Lecture 14. Atomic Spectroscopy
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 continue our exposition of atomic spectroscopy focusing on selection rules, coupling, and terms. Of course, atomic spectroscopy is the basic tool by which everybody learned that the atom was quantized and so it's of very central importance in this whole area of endeavor of quantum mechanics applied to small things. Various sets of transitions are observed. Here is energy level diagram for the hydrogen atom, not to scale because if you do it to scale, the levels crowd together too much at the top and it's very hard to see what's going on. There's a series called the Lyman series in which an electron jumps from the n equals 2 to n equals 1 or 3 to 1 or 4 to 1 and so forth. And then there's another series, the Ballmer series, that's in the visible range in which the electron jumps to the n equals 2 state and then there are other series and they have names according to who worked on them. The main thing to keep in mind here is that any change in the quantum number n in the energy is allowed in principle. The only thing that has to be satisfied is that there has to be conservation of energy and that's taken care of automatically because the photon is emitted with a quantum h nu which is exactly equal to the energy loss of the atom. So energy is conserved. In this diagram, the zero reference energy is at the top and all the energies quoted there are negative being more stable than a proton and an electron at rest at infinite separation. The selection rules are a little bit more complicated, however, and the first thing is that a photon has one unit of angular momentum, h bar. Angular momentum is conserved and so when the electron makes a transition there has to be a change by one unit of angular momentum. So that means that although we draw these things coming down to the one, in fact they're all coming from specific values of L. It's 2p to 1s, 3p to 1s, 4p to 1s and so forth and it's always p. We can't have 2s to 1s because then delta L is zero and that's not allowed. That doesn't conserve angular momentum and likewise we can't have 3d to 1s either because then delta L is equal to 2 and again that doesn't conserve angular momentum. Now if we have a different kind of transition or if more than one photon is emitted that's different because then, of course, that could be allowed because angular momentum could be conserved in a more complicated kind of process. Those kind of processes do happen all the time. They're usually just much slower than the single photon electric dipole allowed transition. If we have transitions to 2p in the Bombers series, for example, they could be from 3s or 3d. If we have transitions to the 2s state then they have to be again from 3p, 4p and so on. And in general, delta L is plus or minus 1 for electric dipole allowed transitions and these are generally the very strongest transitions that you see in the spectrum. Not always, however, if the atom is very heavy like mercury as we'll see in the next lecture, we could have a strong transition that doesn't seem to follow this rule. There are weaker transitions, magnetic dipole transitions and electric quadrupole transitions. We won't touch on those much but you should keep in mind that those can and do occur and therefore you mustn't just assume that this is an ironclad rule. When we use the terms allowed and forbidden in spectroscopy, what we're talking about really is strong and weak. Allowed is the light is green, forbidden is it's red but some people run red lights. Here's a compendium from the NIST database just to show you how accurate and fastidious these spectroscopists have been in cataloging these energy levels. This one has a different energy zero. In this case, energies are quoted from the 1S level of the hydrogen atom being zero and then up from there. But the amazing thing about these numbers is that they have six, seven, eight, and nine digits of accuracy. So you can see that they had to take a lot of precautions when they were doing these. And if you look closely, there are multiple entries. There's 2P and then there's 2S and there's just 2 and then there's 2P again and so forth. And we'll see in a second what these things mean. But you have to keep track of individual substates in these systems and oftentimes you put on a magnetic field to be able to tell which state is which and how they move around to keep track of them. And they go all the way up to 6H. So in the periodic table, we don't go beyond F. But if we go to excited states of atoms, it's easy to go very high and to have very high values of the angular momentum L. If you look closely, you find out that the first 2P state that's quoted is ever so slightly lower in energy than 2S, which is quite a surprise because in other elements, of course, 2S is lower than 2P. That's why lithium is 2S1 and not 2P1, for example. And we don't fill up the 2P until we get over to boron. But in hydrogen, it's kind of an anomaly, one of those 2P states is lower than the 2S. And that was extremely interesting to theoreticians how that could possibly be. And in studying that in great detail, they worked out important clues about actually the residual electromagnetic field that permeates our entire universe like the zero point energy of a harmonic oscillator. You can think of the electromagnetic field as some kind of electromagnetic oscillation. And it turns out that even in a vacuum at absolute