 Greetings, we will continue our discussion on the behavior of phase shifts for large orbital angular momentum quantum number L and examine how the phase shift changes with energy and momentum and then we will discuss other consequences of this situation. Now we already obtained in our previous class the expression for the tangent of the phase shift and we found that the resonance condition is given by this denominator going to 0 this L plus 1 plus a gamma this gamma carrot is the limiting value of the logarithmic derivative as k tends to 0 right. So typically because of this k a to the power 2 L plus 1 all the contributions from higher orbital angular momentum will become less significant. So we have the expression for the tangent of the phase shift and because of this k to the power 2 L plus 1 the contributions of higher orbital angular momentum partial waves will become insignificant unless the denominator L plus 1 plus a gamma approaches 0. So this becomes your necessary condition for resonance and we will have to examine if S wave scattering is the only important contributor or are we going to expect some contributions from some of the other partial waves and this is the resonance condition that we should examine closely so that is what we are going to do now. So this is the necessary condition for resonance and the question is if you do hit the resonance if this denominator really becomes small then what will be the scattering cross section what will be its contribution to scattering from the lth partial wave. So we already know that the total cross section you can write as a sum of partial wave cross sections from different orbital angular momentum quantum numbers right. So you sum over all the values of L going from 0 through infinity and the lth partial wave is just 4 pi over k square 2 L plus 1 and then you have the sin square of the phase shift for that particular partial wave. So that is the contribution from the lth partial wave. So this condition you can see will hit a maximum when the sin square function becomes 1 that is the maximum value it can take and at that maximum value when sin square phase shift is equal to 1 the phase shift will be either half or 3 by 2 and so on. So it will be n plus half times pi so that will be the phase shift at this maximum. So at the resonance the phase shift will be pi by 2 modulo 1 it can be pi by 2 it can be 3 pi by 2 5 pi by 2 and so on. So that is the resonance condition which we have and the minimum value of course will be 0 when the phase shift will become n pi and there is no contribution to scattering by this particular partial wave in that situation. So actually although the expression requires for completeness that we must sum over all the partial waves from 0 to infinity but you really do not have to go to very large numbers so far as the angular momentum quantum number is concerned because if you have a potential with a finite range then even from classical mechanics means this is a semi classical argument it is not very rigorous but it works and it gives you at least an order of magnitude estimate of the maximum number of partial waves that need to be considered in the scattering phenomena because r cross p with the angular momentum so if you have a scattering target over here and you have got a beam which is coming here. So if this beam happens to become not just head on but a little above it so that there is an impact parameter corresponding to the range of the potential then you can see that the r cross p at this point will give you an estimate of the angular momentum. So that is r into p, r into m into v and that will have to be the angular momentum. So the angular momentum of course is quantized so the eigenvalue that you are really considering are eigenvalues of l square which will be h cross square l into l plus 1 but you can approximate that for large value of l, l into l plus 1 will be almost l, it will be almost l square and you can ignore the 1 compared to that square root will give you l itself. So h cross l will become of the order of r into h cross k so that tells you that the maximum value of the orbital angular momentum quantum number that you need to consider will be of the order of k a and it will not really be infinity so you do not have to go to you know very large numbers in most of the physical situations. Now what we are going to discover by doing a small exercise is that this particular condition which we have recognized to be the necessary condition for resonance is completely consistent with the Levenson's theorem it because it corresponds to the condition of getting a bound state for 0 energy for the lth partial wave. So that is something that we would expect and that is what it indeed turns out to be. So we have to examine what is the condition to get a bound state at 0 energy in the lth partial wave. So we do this again by considering a square well potential which is our prototype of our potential there most of the other potentials are more complex but this example does illustrate the key ideas which go into this analysis. Now for this square well potential we have the usual dynamics over here the continuity of the wave function and its derivative is indicated by this logarithmic derivative gamma and this is in terms of the Bessel function and its derivative the argument of which is kappa A and kappa is determined by the depth of the potential which is minus lambda 0 square and it is also determined by the energy which is k square which is h cross square k square by 2 m right. So this is the limiting value which goes in the necessary condition the gamma carrot or the gamma hat is the limiting value as k tends to 0. So as k tends to 0 so this term k square would vanish and kappa will be lambda 0. So that is what you get in the limiting value. What we will do is to make use of the recursion relations for Bessel functions which you will find in