Lecture 12. Spin, the Vector Model and Hydrogen Atoms
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Shaka, Athan J.
 This course provides an introduction to quantum mechanics and principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. It also examines the Schrödinger equation and study how it describes the behavior of very light particles, the quantum description of rotating and vibrating molecules is compared to the classical description, and the quantum description of the electronic structure of atoms is studied. 
 Welcome back to Chemistry 131A. Today we're going to talk about spin, the vector model, and begin to talk about hydrogen atoms, a subject that we'll continue on for a couple of lectures. Recall that when Stern and Gerlach did their measurement, there were two bands that were observed rather than a continuum of silver atoms distributed. And that was interpreted to mean that there were two possible states for the angular momentum of the magnetic moment, the intrinsic magnetic moment of the electron. And that would correspond to M sub S equals to plus one half or minus one half. And the interesting thing is that it was a half rather than integer, which made it seem fairly mysterious at the time. In the vector model of angular momentum, we visualize the angular momentum vector as lying somewhere on a cone. And the Z component we imagine that we've measured and that has a definite value. And we've measured the total angular momentum, that also has a definite value. But the X and Y values are indeterminate for reasons that we'll see in a minute. For an electron then, there are just two such orientations of this particular cone. And we tend to just call them up and down and recall that we don't actually physically think anything is spinning or rotating or there's a little planetary model inside the electron. We have no evidence at all for anything like that. This is just an intrinsic property that there is a magnetic moment and it behaves as if there were an angular momentum and it had the two projections, M sub S equals plus or minus one half. We can do spectroscopy, we can do magnetic resonance, we can make transitions between these two magnetic states. And a lot of information about molecules and photosynthetic systems and various other kinds of things have been gleaned by actually seeing if there is an unpaired electron, what the energy difference is between the two magnetic states and actually getting a spectrum of the possible difference and seeing how that spectrum may change if we irradiate the sample with light or do other things to influence the molecular structure. So for a spin one half, we've got a picture like this of these two cones, one cone up, one cone down, these are the two possibilities and there's a definite value of the z component of the angular momentum, h bar over two and the other possibilities minus h bar over two. For the x and y components though, they're indeterminate and that's why we just draw this cone because that's meant to show that the x and y components could be anywhere on this cone. We'll see why in a second. For a higher angular momentum, we would have more cones and we'll see that in a second. What about the usual conditions that the wave function, whatever it is, be single valued and have at least a second derivative? If we put m instead of e to the i m phi, if we imagine some kind of thing there with m over two, then the problem is that it changes sign. But keep in mind that spin is different because when we derive that e to the i m phi, we were actually had a physical particle on a physical ring and we got those wave functions from the boundary conditions there. But here, we have an observation of a magnetic moment but we don't necessarily know that it corresponds to anything like that. There is no spatial component to spin. It exists just as the magnetic moment of the electron but there's nothing moving around in space that we can ascertain. And the spin part can actually make some calculations quite difficult because, for example, if you have a molecule with an unpaired electron, then it may be that whether the spin is up or down changes the spatial electron density. Why? Because there are a lot of other magnetic particles around the molecule, nuclei, other electrons and so on. And it may be that if the spin has one orientation, the magnetic moment, that it hangs out more over here and if the spin has opposite orientation, that it ends up in a different part of the molecule. And that means that if you're trying to keep track of chemical reactions where electrons can become unpaired or various things can happen, sometimes your calculation gets much more complicated than you would like because you have to know something about the spin or average over it or figure out what's going on. The spin one-half, if you want to think about it, is rather like a Mobius strip in that if I put a twist in a piece of paper, then rather than coming around like a ring and pointing the same way, if I put just half a twist and tape it together, when I come around, I'm actually pointing inside rather than outside and when I come around again, I'm pointing outside, so the wave function repeats when you rotate by 4 pi, that's so-called spinner behavior and usually it's extremely hard to observe but if you have a reference state and then you have a state that you're rotating with respect to the reference and you can do this in magnetic resonance experiments, you can actually see that when you apply a pulse that you actually see the thing change sine if you go around by 2 pi and if you go around by 4 pi, it comes back to the same sine again. Here's a picture for the spatial part. Here are the five components of M sub L for an L equals 2 state which we'll see is called a D orbital and again the cones have to match and the length of L has to also match because we imagine we've measured L, so we know L times L plus 1 and then we've measured the Z component so we know that and you get this series of cones and again the LX and LY components are completely indeterminate but you get these pretty pictures and this is the so-called vector model for angular momentum. We'll see in atomic spectroscopy in a couple of lectures that keeping track of the angular momentum of the electron depending what value of L it has and keeping track of the spin depending whether it's up or down is very important for keeping track of where the spectral lines will appear and to assign the spectrum. To assign the spectrum means that we know the energy level where the electron started and then it falls down to a lower energy level and emits light of a certain frequency that comes out someplace and we know then the energy levels and we know as many quantum numbers about each energy level as it's possible to know. If we know that then we say that we've assigned the spectrum if we can match up everything and then we can infer a lot about the structure of the atom by where these lines actually happen to be like the sodium D lines for example that give the characteristic yellow color of a sodium vapor lamp. Why do we have this uncertainty? Well it turns out that this is another manifestation of the uncertainty principle and at this point rather than just talking about delta P delta X we can make a connection with a much more general idea namely if we're going to make a measurement of one thing and another thing then if we make the measurements in the reverse order the question is whether we get the same result or not. If we get the same result by making the measurements the other way round then we imagine that these measurements are compatible and it's okay. If on the other hand we get a different result like we saw with the coin where if we wanted to see if it was heads or tails or we wanted to see how thick it was that we ended up getting different results at random for whether it was heads or tails after we measured the thickness then they're incompatible and this incompatibility is actually coded in the operators themselves so we don't have to guess about it. We have a mathematical way to analyze what the operators are doing when we can figure out. It turns out that the square of the angular momentum L times L plus 1 or S times S plus 1 and the Z component or any other component but when we're picking a component we pick Z by convention they're compatible. They can both be determined no problem at all no problem with the coin on the edge but if we try to in addition determine in addition to LZ and L squared if we go further and say well I'd like to also know LX and LY as well then things get filed up then what happens is if we measure one of those then LZ gets changed at random and if we go back and measure LZ the other one gets changed at random and no matter how many times we go back and forth we can't get a consistent result because the result we get depends on the order that we measure it in and further more it's random. It doesn't depend like a flip flop or anything like that something orderly that we can predict it's just a random value and the key is whether the operators commute and the commutator really is you put one operator on one side and the other one behind it and then you do it the reverse way you put this one on this side and this one behind it and you put them both on the wave function and then you subtract them and if you subtract them and it comes to zero then they are compatible and you can measure them both as accurately as you want and if they do not then they are incompatible and that means that making one measurement will destroy some