they needed to. All right. Let me just remind you that last time we were talking about the dynamics of molecules. And I think at the end of the hour, we came to the conclusion that if the molecular vibrational state is in its ground state, it's roughly a harmonic oscillator down to the bottom of the well to the extent that the bottom of the well can be approximated by a parabola. There's a harmonic oscillator approximation there. This approximation is obviously only valid when the quantum limit is small, when you're here at the bottom of the well, when you get higher up, it's going to be less accurate. But to repeat, we came to the conclusion that the last time that the measure with this wave packet is an Gaussian wave packet naturally to the ground state, is in the order of magnitude about what I call delta r here, is of the order of magnitude of the mass ratio, the m over mu, to the one-quarter power multiplied times the characteristic length of the system, which is the Bohr radius. And the equilibrium length is also the same order as the Bohr radius. So this is the same order as the equilibrium length of r0. The mass ratio and the question here is the ratio of the electron mass to the reduced mass of the diatomic molecule mu. This reduced mass is the reduced mass of the two nuclei. And therefore, it's a typical nuclear mass. In the case of carbon dioxide, the number turns out to be very roughly about 10 to the minus 4. And this same number very roughly applies to most diatomic molecules, a small parameter problem. So in particular, raising this to the one-quarter power, we get 1-10. And we come to the conclusion that the vibrational amplitude of the diatomic molecules and vibrates is roughly about 10% of the bond length as a rough estimate. And as a result of the diatomic molecule, roughly speaking, does behave as a rigid rotor, which we started talking about before we got in the molecules. All right. Now, by the way, as something else we concluded, this is the frequency of vibrations. We call them the negative p. This is the order of magnitude of the square root of the mass ratio to the one-half power, the square root of the mass ratio times the typical frequency for electron motion, which I didn't put in here. And it's a characteristic of the frequency of the electron motion in time limit to the minus 10. 2 times 10 minus, well, the inverse of this times, 2 times 10 to the minus 17 seconds. And the vibrational period, as I say, is doubted by a factor of 100, meaning that the photons emitted in the vibrational transitions are about 100 times less energetic than those that are emitted in electronic transitions. Electronic transitions correspond typically to the visible light or ultraviolet range. And so the vibrational transitions are near to be at infrared. All right. Now, I think that covers some things that I said in the last lecture. Now, this picture that I've drawn here is actually only for the case where the angular momentum is equal to 0, because I've just drawn the true potential and not included the centrifugal potential. Allow me now to include the centrifugal potential and the two differences that make. So the variable potential, of course, is strictly positive. It's L times L plus 1 h bar squared over 2 q r squared like this. That's just what it is. Now, if we're near the vibrational ground state, then as I just explained, the range of vibrational motion is rather small, roughly small, compared to the overall distance between the two molecules. And if that's so, it's a reasonable approximation to replace the centrifugal potential just by its average value or its value at the middle of the vibrational wave function. In other words, just to replace the r square here by an r naught is reasonable to do that for the ground state vibrational wave functions. And if we do, the centrifugal potential just becomes a constant, which is added to the radial wave equation. Now, that constant won't change the wave functions. It'll just change the energies. It changes the energies by the addition of this constant. And so it now becomes possible to write down rather easily what the energies are. They're a function of both the vibrational quantum number in and the angular momentum.  So this is a very rough expression. It's rough, but it gives you a pretty good idea qualitatively of what the molecular spectrum looks like. All right. Now, if the temperature is not too high, the vibrational motion will be in the ground state. For many molecules, this means ordinary room temperature for diatomics like oxygen and nitrogen and so on. And if that's so, then the vibrational energy is just a constant. And if we're interested in how the energies depend on angular momentum, it just comes from this final term. It has an L dependence, of course. And in fact, what you see is that it's really the energy that's rotational energy of the rigid rotor. We recover our rigid rotor results. The reason this is a mu times r naught squared is the motor of inertia of the rigid rotor. So the molecule does behave like a rigid rotor. And we recover the results, early results, we guessed at regarding a rigid rotor. What about the energy scales of the rotational degrees of freedom? They go as a quadratic function of L, the angular momentum one number. If we take just the lowest one, L equals 0 and L equals 1, let's just take those two, and we look at the number at all times L plus 1, and you get 1 times 0 is 0, and then you get 2 times 1 is 2. Those are the numbers that appear up here. So I'll be to write delta d r o t for the, let's say, the difference in the two rotational energy levels between L equals 0 and L equals 1. If we do, then this is 2 times, for the all plus 1, 2h part squared divided by 2 mu r naught squared. And if we want to estimate this as an order of magnitude, we'll care about the 2s and for the r naught, we'll replace it by a naught squared, because this order of magnitude, that's what that is. Now, this must be proportional to the energy k naught, which is a characteristic energy of electronic transitions. When you just do the algebra, what you'll find is, this is k naught multiplied times the mass ratio of the mass of the electron divided by the reduced mass of the diatonic. And so this is a factor of 10 to the minus 4. What that means is that the typical energy scale for transitions of rotational degrees of freedom is down by another factor of 100 compared to the vibrational energies, which is determined down by a factor of 100 compared to electronic energies. So those three energy scales in the molecule was the rotations, which are much less than the vibrational energy scales, which are much less than the electronic energy scales. And typically, they're about a factor of 100 each. This means that the photons, which are emitted in corresponding transitions, have energies