 We have been talking about central force tension in the case in which the force center has an infinite mass. So the problem becomes essentially a one body problem with the remaining particle. Today I might take one example of that, a fairly obvious example, which is namely one in which the potential is zero. In that case, there is no force center. It doesn't matter at all. So let's talk about the free particle where the potential is zero. In this case, the Hamiltonian, of course, is just a piece further to M. This is really the three-component square. And so to emphasize Hamilton is the square of a vector. If you want to solve it, the free particle Hamiltonian is really an equation for H psi equal to H psi for the free part of Hamiltonian in three dimensions. The most obvious choice of a complete set of commuting observables is the three components of momentum. These, of course, communicate with each other, and they also communicate with the Hamiltonian so that an eigenstate of the PX, PY, and PZ is automatically an eigenstate of the Hamiltonian with an eigenvalue which is a piece square of a 2M. This is really, of course, the same equation as I had over here for the Hamiltonian, except here the Hamiltonian and momentum are considered operators and here they are considered C numbers that are eigenvalues, but it's the same equation anyway. In this case, we know what the wave functions are. The eigenfunctions in momentum in the X representation as a function of X are just plane waves. They're here in the IP dot X over H bar. And if you want to normalize them with the delta function normalization, you need to divide it by 2 by H part of the three halves. So this is the most obvious solution. This is also the solution you get by separating the wave equation in rectangular coordinates. However, it's more in accordance with what we've been talking about recently to think about a different complete set of meeting those other levels. So I have namely one in which we have the Hamiltonian L spread in LZ, because that's what we normally think of in simple force function. And as I've explained in the previous lecture, we get an eigenstate of L spread in LZ automatically by making the total wave function a product of a radial wave function times a YLM. And in which case the radial wave function satisfies one of two different versions of the radial wave equation. Version one is in the original variable R, and version two is in a modified variable and calling F, just a scale factor of R, the radius between them. For the free particle, it turns out that version one is the more convenient version to use. Before I write it down however, let me note something which actually follows fairly obviously from the solution in the rectangular coordinates, which is the energy of the free particle has to be positive or non-negative. This is clear because the eigenvalues of the square and momentum has to be positive. So the same thing, of course, is true regardless of the coordinate system, so it's going to be true in the spherical coordinates as well. But what it means is that we can define an effective wave number for the solution called K, which is the square root of 2MP. E is the positive eigenvalue. And then I mentioned this, the expected is continuous, so there's no quantization. Divided by h part, this gives us a wave number. It allows us to introduce an dimensionless radial variable called rho, which is equal to K times R. And if we make this substitution then into the version one of the regular Schrodinger equation in which the, this is wrong, this is plus the true potential that you are, in which the true potential vanishes, there's only a centrifugal potential, then the regular equation we get in terms of the rho miracle is this. It's minus 1 over rho squared times d u rho rho squared times R plus the quantity which is L times L plus 1 over rho squared minus 1 times R equals 0. It basically, this substitution, rho equals K R, basically gets rid of all physical constants in the version one of the regular Schrodinger equation. It also transforms it into a standard equation of mathematical physics, which is what this is, in which the solutions are the so-called spherical vessel functions. That's to say, one can say that the solution of R as a function of radius R is one true type. The spherical vessel functions are either called J L of rho or else they're called Y L of rho versus the studio convergence. In fact, for a given, see L is a parameter of a regular Schrodinger equation. That's something to keep in mind is really a different radial Schrodinger equation for each value of the angular momentum when you see it here. So the solutions are parameterized by L as well. And this being the second order differential equation, it naturally has two linear and independent solutions, which in mathematics are called as J L and Y L. These are two types of radial vessel functions. Now, if you examine these radial vessel functions in the limit in which rho goes to 0, so it's small rho, you'll find that J L goes like rho to the L power, whereas the Y L goes like rho to the minus L minus 1. And I explained in the previous lecture why this type of behavior of rho to the minus L minus 1 or radius to that power is not acceptable physically. So from a physical standpoint, free particle solutions are only the J L type solutions. And if we do that, then I'll make use of this space over here. If we do that, we now have the solution for the free particle just to plug in the value of J L and what we get is that psi is equal to J L of k rho times the Y L M of theta and pi. And the energy is connected with a k parameter through this substitution we made over here. So I'll just connect you between wave number and energy for free particle. To get a feel for these solutions, allow me to plot the J type spherical vessel functions for you. It looks like this. Let's make this as a function of the dimensionless variable rho and I'll plot J L of rho. For the case L equals 0, it starts off at a constant value, so this is J 