Abramovich and Stegen or even in Arfkan or you know almost any usual book on mathematical physics you will find the recursion relations for Bessel functions and these are the standard recursion relations for Bessel functions in which the derivatives are given in terms of the adjacent Bessel functions. So there are similar relations for the Neumann functions and so on. So you know this is the particular one for the Bessel functions so we make use of the recursion relations for the Bessel functions and then plug it over here to get the expression for the limiting value of the logarithmic derivative gamma in terms of the Bessel function for L minus 1 and L. So these are the Bessel functions which appear in this because in the recursion relations the adjacent values of L they get connected along with the derivatives. So we have this using the recursion relation this form for the limiting value and that gives us for the necessary condition for resonance. Now in this A gamma carrot we put the right hand side from here which we have obtained using the recursion relation and you have A times this is A and then gamma carrot which is this beautiful bracket multiplied by lambda 0 so lambda 0 is here and the other beautiful bracket is here. So now you have got A lambda 0 here and A lambda 0 in the denominator so you can multiply both of these terms by A lambda 0 what happens when you multiply this A lambda 0 this cancels this and you will have L plus 1 here and then minus of L plus 1 here so those will cancel each other and you are left with A lambda 0 times this ratio this product must be of absolute value which should be a small number. So that is now effectively the necessary condition for resonance at low energy. So now we ask what happens when this denominator is small so what is going to happen is that when this denominator is small then you get if you go through this exercise in shifts quantum mechanics which I will not work out then you find that you get a condition for having a bound state. So essentially what is going to happen is that the condition that you get a bound state for L becomes completely equivalent to the necessary condition for resonance which is just the right kind of condition the only difference between this and the case for the partial wave for L equal to 0 is the fact that in the case of L equal to 0 for the S waves what happens is you get virtual bound states you remember that that you had virtual bound states for L equal to 0 when you have the resonance condition at with the depth and the scattering strength parameter was pi by 2 you got virtual bound states whereas in this case for higher orbital quantum numbers you get proper bound states. So it is not like a it is not like a state which is half bound and half not bound which was the case for the L equal to 0 case. So now if you have a weak potential the phase shift will first increase if it is weak if the potential is so weak that it cannot even hold a single bound state then the phase shift will first increase and then since there is no bound state at all it will have to come back and eventually go to 0 but then as you increase the depth of the potential or the range of the potential make the potential stronger we have seen these plots for L equal to 0. So now we are observing that for L equal to L greater than 0 we have somewhat similar plots but then we have some differences also because as the potential becomes stronger then the phase shift increases but does not quite get to pi unless it actually is strong enough to hold a bound state then it flips over and then comes back whereas if the potential becomes stronger still then the once it crosses that threshold then you have a one bound state so that the low energy phase shift will become equal to n pi n is equal to 1 in this case so that it is just strong enough to hold one bound state so the phase shift at low energy limit the phase shift will be equal to pi and then it drops which is given in the curve 3. So this is completely along in consistent with the Levenson's theorem what we will now consider we consider the low energy behavior now we will consider the high energy behavior okay so high energy high with respect to what high with respect to the orbital angular momentum because K and L they come together so any comparison between them tell us that okay if you are much above the L value then you are in the high energy domain so far as this context is concerned so in this case let us set up the relationships for the Schrodinger equation for the inner region you are said the actual radial solution is capital R you set up the differential equation for Y the radial solution being Y over R so you have got these two solutions for the inner region and for the outer region the outer solution will be determined by the scattering phase shift okay and you have got from the continuity of the wave function and its derivative at the boundary this continuity of the logarithmic derivative okay and this must be the same for the inner region as for the outer region so you have kappa square is lambda 0 square plus k square so this is the same kind of parameter that we have considered earlier so what do we get the logarithmic derivative must be continuous and at R equal to A it must be kappa it it will be given by this kappa A so let us take this now I will use these relations which we have obtained earlier this was in unit 1 so we have obtained this relation earlier so I will use that relation directly and consider the high energy limit so the high energy limiting Bessel function values are given by sine of z minus l pi by 2 divided by z for the Bessel function and the Neumann function is a cosine function with the same argument with a 1 over z again but with the minus sign okay so these are the usual relations for the Bessel functions and the Neumann functions which we have to substitute in this relation so let us substitute