Welcome back everyone to Exploring Quantum Physics. Now, were going to complete the solution of hydrogen atom according to class mechanics and this brings in the identification of non-trivial concept of motion. This something that was known for many years by people setting planetary motion, but emerged as something of great importance in the development of quantum mechanics and I think you'll see that it points one to directions that are very fruitfully used when one goes to solving equation for complex systems, well we'll see. Now, just to remind you where we are, we're still working on, wer're leading up to the Bohr model of the atom by solving the classical problem. So, when you do the homework, you'll probably be reminded of some of this material. Again, I just like to keep in view that our objective is to understand real physical systems. And maybe some of you are tiring of seeing these spectral lines, but I never get tired of them. Okay, here's where we left things at the end of the last lecture. We have these two constants in motion, energy and angular momentum and that means we know whatever the initial configurations of the particles are, they move in a plain that is determined by the initial orientation. And then, the motion in that plane is also restricted to have a constant energy. So then basically, there's going to be an orbit in the plane, because you can't have a continuous, well at any given point, there'll be some limits in the motion that are determined by the energy. As we shall see. Okay, now how do we make progress? Well, we started with these two first order equations and motion. Now, we then for the vectors r and p. Then, we took scalar products, you might say r is a scalar, r is the square root in the scalar product. And we found that there was a conserved quantity, the energy that just depended upon the scalar products of the vectors themselves. Then, there was a vector constant of motion naming them r cross p, that's also a constant in time. Well, what other things do we have to look at. Well, from the value of angle momentum, we know that r and p lie in a plane. So, what other independent vectors do we have that might lie in the plane. I think you'll easily see that these two are cross L and p cross L both lie in the same plane as r and p. So, maybe there's some advantage. And they're independent variables. Maybe, there's some advantage to looking at the equations of motion. When framed in terms of these two new factors. Because after all, since they line the same plane, it's quite possible that there's a simpler relationship between all four of these vectors. Than is, that meets the eye right now. Okay, let's see what progress we can make. So here's let's take our cross L and look at it in greater depth, so then we can just say that's, that's simply that. R cross P. Which is then just substituting for r and r dot takes that form. Now, were going to invoke a well known vector identity to this is the so called bac-cab rule. Again, this is going to come up again in the Quantum Theory of Angular Momentum. So, it's a good idea for you to at least be aware that such a thing exists, and then later on we'll see how you'll develop a sort of vector calculus that will help you easily rederive it. So this is how you, when you have two cross products, you can reduce them and simplify them. And so that gives. When one goes through the math, here's this just rewriting that identity. Then from the bac-cab rule, you get something that's proportionate to the position vector and the velocity vector. And now, a remarkable thing happens. When you simplify this and I think it's really worth your writing notes on this or freezing the screen, pausing the screen and taking yourself. This is a very elementary derivation, just a sequence of simple steps, and you see what it gives you is that R cross L is the time derivative of the unit vector times r cubed. Now, what about p cross L? Let's look at its, let's look at the time derivative of P cross L. So that is just, I mean this is an obvious identity since L dot is zero and so we now get. Another cross product for r cross L. So, in other words, well, let's just see where that takes us. So, you see that r cross L Is equal to r cubed times unit vector of r dots. So, this equation implies that r unit vector dot is equal to minus r cross L, divided by r cubed, but you see that's what we have here. So, that means that this vector the sum of p cross L and the unit vector of r is a constant motion. This is often called the Runge Lens vector. There's a Wikipedia article on this that's very, very useful, very readable, very informative but I hope I've given you a path to rederiving it. Actually, when you know that something exists. When you know that there's something involving p cross L the results in the cost of the motion then you should be able to calculate it again for this system. Now, the remarkable thing here we are we have the energy, we have the [INAUDIBLE], we have the Runge-Lenz vector. The remarkable thing is that the, in some sense, the identification of the Runge-Lenz vector solves the problem for us because, look at this. Let's calculate the scalar product of r.A, which is r A Cosine phi if. This is r, and that's A and that's phi. So, first of all we know that A lies in the same plane. As r and p, and so we now know, so that means that the location of r is more or less precisely defined by the angle phi. Okay, but now taking r.A just from here, using this identity, to r A consign phi is this quantity here. Now, here is here is one of these vector identities A dot B cross C. I hope you recognize what to do with that. And if you can recall their identity, you can use it to complete the inline quiz or you can try computing that directly yourself. And in any event, let's have a look at the quiz. Okay, so dealing with this it's A dot B cross c. There's a cyclic identity. That's equal to, let's say we move everything around cyclically. That's equal to C.(A x B) = B dot C. C cross A. And so on. And so in other words, this term just reduces to the square of the angular momentum. And so now, let's see. I'm sorry for making such a mess here. Let's try to get rid of it. It's a very straightforward derivation. Well, now you see we have a very simple equation that relates r and phi. Thusly. So, here is an equation that defines the orbit. It shows r as A rather simple function of the angle phi. And that's, apart from specifying the origin of time or once you have the values of these constants to the motion, things pretty much fall into place. So let's see, we've gone on for quite a while. So one final calculation that we'll do here. We have, this is by the way, an orbit which can be a circle, an ellipse or a hyperbole and the distinction between those depends upon the energies. There's one last calculation that is worth doing and that is to calculate A squared. And I'm going to leave that to you as an exercise. It's worth doing because it's a fairly straight forward calculation and this is, that calculation gives us a relationship between the magnitude of A between the energy and the angle. So in other words, the energy angle, the angle momentum, and the Runge-Lenz vector are each constants in motion but as this equation shows they are not entirely dependent. Okay forgive me but these is literally the last of the in line quizzes. For the moment, and what, what it's going to be used for is to show you the significance of the value of the energy. In other words for this system, as we'll see, the energy can be any, well let's just take the quiz. So I hope you persuaded yourself that the energy of this system can be absolutely anything from between minus infinity and infinity. For example, you can have the momentum vanish at some given time. You just have these god like powers for specifying these conditions. You can keep p at 0 and then bring the particles arbitrarily together, arbitrarily large, negative energy or you can move to some finite value of r and then crank up the value of p to whatever you need. You can make it as positive as you'd like. And once you fix the energy, it stays that way for all time. And now the critical thing to remember, this has to do with some of the readings in the supplementary material would be in the homework. Is that there's a point of separation between positive energies or between negative energies and positive energies. So when the energy is less than 0. All the orbits are elliptical, so these are like the orbits of the planets about the sun. You can get the highly elliptical orbits, their far distance away. Those orbits have negative energies, but they're approaching zero. For e greater than zero, you have a type of trajectory where let's say here's the proton, the electron comes in and just, it just visits once and never leaves. So these two types of classical orbits actually have counter parts on the Quantum Mechanical Spectrum.