0 here, rho. And then it goes down and does some oscillations like this. And the empty oscillations guy's office goes out. For J 1, it rises linearly at first and then reaches a maximum and then oscillates like this, goes on down. For J 2, it's quite valid at first, reaches a maximum and then oscillates on down. I'm probably not drawing this very accurately. But for each increase, so this is the J 1 and this is the J 2. As the L increases, the function lays down ever more flat at the origin. As I explained last time, rises to reach a first maximum. This is really acoustic of the WKB theory. And then after that, there's a oscillation which obviously amounts to a classically forbidden region. The magnitude of the oscillations dies off roughly. It dies off as 1 over rho, 1 over the radius that should go on. And this is the sketch of the, there's just some idea of what these spherical vessel functions look like. I will sketch the Y type solutions. As I mentioned, they're not acceptable as free particle solutions. You have, the particle is free everywhere. But that doesn't mean they're not relevant for quantum mechanics. Because in scattering theory, we have a situation where there's a potential but it has a limited range. So down near R equals 0, the particle is not free. But at a large radius where the potential becomes negligible, the particle is free. And we often need to expand the wave function out of R radius in terms of free particle solutions. Which means, which means spherical vessel functions in YLM. If you do that, then you actually need to Y's. Because out there, as long as you stay away from the origin, that's where the singularity of the Y's is. That's why they're not acceptable if you push all the way down to the origin. But if you don't do that, then they are acceptable. Not only are they acceptable and actually necessary to get a complete set of wave functions in the exterior region. So in other words, they're important in scattering theory. We'll see that come up later next semester when we talk about scattering theory. Alright, so this is the case of the free particle solutions in spherical coordinates. There is, by the way, obviously going to be a transformation. If we have, here is an energy eigenfunction in spherical coordinates with some given energy E. And then over here we've got plane waves, which are also of given energy in the same energy as we make the momentum right. It's obvious that this wave function must be representable as a linear combination of those wave functions. I won't go into that now because it's just special function manipulations to convert one into the other. But we'll actually need that later on in the scattering theory to convert the plane wave solutions into central force type solutions. Alright, now, at this point I'd like to return to this question about the infinite mass of the fourth center, which we needed in order to make this two-body problem into a one-body problem. This is unrealistic. Of course, real force centers are created by particles and other things. So to create a more realistic situation, we have to include the dynamics of the fourth center itself. Let's say there's two of them, they're particles, and that masses M1 and M2. Now we don't have the option of attaching an inertial frame to either particle because if we do it, it won't be inertial. So we need to just choose an arbitrary inertial frame here, x, y, and z. We can call this the lab frame. And then we can draw the radius of vectors in and give us the positions of the two particles called this R1 and R2 vectors like this. And I'll also draw the volume of the two particles. In the case of central force motion, the force between the particles lies along this line, as I drew it. These two particles have a center of mass and it lies somewhere on the line joining them. If the masses are different, it won't be in the center. Let's call the center of mass position, let's give it a name, call it the vector capital R, so it looks like this. Now, this one becomes a two-body problem. And so the Hamiltonian for this is the sum of the kinetic and potential energies. So the kinetic energy has to be the kinetic energies of particle one plus the kinetic energies of particle two, which are expressed in terms of the moment Rp squared over 2 and 1 plus p1 squared over 2 and 1 plus p2 squared over 2 and 2. As far as the potential energy is concerned, we will assume that it's a function only at the distance between two particles, which I'll write this way as the absolute value of the relative vector between the two. So this is the central force motion with two particles now. And we'd like to solve the Schrodinger equation for this. This system has six degrees of freedom. It's the x, y, and z coordinates of both particles. The configuration space is really six-dimensional space now. That means that the wave function, which I'll call capital Psi, is a function of the coordinates of both particles, R1 and R2. So the Schrodinger equation is going to be h-sized into d-sized, which depends on both of these coordinates of R1 and R2. In order to solve the Schrodinger equation, the way to do it is to make a transformation of variables. The transformation is motivated by the physics. We first of all introduce one variable, which is the center of mass, called capital R, as I've done here. The center of mass is just the mass weighted positions of the two particles. So it's M1 times R1 plus M2 times R2 divided by the total mass of the system, which I'll call capital M. So let me put this over here. Capital M is defined in the M1 plus M2 like that. And in addition, let's also introduce the relative position vector between the two systems. Let's say going from M1 to M2 like this. And I'll call that vector lowercase R without any subscripts on it. So lowercase R is defined as R2 minus R1. Now this amounts to an important transformation in the configuration space, as you can see. While I'm writing this down, let me also write down the inverse transformation, which I'll need at this