these values so now you have to put in the value corresponding to kA whereas for gamma you have to substitute the value corresponding to kappa A so you have these values here okay so they are similar relations but then this one is k and this one is kappa and with the ks and the kappas taken care of what have we ignored we have ignored for large z terms of the order of 1 over z square when they come together as additive terms with 1 over z terms so that is the approximation but that is fine that is always approximation we take for the asymptotic limit so here the asymptotic limit is coming because not because of the distance but because of the energy parameter that this is the large energy limit so you have a similar consequence on the values of the Bessel functions so you put in these values okay and get the expression for gamma so these are straightforward substitutions I will not comment on this and now here this angle kappa A minus L pi by 2 is what I write as phi just to make our notation compact okay so I will write this expression in terms of the cosine sign and tangent or cotangent of the angle phi as it the case would be so I introduce the symbol phi just to make our notation a little compact so this is gamma here in terms of phi which is this angle kappa A minus L pi by 2 or in terms of this we can write our earlier result so now the tangent of the phase shift is now given in terms of the trigonometric functions of theta theta is k A minus L pi by 2 and the other angle is phi which is kappa A minus L pi by 2 so these are the two angles which come in our expressions the actual substitutions are straightforward so we have written them in terms of theta and phi and a little bit of substitution as you can see this k cancels this k the 1 over a is in every term in the numerator as well as the denominator so they are going to cancel right so now you get rid of all the 1 over a and you get after these simple substitutions a little bit of rearrangement as you can now recognize by this blue arrow that there is nothing more in it than straightforward substitution so I will not take you through this step by step but you can work it out and you can always refer to the PDF for details so here we are so this is the expression for the tangent of the phase shift and it appears in terms of two angles theta which is k A minus L pi by 2 and phi which is kappa A minus L pi by 2 now I see that there is a k over kappa times tan phi so I introduce this angle epsilon the tangent of which is k over kappa tan phi because that will make it possible to write the tangent of the phase shift in this form so that you can write it as the tangent of a difference of two angles okay so that is the idea so we introduce this epsilon so that you can write the phase shift or the tangent of the phase shift as tangent of another angle and then you can immediately get the two angles you can equate the two angles and you get the high energy limit which is minus k A minus L pi by 2 now this is the one of the two angles this is the theta and then you have got the phi over here but now the phi in terms of phi you have the epsilon so you get this k over kappa kappa minus L pi by 2 right so this is the expression you get for the phase shift at high energy so as energy increases you have this minus k A right so the phase shift will keep continuously decreasing till it becomes 0 okay so that is the high energy behavior now we will now take the example of a square well we have already considered but we have considered square well potentials with finite depth and so on so we will now consider a rigid potential like an impenetrable sphere okay it is a it is sometimes called as a hard sphere you cannot penetrate it because the potential inside in the inner region from 0 to A is infinite so nothing can get it okay it is an ideal situation it is a mathematical construct that the potential is infinite and no scattering no projectile can penetrate that so we will consider this case of hard sphere scattering now in the case of the hard sphere scattering we are not bothered about anything like the shape of the potential or anything because it just does not matter it is infinite so for this inside there will be it will be impossible for the projectile to get in and outside it will be governed by this phase shift what happens at the boundary it must go to 0 because it cannot get in okay so that is the boundary condition because it is hard sphere it is impenetrable and therefore at the boundary r equal to A it must vanish and therefore this J L k A over N L k A must give you the tangent of the phase shift so now we have an expression for the hard sphere phase shift now this is an important result because when we deal with any arbitrary potential we can pretend as if this potential is made up of two pieces one of which is a hard sphere kind of situation and the other contains the dynamics so we will break up the net result into this kind of analysis but that is something that we will discuss in subsequent classes so for the for now we know that the function y must go to 0 at r equal to a which gives us this expression for the tangent of the phase shift and now over here if you now consider the low energy limit then in the low energy limit you know that the basal function what the limiting behavior of the basal function and the Neumann functions are for the low energy limits as Z tends to 0 you can put them in and again you get a same k A to the power 2 L plus 1 behavior from this right so this is the low energy limit for a hard sphere scattering and you can see that it decreases quite rapidly with L again in consistency it is completely consistent with our expectation that higher partial well waves will not matter most of the scattering will be determined just by the L equal to 0 the S waves so this one this is your result which you will remember for L equal to 0 we have found that it