of the information that you gleaned from the previous measurement. Let's just take our operators for position then and momentum. Position remember the x hat operator was just to multiply the wave function by x and the p hat x operator was to take the derivative minus ih bar times the derivative with respect to x. Suppose we want to figure out the commutator then we write the square bracket and we write x comma p and what that means is that means take x px minus px x. So let's do that. The first term is x and then on the right hand side minus ih bar d by dx that's p hat x minus the opposite order which is minus ih bar d by dx of x and what we do then is we put our wave function on the far right of this. We don't insert it where it might seem to go next to the derivative past the x we put it on the far right hand side of these operator equations. Always remember that don't start inserting things in the middle keep them to the right because the way operators work is left to right by convention. Let's apply this commutator to some wave function psi of x that's completely arbitrary we don't care what it is as long as it follows our rules for being continuous and having derivatives and so on. The first term then is quite easy because it's x times minus ih bar times the derivative of psi with respect to x whatever that may be. The second term is now plus ih bar because I have minus minus ih bar times the derivative of the product x psi and so that means I have to take the derivative of x and then times psi plus the derivative of psi times x and if I do that then I find that two of the terms vanish the minus ih bar x psi dx cancels with the plus ih bar x d psi dx but I end up with that third term because of the rules for taking the derivative which is ih bar psi and so the commutator x hat p hat x applied to the wave function psi returns a number which isn't real but it has ih bar in it ih bar psi. Therefore since psi is arbitrary what we say is that the commutator this relation has nothing to do with psi so at the end we throw psi out and we have an operator relation the operator x hat comma p hat x is equal to ih bar and the key is it's not zero. Since it's not zero we conclude that we cannot measure x position and x momentum simultaneously to arbitrary precision and I've written this relationship here on slide 312 this is just an operator relationship you can think of as ih bar times the one operator one hat where the one operator just multiplies psi by the number one and anytime in fact the commutator of two Hermitian operators is nonzero doesn't matter so much what it is it gives you a hint about how it's going to behave but the key thing is is it zero or is it not zero if it's not zero they are not compatible if it is zero they are and in fact we can take the three components of angular momentum L hat z that's remember that these components are just r cross p and that where we take the vector cross product so we can take L hat z that's the angular momentum operator for the z component that's x hat p y minus y hat p x and L x is y hat p z minus z hat p y and L y is z hat p x minus x hat p z if we take these because we now know that x and p x don't commute do y and p x commute yes they do because there's no when I take the derivative with respect to y of x times the wave function it doesn't I don't need to use the product rule because these are partial derivatives and so if there's no y there then I don't bother taking the product rule x is treated as a constant so I can measure the momentum one way and the position another way in principle as accurately as I like I just can't measure them both and I certainly can't measure all three to localize the particle plus no its trajectory if you work this out you can verify on your own that in fact the commutation relations between angular momentum are a little bit more interesting the commutation between position and momentum is just a number an imaginary number but still just a number in fact if you take the commutator of L x with L y what you get is I h bar times L z you not only get a number back but you get another operator and if you take the commutator of L z with L x you get I h bar L y and if you take the commutator with L y and L z you get I h bar L x and when you look at them these things are in a cyclic permutation so if you start with the three operators like this you're allowed to pull this one here put this one here and move that one there and you can do it again and you always get the same result that's called a cyclic permutation and it's very interesting that these cyclic permutations look very much like what a rotation would look like we're going around like this and every time we take the commutator we go around to the other component and in fact there's a very deep connection between the structure of these commutators and the fact that these are in fact referring to a rotation of something that they are angular momentum. Therefore for the spin angular momentum we don't have r cross p we just have this behavior but now all we have to say is we propose that there is such an operator as the spin angular momentum we know that because we have this magnetic moment we know it has the value one half for the spin and that there are two orientations and we just propose that there are three operators we call them S x, S y and S c and that they follow the same exact rules they follow the same exact rules as the orbital angular momentum and that in fact that it's not really the derivatives and so forth that are important but what's more important is whether you follow this particular rule or not if you do you're going to behave like an angular momentum and that's kind of an interesting way to formulate things and in fact as I remark here in more advanced courses that you take on quantum mechanics we just simply propose that there exists there are operators and they follow certain commutation relations and we don't need to specify what they are in concrete terms we don't have to say well this operator is a derivative with respect to x or anything like that we just say x hat p hat x is equal to i h bar anything that follows that is going to behave like position momentum we don't have to write derivatives and various things like that why would we want to do that well there's a kind of a minimalist approach to the theory the less things you have to actually assume that are detailed than the stronger your theory is because it doesn't depend on whether those assumptions are correct if you didn't make them for example implicit in taking the derivative is sort of the idea that space is a continuous thing that there is such a thing as dx and that there's an infinitesimal unit of space we're taking the derivative with respect to it and using these nice stylized functions to represent reality but what if it isn't what if space is digital or sort of like a billboard when you get up close you see little dots and only when you get back away does it look like a continuous picture or a photograph well then you'd have to reformulate your whole theory if it depended on all that kind of thing whereas if you just say well I have got the commutators and that's it you don't have to do anything different because you come up with something else that has the same commutator and then everything else in the theory still works whereas if everything in the theory depends on this linchpin of well there is a derivative and all this stuff that isn't quite true then you're in trouble you have to start over. Okay we're going to talk in a couple lectures about atomic spectroscopy and in all kinds of spectroscopy and rotational spectroscopy and vibrational spectroscopy and atomic spectroscopy you run into this unit of energy that seems a bit odd it's called the wave number and it has a very strange unit because the unit is centimeters to the minus one and that doesn't seem like a unit of energy at all and then why is it centimeters and so I want to take a little bit of an aside here to go into why we use this unit. The main reason is that a wave number is about the right size there's a reason why we have measurements like the foot and stuff like that because it's about the right size to measure things that we encounter in real life and when we're writing down energies and we're writing down transitions and so forth if we express the energy in wave numbers we get numbers like one and ten and a hundred and maybe a thousand but we don't get numbers like ten to the minus thirty four or ten to the plus twenty and chemists don't like big numbers like that because it makes it harder to understand things and talk about it and relate them it's always a little bit more daunting when we've got huge and tiny numbers and we're dealing with them. For atomic and molecular spectroscopy then the relationship between the energy and the wave number is E is equal to H nu that we knew from the photon and we can write that in terms of the speed of light because C the speed of light is the wavelength times the frequency you can see that because if you've got a certain wavelength and the frequency changes then that's how far it moved so that's wavelength times frequency is in fact the velocity of the wave which for light is always C in a vacuum so for the frequency nu we substitute C upon lambda and then we set one over lambda is equal to nu bar and we say that the energy is H C times nu bar and nu bar is the wave number that's what we're quoting when we do that and one wave number or one centimeter to the minus one corresponds to about 30 giga hertz and the reason why is