which are down by the scale of the same ratio. As a result, photons corresponding to rotational transitions in a molecule are down by about another factor of 100 compared to the vibrational energies. That puts them in the far end for rather than the microwave regime. Or to put this another way, if I'm trying to create an energy double diagram, if we say here's 0, then we start off with the vibrational energy levels. That's basically a harmonic oscillator theory. So the first one is 1 1⁄2 h bar made of v. And the next one is a 3 1⁄2 h bar made of v. Like this, these are the vibrational energies. But now, if we have the rotational energies on top of them, it produces a fine structure of levels like this that sit on top of that. It's purely additive, you see, in which the scaling in the little delta energies is about 100 times smaller than the separation between the vibrational. And so this is a typical, what they call, row vibrational spectrum of the molecule. All right. Well, I think this is all I want to say about molecules. If you take a chemistry course, you would have a lot more about them. This is the reason the basic facts about them. And for the problem of the department, no. Now what I'd like to do is to turn to a hydrogen atom as the next example of the central force potential. Potential here is dmr, let's put in a nuclear charge so it becomes minus z squared over r. It's a Coulomb potential. This includes not only ordinary hydrogen, but it also includes other single electron atoms, such as singly ion and isoleum, doubly ionized, lithium, all the way out to uranium, which is 91 times ionized. And they correspond z values 0, 1, 2, 3, up to 92, or more if you want. Actually, in recent years, there's been some experimental interest in how to restrict uranium in other heavy atoms for a test of quantum electron dynamics, where the electric fields are strong. So there's actually some interest in such single electron atoms like this. This would be called hydrogen-like uranium, as I would be described generally. These are hydrogen-like atoms, let us just say, similar electron atoms. Now if we write down Schrodinger equation, and I'll write this down in the radial Schrodinger equation in the second version I presented in the previous lecture, it's going to be minus h bar squared over 2 mu mu is the reduced mass here, d squared f dr squared. And then for the effective potential, there's l times l plus 1 h bar squared over 2 mu r squared for the centrifugal potential minus z squared over r for the true potential, multiply that as an energy times 1. This is the radial Schrodinger equation. And it's the second version here, the f of r. The mu that appears here is the reduced mass of the electron nucleus system. One thing not to be confused about is a mu in a ton of the molecules was large compared to the electron mass, whether they were about 10,000 times larger, because it referred to the reduced mass of two nuclei, two atoms. Here in this context, the mu is the reduced mass of the electron plus a nucleus system. And when you have two masses, one of which is smaller than large, then the reduced mass is almost the same as the small one. So the result is the reduced mass here is very good approximation is the same thing as the electron mass, not much difference. However, we do have the z squared which appears here. Now, this equation has a lot of physical constants in it, and we can get rid of them by introducing the characteristic of distances, energies, times, and so on. Very much as we did above in the case of molecules. So if I start with the distance, for example, let's call the distance scale A. This is a distance scale that can be created on the physical constants in this equation. It's basically the same as the formula of a box. Remember, these characteristic physical quantities came out of using just the three physical constants where e, m, and h, h bar for the electron, and this is the mass of the electron. The only difference here is that instead of the electron mass, we're going to use the reduced mass, which doesn't change things very much. But maybe more importantly, instead of e or e squared, we're going to use z squared. So to get our characteristic quantities in the context of hydrogen, we'll take those values of above, replace them in electron mass m by mu, a very small change. And replace e squared by z e squared. So the A here, as you see from looking up above, becomes h bar squared divided by mu times z times e squared. And we can see that this is approximately the more radius divided by z. So this is just from conventional analysis. But it shows that in a hydrogen like atom that the effective more radius, which I'll call A, as a contrastably more radius defined above, is basically developed by a factor of the nuclear charge. This makes physical sense as the nuclear charge increases. The positive charge is stronger, and it pulls the electrons in closer. But this is in particular how it scales. It scales interestingly proportional with the charge. Likewise, we go into the energy. We get an energy scale, which I'll call capital K. I did it with some K naught above, and then we replaced m by mu. So we get mu that in e squared goes into e to the fourth, goes into z squared e to the fourth divided by h bar squared. And this is very roughly the same thing as z squared times K naught. The only reason it's not exact is because I'm using the reduced mass for the electron instead of the exact mass. But in any case, this shows you that the energies of transitions in the electron hydrogen like atoms goes as the square of the nuclear charge. Already in hydrogen, the 13.6 ionization energy corresponds to a photon in the ultraviolet range of frequencies. And this goes up as K squared if you increase the nuclear charge. So the heavier atoms are talking x-rays, actually quite likely the x-rays for heavy atoms. Let me mention one more thing, which is the velocity that comes to any new velocity. Let's call it v here instead of v dot. It was e squared over h bar up there. This becomes z times z squared over h bar. This is the same thing as z alpha, z times the fine structure constant times the speed of light. And to take two cases, in the case of hydrogen, this is 1 over 137 times the speed of light, which means the velocity of the electron in hydrogen is essentially non-melted. I think it's less than 1% the speed of light. But in the case of uranium, it's 92 over 137, which is about 0.6 times c, which is something like about 60% the speed of light. And so we see that the characteristic velocity of electrons in the ground state and in that electronium is substantially non-melted this state. And as a result, the heavier atoms are not well approximated by the non-melted domestic Schrodinger equation. It's not really a very good first approximation to take for those heavy