goes as minus k A because L for L equal to 0 this 2 L goes to 0 and then you will simply have the tangent of delta 0 going as minus k A now notice that in this case the scattering length which is the ratio of tan delta by k it will go as A and if the scattering length is alpha what does it give us for the scattering cross section it is 4 times pi a square the classical cross section is pi times a square pi a square that is the area of a circular disk of radius a so this is 4 times that and that is that is not surprising because this is not classical this is quantum so obviously you will expect some differences and this is what it is so now you consider the high energy projectiles now when you consider the high energy projectiles we have seen that at r equal to a this function must go to 0 that this is the hard sphere scattering phase shift and the high energy projectiles again you use the high energy limit for the Bessel function and the Neumann functions so you can put these expressions over here for the tangent of this angle and what does it give you for the phase shift itself is this tan inverse of this ratio this ratio itself is tangent of this so you get the phase shift to be minus k A plus half L pi now this is an interesting result because what is happening is that in the 11th sincere we always argued that the phase shift at high energy would go to 0 right that was the reference for the angle that we considered whereas in this case the phase shift keeps becoming more and more negative and it goes to negative infinity okay so this is something which you might think is not consistent with the Levenson's theorem but why does it even have to be consistent with the Levenson's theorem because this is a hard sphere and it does not have any bound state okay so there is no problem with that issue over here so this has got no bound state at all and again you can get the total cross section by carrying out the summation up to the maximum value of L which using our semi-classical argument we know is of the order of k A so for all practical purposes we can take L max to be equal to k A it works in most situations which I have mentioned earlier in various different contexts so this is the cross section that you get partial wave contributions from different values of L you do not have to sum all the way to infinity you just have to go as far as L max which is about k A and what does this sum turn out to be so let us take this term by term so let us write this expansion explicitly because you will find that it sums up to a very compact result which is rather attractive so it is nice to get it in that form so you write all the terms for L equal to 0 this half L pi is the only thing which is going to be different in every term okay you are summing over L going from 0 to L max and this half L pi will go from 0 to pi by 2 and so on right so you get for L equal to 0 a term in sin square k A then for L equal to 1 you have got 3 times you have got a 2 L plus 1 factor here so 3 times sin square k A minus pi by 2 then for L equal to 2 you have got 2 L plus 1 which is 5 times sin square k A minus pi okay for L equal to 2 and that is how you get various terms so you get very different coefficients of the sin square functions 1 over here 3 over here 5 over here 7 over here 9 over here notice the pattern in which the coefficients are coming so that you can generalize it very easily you can go to higher order terms so the maximum value that you have to consider is of the order of k A and at high energies L max can be large so you can sum up to all those values. So this is the pattern that you are getting and this angle which is sin square this is not just theta but it is theta minus pi by 2 so the argument is k A minus pi by 2 but you can write it in terms of sin or cosine of k A itself because sin square of k A minus pi by 2 is nothing but the cosine square of k A so you can simplify this likewise this is sin square of k A minus pi and this simplifies to sin square of k A so all of these angles can be written very simply in terms of sin square or cosine square of k A so this comes out to be sin square k A the sin square k A minus 3 pi by 2 which is here so this term comes out to be cos square k A and then you get terms in cos square k A and cos sin square k A together so you can combine them and accept that sin square theta plus cos square theta is equal to 1 even in atomic physics right so we can simplify this relation and we really get a very simple sum but of course we have to take care of the coefficients here because they are not the same you have got 1 over here 3 here 5 here so keep track of that and we will make use of this relation so let us look at these terms so the next one again I have not written all the details over here but then you will get similar combinations and you will get either sin square theta or cos square theta where theta is k A so these are the terms that you get so this one is cos square k A here you have got sin square k A and here you have got cos square k A and so on so now you write these terms in terms of this angle k A now no matter what value of L is it is always written in terms of sin square or cos square of k A and now you have this factor 3 this is 3 times cos square k A so this 3 I split into 1 and 2 so it is cos square k A plus twice cos square k A so that I can combine this cos square k A with this sin square k A to get unity and then I can combine this twice cos square k A with the twice sin square k A which I am getting from the next term to get a sum a factor of 2 because here I have got 5 sin square k A which I have split into twice sin square k A plus 3 times sin square k A so I take this 2 twice sin square k A with this twice sin square k A and this 3 sin square k A with this 3 cos square k A of the following term so this is just a rearrangement of terms so that I can combine these to get a simple result