zero, in this universe, apparently, we can't have nothing. There's still some zero point energy left over. And that can bounce the electron around. And then the question is, if it bounces a 2S electron around or bounces a 2P electron around, what's the difference? And if you actually follow through the calculation, it's actually very, very beautiful work to follow. And you can actually derive that the 2P should be lower. And we can use the measured energy. Once we have a series of these states, we can use the measured energy, if we're careful, to estimate the ionization potential, also called the ionization energy. And that's just an extrapolation to what the energy would be at n equals infinity. And that's very important to know what that number is. That's one of the main things that's listed in freshman chemistry books in the periodic table to try to account for various trends. If an electron is easy to ionize, and that means that that element's much more likely to give an electron up to another atom than an element that has an electron that's extremely hard to ionize, for example. Let's try a practice problem then and have a look at how we might tackle estimating the ionization energy. Practice problem 18. Consider the following unassigned, that means we don't know n or l, emission lines from a hydrogen arc, all of which terminate in a common level, assign them and estimate the ionization potential or the ionization energy, same thing, of hydrogen. While the hydrogen arc is just we put a big lightning strike, a voltage through, and we excite the atoms and then they emit light. And if they terminate in a common level, that means they're like the other diagram I drew. We have arrows coming down to this, to this, to this, and so forth, always to the same level. And that means the change in energy has to do with the upper level but not the lower level. Of course, in actual fact, you just get all kinds of lines and you have to be very clever to figure out which ones go with which and which go together. There are various tricks that were worked out to do that. So here are the numbers. They all have four or five digits in them, 1.8887 and so forth. And let's just take these four numbers, quoted in electron volts, and let's figure out what the ionization energy should be. The first thing is we have to figure out what common level they terminate in. And if we don't do that, then we aren't going to be able to figure out anything. And so the way we're going to do that is we first write this formula that delta E is equal to the Rydberg constant times 1 over n squared common level minus 1 over n k squared where nk is some higher number, higher up dropping down. We don't know what the common level is, however. And we don't know what the value of r is because it wasn't given. And on an exam, if you aren't given the value of r and a problem like this, that means you're supposed to figure it out without that. So let's try to do that. We don't know n common. We don't know r. We have some values of delta E. And we have to try to figure out if they fit this formula. A very bad way is to just start plugging things in and seeing what happens. A very good way is to proceed systematically and figure out if we can get a straight line plot. Chemists love straight lines. Before computers, straight lines were absolutely essential. And the reason why is that the human eye can distinguish a straight line from any curve. But the human eye cannot distinguish curves that are slightly different. Even if they're statistically significantly different, it's very hard to tell just by looking. And before computers, people had to plot things on paper often and look at them. And in order to tell what's what, if you could organize your equation so that it came into a straight line plot, then you could tell. You could tell if there was a systematic deviation, some kind of slight curvature one way or the other way. You could tell if there was scatter, noise, and so forth. And all those are very important to quantify when you're trying to estimate something. We can cast our equation into a straight line form by letting y equals to delta E, letting x equal 1 over n squared, and then letting the intercept b equals 1 over, rather, r over n common. And we can guess values of nk and see which look good. So we have to assume a value of n common, and then we guess values of nk all greater than that, and we see what looks good. Once the line is straight, then we can extract the slope, y equals b plus mx, and m in this case should be minus r, the Rydberg constant. Once the line is straight, then we can get r. Now the lowest common level could be n equals 1, so in the absence of any other information, we should assume that first and then work our way up. If the common level is very high, really, really high, this would be a disaster because we would take forever. But we know that it can't be that high because we see a couple of electron volts, and so we know it has to be one of the levels further down. If it's very high, then all the transitions would be very, very small numbers because we'd already be most of the way up to the ionization energy. So let's try n common equals 1, and in that case, we could have nk, the upper level, b2, 3, 4, and so forth. If some transitions are missing for some reason, we just didn't see them, they were faint, we made an error, some other reason, then this kind of exercise can get very frustrating because the points