that three times ten to the ten is how fast light goes in centimeters per second so that's the conversion then between wave numbers and hertz and it's much easier to visualize one wave number one unit as a wave number than 30 giga hertz three times ten to the ten per second as a frequency they are the same thing and we can measure the thermal energy and we can see that the wave number is makes sense in that regard so as you may have learned in freshman chemistry the random thermal energy that's around and can be accessed by system at some temperature T in Kelvin of course always in Kelvin is KT that's about the range K is Boltzmann's constant let's figure out then how many wave numbers this is so we take KT let's take 25 Celsius 298.15 Kelvin put in all the constants K 1.38 times 10 minus 23 joules per Kelvin 298.15 Kelvin H 10 to the minus 34 joules times seconds and then for C all we have to do is put it in centimeters per second because we see that the joules go away the per second goes away the Kelvin's go away and so if we just put C in centimeters per second rather than meters per second we get the answer in inverse centimeters and what we find then is random thermal motion is about 207 wave numbers therefore if we've got energy levels that are spaced less than 207 wave numbers then just things knocking around can knock stuff up into these higher levels because there's plenty of energy around to do that you can think of the 207 as if there's lots of hundred dollar bills on the ground just lying around randomly everywhere then it's pretty easy to buy a cup of coffee if that's the way it is if it's very cold there's no energy there's no money anywhere there's a couple of cents there's no way you're going to buy a cup of coffee so you can't make any kind of molecular transition in that case. Electronic excitation of atoms is usually much higher than this number 207.2 wave numbers and that makes sense because materials only glow let's say red hot so we know if they're glowing they're emitting light that electrons are making transitions in the material but that only happens if it's red hot if it's actually very very hot it's not going to happen for any kind of metal or material like that at 205 Celsius they don't emit light and so that explains why the numbers we're going to get for electronic transitions in atoms are very very much higher numbers than that and that also explains for example why infrared light just heat as long as you don't get too much of it because the photons are low doesn't do anything to chemical bonds and neither does the radio waves or microwaves from your cell phone regardless of what people say to the contrary who don't know anything about molecular structure or radiation whereas UV light if we figure out how many wave numbers that has that's different that has the potential to break some bonds and that's why we can use it to sterilize water when we go camping because if we stick in an intense LED UV source it breaks bonds in all the bacteria that are swimming around in the water and of course they're very minute amounts but the key is when you drink it are they alive and then they start doubling and quadrupling and so forth and make you very sick or are they there but they're dead because their bonds have been broken by UV light same thing with your skin you get too much you get sunburned it's very bad. Now in atomic spectroscopy it was experiment first and then theory next in fact most of the time it's experiment first and then theory comes in afterwards once you know the answer then you can appear to be quite smart it's only very rarely let's say with the observation of mercury by prediction by special relativity and a few cases like that where the prediction was made and then the experiment was done and it actually agreed with the prediction because the theory was so deep and in this case a dedicated spectroscopist Rydberg a Swedish man found that in 1890 that the lines the emission lines from a hydrogen arc so what's a hydrogen arc well I can take hydrogen gas that's a diatomic and I can like put a lightning strike through it boom just high voltage and two things will happen the bond will break I got so much energy there I got a cascade of electrons slamming in knocking things up and then the isolated hydrogen atoms then will have electrons in very high orbitals a random number and depending how big I make the arc and then they start emitting light and then I record all the light they emit and I have a look at it and quite how he defined this is a mystery but he must have been very very good with numbers and some people are very good with things like that and others are not so strong and he looked at where these lines were and he worked out that they followed this relationship that the wave number of the line appeared to be a constant times the difference in 1 over n2 squared where n2 is an integer minus 1 over n1 squared for example it could be 2 squared is 4 and 1 squared is 1 and so you get that one and then there's 3 and 2 and so forth and he saw all these things and this number R sub H the Rydberg constant referred to hydrogen was this number 109,677 wave numbers why it followed that was a mystery but it was interesting that it followed that and what was even more interesting especially now quantization quantum mechanics not continuum mechanics is that these numbers were integers that was a crucial observation because now this is a simple thing hydrogen one proton one electron and it's giving you this clue that there are numbers and they're integers that are dictating where the light comes out if you just take differences between them you can work all these states out it would make sense then if you get all these combinations 1 over n squared minus 1 over m squared that the energy levels themselves go like 1 over n squared and then could be 1 that would be down at the bottom 2, 3 and so forth and maybe up to infinity and the zero of reference here for this kind of atomic system and all these systems in general is that you take the electron and the proton and you move them apart so they're infinitely far apart and you leave them at rest if they're infinitely far apart their electrostatic energy is zero and if they're at rest their kinetic energy is zero that imaginary reference state you call zero and then you compare what the energy is as the atom and of course the energy is going to be negative in that case because the way to think about it is if I'm stuck down here where they've attracted each other and I want to pry them back out to here I've got to put in positive energy so if I go the other way from the reference I had to go down in energy I couldn't go up in energy or I'd be getting energy back when I went off to infinity therefore we can write the Hamiltonian and the wave function for the hydrogen atom we've got the Schrodinger equation time independent Schrodinger equation we've got the clue from the experiment as to what happened and so we'll write down a Hamiltonian and it has the kinetic energy of the nucleus which in this case is a single proton it's got the kinetic energy of the electron and it's got the potential energy between them so I've written this here H hat is minus H bar squared over two times the mass of the electron times the second derivative with respect to where the electron is minus H bar squared over two times the mass of the nucleus times the second derivative with respect to where the nucleus is minus E squared over 4 pi epsilon not r where r is a special thing it's the distance between them now suppose we solve this just written like this what would the wave function depend on well it would depend at the minimum on six coordinates because it would depend on where the electron is in x y and z and then it would depend on where the nucleus is in x y and z that's six things that I have to put in to get a value for the wave function now how do I plot that there's no way I can plot that because I can't see what I'm doing it's as if I make a value I have six axes all I can do is make like contour plots and make fixed certain values and then cut through just like we do with a mountain range when we make a contour plot we turn a three dimensional thing into a map that's flat and we draw contours on it but here I'd have to fix a bunch of these guys and then let two of them vary and then plot the wave function and it's really you're really blinded it's like being stuck between two tall buildings and you just can't see where on earth you are in the city because you're just going down this narrow alley and so this is completely unworkable and as you get more particles involved the real wave function itself is just impossible to understand what it actually looks like and it's extremely hard to calculate what it is as well since the wave function gives us the most information possible most of the time we don't need to know everything about it we just need to know enough to be able to calculate what we want to calculate and there's a trick here I'd like to go through to show you how we get rid of three of these coordinates so that we just have three things and then we can make these plots in 3D space where we use transparency and shapes to indicate a certain percentage chance that the electron is in a certain area. So the first thing we do is our potential depends on just the difference in position between the two things so that's a clue that trying to specify the exact position of the nucleus in XYZ and the electron in XYZ is kind of a waste of time and I need a trick so that what I'm looking at is the nucleus