moment. R1 is equal to the center of mass position capital R minus M2 over capital M times the relative vector R. And R2 is the center of mass position capital R plus M1 over capital M times the relative vector R like this. You can easily see for yourselves that if you let capital M1, let's say, go to infinity, you go back to the earlier situation we were talking about before, where one of the masses was infinite. Alright, so this is the transformation on the position coordinates forward and backwards to go to the opening coordinate system. We like to transform the Hamiltonian over to the new coordinates. The first obstacle we run into is the Hamiltonian has momentum in it, as well as positions. So how do we transform the momentum, P1 and P2? Well, we'll do it like this. In the configuration representation, P1 is the differential operator minus i h bar e dr1. And P2 is similarly minus i h bar e dr2. That's what it is in the position representation acting on wave functions like this side here. Now, let's define a momentum called a capital P, which is to be conjugate to the center of mass flow in capital R. And it will be defined as minus i h bar d d capital R. Let's make that definition. Well, this can be expressed in terms of the other two momentum, but it's just by using a chain rule. This becomes minus i h bar multiplied times d dr1 with respect to capital R times d dr1 plus d dr2 with respect to capital R times d dr2. These are really vectors here. So to get it right, you really need to have a, so this thing is really a matrix. You can track the r1 there with the r1 there. If you wanted to put i's on it like this and an i there that introduces some on i, I can do that to clarify what this means. But you can also think of this as a matrix, multiple final vector. Well, in any case, these derivatives, d r1 with respect to R, are easily evaluated by looking at these formulas. d r1 with respect to capital R, you see, is just unity over the identity matrix. So this thing just becomes the identity matrix. And this thing here, d r2 with respect to capital R, it's the same story. That's also the identity. And so the result means this turns into minus i h bar simply times d dr1 plus d dr2. This is just using the chain rule. However, we now see that's the sum of p1 plus p2. And so we get a result which is nice enough that I'll put it on the line by itself. This is that the momentum conjugate in the center of mass transition is actually the sum of the momentum of the two individual particles. Now, let's do something similar for the momentum I'll call the lowercase p, which is conjugate to the relative position. That's the same thing as we define that as minus i h bar and d d r with respect to the relative position. And now you can go through the same kind of a chain rule here. And the difference is that the matrices that appear here are going to be the derivatives of r1 and r2, not with respect to capital R, but rather with respect to small r. So we have coefficients minus m2 over m and plus m1 over m. The result is that the momentum conjugate to the relative position is minus m2 times p1 plus m1 times p2 divided by the total mass dealt. And this then gives us the transformation of the momentum. Work out this way. Now, once we have this, then we can solve these equations for the old momentum to the function of the new. It's just algebra to do it. You'll find that p1 is equal to m1 over capital M times the capital P and then it's minus of the lowercase p. And then you find that p2 is equal to m2 over capital M times capital P plus lowercase p. These are the inverse transformations. And so now we've got a complete transformation between old variables, old position and momentum variables, r1, r2, p1, and p2. And new position momentum variables, capital R, center of mass and relative position and momentum. And having that, we can now take the Hamiltonian and transcribe it into the new operators. Before I do that, let me mention that the commutations work by looking at the commutation relations of the new variables amongst themselves. In the first place, we have the commutation relation of capital R i with capital P i. These are the center of mass position momentum is minus i h bar times delta i j. The reason for that is that we define the capital P as this derivative with respect to R and some usual rules of computing the commutator's work out in the usual way. Likewise, this is also equal to the commutator of the components of the relative position and momentum. Lowercase r ij, lowercase r i with lowercase pj is the same commutator. This should have been a plus by the way, plus i h bar delta i j. These are the main commutators here. And then all other commutators that you can make by combining these four sets of variables are equal to zero. So it is the same structure of commutation relations we had initially when we r 1, r 2 and p 1 and p 2. The common structure of commutation relations is preserved. This is what you call a canonical transformation in classical mechanics. You could call it that in quantum mechanics too. We saw another example of canonical transformation dealing with the probably charged particle in the uniform magnetic field. This is actually a little bit simpler here. But in any case, one of the consequences of this is that any function of the center of mass variables, the capital variables, will commute with any function of the relative variables. So that's a simple rule that's easy to remember. Now, having that transformation, let's now transform the Hamiltonian to the variables. And if we do that, it's some algebra to transform the kinetic energy. We're just plugging these formulas here into that Hamiltonian out at the top of the board. And what you do, here's what you find, is the Hamiltonian now becomes the capital P squared over twice capital M, the square of the total limit of the system divided by twice the total mass, plus lowercase p squared over twice mu, or mu is the deduced mass, plus the potential energy, which is B and B of the magnitude of r 1 minus r 