don't appear to fit, and if you have to start deleting values, then it's very tedious, but in any kind of exam problem, we'll never have something like that. But in the lab, for various reasons, it can be that some transitions are dark, something happens, there's another pathway where the electron can go, and it's very hard to see that, and sometimes it takes a lot of patience to be able to work out what's going on. So let's plot these. I drew up a table, I have the delta E values for y, and then I have the 1 over nk squared, and for 2, it's a quarter, for 3, it's a ninth, and so forth as we go up. And if we plot those, which I've done here, well, first of all, we notice that the slope is negative, excuse me, and that's, of course, exactly what we expect because the slope should be negative r, and r is a positive number, so that part's okay. And let's plot them now, and here I've drawn a plot. Now, there are four black dots on this plot on slide 370, and if you just draw a line through the four black dots by linear regression, you will get a pretty good fit. In other words, there is a line that misses them all, but is pretty good, and so you have to decide if that kind of line is going to be good enough for what you're doing, and in this case, it's not nearly good enough. And what I've tried to do is emphasize that by first drawing a line through the first two points on the left, the red line, that has a certain slope, and keep in mind that these points have very, very, very tiny errors. So although I've put these black dots so that you can see them on the slide, the points themselves are very, very accurately determined. We draw that red line, and then we compare that with the slope from the last two points, the blue line. We only have four points, so we have to be kind of careful, and we plot that, and when you do it that way, what you see is that the blue line has a much shallower negative slope than the red line, that it is quite a bit different, and so there, and if you took the center two points, it would have a slope in between, and what that means is that this is curved. It's not n equals, n comma equals 1, and this is not good enough, but you could easily assume it worked good enough if you're used to other kinds of data that isn't this well determined. In any case, we're going to discard that and say, well, n comma equals 1 doesn't seem to work. Let's go up to the next level. So this is not close enough. If we assume the next level up, n equals 2 for the common level, and then we have 3, 4, 5, and 6 for nk squared, then we get 1 ninth, 1 sixteenth, 1 twenty-fifth, 1 thirty-sixth. We put those values in and we get this table that I've shown on slide 371, and using this data, we get the plot on the next slide. And now the difference is it's dead on. So I put one blue line and boy, it goes right through all those points, perfectly through them, nice and straight. So there's quite a bit of difference between the two when you view it this way. And if you pick, I'll let you try it on your own. If you pick the common level equals 3 as the terminating level and do the same exercise and plot it, you will see that it is significantly curved again. But not by much. It's curved, but you have to have a critical eye in order to see it, and you have to understand how accurately these plots should come out. If you do a linear regression on this and you take the slope, then the slope M is equal to minus 13.5984 electron volts. That's R, and that's the ionization energy, because if we put N equals 1 in the formula and N equals infinity, the ionization energy is just R. And therefore, the ionization energy for hydrogen, according to this analysis with those four points, should be 13.5984 electron volts. If you go back to the NIST compendium of data, what you'll find out is that we did pretty well. That comes out very close to the exact value of the ionization energy for hydrogen. And doing that with just four points is pretty good, because we didn't assign very many of the transitions. It's especially important to note that it's not enough to fit the data if you don't do a statistical analysis of the fit. If you just get a fit, it can look great, and it can be completely wrong, and you have to do a statistical analysis based on what you think the errors in the data points might be and how well it fits. If you don't bother to do that, and most of the time people don't, they just do some fit, it looks good, they assume it's right, and they quote an answer, you can get really crossed up if you have systems that are in fact very subtle, the curvature subtle, and the data is very, very highly precise. So in these cases, if it's not right on, then your model isn't correct. Keep that in mind when you're doing these kinds of problems. Okay, now I want to talk about coupling schemes, and this has to do with why there were two values of 2P in the NIST database, there was one of them, and then another one, and you think, well, an electron can be in 2P, it can be in 2P, what's the difference? Is it 2PX2? No, it's not, that's not the difference. The difference is that the electron itself has a magnetic moment, and we have to take that into account when we have a nonzero value of L. Well, kind of, in order to talk this through, we'll have to assume a kind of a classical model of the atom, but this is just a heuristic way to kind of look at what's going on. We don't actually believe that something's orbiting around, but