and then I can pretend the nucleus is at zero and then it's out of the picture and the way to do that is to change our coordinates so that one system is the center of mass one system is wherever the proton is and the electron is I take their center of mass which is very close to the proton and I call that big M and the other part is called the reduced mass which should be familiar from vibrational problems and that's given the symbol usually mu and that's the mass of the nucleus times the mass of the electron over the mass of the nucleus plus the mass of the electron and when the nucleus is heavy mu is pretty much the mass of the electron. The mass big M the center of mass just drifts around as a free particle because there is no potential that refers to the coordinates of the center of mass the potential only refers to the difference between the two particles not to where they are in the universe as the center of mass and we know the solution for that that's just our plane waves e to the I px upon h bar or I've written it here as psi for this big M the center of mass is equal to some constant times e to the minus I k dot r and so that's done now all we have to do is set a derivative with respect to that that's all kinetic energy all the center of mass can have is kinetic energy it can't ever have any potential energy in this kind of problem the other part which separates only depends on the difference between the two particles not not their absolute coordinates and we can write that then as minus h bar squared over two times mu times del squared where now del is referring to the difference just the difference what's the difference between x between the nucleus and the electron y and so forth and then this is a dodge usually we aren't interested in the center of mass motion we know the things going to drift around in fact we try to design experiments where things are either very cold or very quiet so that things aren't drifting around too much because if things drift around we get slight frequency shifts just like a train whistle coming toward you or going away and usually if we're trying to do an accurate measurement we don't want that stuff we'd like to know what the train whistle is when the train is stationary mathematically though this lets us just forget about the center of mass we just take the coordinate of the nucleus and we just artificially totally artificially assume it's fixed we know it can't be fixed because of the uncertainty principle but we just assume that it is for the purpose of the calculation and then if we're really doing something detailed we add the center of mass calculation and later now then this is great because we're down to one derivative and it's a three dimensional problem but the potential has spherical symmetry and therefore we just imagine the nucleus at zero zero zero the electron then is the spatial variable and we just go forward and as I remarked of course in real fact the nucleus can't literally be fixed we go ahead then we know how to transform the d by dx squared d by dy squared so forth into spherical polar coordinates we get the partial derivative the second partial derivative with respect to r plus two over r times the derivative with respect to r plus one over r squared times the same thing we had before for the particle on a sphere recall that before we got to this point and then we just said well r is fixed let's look at theta and phi now we've got r in there but we've got the other thing where we already know what that part does and so all we have to really figure out is what does this part r do where instead of being fixed it has this electrostatic potential between the two particles that only depends on r but not on theta and phi so as I remarked we've solved that part of that problem on a sphere no big deal and all we do then is we just assume like we always did that the wave function whatever it is is a product and it's a part in r that I'm going to call big r of r and it's a part in theta and phi that I'm just going to call y because that's the conventional term for the spherical harmonics ylm and you can verify on your own as an exercise that if you make this substitution into the Schrodinger equation that it will separate you'll get two parts one part that depends only on theta and phi the other part depends only on r therefore you can do them separately and you get the following two equations you get minus h bar squared over 2 mu y times what we call lambda squared which was all the sine theta and d by d phi times y is a constant and the other part we get is minus h bar squared over 2 mu big r times r squared d squared big r dr squared plus 2 r d big r dr plus v r squared minus e r squared and that should be a minus whatever the other one constant was so that the two of them together add up to zero but they're separately constants once you pick one as a number the other is the same number with the opposite sign we already know the top part is just the particle on a sphere so good good thing we did that we know that h bar minus h bar squared over 2 mu times lambda squared y is equal to h bar squared over 2 mu times l times l plus 1 times y and so the constant is h bar squared over 2 mu times l times l plus 1 that's the constant we need and the second equation we can simplify instead of dealing with big r let's make a substitution little u of r is equal to r times big r and let's see if we can let's do this together as a practice problem now and see if we can come to some conclusion so here's practice problem 15 show that the differential equation simplifies substantially if we express it in terms of u rather than in terms of big r what is the connection with the classical case okay well we're going to write u is equal to little r times big r and we're going to have to use the product rule to take derivatives so that being the case du dr is r d big r dr plus r times the derivative of r with respect to r which is 1 and we can take the second derivative we take the derivative of the derivative and we just follow the same thing the first thing is r d squared big r dr squared plus dr over little dr plus dr over little dr and then we get two terms out the end r d squared big r dr squared plus 2 times the derivative of big r with respect to r then we just solve for the second derivative of big r and what we get is 1 over r d squared u dr squared minus 2 dr d big r dr and if we substitute that into our original differential equation substitute the second derivative of r then it turns out that minus 2 d big r d little r cancels the other term that I had perfectly and we find the following then that we've got the sum of these terms plus v r squared minus e r squared is equal to the constant which is minus h bar squared over 2 mu times l times l plus 1 after we cancel the terms here's what we get we get minus h bar squared over 2 mu times little r over u times little r times the second derivative of u with respect to r squared plus v r squared minus e r squared is equal to this constant if we divide by r squared and multiply by u we can simplify that and we can finally write the equation in this form that I've written it at the bottom where we have the second derivative of u with respect to r squared that looks like a potential that looks like a kinetic energy excuse me in one dimension then we have minus e squared times u of r that's the potential e squared over 4 pi epsilon not r times u of r that's the potential then we have another term times u of r which is h bar squared over 2 mu r squared times l times l plus 1 this whole thing then should equal e times u of r and that's just a one dimensional problem now on the interval r is equal to 0 to infinity so rather than minus infinity to infinity like x slightly different at 0 to infinity and we can write it simply as a pseudo kinetic energy operator operating on this wave function u of r plus v effective on u of r and v effective is the real electrostatic potential which has a 1 over r dependence plus h bar squared l times l plus 1 over 2 mu r squared and that looks just like a centrifugal term if I have something moving in a circle it has an energy l squared over 2i and that's exactly what we would expect if we have non-zero l so it has angular momentum then we have to add the energy of the angular momentum before we solve it for u so that's the nice simplification we still have to solve for u and then we have to solve for r in order to see what kind of electron density we expect to find as a function of little r which is the separation so that's what we'd like to know is if we've got a hydrogen atom where's the electron hanging out and what's the chance of finding it here or here and so forth the case when l is not equal to 0 is harder for us to do why because while we have that kinetic energy term and the potential energy 1 over r well we still have to figure out about that because last time we either had no potential energy for the particle in a box or 1 half kx squared for the harmonic oscillator now we've got something a little bit more challenging 1 over r in particular you might worry a little bit that when r is 0 what's that blows up and so maybe it's going to be very difficult to solve the differential equation it turns out that's not such a big worry and then we've got another term this l times l plus 1 upon r squared and so that adds another thing in so now we have a 1 over r and a 1 over r squared and we have to find the function that does that we're going to explore that how to do that step by step in the next lecture where we'll talk about these radial distribution functions.