2, which is the magnitude of the relative position vector, which I'll call r like this. The yield here of the reduced mass is defined this way. It's 1 over 2, it's 1 over 1, it's 1 over 22. That's what you call a harmonic mean of the two masses of the two bodies. This just comes from the algebra substituting there. I mentioned one thing about reduced mass. If you're in a situation where one particle is much more massive than the other one, then the reduced mass is approximately the same as the lighter of the two particles. So for example, in hydrogen, when the proton is much heavier than electron, the reduced mass is almost the same thing as the electron mass. However, if two masses are equal, when the reduced mass is equal to the mass of unit particle divided by 2, it becomes comparable to the masses. Anyway, just some rules about reduced masses. So we have two terms here. Let's call it the first term. The first term depends only on the center of mass variables, so let's call it HCM. And then the second term depends only on the relative variables. Let's call it H-relative. So the total Hamiltonian breaks up into a center of mass, a term plus a relative term. Moreover, these two terms commute with one another because they are functions of the different sets of variables. And that means that possessed simultaneous eigenstates. You can diagonalize these two things separately. And once you've done that, you'll automatically have an eigenstated H because H commutes with these two as well. So this is a convenient decomposition. As far as the wave function is concerned, we started out by writing it as a function of R1 and R2. But by doing the change on position variables, which is given right up here, we can re-express it as a function of the center of mass core against the relative coordinates. And if we do this, then the HCM acts only as the capital R, or the H-rel acts only as the lowercase r. And the result is the Schrodinger equation is separable. And it means that solutions exist in the form of a product. Let's call it capital 5 and capital R for the center of mass part and lowercase i of lowercase r for the relative part. I don't mean that every solution for Schrodinger equation has this form, but I mean there exist eigenfunctions of this factored form. If you put this in here, you'll find that just by applying HCM, which acts in the first factor and H-rel acts in the second factor, what you'll find is that you have these breaks up into two eigenvalue problems. HCM acting like 5 is equal to the center of mass energy multiplied by, and you apply the H-relative acting on the side, because that's the relative energy multiplied by 2 times 2 eigenvalue problems like this. And if you solve these, then the total energy of the entire system, the entire Hamiltonian, is just the sum of the center of mass plus the relative energies. The center of mass Schrodinger equation is particularly simple because it's a free particle problem. This corresponds to the classical fact that the center of mass of a system like this moves on a straight line of constant velocity, and so it appears here also. And thus the solutions for these wave functions capital 5R could be taken to be any free particle solution. We could, for example, take it to be a plane wave. This is probably the simplest solution. You need to be on a capital P, got into the capital R over H-bar, and then divide it by 2 by H-bar to the 3-has if you want to normalize it. This is getting the plane wave solutions to the center of mass motion. But of course we can also do this in spherical coordinates in which case you'd have spherical vessel functions. You get to choose how you want to solve this free particle problem. Having done that, we turn our attention to the relative Schrodinger equation, which is down here. And what we see about the relative Schrodinger equation is that it's an effective one-body problem that has the same form as the one-body problems we considered earlier when we considered the force centers having an infinite mass. The only difference is that the variables in question and the positions in moment and in coordinates are no longer the coordinate in particular is no longer the coordinate of the particle relative to an inertial frame. Instead, it's the relative position between the two masses. And in fact, to refer back to these equations up here, you see it's actually a function of the positions of both particles. Likewise, the momentum which occurs here is also a function of the momentum of both particles. It's worth all remembering that. If you're dealing with hydrogen, for example, you're talking about the radius, this is really the distance from the proton. And as operators, they are, they are operators that satisfy the usual operator, the polar commutation relations. But physically, they're actually functions of the positions and momentum of both the electron and the proton. Well, in any case, mathematically, it has the form now of a one-body problem. Oh, another important change is that the mass of the particle, which is moving in the field, is not ever replaced by the reduced mass. Except for those changes, it becomes a one-body problem, just like we were talking about in the movie. So solving one-body problems is enough to solve two-body problems when the force is the central force. All right. Plus a three-particle solution for the center of mass, which is pretty easy. Okay. So now what I'd like to do is to turn to some examples of central force motion. Let me begin with a simple example, which is that of a rigid rotor. I'll talk about it classically first. So we get some ideas about it in a picture of it. A rigid rotor we can think of as being two masses, let's call it L1 and M2, which are joined by a massless rigid rod. In other words, it has a fixed length, let's call it R0 connected with 2. This is obviously an idealization because there are no rigid rods in nature. Let's say somewhere here is the center of mass positioned like this. And if we compute the moment of inertia of this rod and the