 So we have the second guest lecture today by Professor Steve Manson, and today he will talk about collisions, quantum collisions. Okay, thank you. Today's talk will be a little bit different. Today's class will be a little bit different in the sense that it will be much more quantitative. What I want to talk about today is charged particle collisions, particularly with atoms, although what I say is going to be fairly general and can be applied to molecules as well, but I will aim at the demonstration of what goes on in a simple system. Now, why are we interested in charged particles impinging on atoms? Well, for a number of reasons. First of all, that and photo ionization are probably two of the major processes that go on in the universe. The universe is much bigger than a few miles above the crust of the Earth. Most of the universe has lots of charged particles coming out and interacting with all kinds of things. And so, to understand what goes on, we need to understand charged particle collisions. Furthermore, there's all kinds of applications. Fluorescence lighting, for example, charged particles that ionize and excite and the resulting radiation is what you see. Well, actually, it's a lot of the radiation you don't see, but never mind that. However, unlike photons, charged particles have a very strong interaction with the targets. And that's something that needs to be looked at carefully. So, what I'd like to do is start out with a relatively simple problem. A structuralist charged particle interacting with a hydrogen atom in its ground state. And that's, when I say structuralist, what I mean is a bare nucleus. Because I'm not going to get into that today, supposing you have a collision where the projectile brings in its own electrons, that makes it much more complicated. We're not going to talk about it today. What I want to talk about is complicated enough. And aside from doing the derivation of what the probability, what the cross-section looks like, what I want to focus on is the approximations made and what the implication of the approximations are. Because, you know, it's easy to do the mathematical derivation. I mean, we have, you know, various packages that essentially do it for you, but they won't interpret it for you. And that's the important thing. One of the things we do as physicists is you take a problem and you say, okay, what methodology shall I use to solve this problem? In other words, what methodology should I use that has the essential physics of the particular problem? That's what you have to worry about. And in order to know what has the essential physics, you have to know A, the essential physics of the problem, and B, the essential physics of the method. I'm not going to talk about the problem now. I'm going to talk about the method right now. All right, let's see, chalk. Scattering, I mean, I am told that you have sort of a background in at least elastic scattering. We know that the wave function, let's say just for potential scattering, the elastic scattering of a particle off some center of force, goes as r goes to infinity. That is, get very far away as e to the ikz plus f of theta and phi e to the ikr over r, where my force is centered at r equals 0. If this is the asymptotic form of the wave function, asymptotic form, gets much messier if r does not go to infinity, then the differential scattering cross-section is equal to, ooh, that's pretty good. So that doesn't sound too hard. You get a wave function that has that asymptotic form, and then you just pull out that f and you take the absolute square, oh, we can do that, f star times f, you know, where pros are dealing with complex things, and get, well, not quite that easy. The way you get this is by looking at the probability current density for the incident wave, that's just, you know, a plane wave coming in, and the scattered wave. And it turns out that, you know, the probability current density, you just get from the continuity equation for the Schrodinger equation. It relates the time derivative of psi star psi, or which is the probability per unit value, to the probability current density. Actually, if I call rho as psi star psi, then rho dt plus del dot s is equal to zero. That just says that the total probability of all space remains constant. That's all that that is. But this thing here is the so-called probability current density. Very often students look at that and say, oh, my God, that's so complicated. But it's really quite trivial. If it's a charge particle and you multiply by the charge q, and I put it in here, and I put it in here, then this is just the charge density. That's nothing. And this is the current density. You know that. So these are, so without the q, these are just general concepts that we use in quantum mechanics which represent the same kind of thing except probability. Charge density sounds so easy. And probability density sounds so hard, but it's really not. OK. So you do this, and actually what you get for the, if I call this the incoming wave and the scattered wave, what I get is that s incoming is equal to, well, all I care is the magnitude, is h bar k over m, and s scattered as r goes to infinity. It's actually a