as we get we know what the resonant condition for L greater than 0 will be for L equal to 0 we already discussed. We know that what approximations we have made because over here we have ignored terms of the order of k square a square now you can again find some common terms and simplify this. So, there was an extra factor of L which was needless so we can get rid of it. So, we already know that the S wave scattering is the most important one because you see that the phase shift goes as 2 L plus 1 and as L increases it will become smaller and smaller. So, that is the importance of the centrifugal barrier effect unless of course, you have a resonance. So, all this is fine, but when you do have a resonance which you may have because depending on the range of the potential and the logarithmic derivative you may hit a value of L that for some partial wave you may hit a resonance and the necessary condition for resonance now is that the denominator goes to 0 or it becomes much smaller than 1 that is now the condition for resonance. So, the resonance in the Lth partial wave will take place when you have this condition satisfied that this denominator is much smaller than 1 and this is a good approximation the only thing is it ignores is terms of order k square a square. So, all the contributions from higher partial waves can be known you can work only with S waves except when the resonance condition is satisfied and we have found what the resonance condition is that this modulus of L plus 1 plus a gamma must be lot less than 1. If there is any question I will be happy to take otherwise we continue from here in the next class.