 Welcome back to Chemistry 131A. Today we're going to talk about spin, the vector model,  and begin to talk about hydrogen atoms, a subject that we'll continue on for a couple  of lectures. Recall that when Stern and Gerlach did their measurement, there were two bands  that were observed rather than a continuum of silver atoms distributed. And that was interpreted  to mean that there were two possible states for the angular momentum of the magnetic moment,  the intrinsic magnetic moment of the electron. And that would correspond to M sub S equals  to plus one half or minus one half. And the interesting thing is that it was a half rather  than integer, which made it seem fairly mysterious at the time. In the vector model of angular  momentum, we visualize the angular momentum vector as lying somewhere on a cone. And the  Z component we imagine that we've measured and that has a definite value. And we've measured  the total angular momentum, that also has a definite value. But the X and Y values are  indeterminate for reasons that we'll see in a minute. For an electron then, there are  just two such orientations of this particular cone. And we tend to just call them up and  down and recall that we don't actually physically think anything is spinning or rotating or  there's a little planetary model inside the electron. We have no evidence at all for anything  like that. This is just an intrinsic property that there is a magnetic moment and it behaves  as if there were an angular momentum and it had the two projections, M sub S equals plus  or minus one half. We can do spectroscopy, we can do magnetic resonance, we can make  transitions between these two magnetic states. And a lot of information about molecules and  photosynthetic systems and various other kinds of things have been gleaned by actually seeing  if there is an unpaired electron, what the energy difference is between the two magnetic  states and actually getting a spectrum of the possible difference and seeing how that  spectrum may change if we irradiate the sample with light or do other things to influence  the molecular structure. So for a spin one half, we've got a picture like this of these  two cones, one cone up, one cone down, these are the two possibilities and there's a definite  value of the z component of the angular momentum, h bar over two and the other possibilities  minus h bar over two. For the x and y components though, they're indeterminate and that's why  we just draw this cone because that's meant to show that the x and y components could be  anywhere on this cone. We'll see why in a second. For a higher angular momentum, we  would have more cones and we'll see that in a second. What about the usual conditions  that the wave function, whatever it is, be single valued and have at least a second derivative?  If we put m instead of e to the i m phi, if we imagine some kind of thing there with m  over two, then the problem is that it changes sign. But keep in mind that spin is different  because when we derive that e to the i m phi, we were actually had a physical particle on  a physical ring and we got those wave functions from the boundary conditions there. But here,  we have an observation of a magnetic moment but we don't necessarily know that it corresponds  to anything like that. There is no spatial component to spin. It exists just as the magnetic  moment of the electron but there's nothing moving around in space that we can ascertain.  And the spin part can actually make some calculations quite difficult because, for example, if you  have a molecule with an unpaired electron, then it may be that whether the spin is up  or down changes the spatial electron density. Why? Because there are a lot of other magnetic  particles around the molecule, nuclei, other electrons and so on. And it may be that if  the spin has one orientation, the magnetic moment, that it hangs out more over here and  if the spin has opposite orientation, that it ends up in a different part of the molecule.  And that means that if you're trying to keep track of chemical reactions where electrons  can become unpaired or various things can happen, sometimes your calculation gets much  more complicated than you would like because you have to know something about the spin  or average over it or figure out what's going on. The spin one-half, if you want to think  about it, is rather like a Mobius strip in that if I put a twist in a piece of paper,  then rather than coming around like a ring and pointing the same way, if I put just half  a twist and tape it together, when I come around, I'm actually pointing inside rather  than outside and when I come around again, I'm pointing outside, so the wave function  repeats when you rotate by 4 pi, that's so-called spinner behavior and usually it's extremely  hard to observe but if you have a reference state and then you have a state that you're  rotating with respect to the reference and you can do this in magnetic resonance experiments,  you can actually see that when you apply a pulse that you actually see the thing change  sine if you go around by 2 pi and if you go around by 4 pi, it comes back to the same  sine again. Here's a picture for the spatial part. Here are the five components of M sub  L for an L equals 2 state which we'll see is called a D orbital and again the cones have  to match and the length of L has to also match because we imagine we've measured L, so we  know L times L plus 1 and then we've measured the Z component so we know that and you get  this series of cones and again the LX and LY components are completely indeterminate  but you get these pretty pictures and this is the so-called vector model for angular  momentum. We'll see in atomic spectroscopy in a couple of lectures that keeping track  of the angular momentum of the electron depending what value of L it has and keeping track of  the spin depending whether it's up or down is very important for keeping track of where  the spectral lines will appear and to assign the spectrum. To assign the spectrum means  that we know the energy level where the electron started and then it falls down to a lower energy  level and emits light of a certain frequency that comes out someplace and we know then  the energy levels and we know as many quantum numbers about each energy level as it's possible  to know. If we know that then we say that we've assigned the spectrum if we can match  up everything and then we can infer a lot about the structure of the atom by where these  lines actually happen to be like the sodium D lines for example that give the characteristic  yellow color of a sodium vapor lamp. Why do we have this uncertainty? Well it turns out  that this is another manifestation of the uncertainty principle and at this point rather  than just talking about delta P delta X we can make a connection with a much more general  idea namely if we're going to make a measurement of one thing and another thing then if we  make the measurements in the reverse order the question is whether we get the same result  or not. If we get the same result by making the measurements the other way round then  we imagine that these measurements are compatible and it's okay. If on the other hand we get  a different result like we saw with the coin where if we wanted to see if it was heads  or tails or we wanted to see how thick it was that we ended up getting different results  at random for whether it was heads or tails after we measured the thickness then they're  incompatible and this incompatibility is actually coded in the operators themselves so we don't  have to guess about it. We have a mathematical way to analyze what the operators are doing  when we can figure out. It turns out that the square of the angular momentum L times  L plus 1 or S times S plus 1 and the Z component or any other component but when we're picking  a component we pick Z by convention they're compatible. They can both be determined no  problem at all no problem with the coin on the edge but if we try to in addition determine  in addition to LZ and L squared if we go further and say well I'd like to also know  LX and LY as well then things get filed up then what happens is if we measure one of  those then LZ gets changed at random and if we go back and measure LZ the other one gets  changed at random and no matter how many times we go back and forth we can't get a consistent  result because the result we get depends on the order that we measure it in and further  more it's random. It doesn't depend like a flip flop or anything like that something  orderly that we can predict it's just a random value and the key is whether the operators  commute and the commutator really is you put one operator on one side and the other one  behind it and then you do it the reverse way you put this one on this side and this one  behind it and you put them both on the wave function and then you subtract them and if  you subtract them and it comes to zero then they are compatible and you can measure them  both as accurately as you want and if they do not then they are incompatible and that  means that making one measurement will destroy some of the information that you gleaned from  the previous measurement. Let's just take our operators for position then and momentum.  