rotation is about the center of mass, what you'll find is that it's the reduced mass mu multiplied by the square of the distance between two particles. That's a classical calculation of what you do to see if this is true. Also, the motion, as we know, is going to be just rotating. So the two particles rotate about the common center of mass and they're going to do so in a circle, which therefore lies in a plane. The angular momentum factor will be perpendicular to the plane. And so will the angular velocity be in the same direction. And so, just by elementary mechanics, the energy in the system is going to be one half of the moment of inertia times the square of the angular velocity. We also know the angular momentum is the moment of inertia times the angular velocity. And if you're writing energy in terms of L instead of omega, it turns into an L squared divided by twice the moment of inertia. Now, here's an interesting fact. If we take this expression for the moment of inertia in terms of the reduced mass and plug it in, we get L squared over twice mu times r not squared like this. And what this shows you is that this, as you see, is actually the centrifugal, this is actually the centrifugal potential. I remind you the centrifugal potential is physically a part of the kinetic energy. In fact, physically, it's the angular part of the kinetic energy. For a rigid rubber, there is no radial part because there's no radial motion. So this is all the kinetic energy there is. And I'm not including any potential here. So in fact, that's the total energy. It's the effect that comes to the Hamiltonian in the system. All right. Now, the configuration of the classical rigid rubber is determined by, let's say, a unit vector that points from, let's say, let me swap these things out one and two so that I have a vector that points from one to two going out like this. And this has the direction between which we specify the spherical coordinates by theta and phi. Basically, if you tell me what theta and phi are, it's functions of time. You've got the complete evolution of the rigid rubber. This suggests that when we go to quantum mechanics, that the wave function should be a function of the spherical angle stated in phi. It should be a wave function on the unit sphere. Likewise, this expression, either one of these two expressions with the classical energy, suggests that the quantum Hamiltonian should be elsewhere over twice the moment of inertia, where L is now the angular momentum operator, which we know how it acts on functions on the unit sphere. Particularly, it brings out, well, we know the eigenfunctions of YLM. So the energy eigenfunctions should be YLMs of theta and phi. And if we let this Hamiltonian act on that, then the L squared is going to bring out L times L plus 1 h bar squared. So this is the energy, then, that the energy depends on the quantum number L and it's L times L plus 1 h bar squared divided by 2i, which if I replace i by a reduced mass R0 squared, you can see, again, it's really the same as the centrifugal potential frozen at the radius R0, the fixed radius. Okay, so this is some of the quantum mechanics of the rigid rotor. There's a degeneracy you can see because the energy doesn't depend on the magnetic quantum number. This is the 2L plus 1, this is 2L plus 1 called the degeneracy because of the possible values of the magnetic quantum number. And the usual reason is that the magnetic quantum number indicates the orientation of the system and the energy doesn't depend on that. All right. Now, the rigid rotor is an idealization and we've been making a lot of guesses in terms of writing down these quantum mechanics relations. They're kind of obvious, but they're still guessing. If you wanted to make this more rigorous, you'd have to make the rigid rotor as a kind of a limit of a central force problem in which you end up finding potential and it keeps the two particles at very close to a constant radius. I'm not going to bother you because it's not what I want to do today. But I just want to say that this involves some guesses to what we've done. Yes. So when you go down like this, you have to do know that it's dependent on the size of the atom, 8 and 5. But why do you know that it's the YLNs? Because the energy eigenfunctions, the eigenfunctions of L squared are the YLNs. Okay. And then how can you say or how do you know that it's independent of N? Because the energy only depends on the L1 number and not on the N1 number. Because the Hamiltonian doesn't involve the ALZ. That's because it's rotationally invariant. This is the normal situation with rotationally invariant Hamiltonians. All right. Now, rigid rotors are idealizations of non-existent nature, but there are biotomic molecules which exist in nature that provide nearly the same features as a rigid rotor. So I'm going to tell you about a diatomic molecule as examples of central force motion. So a particular diatomic molecule, take a specific example, look at the carbon monoxide molecule. The mass of the carbon, well, first of all, the atoms, the carbon monoxide atoms are composite particles. They're not elementary particles. That's because they're made up of sub-components electrons and nuclei and so on. But it is possible in quantum mechanics to treat composite particles as if they were single particles under certain circumstances. I won't go into that now, but we'll just go ahead and do that. As far as carbon atom is concerned, the mass of the carbon is about 12 times the mass of hydrogen. The mass of the oxygen is about 16 times the mass of hydrogen if we take the normal isotopes. The mass of hydrogen is about, something like 1800. I'll make it very roughly about 2,000 times the mass of the electron. Let me also talk about the reduced mass of the carbon oxygen system, which I'll call u. This is about 7 times the mass of the hydrogen if you work it out. Momentary net by 2,000, we get about 14,000. I'm going to make a very rough estimate of this and just replace it by about 10 to the fourth times the mass of