very complicated expression, but most of it goes away as r goes to infinity. You get h bar k over m f absolute square over r squared. And the cross section, the differential cross section, is almost a ratio of these two things, but you have to do one more thing you have because it is, there is in the definition of the cross section, on the bottom it's a number of particles, or particles per unit area, or probability per unit area. On top it's probability, well, with the unit time in there as well. On top it's the same thing, but per unit solid angle, and so you have to get per area per solid angle, that gives you r squared, and that's why everything just cancels out, and you just get the f absolute square. Again, I'm assuming that this is review, I hope it's review, because this is hardly a derivation. I'm just trying to remind you of some things that I hope you already know. Okay, now let's get to our problem. What we want to try to solve is a charged particle striking a hydrogen atom, and what I particularly, you see this is potential scattering, just get elastic scattering, what I particularly want to focus on is inelastic scattering, where it gives some energy to the hydrogen atom, and let's say if the hydrogen atom starts in the ground state, it ends up excited. Okay, well, how do we solve it? Well, first we've got to write down the equation that we want to solve, and that is the Schrodinger equation, which I'm going to write in this form. Remember, now we have two electrons, the electron coming in and the hydrogen atom electron. So, notice I multiplied the Schrodinger equation by a minus sign, because I always like the first term to be positive. I just feel better like that. It's like in mathematics having an eye on the bottom. I never like an eye on the bottom. I always like to get rid of it and put the eye on top. I know it's a quirk. Anyway, 1 squared plus del 2 squared plus e squared over r1 plus e squared over r2 minus e squared over r12 psi, I should make this a capital psi of r1 and r2 equals 0. No, it isn't. I forgot something. It didn't die. I'll squeeze it in here. That's important. Move everything over to this side. This is simply the Schrodinger equation for two electrons. Now, I've got to solve that with the right boundary conditions. We'll come in a moment to what the right boundary conditions are. But now, I'm going to make my first approximation. You know that electrons are identical. So we have no way of knowing if this happens or if this happens. However, if my electron is very fast compared to the velocity of the bound electron, the chances of exchange are very small. So my first approximation is no exchange. And that implies that this will be reasonable as long as the incident electron is going much faster than the, or has much higher energy, is another way of saying it, than the atomic electron. Okay, that's it. You see, saying no exchange or without exchange doesn't carry with it immediately the idea of where that approximation is good. That's, now, this is the point that I want you to think about. So, if I am sending in a 3EV electron, or say a 10EV electron, when the hydrogen atom is bound by 13.6EV, this is a crummy approximation. I'd have to put exchange in. Now, putting exchange in is not terribly, terribly difficult, but it just makes the equations much messier, and it's harder to see what's going on. That's why I like to leave it out to begin with. All right, now, it's a partial differential equation. Again, two ways we know of solving partial differential equations. Separation of variables, we don't know how to do this. Guessing, I don't have any good guessing. We have to approximate. Okay, now one of the postulates of quantum mechanics is that you can expand any wave function in any complete set. Right? Unfortunately, complete sets in this case are infinite. So, we take a finite set. However, let me just write it down in principle what we would do. I take psi of r1 and r2. I'm going to write it as a sum. Remember, it's a two-particle, so I'm going to do it in two particles, and I'm going to write it as psi n of r1, actually of r2, I think I did, yeah, fn of r1. Now, since I'm not allowing exchange here, r1 is, I'm going to use the incident particle and r2 for the bound electron. Because without exchange, they become, in principle, distinguishable. Okay, I'm free to pick my set. Since I know that I'm dealing with a hydrogen atom, I'm going to pick these to be the eigenfunctions of the hydrogen atom, those you know. And these, I'm going to leave for the moment. These are the functions representing the continuum electron, the incident and scatter electron. Before I can begin to even approximately solve a partial differential equation, unlike an ordinary differential equation, there's something crucial that I need. Boundary conditions. You remember back to when you took a course in the mathematical methods of physics. You can't even begin to solve a partial differential equation even approximately unless you pick boundary conditions because you get different solutions under different cases. And you see it's easy when it's a bound state. Boundary conditions, everything goes to zero and infinity. Bang, that's