anyway, if L is not equal to zero, we can think that something is orbiting around in some sense. If I'm an electron and I'm orbiting around the proton, or the nucleus, if it's a bigger atom, and I go on the electron, so I'm on the electron, boy, is that a wild ride, what I see is I see the proton going around me, because I'm the fixed guy, and the proton's whizzing around me like that the other direction. But the proton has a charge, and so what I see is I see a charge whizzing around me in some kind of a pattern, like a circle or something, and that to me as the electron looks like a current loop. It looks like an electromagnet, but as an electron, I have an intrinsic bar magnet, the spin that we discovered from the Stern-Gerlach experiment, and therefore, my own bar magnet can either be aligned with this magnetic field that I appear to feel, or it can be the other way, and those two could have slightly different energy because there's some magnetic energy. We didn't take that into account in the Hamiltonian when we wrote it down, but in fact, we know that there is, and if we were more sophisticated, we could take it into account. And therefore, those two orientations there pretty much is the resolution of why there's two P levels. They're very close because this magnetic orientation of the electron either being aligned or against this field that the proton appears to induce is very tiny compared to the electrostatic interactions with the potential. And so, it's out in several digits out. But as we'll see, we'll do a calculation probably with sodium, I think this magnetic field that this nucleus appears to generate at the electron as measured by the difference in the energies is huge. And it's bigger than almost any magnet we can make in the lab. So, it has a very big effect because things are moving so fast and it's close by, so it has a very large effect on the electron compared to the kinds of magnetic fields that we can generate in the lab. So, quite a bit bigger than 1 Tesla. 1 Tesla would be the kind of magnetic field that if you went in for an MRI in a whole body scanner and you had a state of the art annular magnet, superconducting magnet, you might have 1 Tesla in there. And it's very much bigger, of course, than any kind of normal magnet that you would ever find. So, this kind of interaction between the spin magnetic moment and the orbital motion is called spin orbit coupling because it's a magnetic interaction between the apparent magnetic field of the nucleus and the intrinsic magnetic field of the spin. Whenever there is more than one electron, the electrons themselves can interact magnetically. They're little bar magnets. They can either align with each other or they can not align. If it turns out, we'll see it when we get to it that if they're in the same orbital, they have to be paired. But if they aren't in the same orbital, if there's an excited electron, there's another one here, then they can have any kind of orientation, either for or against. And if we have an open shell atom with more than one electron, then these spins can add up and we first add up all of the intrinsic bar magnets to give what's called the total spin angular momentum, S. And if there's electrons with nonzero L, then there's orbital angular momentum. If we have more than one of those, we add all those up, and I'll explain how in a second, to give the total orbital angular momentum big L. And then L and S, again magnetically, interact to give the total electronic angular momentum for the atom, which is given the symbol J, L and S, added together. If you take it one stage further, sometimes the nucleus, for example, proton, has a magnetic moment too. And it's been one half like the electron. And in that case, it can either be aligned or not aligned. And we have to add its angular momentum. It's magnetic interaction with J. And in that case, we have to invent a new letter and that new letter is F. We'll usually quit at J, but keep in mind that nuclei do have magnetic properties as well. Their spin is not always one half like the electron. A deuteron has spin one. Chlorine has spin three halves. And if you look very closely at electronic transitions, you actually have to be very careful to take into account the properties of the nucleus to explain what you see. If you try to leave that out, then they're missing pieces out of the puzzle and you don't appear to get the right answer. Here on slide 376 is an image adapted from Wikipedia which shows how you can take these two cones, a big cone for L. L is pointing somewhere. And another cone for S. And you can combine them to get a specific value of J. And then there are 2J plus 1 substates, magnetic substates of J, which are the projections onto the Z axis of the total angular momentum. How do we add these things up to get big L? Well, there's, you, in order to really do this correctly, you have to do a little bit of work with the theory of angular momentum. But as I will explain as we go along, it makes sense that if I've got two electrons, let's say, and they have orbital angular momentum L1 and L2, then I could add them up. They could be parallel and that would be L1 plus L2. So that would be the maximum if everything were co-linear, everything were the wind blowing from behind. And then they could be out of alignment. And because things are quantized, then the next value is L1 plus L2 minus 1. And the worst case is when they're completely misaligned. But the total