the electron. I need to eat off the lowercase sample standard of the mass of the electron in the following discussion. In the first place, there is a large disparity between the reduced mass of the system and the electron mass. It's a ratio of 10 to the fourth. Let's give this already some small parameters, which are all right like this. First of all, u is the reduced mass of the carbon-oxygen system. If I divide this by the electron mass, I get, let me do it the other way, put the small number on top, and then divide it by a mu. This is about 10 to the minus 4, a small parameter. Another small parameter that we'll run into is the square root of this, as we'll see. This is about 10 to the minus 2, and yet another small parameter will be the fourth root of this, which is about 10 to the minus 1, or about 10%. We'll see these small parameters occurring. We're going to treat the carbon-oxygen molecule as a central force problem, where the two atoms interact by means of a central force potential. I'm going to tell you something now about the potential between the two. I'll start by drawing a picture of it. If we draw the potential like this as a function of radius r, and here's v of r, so this is the carbon-oxygen potential. It has a gentle structure like this. It rises very sharply. It's small-radiated. It decreases down and creates a well, and then the well has a shallow tail that goes out like this. In fact, it was a curve of the same shape, which applies to essentially all biotomic molecules. The physics of this is that at large distances out here, the two atoms, the electron clouds and the two atoms polarize each other to create dipole moments. It's a dipole interaction. It's only a prosthetic interaction between the two atoms, and that's what causes the attraction that pulls them in, just like when you're pulling your hair up. However, at short distances, we start to feel a repulsive force, and that arises when the two electron clouds start to overlap substantially, because you're trying to force electrons into the same region of space, just like taking a particle in a box and keep on adding more and more electrons. They have to go on the higher energy levels. And so the repulsion which occurs here is in the following sense is being due to the pressure of a degenerate electron gas that's being squished into a smaller and smaller volume. And it's a competition between the two these two effects, which gives you the well of the bottom. Now, there's no simple analytical formula for this curve, so I don't even attempt to write one down, but there are some obvious parameters. One of them is the radius of the equilibrium position. Obviously, this is the equilibrium position, and the carbon and oxygen molecule can undergo vibrations around this equilibrium position. This is the potential in which it's moving. So one obvious parameter when it's in dotted lines here is the radius which I'll call r0 out to the middle of the potential well. Another obvious parameter is the depth of the well which I'll call v0 like this. In the case of the carbon and oxygen molecule it turns out that v0 is about 11 electron moles and it turns out that r0 is about 1.1 angstroms. Other molecules have different numbers but very roughly they're all in the same order of magnitude. There's some variations in it, but very roughly this is fairly typical of numbers. All right. Now to understand this in a slightly more quantitative way there's no, as I said, analytical formula for this. So I'm going to rely on order of magnitude estimates using some dimensional analysis. As I pointed out the forces between the atoms are determined by the properties of the electron clouds alone, whether it's or dipole-dipole interactions. Now the physics of the electron clouds is governed by three fundamental constants. The charge of the electron, the mass of the electron, and the plane's constant h bar. The speed of light is not near because the forces between the electrons, the electrostatic forces are electrostatic and not electromagnetic. Magnetic forces are not important. There's no seclusion to one's law. Another thing is that the velocities of the electrons are essentially non-wilderistic but very small comparisons to the speed of light. So they're already relative distinct effects either and that's why, at least not the lowest order. And so that's why c is not here. So given that we've got just these three constants to work with it becomes possible to take these two dimensional analysis and to construct a fundamental length a fundamental distance time, the length is the same as distance, length, time, energy, etc. And if you do this here's what you get. So let me call the fundamental length let's call the fundamental length a0 and this turns into h bar squared divided by e squared. And if you work this out the variably it's about 0.5 angstroms. This is the only this is the only quantity of distance you can work out by using these three physical constants. And as far as energy goes there is let me call it kgon as the energy this is contributed to the force divided by h bar squared and this turns out to be about 27 electron volts. If we talk about velocity it's called v0. This is equal to e squared over h bar. I'm going to take this e squared over h bar and write it as e squared over h bar c times c. I said a moment ago the speed of light is not on the list but I have modified it and modified it so c's really not there even now. But it's still useful to write it that way because e squared over h bar c is the fine structure constant and you can see that this turns into alpha times c which is the same thing as 1 over 137 times the speed of light. So this confirms what I said earlier that the typical velocity of the electron is small compared to the speed of light it's less than 1 percent of the speed of light. It's fast and it's not relativistic. There's a frequency that comes in fundamental frequency and this is basically the energy divided by h bar so this is m e to the fourth over h bar q the reciprocal of this gives