Position remember the x hat operator was just to multiply the wave function by x and the  p hat x operator was to take the derivative minus ih bar times the derivative with respect  to x. Suppose we want to figure out the commutator then we write the square bracket and we write  x comma p and what that means is that means take x px minus px x. So let's do that. The  first term is x and then on the right hand side minus ih bar d by dx that's p hat x minus  the opposite order which is minus ih bar d by dx of x and what we do then is we put  our wave function on the far right of this. We don't insert it where it might seem to  go next to the derivative past the x we put it on the far right hand side of these operator  equations. Always remember that don't start inserting things in the middle keep them to  the right because the way operators work is left to right by convention. Let's apply this  commutator to some wave function psi of x that's completely arbitrary we don't care  what it is as long as it follows our rules for being continuous and having derivatives  and so on. The first term then is quite easy because it's x times minus ih bar times the  derivative of psi with respect to x whatever that may be. The second term is now plus ih  bar because I have minus minus ih bar times the derivative of the product x psi and so  that means I have to take the derivative of x and then times psi plus the derivative  of psi times x and if I do that then I find that two of the terms vanish the minus ih  bar x psi dx cancels with the plus ih bar x d psi dx but I end up with that third term  because of the rules for taking the derivative which is ih bar psi and so the commutator  x hat p hat x applied to the wave function psi returns a number which isn't real but  it has ih bar in it ih bar psi. Therefore since psi is arbitrary what we say is that  the commutator this relation has nothing to do with psi so at the end we throw psi out  and we have an operator relation the operator x hat comma p hat x is equal to ih bar and  the key is it's not zero. Since it's not zero we conclude that we cannot measure x position  and x momentum simultaneously to arbitrary precision and I've written this relationship  here on slide 312 this is just an operator relationship you can think of as ih bar times  the one operator one hat where the one operator just multiplies psi by the number one and  anytime in fact the commutator of two Hermitian operators is nonzero doesn't matter so much  what it is it gives you a hint about how it's going to behave but the key thing is is it  zero or is it not zero if it's not zero they are not compatible if it is zero they are  and in fact we can take the three components of angular momentum L hat z that's remember  that these components are just r cross p and that where we take the vector cross product  so we can take L hat z that's the angular momentum operator for the z component that's  x hat p y minus y hat p x and L x is y hat p z minus z hat p y and L y is z hat p x minus  x hat p z if we take these because we now know that x and p x don't commute do y and p x  commute yes they do because there's no when I take the derivative with respect to y of  x times the wave function it doesn't I don't need to use the product rule because these  are partial derivatives and so if there's no y there then I don't bother taking the product  rule x is treated as a constant so I can measure the momentum one way and the position another  way in principle as accurately as I like I just can't measure them both and I certainly  can't measure all three to localize the particle plus no its trajectory if you work this out  you can verify on your own that in fact the commutation relations between angular momentum  are a little bit more interesting the commutation between position and momentum is just a number  an imaginary number but still just a number in fact if you take the commutator of L x  with L y what you get is I h bar times L z you not only get a number back but you get  another operator and if you take the commutator of L z with L x you get I h bar L y and if  you take the commutator with L y and L z you get I h bar L x and when you look at them  these things are in a cyclic permutation so if you start with the three operators like  this you're allowed to pull this one here put this one here and move that one there  and you can do it again and you always get the same result that's called a cyclic permutation  and it's very interesting that these cyclic permutations look very much like what a rotation  would look like we're going around like this and every time we take the commutator we go  around to the other component and in fact there's a very deep connection between the  structure of these commutators and the fact that these are in fact referring to a rotation  of something that they are angular momentum.  Therefore for the spin angular momentum we don't have r cross p we just have this behavior  but now all we have to say is we propose that there is such an operator as the spin angular  momentum we know that because we have this magnetic moment we know it has the value one  half for the spin and that there are two orientations and we just propose that there  are three operators we call them S x, S y and S c and that they follow the same exact  rules they follow the same exact rules as the orbital angular momentum and that in fact  that it's not really the derivatives and so forth that are important but what's more important  is whether you follow this particular rule or not if you do you're going to behave like  an angular momentum and that's kind of an interesting way to formulate things and in  fact as I remark here in more advanced courses that you take on quantum mechanics we just  simply propose that there exists there are operators and they follow certain commutation  relations and we don't need to specify what they are in concrete terms we don't have to  say well this operator is a derivative with respect to x or anything like that we just  say x hat p hat x is equal to i h bar anything that follows that is going to behave like  position momentum we don't have to write derivatives and various things like that why would we  want to do that well there's a kind of a minimalist approach to the theory the less things you  have to actually assume that are detailed than the stronger your theory is because it  doesn't depend on whether those assumptions are correct if you didn't make them for example  implicit in taking the derivative is sort of the idea that space is a continuous thing  that there is such a thing as dx and that there's an infinitesimal unit of space we're  taking the derivative with respect to it and using these nice stylized functions to represent  reality but what if it isn't what if space is digital or sort of like a billboard when  you get up close you see little dots and only when you get back away does it look like a  continuous picture or a photograph well then you'd have to reformulate your whole theory  if it depended on all that kind of thing whereas if you just say well I have got the commutators  and that's it you don't have to do anything different because you come up with something  else that has the same commutator and then everything else in the theory still works whereas  if everything in the theory depends on this linchpin of well there is a derivative and  all this stuff that isn't quite true then you're in trouble you have to start over.  Okay we're going to talk in a couple lectures about atomic spectroscopy and in all kinds  of spectroscopy and rotational spectroscopy and vibrational spectroscopy and atomic spectroscopy  you run into this unit of energy that seems a bit odd it's called the wave number and  it has a very strange unit because the unit is centimeters to the minus one and that doesn't  seem like a unit of energy at all and then why is it centimeters and so I want to take  a little bit of an aside here to go into why we use this unit.  The main reason is that a wave number is about the right size there's a reason why we have  measurements like the foot and stuff like that because it's about the right size to  measure things that we encounter in real life and when we're writing down energies and we're  writing down transitions and so forth if we express the energy in wave numbers we get  numbers like one and ten and a hundred and maybe a thousand but we don't get numbers  like ten to the minus thirty four or ten to the plus twenty and chemists don't like big  numbers like that because it makes it harder to understand things and talk about it and  relate them it's always a little bit more daunting when we've got huge and tiny numbers  and we're dealing with them.  