angular momentum L has to be a positive number because it refers to L squared. And so we terminate the series at the absolute value of L1 minus L2. We say absolute value because we don't know whether L1 is bigger than L2 or less than L2. And so we just put the absolute value there to terminate the series. It terminates at a positive number, which could be zero because L1 and L2 could both be one, for example. This series is called the Klebsch-Gordon series. And the same algebra exactly applies to the total spin angular momentum, big S. Big S can be S1 plus S2 or S1 plus S2 minus 1, blah, blah, blah, down to the absolute value of S1 minus S2. And we can see this in a vector picture keeping in mind that the interaction between these guys is magnetic in nature. And so it's the magnetic little bar magnets that are doing the talking here. And then they're adding up in certain ways to give these overall angular momentum that we observe. Let's, for an example, let's just take two electrons and let's couple the two spin one-halves. We've got the two bar magnets here. Let's couple them together. They're interacting. How could they be? Well, there are four possibilities. They could be like this, like this, like that, or like that because each electron can just be up or down. So I've quoted that in a notation which you may see in a more advanced course and these are called KETs. But for our purposes, we don't need to know what they're called. They just have the arrows in them. They're either two up, up down, down up, or down down. Those are the four possibilities. And we have to see how these can give an overall spin, big S, that's not going to be one-half like the either spin of the electron because the Klebsch-Gordon series says, well, big S could be S1 plus S2, that's one, or it could be S1 plus S2 minus one, that's zero, and that's the absolute value of S1 minus S2 since they're both spin one-half. So we could only have big S be one or zero when we couple the spins together. So let's look at these then from a more pictorial view in a vector diagram. But here's a very nice picture on slide 379. One side in blue is called the singlet. We'll see why in a second. The other side in red is called the triplet. The singlet has one state. The triplet has three states. That part of it makes perfect sense. And we can see I've written these kets next to each orientation. For the triplet, they could be both parallel, both up. They could have one up and one down, but pointing the same direction like that, okay? So this would be an S equals one, S equals zero, and then they could both be pointing like this, that's S equals minus one, M sub S, excuse me. Those are the three values of the M quantum number that would go with an S equals one state, M sub S, big S equals plus one, zero, or minus one. And then the singlet has just M sub S equals zero, because S is zero, so that can be the only value of M sub S. And for that one, we draw it slightly differently. We draw it, instead of drawing it like this, we draw it like that. So we've got one up and the other pointing the other way. They add to zero, but they're kind of like the two guys in the Star Trek time tunnel, they're just stuck battling each other in this state, absolutely anti-symmetric. Instead of calling one of them up, down, and the other one down, up, you may notice that I have this root two over two down, up plus up, down for the triplet, and root two over two up, down minus down, up for the singlet. Why is that? That we have to have them. Well, we know of course that since the electron can go through both slits, it's perfectly feasible for quantum systems to decide that they want to exist in a superposition of what we might think of as the most concrete things, namely, well, it's either down, up or it's up, down. Well, not in a quantum system, it could be 50% of that and 50% of the other one together. That could be the solution. We see then we've got four states and they break apart into three and one, and that's a common pattern because the odd numbers add up to a perfect square. And so if we have a certain angular momentum choice, three here and three here, we get nine and we can write that as a sum of one, three and five. And that's basically what the Klebsch-Gordon series says. It turns out that neither of the states up, down or down, up will do and the reason why is because these guys either in the triplet or the singlet or in other systems, they have to have the same stripes. They have to have the same characteristics with respect to swapping the particles. The particles are identical, so which one we call one and two is unimportant. But if we have two of them up and we swap them, we get the same thing. And if we have two of them down and we swap them, we get the same thing. But if we then say, well, I think the middle one then for the triplet is up, down and we swap them, we get a new guy, we get the other one, which wasn't part of the mix and we can't have that. We have to, when we swap them, we have to get the same thing because it's part of the same series. It has to have the same symmetry as the other ones. Otherwise it doesn't work. And therefore, we have to take a combination between up, down and down, up such that when we swap them over, we get the same thing as the M equals zero state of the triplet. It's, unless you've seen how to do this, it's not, not quite so easy to figure out how you might do it. But let's explore this symmetry by artificially