us a fundamental time t zero which is 1 over omega zero which is equal to h bar q over m e to the fourth and this is about 2 times 10 minus 16 seconds it's a pretty it's a pretty small amount of time okay so these are these are the quantities which enter now these fundamental units of physical units of distance energy and so on in fact they're exactly the same ones that arise in the hydrogen atom and the reason is the same is you've only got these three physical constants to work with so for example this a zero is the same as the Bohr radius over here is actually twice the ionization potential of hydrogen which is 13.6 electron volts this velocity d zero is basically the velocity of the electron in the ground state of hydrogen and so on so these are basically hydrogen numbers that are used here so just on the conventional analysis along we would expect that the depth of this well the carbon oxide molecule as well as the radius should be in the same order magnitude of these numbers given here if you look you see that p zero the depth of the well is something like 40% of k zero so it's not too badly off and you see that the radius is about twice, a little more than twice the Bohr radius so within a factor of two or so actually not too bad it gives you the right order of magnitude you know if you want to change the problem around the management you may need a molecule like carbon oxide but instead of using electrons let's say mu minus some muons which have the same charges of electrons that are heavier and you want to know how big would the molecule be the dimensional analysis would just tell you that you just need to replace the mass of the electron but the mass of the muon and the answer will come out now that means you'd like to do some things because it would be faster and so on and still you need to check to see that it's still not all the distance but roughly it would be the correct answer nobody makes muonic molecules like that there's too many muons required but they do make muonic molecules that have one or two muons replacing electrons and there's actually quite some interest in that alright there's no such thing as muonic carbon and oxide as far as I know alright alright so so this is the dimensional analysis now given this we can start to do some other things alright you can see that the that the vibrations of the molecule are going to be represented classical by motions in this well and therefore quantum mechanically they'll be quantized and there'll be energy levels in this well these will be the vibrational energies also you can see that if we approximate the bottom of the well by parabola that the ground state and hopefully the first few excited vibrational states can be roughly described as a harmonic oscillator so let's talk about this harmonic oscillator let's make a harmonic oscillator approximation to this well that would mean that there's an approximate potential here we look like this there would be one half the mass mu which has to be the reduced mass of the carbon monoxide system which is like put up there it's 10 to the 4 times the electron grass times the vibrational frequency square times the radius minus the equilibrium radius square this is the harmonic approximation now we know what the reduced mass is but we don't know what the vibrational frequency is that depends on the curvature at the bottom of the well but it's possible to estimate the vibrational frequency and we do this in the following way we'll say that as a as a guess or a rough order of magnitude we'll say that when the when the distance between the atoms moves a distance which is comparable to A0 which is comparable also to R0 and that far we must be up to the point where we're at the association energy so in other words let's do this let's equate one half mu of a mega vibrational square times A0 square to the square of the deviation away from equilibrium equals A0 which is an estimate of the dissociation energy of V0 and I'll just ignore factors of 2 and as long as I ignore factors of 2 let's put away the one half here as well again it's just a potential analysis but you see it now allows us to solve for a mega V in terms of things we know so we find a mega V then it's equal to B squared it's equal to 1 over A0 times the square root of K0 divided by the reduced mass mu like this now let's plug these things in this becomes equal to H bar squared over MB squared or excuse me it's 1 over A0 it's equal to B squared over H bar squared and then we get the square root of K0 over mu so there's the K0 so this becomes M over mu times E to the 4th over H bar squared and so you can take the square root of E to the 4th over H bar squared and you get E squared over H bar multiplied here so you get M E to the 4th over H bar Q times the square root of M over mu but the M E to the 4th over H bar Q is the polynomial frequency we have here since the left hand side is dimensions of frequency the right hand side has to be equal to this omega not times something dimensionless and in fact here's what we find so we find the mega V is equal to the square root of M over mu times the omega not so this is an important relation the square root of M over mu is the square root of the ratio of the mass of the electron of the mass of the nucleus which is really the reduced mass of the 2 in the eye which as you see is about 10 to the minus 2 roughly so this means that the vibrational frequency is about 10 to the minus 2 times omega not and this basic factor of 10 to the minus 2 roughly speaking applies to most diatomic molecules some variations but that's the basic idea it's a rough order of management this is how it works this frequency of omega not as I mentioned earlier is the same as the orbital frequency of the electron and hydrogen this typical orbital frequency of the electrons in the carbon and oxygen molecules and so H bar times this is the typical energy of a photon which is emitted in an electronic transition in an atom these electronic transitions as you know are in the optical or also in the ultraviolet