atoms. We'll develop this de-quantum mechanics in the second semester. But for now, since we're sticking with the non-melted domestic Schrodinger equation, our results will apply mainly for the atoms with not too large a value of c. All right. So this is just a dimensional analysis. Now, if we make a change of variables in a radial Schrodinger equation, let's say right r tilde is equal to r divided by this a. And let's say e tilde is equal to the energy divided by k, then it has the effect of getting rid of the physical constants. So we end up with an equation like this. It's minus 1 half d squared 1 half of d r tilde squared. It's exactly the same structure, l times l plus 1 divided by twice r tilde squared minus 1 over r tilde for the potential now is equal to e tilde times f. As you see, it just gets rid of the physical constants. Now, to proceed with this equation, or either version of this equation, it helps to get, before doing any math, to get some qualitative idea of what the energy levels will look like. We can do this just by looking at the potential energy curve. Let's take the potential energies. There's more than one, so I won't say which it is. If we take the true potential, this is the Coulomb potential, and it looks something like this. This is minus z squared over r. The potential goes to minus infinity as r goes to 0. And it goes to 0 as r goes to infinity. If we take this integral potential, it looks like this.  This goes to plus infinity as r goes to 0. And it goes to 0 as r goes to infinity, kind of the opposite of the true potential. At small distances, the r squared diverges more rapidly than we are. And so the positive, centrifugal potential dominates. So the true potential is positive at small radii. At large radii, both curves are going to 0, but the centrifugal potential goes to 0 faster, so the negative true potential dominates at large radii. So the curve has to go negative. So somewhere it must go negative, and then it has to, as some told out, go to 0 like that. This is what the U of r curve, the effective potential has to look like. So in other words, it has a well like this. This well will support bound states. And the result is from a diagram like this, you can see without much trouble that the hydrogen like atoms will have some number of discrete bound states at negative energy. But you can also see that when the energy is positive, that there is any number that can actually continue a positive energy theory. You can send in a particle, some energy will bounce off this effective barrier and come back up. And so the spectrum of the hydrogen atom, hydrogen like atom, in a complex energy plane, which is where you draw spectra, is first of all, there's a continuous spectrum going from d equals 0 to d equals infinity. And then you've got some number of discrete states at negative energies like this. Actually, it turns out that there's an infinite number of bound states. And they accumulate like this. There's an accumulation point as the energy goes to 0. So a more accurate picture for the bound states is this dot here, and then a dot here, and then a dot here, and then an infinite number of dots accumulate on the hydrogen after which just becomes a continuum. All right. So this is the qualitative features of the hydrogen spectrum. In this course, we're mainly going to concentrate on the bound states. They're simpler mathematically. But you should keep in mind that there are also these positive energy continuum states. And in particular, if you want to have a resolution of the identity using hydrogen atom wave functions, you better include the continuum because otherwise you don't get the identity just by sending over the bound states. Now, in talking about the bound states, then it will help to do some further coordinate transformations on the already-scaled Schrodinger equation there. Let's introduce a radial variable on all row. This is a convenient definition that it's too hard to know before I do that. So again, do something else. Let's take the case where the energy is negative. So we're talking about the bound states. So let's define a quantity I'll call nu to be equal to 1 over the square root of minus 2 times e tilde. This is the scale of energy. You referred to my notes here. And yes. And let's also introduce a radial variable row, which is defined as twice r tilde divided by nu. E tilde is negative for the bound states. So nu is a positive number. Anyway, this is a convenient change of variables. And if you do this, then you get for the gradient v squared after rho squared plus minus l times l plus 1 over rho squared plus nu over rho minus 1 quarter after 1 half equal to 0, which is the main reason. This is convenient as it puts it into a more or less standard form that you find in Wilson's special functions. In any case, so when you apply standard techniques of differential equations to this equation, in particular, to look for solutions which are normalizable, so they vanish in infinity, what you find is the index nu here, which is, effectively, just a function of the energy. You find the index nu must be equal to the integer in order to get a normalizable solution. So let me just call it n, replace it nu by n just to emphasize that this is an integer. You also find that this takes on the values l plus 1 and l plus 2 and so on all the way out to infinity. I'll remind you that the radial Schrodinger equation is parameterized by the angle momentum one number l. So in effect, it's a different radial Schrodinger equation for each value of l. And the n here, which is a label of the eigenfunctions, starts at l plus 1 and it goes on out. As far as the energy eigenvalues themselves are concerned, the etildes depend only on this one number n in our minus 1 over 2 n squared.  We'll multiply this by k, which up here is our characteristic energy for the system. But they go with minus 1 over 2 n squared. As far as the wave functions themselves are concerned, they depend on the n and l. And as a function of rho, they look like this. Their rho to the right rho to the l plus 1 power multiply times e to the minus rho over 2 times the associated Laguerre polynomial to l plus 1 upstairs to e plus l downstairs on the rho. And this is the associated Laguerre polynomial. So there's a lot of technical mathematics that's involved in it. And let's straight forward to setting up these results. I think in terms of what's useful to remember, it's useful to remember that we have this exponential factor here minus rho over 2 if you follow back the definitions. This is in terms of e to the minus r divided by n times a. The a is the effective for a radius. And n is the principal quantum number, which is then multiplied times a polynomial n radius r. It's much better remembering that the hydrogenetic wave functions have that form. All right. I didn't