no problem. But here, everything does not go to zero. And what does the asymptotic form of this wave function have to look like? Well, this tells me something about that. It's equal to, or it goes to, what? To the ik dot r1 times psi zero of r2. This is a plane wave in the ground state of hydrogen atom. That's what you had at infinity. Plus where, see, what does this represent? From what we talked about here, this represents the hydrogen atom in various states and waves coming out. However, the k is different. In other words, here, k was, or k squared, was 2mE over h bar squared. That's the energy of the incident electron. Here, when the electron has lost some energy, kN squared is equal to 2mE minus En over h bar squared. The excitation energy to a particular state of the hydrogen atom. I mean, so this is just definition so far. But now, let me mention something else. Supposing the hydrogen atom was not in the ground state. Supposing it's in some excited state. And the electron comes in. Could it knock it down? And the electron come out with more energy? Absolutely. That's called a super elastic collision and just falls automatically out of this formalism. However, in this case, it's just a little bit easier to think of it in the hydrogen, in the ground state, but it could be in any state. And you could add or subtract energy. And this actually happens not that much. The reason is that usually excited states decay fairly quickly. So in order to see this kind of thing, you have to have a really, really high density of charged particles or everything will decay. Or you have to have a metastable state. That'll work. OK, so you need boundary conditions like this. So that tells me what the FNs are. In other words, F0 has this in it, plus psi n, psi 0, e to the i. But all the others just have this. And since I want to focus on inelastic collisions, Fn goes as r goes to infinity, e to the i, kn r1 over r1, psi n of r2. And now I have a boundary condition or boundary conditions. Oh, by the way, there's one more thing that I want to mention. Let's assume that I can solve this. A really bad assumption, but let's assume that I can. Then from what we did before, d sigma n, d omega, be equal to what? Would it be, I'm sorry, I forgot something here. Would it be, and I forgot it here too. That's important. I mean, so it looks like the elastic scattering case. Is it just the Fn squared? No, why not? Because remember when I wrote down what the probability current density was, it had this h bar k. Here I have the h bar kn in the ratio. I still have the h bar k for the incident wave. So what you get is kn over k Fn absolute squared. Now I've got to bite the bullet and actually calculate this Fn, which is not easy. Let's say, how am I going to do it? Well, let me take this form, plug it in here. You see the reason I do it this way, I pick this fixed and this I haven't fixed. Because I know first of all asymptotically what the wave function is going to be, and asymptotically I'm going to have these by themselves here. And this also, why don't I just take something like plane waves or because I can't take an infinite set, because I know that I'm not going to be able to do an infinite set. So I want to determine these to get the best I can with a finite set. And so let me get the equations now that govern Fn. So what I do, I simply take this and plug it in here. So far there were no approximations except for the exchange business. All right? I'm going to plug in. But before I do that I want to note something. I want to note that I know what equation that is a solution to. It's hydrogen. So h bar squared over 2m del 2 squared plus En minus E squared over R2. Right? I mean, that's just a rendering of the Schrodinger equation for the hydrogen atom. OK? Now, here's what I'm going to do. I am going to plug this into here and then multiply by R2 star and integrate. It means the same thing. I like to do it this way because there's less writing. It means exactly the same thing. Let's see what happens then. What I'm going to get is sigma over En psi n star of R2. Make sure I've done this right. Well, no. I'm just now plugging that in and multiplying times h bar squared over 2m del 1 squared E minus En. By the way, I've made a mistake here. That's a plus sign, isn't it? Plus E squared over R12. Yeah, because if I multiply everything by minus 1, that gets multiplied also. fn of R1 is equal to what's on the other side. Well, let me write what's on the other side this way. I'm moving some stuff to the other side. It's E squared over R12 minus E squared over R1. See, all I have done here is I've taken part of this and plugged in here, and that's how I got this and this together on this to give me the En. That's the way I got the En here. That's R12, and that is R1. No, that's gone. That's what it looks like. I'm sorry. The R2 is gone from here, and here's everything else you see. Okay, to get rid of this now, I'd have to be multiplied by the star here. Okay, now I multiply by psi n star and I integrate this just gives me 1, and I get what? I get h bar squared over 2m del 1 squared plus E minus En over R1 is equal to the integral of, I multiply one side by psi n star. I