regime the 13.6 electron volts for hydrogen and the ionization energy is in the ultraviolet some of the other transitions to hydrogen are in the visible part of the spectrum but what this shows is the vibrational energies are downed by a factor of about 100 the vibrational frequency is the F4H bar of omega which is the H bar of omega B which is the delta E on a vibrational transition of the molecules downed by a factor of 100 it's a square root of mass ratio what that means is the vibrational transition of the frequencies or the photons that are emitted and absorbed and changing the vibrational state are not in the optical they're down in the near to middle infrared this is important for example in the greenhouse effect optical radiation in the sun well the sun puts out most of its power in the optical range of frequencies and the atmosphere is obviously transparent to the light but when the light hits the ground it warms up the ground with something like 300 degrees Kelvin which then radiates how a formal radiation is about that temperature it's not exactly a black body but it's the idea that's the right temperature range and that as it turns out that's in the infrared so the radiation that has to cool the air off get back out of the space has to pass back through the atmosphere but as it does so it's in the right frequency range of the vibrational transitions of molecules now part of the monoxide is not one of the greenhouse gases this part of the monoxide is the usual culprit that's a triatonic molecule which I know I want to talk about but still the basic physics applies that it's vibrational transitions of those molecules is what is it swerving those infrared photons and then they get reemitted as well in the atmosphere of course nitrogen and oxygen are also diatomic molecules in the atmosphere and they also have a vibrational spectrum it's very similar to this there's a question about why those infrared photons don't interact with them in other words why are oxygen and nitrogen greenhouse gases as well the reason for that is those are what you call homonuclear diatomic molecules those atoms are identical they're identical the result is they don't have any electrical dipole moment there's no charge separation carbon monoxide has got a nice it's one way or another I'm not a chemist so I don't know which way it is but there's a charge separation on it and this produces a dipole moment which strongly interacts with radiation at the right frequency with O2 and N2 that doesn't happen because they're neutral that's why the O2 and N2 are not greenhouse gases so the fact that although the N2 and O2 have vibrational transitions in the same range as just CO2 but it's the other fact that there's no interactions between the photons they don't interact with the electromagnetic field but the infrared photons produces an electric field that vibrates at the right frequency to stimulate the bonds the nuclear bonds of molecules such as CO2 alright so that's a little bit of the physics of the greenhouse effect the I mean if nitrogen and nitrogen were greenhouse gases it would be a trouble because it would be like Venus which is instead of most of the CO2 so it's really hot okay alright so this story's not quite over yet there's other interesting things we can do if we want to calculate the number of vibrational eigenstates for dissociation one of the ways of doing that is just to take the dissociation energy which is v0 here which we'll estimate as an order of magnitude by k0 so let's take k0 and then let's just divide it by h bar we'll make it v which is the vibrational energy so this is the separation between the vibrational energy levels if you work this out which you find the square root of u over n it's the inverse of that square root so it's an order of order of about 100 and the result is the typical diatomic molecules have about 100 vibrational states before they break and dissociate alright now this potential that I drew here is only the true potential and I've been ignoring the reciprocal potential before I get to that though there's one more thing to say the vibrational energy the ground state vibrational energy is of course one half h bar who may not be the vibrational frequency and from many molecules this is common to somewhat argument in the room temperature so at ordinary temperatures the diatomic is common for diatomic molecules if we mostly in the ground state vibrational ground state well perhaps it's just in the low line problem numbers, vibrational ground state, you get an idea of how big the vibrational amplitude is so that's fairly easy, we go back to the so I call this delta r here it's called this delta r as the width of this it's a Gaussian wave pattern you see the width of this Gaussian wave pattern what is it equal to? well if we go to the the story of harmonic oscillators you look in the notes on harmonic oscillators and you look at the characteristic length of scaling for harmonic oscillators it's sort of a h bar over h omega if you look in the notes you see that and this delta r is really the same thing except the h bar is the same but the mass is replaced by the reduced mass and the omega becomes the vibrational frequency omega v like this now this has dimensions of length so it's got to be proportional to our a0 times the dimensional slumber and if you work it out that dimensional slumber is the mass ratio n over u to the one quarter power times a0 which is very roughly something like 10% of the more radius and so very roughly the idea is that in the ground vibrational state the molecule is undergoing zero point oscillation which is magnitude is about 10% of the equilibrating radius and in that sense it is very close it is roughly a rigid roller it's got some small vibrations the radius doesn't change very much and I see I didn't hear the bell it's so interesting well it's too bad I've just got a couple more comments to do to finish this story of molecules and then we'll go on to a hydrogen atom which we'll talk about in a minute maybe that's all