normalize this either. That's a further calculation one needs to do. All right. Now, as I keep saying, a regular Schrodinger equation is parameterized by l. The effective potential, which is the sum of centrifugal and true potentials, is therefore different for each different value of l. For most central force problems, this means that the different radial Schrodinger equations that you get for different values of l effectively have different potentials that don't even know about one another. It's almost as if you're choosing random potentials for each different l. And as a result, the energy level is a typical problem for different values of l. There's no relation one to another. In particular, it's not likely that it's a degeneracy. What this does do is it motivates us to organize energy levels both by the angle of the quantum number as well as by this n-quantum number. The n-quantum number, by the way, is called the principal quantum number in the case of hydrogen, just because of the name. So let's make a table of which we have l going across the top. Of course, the values are 0, 1, 2, 3, and so on, not like this. Speaking of l values, there's some ancient terminology that has historical meaning only in which these different l values are given l-thetic letters. They go S, P, V, F. And then after that, you go through the l-thetic, G, H, I. And I guess at some point, you run into P, and you have a skip, but that doesn't happen in real atoms. So anyway, this is the old terminology for describing the different negative meta-values. It's unfortunately just to have to learn it. If you want to be able to talk to time and physics, you should have to know what those symbols mean. And then on this side, let's put the energy levels coming up and down. And as I explained, the energy levels are basically minus 1 over 2 in square. Basically, it's multiplied by the characteristic energy. So if I measure things in terms of the characteristic energy, then n equals 1, and I didn't say this either, because we're in runs of 1, 2, 3, and so on. That's the range of the principle of 1 numbers. The ground state is that n equals 1, which is at an energy of minus 1 half in terms of these units capital K. And because n starts at l plus 1, if l is 0, n equals 1 is the first in state like this. Let's do the l equals 0 column here in the S waves. The next one up is most of the way to the top. This is n equals 2, which is at minus 1 half in quarter of 2 square. It's 3 quarters of the way up. So this is the n equals 2 level. And somewhere up above that, we've got n equals 3, which is at minus 1 half of 1 over 3 square minus the 9th level, we're speaking. And so you can see what happens is that these accumulate like this when you get an infinite number of views. If we move on to the l equals 1 radial equation, the principle of model number n starts with l plus 1, which is 2. So the first model is right here. This is an n equals 2 level, but then you have n equals 3, 4, 5, and so on. If you go to the l equals 2 or d states, then the first n value is n equals 3, and they accumulate up like this.  So these are the energy levels. They organize both by the principle of model number n, as well as the angle of n. This is a typical way of organizing energy levels in atomic physics and many other places, as well, using these two quantum numbers. This kind of a diagram, by the way, especially if you draw on the transitions, is called a Rotary and Diagram, which is named for. Now, the levels that are shown here are given in designations that indicate both the principle of quantum number, as well as the angle of momentum. So for example, this one is called the 1s. That means n equals 1, and s means l equals 0. This is the 2s, this is the 3s, and so on. This level's called the 2p. That means n equals 2, and l equals 1. Here is the 3p, here is the 4f, and so on. Here is the 3p, here is the 4p, and so on. The levels are labeled like that. I'm calling these levels, but actually what they represent are pairs of quantum numbers, n and l. There is another quantum number in the wave function, which is the magnetic quantum number. This enters because the total wave function of the dimensional space depends on all three quantum numbers, psi, n, l, m. It's equal to the radial wave function. All right, so the radial wave function is that this f is equal to r times capital R. So it's the radial wave function is a function of n and l of r times the ylm, which depends on theta and phi, even with the angular dependence. And in particular, the magnetic quantum number appears when you talk about the angular dependence. The energy doesn't depend on the magnetic one, but it depends on the energy's independent orientation. And the result is that the different all levels are two-all plus one-fold degenerate, and if I make those numbers, when we cross here, that becomes 1, 3, 5, 7, 9, and so on. Those of them, those of them will be, that's the order, the order of degeneracy of the levels that are going here. 2s, for example, is non-degenerative. 2p has got a 3-fold degeneracy. 3d has got 5-fold and so on across like this. Now, I mentioned several times that the different radial Schrodinger equations have different effective potentials. For most problems, those different effective potentials means that the spectra of those radial wave equations don't talk to them or they don't know about them. That means most likely the levels in one of these columns is not going to coincide with the levels in the other columns. Thus, there tends to be, for a randomly chosen radial potential, potentially no degeneracies amongst these different columns. What I want to point out is that that's not the case for the Coulomb potential, the hydrogen atom Coulomb potential, because, in fact, they have exact degeneracies running across. That's because the energy levels depend on the principle 1 and the n, and they do not depend on the angle of any quantum number. And this is not generic. I guess I've erased out my molecule energy levels, but they did depend on both n and l. There was a harmonic oscillator part, which was n plus 1 half h bar omega. That's the omega 3. That's the vibrational part.  