got to multiply the other side, psi n of R2 E squared over R12 minus E squared over R1 psi of R1 and R2 dr2. Move it here so you can actually see. And what this is, it's basically just a transformation of this equation that I'm erasing. Now, multiplying through by 2m over h bar squared, then this becomes what I call Kn squared. What I get is then del 1 squared plus Kn squared equals 2m over h bar squared integral E squared over R12 minus E squared over R1 psi n star of R2. Now again, this is simply a transformation of the Schrodinger equation. Nothing else. It leaves something to be desired if you actually want to solve it from the following point of view. First of all, there's an infinite number of these terms. That already is a downer. Secondly, in order to solve the equation, you need to know the right-hand side and the right-hand side has what's unknown to begin with. So what I have is exact, but in this form it's not very helpful yet. So what can I do? Well, what I can do is say, all right, let me take some zeroth order approximation here. Now we've got a first order approximation. We can go back and I can do an iteration and I can do this approximation at various levels. For example, remember the expansion for this, which I just erased. Well, I can pick instead of an infinite number of terms, a finite number of terms. That will help a lot. You know, so if I take capital psi as sigma n equals 0 to some capital N, infinity of psi n of r2 fn of r1, then if I plug this in here, I will have on this side the fn's. Here are the fn's and you can actually, in principle, solve this. It's messy, but it's solvable. When I take those terms, a bunch of terms there, that is known as a close-coupling expansion. You can't solve this analytically, but you can do it numerically. You've got to do it by iteration. But it is solvable. By doing this, by taking a finite number of terms, you're saying that the higher part for your purposes is unimportant. And one has to consider under what conditions the higher part will be unimportant or at least not affect the part that you're looking at. It could even take something more. Supposing, in lowest order, what will happen? The electron will miss. So if there were no interactions at all, what would psi be? Psi would just be e to the ik dot r 1 psi 0 of r2. Ooh! That looks like something I just might be able to deal with. So my philosophy is this. I'm going to start with this as my 0th order wave function. I'm going to plug that in here, and I'm going to get df n's out. Then I can get a new psi, the next order, plug that in. This, in a general sense, is known as the Born approximation. It's not the usual way of deriving it. But I like this way because you can see exactly the approximations that are being made. So you are not approximating no scattering. That's my 0th order approximation, and I'm going off. And in principle, if I carry this out to infinite order, it'll be exact. I won't carry it out to infinite order though. So, I take this and plug it in. Now, again, let us think about this. Under what conditions is this a reasonable approximation? That's very important to think about. Well, I'm guessing because I know something about how physics works. The longer time the incident electron spends around the hydrogen atom, the more likely it is to do something. And I'm starting off with the 0th approximation that it does nothing. So how do you get it to spend less time? Have a fast collision. No, a high energy collision. So this, to just think about the physics of the situation, is a high energy phenomenon. A high energy approximation. This should be good at high energy. All right. It's still non-trivial to do. My equation becomes what? Del squared plus kn fn of r1 is equal to, that's h bar squared 2m over h bar squared the integral e squared over r12 e squared over r1 e to the i k dot r1 I'm plugging this into that phi n star of r2 0 of r2 Is that everything? Make sure, yeah. d r2. Now I'm not going to make another approximation yet, but I'm going to restrict myself. Consider specifically inelastic collisions. What do inelastic collisions mean? In this case, n is not equal to 0. Because if n equals 0, then it stays in the ground state, and that's okay, this is important. Because look at this integral over r2. With this term here, we have if I forget about this term, all I have, these are the only two things I've depended on r2. When I integrate over r2, that gives 0. Because the wave functions for different states of the hydrogen atom are orthogonal. So, but n not equal to 0, I can forget about that term. All right? So then, l squared is kn squared of r1 is equal to 2m h bar squared the integral of e to the i well, I'm going to pull out the e squared as well. e to the i k dot r1 I'm going to make sure I'm not making any mistakes. Yeah. Over r12 psi n star of r2 psi 0 of r2 dr2. Believe it or not, we can actually solve this equation. You use what is called the Green's function of this. I don't have time to go into deriving the Green's function, but the actual solution then fn of r1 is equal to 2 e squared over h bar squared the integral e to the i kn uh or let me write this as r, I get too many r dot r1 to