coloring the arrows, red and blue. That way we'll keep track of them. When we swap them, we'll see what we get in terms of whether we get the same thing or not. At the top here on slide 381, I've shown a picture with the coloring in. We start with, we know we can't have one of them because it switches to the other. So we know we've got to have both of them and we know we've got to have 50% of each. Let's forget about the root 2 over 2. That's the normalization constant just to keep the probability of being in some state equal to 1. And let's just take the combination of adding them. Let's take up in red, down in blue, plus down in red, up in blue. And then let's swap the positions of the two arrows. What we get then in the first one, where we're up in red and down in blue, if we swap them, we get down in blue, up in red. Now in actual fact, the red and blue don't matter, but it lets us keep track of it. So we get down up. The other one, which was down in red, up in blue, we get up in blue, down in red. And then if we just change the order, we find we get up in blue, down in red, plus down in blue, up in red. That's what we started with except the colors are swapped. But the colors don't matter because the electrons are identical particles. We're just coloring them so that we can keep track of what we're doing, otherwise we can't tell what we've swapped. And the key is that we get plus one. We get the same thing. The eigenvalue for this swapping is plus one. So that's kind of a parity value. Plus one for swapping this way, plus one for this one, plus one for swapping that way, they all go together. All three of them are birds of a feather. If one of them is plus, then the other one to be orthogonal can't be plus as well. And therefore it has to be minus. And if we take that combination, which I've shown in the second line here, and we take up down, minus down up, and we swap them, we get down up, minus up down. And that's equal to minus one times what we had before, because of the fact that they're opposite. And what that means then is that the singlet has different symmetry than the triplet. And so although they both have M equals zero states, the states are different because one is symmetric under exchange and the other is anti-symmetric under exchange. In more complex systems, you have to look at the symmetry of each of the states and decide if it fits in or not. And there are well-documented procedures for doing exactly that if you have to. So whether you're symmetric or anti-symmetric under exchange is a very important property. It's not just trivia. Okay, suppose then we have these values. We've seen we can take S2S1 half, spin one half particles. We can get S equals one, S equals zero. We can do the same thing with L, and we can follow that through the same way. Now we've got big L and big S. How do we know what values of J will result? Big L is something, could be two. Big S is something, could be one or three halves, depending how many particles we have. And we want to figure out what values the total angular momentum of the atom J can be. Well, the answer is it's pretty much the same thing again because angular momentum always adds in exactly the same way. Doesn't really matter what it is. Well, if L and S happen to align, that's the maximum of value of J, L plus S. And then if they don't, then we decrement by one quantum, and we have L plus S minus one and so forth. And we go down to, again, the absolute value of L minus S because we don't know whether L or S is bigger. But we know that J referring to the square of the angular momentum or the square root of the square of the angular momentum, if you like, has to be a positive thing. And so we can't have anything go negative. And therefore, we get another Klebsch-Gordon series, L plus S, L plus S minus one and so forth. And therefore, what this means is that if you have a certain configuration, an electronic configuration of an atom that's open-shell that has several electrons, depending on the details of how these magnetic moments are interacting, and since these energies are small, then if I have something like a blast, like an arc, an electric arc or something like that, they're all going to be there. That's not like one of them is way high compared to the others. If there's enough energy to populate the excited state, there's probably enough energy to populate all the possibilities randomly, statistically, just depending on how many possibilities there are for each state. And therefore, we need some way to keep track of these things in order to figure out what kinds of emission lines we're going to see in the spectrum and to assign it. And the way we do that is with something called a term symbol. The term symbol summarizes not only what state the atom is in, but tells you what L, S, and J are. And the term symbol has the structure 2S plus 1 on the left, big L, which is like a letter like S, P, and so forth, just like for the electron itself, little S, little P. This is big L, the total angular momentum, orbital angular momentum. And then J as a right subscript to tell you which way S and L are adding up to give the result for the total angular momentum of the atom. And in order to figure out what kinds of transitions you're going to see, you have to know what the term symbol is. That's by far the best and most concise way to look at things. And therefore, you have to know what these are. 2S plus 1 is called the multiplicity. And if S