That's the angular part. And so the molecule shows you what is typical for central force problems. The energy depends on both of those ones. But to go back to the case of hydrogen, there's something very special about hydrogen that it does not. It only depends on n and l and l. Well, one question is, why is this so? Why is this true for hydrogen? The reason is that hydrogen has extra symmetry that goes beyond the rotational invariance of the system. If we had no degeneracies amongst these different columns, which is a typical case, then the only degeneracy you would see would be the degeneracy from the magnetic quantum numbers, which is due to rotational invariance. That's the 2l plus 1 full degeneracy. That's what you get for any system that's rotational invariant. The hydrogen has extra degeneracy. The degeneracy of energy levels and quantum mechanics corresponds to symmetry. I can't go into this in too much detail, but it's a fact. And the fact that hydrogen has extra degeneracy, the only what's expected on the basis of rotational invariance alone, means that the hydrogen Hamiltonian has extra symmetry that goes beyond the rotational symmetry in three-dimensional space. In fact, it turns out that it's possible to model or to make a mathematical representation of the hydrogen atom in a four-dimensional space. This is not space time. It's just a mathematical space, but it has an extra dimension. And you find that the hydrogen atom Hamiltonian is invariant of the rotations in the full four-dimensional space. These are SO4 rotations. This is what lies behind the extra degeneracy that appears here. I won't lecture on this any further than just to make a mention of that. But if you look in Schiff's book, he has actually a nice presentation of this SO4 symmetry that's pretty easy to understand. And also, it's getting up on that subject. However, the fact remains that we have this extra degeneracy in hydrogen. So let's calculate what it is. In particular, what is the degeneracy of one of these n-levels? The energy depends only on n. So what is it? So the degeneracy, so if we fix the value of n, see earlier I said it would be. Earlier I said that we fixed it out. That means we're solving a particular rate of wave equation where n starts at l plus 1 and goes to infinity. That's like looking at one of these columns and starting from the bottom and running out. But suppose we fix the value of n and ask what l values are allowed. We fix one of these rows and run across. Well, you can see that the allowed l value then goes from 0, 1 all the way up to n minus 1. This is the range on l values for a fixed value of n. For example, when the n equals 2 level, we have l equals 0 and l equals 1, or 0, 1, 2, and n equals 2 levels. So the order of the degeneracy then is the sum of the allowed l values, namely l equals 0 up to n minus 1, of the degeneracy of each l value, which is 2 l plus 1. And this stuff can be done when the answer is n squared. The degeneracy of the hydrogen levels is a function of n is n squared. For example, for n equals 2, you have one level of the 2s plus three levels of the 2d in a total of 4, which is 2 squared, the example of this. All right, this is the extra degeneracy of hydrogen. I'll come back to this extra degeneracy in a moment, but I'd like to provide you now with a rationale for this rule and the whole quantum numbers. For a fixed value of n, as I say, for a fixed value of the energy, the angle of the quantum number has a maximum of a goes from 0 up to maximum, which is n minus 1. I'd like to show you that something like this occurs also in classical mechanics, and it gives you some classical intuition through what these quantum numbers mean in the case of hydrogen. So allow me to talk. The classical mechanics, the two-well problem, or a Kepler or one-over-R potential for just a moment. So let's say we had classically a potential d of R, which let's write it as minus k over R, or k is some constant. Of course, our problem, k, is the same things as z squared for our common problem, but we can apply this to the planets, too. The orbit says you know our ellipses, and I plot them in an x, y plane with the core center of the origin, typical argument like this, the core centers at the focus. The ellipses described by two parameters, a and e, a is the semi-major axis, so that going from one side of the ellipse to the other is twice the semi-major axis. This is going a long way. It's what's called a semi-major axis, semi-minor axis, goes the other way. And e is the eccentricity, so a is the semi-major axis. And e is the eccentricity of the ellipse. This is in the range 0 is less than or equal to k is less than or equal to 1. The eccentricity of 0 corresponds to the circle, and the eccentricity of 1 corresponds to ellipses, and it's smashed so flat, it comes just a straight line orbit. To give you different cases of this, if we take the case of eccentricity equals 0, we've got a circular orbit like this. This is equal 0. I increase the eccentricity. You get orbits like something like this. If you increase the eccentricity all the way down to its maximum value of 1, you get a needle-like orbit that goes back and forth like this. This is e equals 1. This is an eccentricity half way between. The energy of the angular momentum of the classical orbit are functions of the semi-major axis and eccentricity. The energy of that is equal to minus k over twice a. It depends only on the semi-major axis and not on the eccentricity. The angular momentum, however, is the square root of mkA times 1 minus e squared, so it does depend on the eccentricity. In fact, you can see that l of m has a minimum when the eccentricity is equal to 1 because that makes this equal to 0. l of m is equal to 0. This is the eccentricity of 1. That's this needle-like orbit here. And l is a maximum when the eccentricity is 0. And in that case, it's just a square root of mkA. This is the case of e is equal to 0. The equal 0 is a circular orbit. So the angular momentum ranging from 0 to its maximum is three orbits that look like this. Here what I'm doing is I want to make it look like a quantum problem. I want to hold the energy fixed to look at the range of angular momentum. So if we hold the energy fixed, it's the same as holding the semi-major axis fixed. The semi-major axis of the circular orbit is the same as the radius of the circle. When we squish it out to make an ellipse, we have to keep the semi-major axis the same. That means the diameter of the circle has to be the same thing as the twice the semi-major axis of the ellipse, or even the case of the needle has to be the same. So therefore, the needle goes from 0 to twice the radius of the circular orbit. And that's the range on these orbits that look like this. And so you see that even in classical mechanics, the angular momentum goes from 0 to its maximum in terms of energy. In fact, we read it out more explicitly. If I solve the semi-major axis in terms of the energy, the ellipse turns into a k k squared divided by minus twice the energy. This is of course the negative energies of the supplies. And that's your maximum angular momentum. But the way this manifests itself in quantum mechanics, one of the ways of connecting this classical picture with quantum mechanics, how do you connect these different types of orbits for a fixed value of energy with the wave functions that we'll see in quantum mechanics? Let's just take the case of the n equals 3 levels here. So there's a 3S, 3P, and 3E