make sure I'm getting this right psi n star of r2 psi 0 r2 dr2 dr1 and this is the exact solution of that. You see, I have this Green's function integrate over it and then just everything else here. Fortunately, I know something about how this works and I can take the asymptotic form of it because for large enough r r is greater than r this just becomes r and this becomes a little bit more complicated but when we put it all together what I get is fn of r is equal to 2 psi n I'm sorry but for large r why don't I just actually solve for fn why do I want it for large r because remember it's for large r that it has the form fn times e to the iknr over r and all I'm looking for is that little f I mean I could do a lot more but it's much harder so this is just the asymptotic form for large r which I take for large r here as I go along I lose things like dh1 which happens when you get old the double integral of e to the ik k minus kn dot r over no I'm sorry not over anything actually it's a minus sign here let me rewrite this a little bit because I didn't give myself enough room it's actually the Green's function has a minus sign there too actually and there's something there's something funny yes there is something very funny I lost that too okay r12 psi n star of r2 psi 0 of r2 r1 dr2 times to the ikr we're going this way particularly because then this is my scattering amplitude the fn and if I can calculate that I can calculate the cross-section ooh and I only have 5 minutes to do it what I do is I change I make a change of variables I change the variable well first of all I define capital K that's generally called a momentum transfer although it's actually h bar capital K which is the momentum transfer because h bar K is the momentum of the incident particle h bar K n is the momentum of the scattered particles so that's the difference that just makes things a little bit easier and then I change variables for r1 I change it to r12 it's a simple change the Jacobian of the transformation is 1 I would have done it but I don't have time to at the moment but if I but in any case just looking at this then the differential scattering cross-section is what? it's K n over K times this absolute square which is 4 pi squared m squared e to the fourth over h bar to the fourth double integral K p dot r psi zero of r2 psi n star of r2 dr1 dr2 squared now what I was saying here is that let me just look at a piece of this the integral of one sorry that's I've lost something somewhere yes I have that's r1 I'm integrating over that so 1 over r12 e to the i K dot r1 psi zero of r2 psi n star of r2 dr1 I can make that that substitution here and I simply get here I get 4 pi over K squared e to the i actually minus iK dot r2 and psi zero of r2 psi n star of r2 um yeah I'm sorry this so that my square of my matrix element absolute square or just the absolute value of the matrix element then is 4 pi over K squared the integral of now we're going to write it in a particular way now psi n star of r2 e to the minus i dot r2 psi zero dr2 so look at what that means I have reduced this whole mess to a matrix element between the ground state of the hydrogen atom and the excited state of e to the minus K dot r2 if you look this up you will sometimes see it as e to the plus e to the iK dot r2 but it doesn't really matter because you're taking the absolute square so you can do it either way in other words then if I um if I call this epsilon n zero of K the momentum transfer what I'm going to get is the sigma the omega is equal to a whole bunch of constants for pi the K n's epsilon n zero I'm going overtime but I won't go overtime much I promise there are various properties associated with but the crucial thing here is that the cross section depends only well not only it depends on the incident energy because you have this K n over K here but aside from that it depends only on this momentum transfer this is a characteristic of the Born approximation only on the momentum transfer so you have taken a very messy problem and said okay I take this plane wave I take the matrix element this integral now supposing you have something more complicated than a hydrogen atom well whatever the wave functions are this by the way this matrix element is very easy to take what you do is you expand this in a series just like the series that you expanded in for a plane wave of scattering and since these things have spherical harmonics associated with it you only get a few terms and this is a very quick and dirty view of how you do charge particle scattering now all this work is just for doing it in the Born approximation although what we really did is called the first Born approximation because what you can actually do is you actually get you can solve this you can do this integral numerically if you have to and you can iterate and get the second Born approximation etc it's very messy but it's perfectly doable and in the Born approximation the first Born approximation which is applicable again for high energy collisions you can get very very good answers when I say very good answers it agrees rather nicely with experiment in many many cases and I guess since I've gone over time I should stop here