and what you see is that we are simply adding 1 plus 2 plus 3 plus 4 plus 5 and so on up to L max and that sum you can easily get to L max into L max plus 1 by 2 so that is the compact expression for the scattering cross section and you find that it is determined only by this upper limit in this particular case so this is the case that we have considered and when L is large then you can approximate this to be L square by 2 if you ignore this 1 compared to L max because L max is large it can be 10 it can be 20 so you get the high energy cross section and in this case now if you put this equal to L square over 2 what do you get twice pi A square so this is half the previous result this is twice pi A square the previous result was 4 times pi A square and now we can consider the resonances so once again we go back to this expression so this is a common expression that we are going to use throughout our analysis that is the basic relationship to discuss the phase shifts but what we will do is we can use any pair of base functions our base pair has been the Bessel function and the Neumann function we can also use the Hankel functions the Hankel functions of the first kind and the Hankel functions of the second kind because they are made up of the Bessel and the Neumann functions and you get an alternate pair of bases so that gives some advantage so you have got the Hankel functions of the first kind which is just the Bessel function j plus i times n so you got the Bessel function plus i times the Neumann function in the Hankel function of the first kind in the Hankel function of the second kind it is the Bessel function minus i times the Neumann function so in terms of the Hankel functions now we can rewrite the expression for the tangent of the phase shifts because the Bessel functions and the Neumann functions can be written in terms of the Hankel functions backward so you substitute all that and then you get the phase shifts and you get an expression for the phase shift which is now given completely in terms of the Hankel functions so this is again straightforward substitution which I will let you work out this is the result that you get that the scattering phase shift is given by this expression here. Now the reason to put it in this form is will become clear perhaps in the next class or maybe in the next one or two classes because what it allows us to do is to introduce two different angles and you can write the phase shift the net scattering phase shift is delta but you can write this scattering phase shift in terms of two angles you can write the scattering phase shift as a sum of two different angles one is xi and the other is rho so we introduce these two angles and one is defined in terms of the ratio of the Hankel function of the second type to the Hankel function of the first type which you see is factored out in this term already right and then you have got the other factor in terms of which we will write another angle which is rho so that the scattering phase shift can be written in these two parts and then it turns out that these two parts refer to Hartzware scattering and the other will be given by the actual dynamics of the potential and that is the one which we will be which will be of central interest in considering resonances because when you have a resonance the other term which is rho which gets added to xi is the one which will vary rapidly at the resonance and it is this rapid variation in the resonance region which is going to be the subject of our discussion in the next several classes. So, this is the phase shift in terms of the Hankel functions we have introduced these two angles now one is xi and this xi is written in terms of this ratio of the Hankel function of the second type to the Hankel function of the first type together with a minus sign. So, this e to the 2i xi l is now defined this is the defining relation for this particular phase shift this is the part of the phase shift. So, this is one part of the phase shift and then there is the other part you will see immediately that the Hartzware scattering phase shift which we obtained earlier was nothing but this it was exactly this. So, we have factored out from the total phase shift one part which is coming which can be attributed completely to the Hartzware component. So, it is not that there is a physical Hartzware which is sitting over there what this analysis is letting us do is that the net scattering phase shift which is determined by the real potential by the real dynamics of that potential. So, that real potential generates a result a resulting scattering phase shift which you can write as a sum of two terms one of which can be attributed to Hartzware scattering because the ratio that you get which in terms of which you define the phase shift xi gives you for its tangent nothing but the same ratio that you get for scattering by a impenetrable sphere. This is just the ratio of the Bessel function to the Neumann function. So, this result we have from an earlier discussion that this is the Hartzware scattering phase shift and now what we will do in subsequent classes is to focus our attention from the Hartzware part to the other part which is the row. So, there are two contributions to the net scattering phase shift one is what we will call as the Hartzware component and the other is the dynamical component and the dynamical component in the absence of resonances will vary smoothly with energy or it will have a smooth variation with respect to k but at resonance it will change very rapidly. So, these are the resonances which are of great importance for our study in collision phenomena. There are two kinds of resonances you have got the shape resonances and you have the phono of fresh back resonances. So, these we shall discuss in the next few classes. So, if there is any question I will be happy to take otherwise we take it from here in the next class.