going from minimum at s equals 0. That's the needle-like orbits to the maximum, which is the circular orbits. What do they look like? Now I'll show you the sketch and plot. We can make a plot of this first start with the first start of the needle-like orbit. Let's take the 3S, take its absolute value squared, so it's really a probability, as a function of radius. And you plot them in a little something like this. There's a first component, and there's a zero in the bigger component, and there's a bigger component, and that's the end of it. That's three points like that. There's two nodes if you are a plot, two nodes of function if I plot the function instead of the square. If you plug in the values for the energy and calculate the classical semi-major axis, you'll find that it's about halfway in this function like this. This is a classical semi-major axis. And in fact, this corresponds to the needle-like orbit right here. Now, if I then plot the next one over, which is the F3P, where L is equal to 1 squared, now the wave function moves in from the origin. It only has two bumps, and the maximum bump has now moved, and both ends moved in, and one bump disappears. So it starts to look like this. And this corresponds to, you see, now there's an inner radial turning point, which is not zero. That's why this first bump is moved outwards. And there's an outer radial turning point that's not as far out. It's moved inwards. That's why this bump is moved inwards. And then finally, I plot the last one, which is the F3SPD. This is the maximum angular momentum corresponding to the circular orbit. We've got a single bump like this, which is concentrated closely at the value of the classical radius. That's the circular orbit. So it's worthwhile remembering this. The S wave, you see, go all the way down to the origin, whereas the states of high angular momentum, the circular orbits, stay away from the origin. In fact, they fly off. They have way small values at the origin. I hope you remember that I said that the greater wave function of RL of R goes as R to the power here, the origin. And that's part of what you're seeing here. This wave function is lying down ever more flat near the origin as L increases. All right, so that's just some pictures held with a two-day understanding of quantum mechanics, the quantum mechanics of hydrogen. Now, allow me to go back to this question of channel which is I explained. This extra degeneracy goes beyond the rotational variance, which is manifested in this diagram by these energy levels being the same as the move across a and a many values. This is a special property, the Coulomb potential. And if we take a mega perturbation of the Coulomb potential, even while making a central force problem, remaining in the framework of the central force problems, let's make some perturbations on this and see what happens. We'll see what happens is that the energy levels for different L's would have given in to move up and down. In other words, it breaks as a degeneracy. I'll show you a couple of physically interesting ways in which that may take place. One of these is called the volume effect. The volume effect, and by the way, as I say, this will still be part of a central force potential. I won't change the, it'll still be rotationally varied potentials. Here's the physics of the volume effect. It has to do with the fact that the proton in the face of hydrogen, or the nucleus for a higher hydrogen like atoms, it is not really a point charged. The proton is a composite particle and has an internal structure. And when you have a heavier nuclear, either made up of protons and neutrons. And so the Coulomb potential, minus z squared over r, is actually not valid once you get to a radius which is inside the nucleus. So allow me to make a plot here of potential to be as a function of radius. And let me just, this is very schematic. Let's go on, I've got a line here for let's say r equals capital R, which is the nuclear radius. Now the Coulomb potential does this kind of thing. Of course it's going down like this. But when you come inside the nuclear radius, then you see it's not a point charge, but a charge distribution. And we round this potential out and we'll do something like that. So compared to the Coulomb potential, there's, this can be regarded as a perturbation, which is the difference between the Coulomb potential valid everywhere and the actual one which smooths out. The true potential is not as negative as the Coulomb potential where you get the small radii. Now, given the fact that it's the S-waves with zero angular mental waves, are the wave functions which are most, which are largest near the origin, remember the radial wave function, the r, r1, s, the rs, s-waves, rl, or l equals zero, the radial wave function is a constant in there and then it gradually dies out. And whereas the l equals one looks like this and so on, it's the s-waves or l equals zero waves that have the largest values near the origin. The size of the nucleus in the case of hydrogen is very small compared to the size of the atoms. In fact, they're 10 to the fifth. And as a result, the perturbation which is produced by this effect is really very tiny in the case of hydrogen, almost always negligible. But for heavy atoms it becomes more important because the wave functions shrink towards the, or pull them in tighter towards the nucleus and also the nucleus grows in size. And so for heavy atoms the volume effect actually does become important. To go off with a slight little tangent, let me say something about the nuclear size and give you a basic fact, the nuclear physics. Nuclear physics, of course, the completely oriented for the neutrons and protons. I think Z is the usual symbol for the number of protons and N is the symbol for the number of neutrons what they usually do in nuclear physics. And A is defined as the sum of two of the number of nucleons plus Z. The basic fact that I wanna mention is that the radius of the nucleus goes as A to the one-third power. And this is because the nucleons interact with one another by only a short range of interaction. So they more or less have to be touching. And if you try to push them too close together there's a stronger pulse rate from the quality principle. And the result is that they can collect little hard balls, hard spheres. And that's why the radius goes as A to the one-third. This is quite different in the case of atoms. The radius of atoms is not as low as Z to the one-third. That's the roller dependence. In any case, this is basically just counting the volume occupied by the nucleus with the neurons. But as you see, if you go to Uranium where A is 238, the nucleus is bigger and the wave function is smaller. Anyway, the volume effect becomes more important for heavier atoms. But to go back to the point I wanna make about