is 0, it's called singlet. Because there's one line, one state there, like we saw with the S equals 0 for the two electrons. S equals 1 is called triplet. S equals 1 half, if we have that. That's called doublet and so on. We can have all the way up. And let's try an example with this then and see how we might use this symbol to figure out. So let's take, here's practice problem 19. Let's look at the sodium emission spectrum. The ground state of sodium is got a 3S electron outside a closed shell. The first excited state is 3P because in the sodium electron, 3P and 3S are quite a bit far apart. And so that turns out to be a transition that has visible light. It's not very tiny transition energy like radio waves. So we can see that easily in the sodium spectrum. The question is, if we have the electron in the 3P state outside a closed shell, what terms can arise? In other words, what are the possibilities for these term symbols? Which way can L, S and J add up? Well, for the P state, little L is 1. There's one electron and therefore big L is 1 because the Klebsch-Gordon series terminates. Big L is little L in this case because there's just one electron. Why don't we include all the other electrons in sodium? They're inside a closed shell. And inside a closed shell, the orbital angular momentum is zero and the spin angular momentum is zero. And so inside a closed shell, you just forget about it, throw it away. You don't want to include all those or you'll spend the rest of your life playing around. And there's one electron. So S is 1 half and therefore the two possible values of J are 1 plus a half and 1 minus a half, absolute value. That's it. Two possible values of J are 3 halves and 1 half and therefore the two terms are 2S plus 1, 2 times a half plus 1 is 2, that's doublet P, 3 halves and doublet P, 1 half. Those are the two terms that can arise. So as I said, closed shells ignore them. They have a total angular momentum zero. The two terms have close but not identical energy. Both of them, it turns out, we'll see, can emit a photon to the ground state, 3S. And therefore we get two closely spaced lines which are called the sodium D lines, which are quite famous because they were studied very early. In fact, even before the electron, it was figured out that the electron existed as a particle by Thompson. The year before, Zaman was looking at sodium and putting on magnetic fields and looking at how these lines changed and working out a ton of information from these yellow lines, which were very easy to excite, for example, with sodium vapor with a flame. And that's, of course, the sodium, the flame test to see what kind of element it is. You put a flame on and you look at the color and you can tell if it's yellow, it's sodium. These are also the characteristic yellow color of the fog lamps you see down toward the beach in places like San Diego or at least you used to. They might not be so energy efficient now. But the yellow color was thought to be much better when you had fog because if you have white light, sometimes what you see with very strong white light is you just see the fog. You don't see through the fog. You instead light up the little droplets and you get a lot of reflection from the fog so you don't see the pedestrian. Whereas if you have the yellow color without the white, which is why you use the sodium because you have these two lines, then you just see the pedestrian and then you put on the brakes. The term symbol gives us a really quick way to tell whether a transition is quote allowed or forbidden according to the electric dipole selection rules. In light atoms, L and S are still pretty good quantum numbers. The reason why they aren't perfect quantum numbers is that L refers to orbital angular momentum that came from hydrogen. In hydrogen, everything is spherically symmetric because there's that one proton and one electron. But in other atoms, there are other electrons and they muck things up. And what that means is that L is not quite so good because things are wobbling a little. And so although, yeah, it's going around, but there's a little bit of wobble and then there's S in there and these magnetic interactions with more than one particle. And so they're 90% good but not 100% good anymore and that means that you can get some forbidden transitions and if you look closely, you'll see some weak things that you can explain by that. But one thing we know for sure if we're considering it's the electric part of the light wave that's moving the electron around. The electric part of the light wave can't change the magnetic bar magnet of the electron. And therefore S, big S, can't change in the transition. So delta big S has to be zero. And because the photon has one unit of angular momentum, delta big L has to be plus or minus one. And putting those together, we can get the delta J is zero or plus or minus one. However, if J starts at zero, we can't terminate a J equals zero because then it's impossible to satisfy the delta L selection rule as well. So in that case, we get shut down. So those are the electric dipole selection rules in terms of the term symbol. Delta S is zero, delta L plus or minus one, and delta J is zero plus or minus one. We'll explore in the next lecture how we can use this notation to analyze some of these spectra.