For atomic and molecular spectroscopy then the relationship between the energy and the  wave number is E is equal to H nu that we knew from the photon and we can write that  in terms of the speed of light because C the speed of light is the wavelength times the  frequency you can see that because if you've got a certain wavelength and the frequency  changes then that's how far it moved so that's wavelength times frequency is in fact the velocity  of the wave which for light is always C in a vacuum so for the frequency nu we substitute  C upon lambda and then we set one over lambda is equal to nu bar and we say that the energy  is H C times nu bar and nu bar is the wave number that's what we're quoting when we do  that and one wave number or one centimeter to the minus one corresponds to about 30 giga  hertz and the reason why is that three times ten to the ten is how fast light goes in centimeters  per second so that's the conversion then between wave numbers and hertz and it's much easier  to visualize one wave number one unit as a wave number than 30 giga hertz three times  ten to the ten per second as a frequency they are the same thing and we can measure the  thermal energy and we can see that the wave number is makes sense in that regard so as  you may have learned in freshman chemistry the random thermal energy that's around and  can be accessed by system at some temperature T in Kelvin of course always in Kelvin is  KT that's about the range K is Boltzmann's constant let's figure out then how many wave  numbers this is so we take KT let's take 25 Celsius 298.15 Kelvin put in all the constants  K 1.38 times 10 minus 23 joules per Kelvin 298.15 Kelvin H 10 to the minus 34 joules  times seconds and then for C all we have to do is put it in centimeters per second because  we see that the joules go away the per second goes away the Kelvin's go away and so if we  just put C in centimeters per second rather than meters per second we get the answer in  inverse centimeters and what we find then is random thermal motion is about 207 wave  numbers therefore if we've got energy levels that are spaced less than 207 wave numbers  then just things knocking around can knock stuff up into these higher levels because  there's plenty of energy around to do that you can think of the 207 as if there's lots  of hundred dollar bills on the ground just lying around randomly everywhere then it's  pretty easy to buy a cup of coffee if that's the way it is if it's very cold there's no  energy there's no money anywhere there's a couple of cents there's no way you're going  to buy a cup of coffee so you can't make any kind of molecular transition in that case.  Electronic excitation of atoms is usually much higher than this number 207.2 wave numbers  and that makes sense because materials only glow let's say red hot so we know if they're  glowing they're emitting light that electrons are making transitions in the material but  that only happens if it's red hot if it's actually very very hot it's not going to happen  for any kind of metal or material like that at 205 Celsius they don't emit light and so  that explains why the numbers we're going to get for electronic transitions in atoms  are very very much higher numbers than that and that also explains for example why infrared  light just heat as long as you don't get too much of it because the photons are low doesn't  do anything to chemical bonds and neither does the radio waves or microwaves from your  cell phone regardless of what people say to the contrary who don't know anything about  molecular structure or radiation whereas UV light if we figure out how many wave numbers  that has that's different that has the potential to break some bonds and that's why we can  use it to sterilize water when we go camping because if we stick in an intense LED UV source  it breaks bonds in all the bacteria that are swimming around in the water and of course  they're very minute amounts but the key is when you drink it are they alive and then  they start doubling and quadrupling and so forth and make you very sick or are they there  but they're dead because their bonds have been broken by UV light same thing with your  skin you get too much you get sunburned it's very bad.  Now in atomic spectroscopy it was experiment first and then theory next in fact most of  the time it's experiment first and then theory comes in afterwards once you know the answer  then you can appear to be quite smart it's only very rarely let's say with the observation  of mercury by prediction by special relativity and a few cases like that where the prediction  was made and then the experiment was done and it actually agreed with the prediction  because the theory was so deep and in this case a dedicated spectroscopist Rydberg a  Swedish man found that in 1890 that the lines the emission lines from a hydrogen arc so  what's a hydrogen arc well I can take hydrogen gas that's a diatomic and I can like put a  lightning strike through it boom just high voltage and two things will happen the bond  will break I got so much energy there I got a cascade of electrons slamming in knocking  things up and then the isolated hydrogen atoms then will have electrons in very high orbitals  a random number and depending how big I make the arc and then they start emitting light  and then I record all the light they emit and I have a look at it and quite how he  defined this is a mystery but he must have been very very good with numbers and some  people are very good with things like that and others are not so strong and he looked  at where these lines were and he worked out that they followed this relationship that  the wave number of the line appeared to be a constant times the difference in 1 over  n2 squared where n2 is an integer minus 1 over n1 squared for example it could be 2  squared is 4 and 1 squared is 1 and so you get that one and then there's 3 and 2 and  so forth and he saw all these things and this number R sub H the Rydberg constant referred  to hydrogen was this number 109,677 wave numbers why it followed that was a mystery but it  was interesting that it followed that and what was even more interesting especially  now quantization quantum mechanics not continuum mechanics is that these numbers were integers  that was a crucial observation because now this is a simple thing hydrogen one proton  one electron and it's giving you this clue that there are numbers and they're integers  that are dictating where the light comes out if you just take differences between them  you can work all these states out it would make sense then if you get all these combinations  1 over n squared minus 1 over m squared that the energy levels themselves go like 1 over  n squared and then could be 1 that would be down at the bottom 2, 3 and so forth and maybe  up to infinity and the zero of reference here for this kind of atomic system and all these  systems in general is that you take the electron and the proton and you move them apart so they're  infinitely far apart and you leave them at rest if they're infinitely far apart their  electrostatic energy is zero and if they're at rest their kinetic energy is zero that  imaginary reference state you call zero and then you compare what the energy is as the  atom and of course the energy is going to be negative in that case because the way to  think about it is if I'm stuck down here where they've attracted each other and I want to  pry them back out to here I've got to put in positive energy so if I go the other way  from the reference I had to go down in energy I couldn't go up in energy or I'd be getting  energy back when I went off to infinity therefore we can write the Hamiltonian and the wave  function for the hydrogen atom we've got the Schrodinger equation time independent Schrodinger  equation we've got the clue from the experiment as to what happened and so we'll write down  a Hamiltonian and it has the kinetic energy of the nucleus which in this case is a single  proton it's got the kinetic energy of the electron and it's got the potential energy  between them so I've written this here H hat is minus H bar squared over two times the  mass of the electron times the second derivative with respect to where the electron is minus  H bar squared over two times the mass of the nucleus times the second derivative with respect  to where the nucleus is minus E squared over 4 pi epsilon not r where r is a special thing  it's the distance between them now suppose we solve this just written like this what  would the wave function depend on well it would depend at the minimum on six coordinates  because it would depend on where the electron is in x y and z and then it would depend on  where the nucleus is in x y and z that's six things that I have to put in to get a value  for the wave function now how do I plot that there's no way I can plot that because I can't  see what I'm doing it's as if I make a value I have six axes all I can do is make like  contour plots and make fixed certain values and then cut through just like we do with  a mountain range when we make a contour plot we turn a three dimensional thing into a map  that's flat and we draw contours on it but here I'd have to fix a bunch of these guys  and then let two of them vary and then plot the wave function and it's really you're really  blinded it's like being stuck between two tall buildings and you just can't see where  on earth you are in the city because you're just going down this narrow alley and so this  is completely unworkable and as you get more particles involved the real wave function  itself is just impossible to understand what it actually looks like and it's extremely  hard to calculate what it is as well since the wave function gives us the most information  possible most of the time we don't need to know everything about it we just need to know  enough to be able to calculate what we want to calculate and there's a trick here I'd  like to go through to show you how we get rid of three of these coordinates so that we  just have three things and then we can make these plots in 3D space where we use transparency  and shapes to indicate a certain percentage chance that the electron is in a certain area.  So the first thing we do is our potential depends on just the difference in position  between the two things so that's a clue that trying to specify the exact position of the  nucleus in XYZ and the electron in XYZ is kind of a waste of time and I need a trick  so that what I'm looking at is the nucleus and then I can pretend the nucleus is at  zero and then it's out of the picture and the way to do that is to change our coordinates  so that one system is the center of mass one system is wherever the proton is and the electron  is I take their center of mass which is very close to the proton and I call that big M  and the other part is called the reduced mass which should be familiar from vibrational  problems and that's given the symbol usually mu and that's the mass of the nucleus times  the mass of the electron over the mass of the nucleus plus the mass of the electron  and when the nucleus is heavy mu is pretty much the mass of the electron.  