breaking the degeneracy, you can see that the S-waves, L equals zero, volume effect has been the most important for those. And in fact, it tends to be negligible for all of others. So what it'll do is it'll take, for example, this 2S model and raise it. It raises it because the potential is not as negative as the calculation is using. Then in particular, and this is a splitting between the 2S and the 2P. So that's one example of a physical effect which breaks this special degeneracy control Let me show you another example of a physical effect that does the same thing. In this case, more mathematically. This has to do with the ultrali atoms. The ultrali atoms, of course, are starting with lithium, there's sodium, potassium, probation, cesium, and pranzium if you wanna deal with a radioactive element. Those are the six. The ultrali atoms are characterized by having a single electron outside the core composed of a closed sub-shells. It's really a multi-electron problem. These are not simply analyzed in detail, but there's a crude model where the ultrali atoms, it's often times useful. And that is to imagine that you've got a core which is rotationally symmetric and it's made of the inner core electrons which produce some density. Some density grows a function of the radius R as soon as it's rotationally varying. And then out here is a single electron which is the valence electron moving in this simple force potential. Well, this row of R is the density of the inner core electrons. So if we integrate this zero to infinity four pi r squared dr times row of R, that's a total amount of charge contained in the core, it's gonna be Z minus one times E with overall minus sign because there are electrons for usefulness. Of course the nucleus has a charge of plus E. And so if you're well outside the core, this valence electron is gonna see the same electric field that it would in the case of hydrogen, it's a single charge. However, if the electron penetrates into the core, which it will do in the case of the S wave, which go all the way down to the origin, remember they see high angular momentum like the circular orbits. So for high angular momentum, this electron is gonna be moving out here in circular orbit and won't really see the core very much. But from lower angular momentum, it will penetrate in and as it does, it's less and less screaming in the core in fact it may come through S wave, so it'll come all the way down to the nucleus and see the full one screen charge of the nucleus which is plus ZB. So this is a rather dramatic breaking of the coolant potential. It's only the coolant potential outside of the core. Given an example of the effectiveness, let's talk about the spectrum of sodium. And here I'll do this again at one of these diagrams where we have SPGF for angular momentum going like this. The ground state of sodium is at a 3S level. It's an N equals three, L equals zero level. It's not a 1S as in hydrogen because the 1S and the 2S and 2P levels are what make up the core. The valence of electron has to go to different states so it has to go to the 3S. But in comparison to hydrogen, the 3S level is right up there, you see that's where it is. And it's degenerate to 3P and 3D. Well, in sodium it's quite different. The 3P levels are higher than the 3D levels are higher like this. But in fact the splitting is not small. The difference in the energy between the 3S and the 3P is an optical transition. This is the yellow-sodium deline which is what makes fires yellow. It's the very bright line from the sodium spectrum. So it's in the, this is a measurement of several electron volts, this energy difference here. It's also easy to see why the energies increase with increasing angular momentum. Because as the angular momentum increases the organs become more and more circular. And so this valence of electron is seen less and less at the core. Because the S-ways penetrate down to where the core is and they see the full incident charge in the nucleus so they're pulled in more tightly. So they're more tightly bound in that lower energy. This is exactly what's going on in these energy valves, I think that's so important. All right. So those are two physically interesting ways breaking the special degeneracy which holds in the case of hydrogen. Now this model of hydrogen that I've been talking about so far is what I would call the electrostatic spinless model of hydrogen. It's electrostatic because the Coulomb potential is based on Coulomb's law, which of course is electrostatics. We know that electrostatics is only an approximation to the full electromagnetic phenomena. We know that in the case of hydrogen there will be electromagnetic phenomena. There's going to be effects of retardation, and all the visiting effects that the electron has some velocity. In addition, there are magnetic effects coming from moving charges. The spin of the electron interacts in particular, the spin of the electron interacts with magnetic fields. And the result is that when we include the electron spin in the degrees of freedom description of the hydrogen, there are effects that couple together the Gordel and spin degrees of freedom giving rise to what's called the fine structure of hydrogen. These can be regarded overall as relativistic effects, so maybe better you could think of them as spin plus relativistic effects. Although in a sense then it is by itself a relativistic effect. So we'll go into that next time and talk about how we include spin, and in particular the problem in addition to angular momentum. So that's all. Remember, if anybody wants to borrow


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But first let's talk about the man who came, who saw and who conquered and then he gave it all away. Ramesh Johan is credited with building the Erated Drinks market in India. More than two decades ago at a time when Thumbs Up had more than 70% market share in the Indian soft drinks business, Johan had actually exited. He sold it with top brands like Citra, Goldspot, Mazda and Limca, all of which were then Pali products and all of them were sold to the mighty Coca-Cola. 23 years later he's back in the game, spitting his Bistlery Pop range of fizzy drinks against the might of the global giants, many of whom of course are his old rivals. So how is he likely to fare? That's a top story. He rose like a phoenix despite selling his well-entrenched range of carbonated beverages to Coca-Cola in 1993, tirelessly rebuilding his fortunes around Bistlery. But the question is can the cola king of the 90s recreate his magic now that he's re-entering the 14,000 crore soft drinks market with his new range of aerated beverages called Bistlery Pop? It's not the best phase for the carbonated drinks industry which is witnessing a very sluggish growth. Even Coca-Cola was supposed to shut down three manufacturing units due to lack of demand. The increasing health consciousness in the country and the growth of fruity drinks is clearly taking the fizz out of these aerated beverages. Can Ramesh Chauhan change the dynamics of this industry? One he himself was skeptical of entering six months ago. Good point. See, we are not making a cola. What we are making is a spicy cola. Now that is a variant which consumers according to us would love it because we don't like anything straight. We like everything with some masala on it. Good fruits is there. You put some chaat masala. So, here is an opportunity of giving people something different. See none of our four flavors are like any other flavor which is there in the market. And we are looking for a excitement among the consumers who are bored with the same old cola, same of orange, same old lemon lime. Even our Limonata doesn't taste like Sprite or 7up. Somebody says it tastes like Citra. I said you seem to have a damn good memory. Something which was discontinued how many years ago. Interestingly, Chauhan's re-entry into the soft drinks market comes at a time when Coca-Cola, the number one player in the market today, has just launched its fruit with fish drink called Fanta Green Mango. As Bisturi Pop prefers to counter Fanta Green Mango in the market, it will be interesting to find out who eventually has the advantage in this game. My approach to more players is welcoming because I believe that because the per capita consumption in the carbonated space is low, more players coming in creates more awareness around the category, creates more zing around the category and everybody benefits. So I do not feel threatened by the entry of any new player. I believe there is a room for everybody to prosper and grow. Post acquiring Parle beverages in 1993, Coca-Cola fizzled out some of the brands nurtured by Chauhan like Goldspot and Citra to make way for Coca-Cola's international brands like Fanta and Sprite. But now more and more companies are creating innovative drinks to suit Indian tastes. Another competitor PepsiCo is localizing its 7up brand by launching a new flavor, 7up Mimbo's Masala Soda. But Bisturi feels it clearly has an advantage here. We have definitely a bigger advantage because we've not been in this for a long time. I feel like it's more exciting to see a company that has been doing water now suddenly come into carbonated soft drinks I think would create more of a buzz. In terms of sales figures we'll have to wait and watch. We have made a conscious effort that our cola contains no caffeine and no phosphoric acid which is what is very unhealthy for you and which most colas will contain. So we are taking the steps that we can in making sure that our products are good. But yes the trend is definitely going towards a healthier lifestyle and a healthier diet. But so far the response has been really great towards Bisturi Pop and towards all our flavors so I'm quite excited to see how this is going to progress. While both Bisturi Pop and Fanta Green Mango campaigns are yet to hit TV screens one would expect some epic advertising as has been the trend in this category. The addition of Bisturi to the carbonated beverages market will definitely open up a new frontier in the existing cola adverts between Coca Cola and PepsiCo. It's interesting that Bisturi is coming out with its own cola and I think the entire suffering market is going to heat up during summer. I'm not sure whether Bisturi will be able to match the width and depth of distribution which Coca Cola company has today. But who knows what happens in the future. So it's going to be fun in the advertising business because there's going to be an advertising bar once again. The consumers are going to probably have a good time finally and that's what marketing is all about. Bucking the trend Chauhan plans to launch a joint campaign for all the four drinks as opposed to giving them separate identities in the first leg of the advertising journey. It will largely be focused on the digital medium and OOH the creatives for which are being handled by Soho Square and Foxy Moron. The founders of Thumbs Up have created products which will now stand in direct competition to their iconic brand currently owned by Coca Cola. I'm a very small time player in the carbonated beverage and our infrastructure and in terms of facilities for carbonated beverage is quite low. So we are not going to be market shaker. We are going to try and create niche market and unique flavors that too big and way ahead of us in terms of yours. So to think that way would be disastrous for us. See you look for opportunities and you look for your strengths. Now we feel that we have got a good distribution setup and managerial setup to handle soft drinks because earlier Mr. He was very small. Now with what Mr. He is today I think we can proudly say at many places we outsell Coke and Pepsi but that aside you have to be ready for it to enter into a field and you don't enter a field as a defense mechanism. At a time when the two top beverage makers in the country Coca Cola and PepsiCo are in the pincer grasp of local soft drink makers the entry of Bistleri International is guaranteed to stir up the marketplace. While the local players have cornered around 10 to 15% of the market with lower priced products the question is can Ramesh Shahan and his competitively priced Bistleri pop wean them away or does he intend to play a bigger game and cannibalize on the market share of the cola giants one just has to wait for summer to begin and a thirsty market. Right so while Ramesh Shahan is back with all that fears the question is can Bistleri pop turn out to be the big disruptor for the Indian cola market that's a space we'll be watching very closely but right now we take a quick break when we return we talk to the man who runs the Polish cosmetics brand Inglot chairman Dr. Inglot is joining us in the studio here to talk us through how they're planning to ramp up the brand's presence in India that's coming up in a moment to stay with us.


We want to pay tribute to a former Yankees, Phil Linz. On Thursday, it was reported that Linz had passed away at the age of 81. He only played four seasons with the Yankees, but he was a member of their 1962 World Championship Club as a rookie. And two years later, he earned a permanent spot in franchise lore. During a rough patch in 1964, Linz infuriated manager Yogi Berra, when after being goaded by Mickey Mantle, he loudly played a harmonica on the team bus following a tough loss in mid-August. But the Yankees went 30 and 13 over the remainder of the season to win the American League pennant, and the harmonica incident was never forgotten. The Yankees traded Linz to the Phillies after the 1965 season, and the last three years of his career were split between the Phillies and the Mets.