The mass big M the center of mass just drifts around as a free particle because there is  no potential that refers to the coordinates of the center of mass the potential only refers  to the difference between the two particles not to where they are in the universe as the  center of mass and we know the solution for that that's just our plane waves e to the  I px upon h bar or I've written it here as psi for this big M the center of mass is equal  to some constant times e to the minus I k dot r and so that's done now all we have to  do is set a derivative with respect to that that's all kinetic energy all the center of  mass can have is kinetic energy it can't ever have any potential energy in this kind of  problem the other part which separates only depends on the difference between the two  particles not not their absolute coordinates and we can write that then as minus h bar  squared over two times mu times del squared where now del is referring to the difference  just the difference what's the difference between x between the nucleus and the electron  y and so forth and then this is a dodge usually we aren't interested in the center of mass  motion we know the things going to drift around in fact we try to design experiments where  things are either very cold or very quiet so that things aren't drifting around too  much because if things drift around we get slight frequency shifts just like a train  whistle coming toward you or going away and usually if we're trying to do an accurate  measurement we don't want that stuff we'd like to know what the train whistle is when  the train is stationary mathematically though this lets us just forget about the center  of mass we just take the coordinate of the nucleus and we just artificially totally artificially  assume it's fixed we know it can't be fixed because of the uncertainty principle but we  just assume that it is for the purpose of the calculation and then if we're really doing  something detailed we add the center of mass calculation and later now then this is great  because we're down to one derivative and it's a three dimensional problem but the potential  has spherical symmetry and therefore we just imagine the nucleus at zero zero zero the  electron then is the spatial variable and we just go forward and as I remarked of course  in real fact the nucleus can't literally be fixed we go ahead then we know how to transform  the d by dx squared d by dy squared so forth into spherical polar coordinates we get the  partial derivative the second partial derivative with respect to r plus two over r times the  derivative with respect to r plus one over r squared times the same thing we had before  for the particle on a sphere recall that before we got to this point and then we just said  well r is fixed let's look at theta and phi now we've got r in there but we've got the  other thing where we already know what that part does and so all we have to really figure  out is what does this part r do where instead of being fixed it has this electrostatic potential  between the two particles that only depends on r but not on theta and phi so as I remarked  we've solved that part of that problem on a sphere no big deal and all we do then is  we just assume like we always did that the wave function whatever it is is a product  and it's a part in r that I'm going to call big r of r and it's a part in theta and phi  that I'm just going to call y because that's the conventional term for the spherical harmonics  ylm and you can verify on your own as an exercise that if you make this substitution into the  Schrodinger equation that it will separate you'll get two parts one part that depends  only on theta and phi the other part depends only on r therefore you can do them separately  and you get the following two equations you get minus h bar squared over 2 mu y times  what we call lambda squared which was all the sine theta and d by d phi times y is a  constant and the other part we get is minus h bar squared over 2 mu big r times r squared  d squared big r dr squared plus 2 r d big r dr plus v r squared minus e r squared and  that should be a minus whatever the other one constant was so that the two of them together  add up to zero but they're separately constants once you pick one as a number the other is  the same number with the opposite sign we already know the top part is just the particle  on a sphere so good good thing we did that we know that h bar minus h bar squared over  2 mu times lambda squared y is equal to h bar squared over 2 mu times l times l plus  1 times y and so the constant is h bar squared over 2 mu times l times l plus 1 that's the  constant we need and the second equation we can simplify instead of dealing with big r  let's make a substitution little u of r is equal to r times big r and let's see if we  can let's do this together as a practice problem now and see if we can come to some conclusion  so here's practice problem 15 show that the differential equation simplifies substantially  if we express it in terms of u rather than in terms of big r what is the connection  with the classical case okay well we're going to write u is equal to little r times big  r and we're going to have to use the product rule to take derivatives so that being the  case du dr is r d big r dr plus r times the derivative of r with respect to r which is  1 and we can take the second derivative we take the derivative of the derivative and  we just follow the same thing the first thing is r d squared big r dr squared plus dr over  little dr plus dr over little dr and then we get two terms out the end r d squared big  r dr squared plus 2 times the derivative of big r with respect to r then we just solve  for the second derivative of big r and what we get is 1 over r d squared u dr squared  minus 2 dr d big r dr and if we substitute that into our original differential equation  substitute the second derivative of r then it turns out that minus 2 d big r d little  r cancels the other term that I had perfectly and we find the following then that we've  got the sum of these terms plus v r squared minus e r squared is equal to the constant  which is minus h bar squared over 2 mu times l times l plus 1 after we cancel the terms  here's what we get we get minus h bar squared over 2 mu times little r over u times little  r times the second derivative of u with respect to r squared plus v r squared minus e r squared  is equal to this constant if we divide by r squared and multiply by u we can simplify  that and we can finally write the equation in this form that I've written it at the bottom  where we have the second derivative of u with respect to r squared that looks like a potential  that looks like a kinetic energy excuse me in one dimension then we have minus e squared  times u of r that's the potential e squared over 4 pi epsilon not r times u of r that's  the potential then we have another term times u of r which is h bar squared over 2 mu r  squared times l times l plus 1 this whole thing then should equal e times u of r and  that's just a one dimensional problem now on the interval r is equal to 0 to infinity  so rather than minus infinity to infinity like x slightly different at 0 to infinity  and we can write it simply as a pseudo kinetic energy operator operating on this wave function  u of r plus v effective on u of r and v effective is the real electrostatic potential which  has a 1 over r dependence plus h bar squared l times l plus 1 over 2 mu r squared and  that looks just like a centrifugal term if I have something moving in a circle it has  an energy l squared over 2i and that's exactly what we would expect if we have non-zero l  so it has angular momentum then we have to add the energy of the angular momentum before  we solve it for u so that's the nice simplification we still have to solve for u and then we have  to solve for r in order to see what kind of electron density we expect to find as a function  of little r which is the separation so that's what we'd like to know is if we've got a hydrogen  atom where's the electron hanging out and what's the chance of finding it here or here  and so forth the case when l is not equal to 0 is harder for us to do why because while  we have that kinetic energy term and the potential energy 1 over r well we still have to figure  out about that because last time we either had no potential energy for the particle in  a box or 1 half kx squared for the harmonic oscillator now we've got something a little  bit more challenging 1 over r in particular you might worry a little bit that when r is  0 what's that blows up and so maybe it's going to be very difficult to solve the differential  equation it turns out that's not such a big worry and then we've got another term this  l times l plus 1 upon r squared and so that adds another thing in so now we have a 1 over  r and a 1 over r squared and we have to find the function that does that we're going to  explore that how to do that step by step in the next lecture where we'll talk about these  radial distribution functions.
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