 Welcome back to Chemistry 131A.  Today we're going to continue our exposition  of atomic spectroscopy focusing on selection rules,  coupling, and terms.  Of course, atomic spectroscopy is the basic tool  by which everybody learned that the atom was quantized  and so it's of very central importance in this whole area  of endeavor of quantum mechanics applied to small things.  Various sets of transitions are observed.  Here is energy level diagram for the hydrogen atom, not to scale  because if you do it to scale, the levels crowd together too much  at the top and it's very hard to see what's going on.  There's a series called the Lyman series  in which an electron jumps from the n equals 2 to n equals 1 or 3  to 1 or 4 to 1 and so forth.  And then there's another series, the Ballmer series,  that's in the visible range in which the electron jumps  to the n equals 2 state and then there are other series  and they have names according to who worked on them.  The main thing to keep in mind here is that any change  in the quantum number n in the energy is allowed in principle.  The only thing that has to be satisfied is  that there has to be conservation of energy  and that's taken care of automatically  because the photon is emitted with a quantum h nu  which is exactly equal to the energy loss of the atom.  So energy is conserved.  In this diagram, the zero reference energy is at the top  and all the energies quoted there are negative being more stable  than a proton and an electron at rest at infinite separation.  The selection rules are a little bit more complicated, however,  and the first thing is that a photon has one unit  of angular momentum, h bar.  Angular momentum is conserved and so when the electron makes  a transition there has to be a change  by one unit of angular momentum.  So that means that although we draw these things coming  down to the one, in fact they're all coming  from specific values of L. It's 2p to 1s, 3p to 1s,  4p to 1s and so forth and it's always p. We can't have 2s  to 1s because then delta L is zero and that's not allowed.  That doesn't conserve angular momentum  and likewise we can't have 3d to 1s either  because then delta L is equal to 2 and again  that doesn't conserve angular momentum.  Now if we have a different kind of transition or if more  than one photon is emitted that's different because then,  of course, that could be allowed  because angular momentum could be conserved  in a more complicated kind of process.  Those kind of processes do happen all the time.  They're usually just much slower than the single photon  electric dipole allowed transition.  If we have transitions to 2p in the Bombers series,  for example, they could be from 3s or 3d.  If we have transitions to the 2s state then they have  to be again from 3p, 4p and so on.  And in general, delta L is plus or minus 1  for electric dipole allowed transitions  and these are generally the very strongest transitions  that you see in the spectrum.  Not always, however, if the atom is very heavy like mercury  as we'll see in the next lecture,  we could have a strong transition  that doesn't seem to follow this rule.  There are weaker transitions,  magnetic dipole transitions  and electric quadrupole transitions.  We won't touch on those much but you should keep in mind  that those can and do occur and therefore you mustn't just assume  that this is an ironclad rule.  When we use the terms allowed and forbidden in spectroscopy,  what we're talking about really is strong and weak.  Allowed is the light is green, forbidden is it's red  but some people run red lights.  Here's a compendium from the NIST database just  to show you how accurate and fastidious these spectroscopists  have been in cataloging these energy levels.  This one has a different energy zero.  In this case, energies are quoted from the 1S level  of the hydrogen atom being zero and then up from there.  But the amazing thing about these numbers is  that they have six, seven, eight, and nine digits of accuracy.  So you can see that they had to take a lot of precautions  when they were doing these.  And if you look closely, there are multiple entries.  There's 2P and then there's 2S  and there's just 2 and then there's 2P again and so forth.  And we'll see in a second what these things mean.  But you have to keep track of individual substates  in these systems and oftentimes you put on a magnetic field  to be able to tell which state is which  and how they move around to keep track of them.  And they go all the way up to 6H.  So in the periodic table, we don't go beyond F. But if we go  to excited states of atoms, it's easy to go very high  and to have very high values of the angular momentum L.  If you look closely, you find out that the first 2P state  that's quoted is ever so slightly lower in energy  than 2S, which is quite a surprise  because in other elements, of course, 2S is lower than 2P.  That's why lithium is 2S1 and not 2P1, for example.  And we don't fill up the 2P until we get over to boron.  But in hydrogen, it's kind of an anomaly,  one of those 2P states is lower than the 2S.  And that was extremely interesting to theoreticians  how that could possibly be.  And in studying that in great detail,  they worked out important clues about actually the residual  electromagnetic field that permeates our entire universe  like the zero point energy of a harmonic oscillator.  You can think of the electromagnetic field as some kind  of electromagnetic oscillation.  And it turns out that even in a vacuum at absolute zero,  in this universe, apparently, we can't have nothing.  There's still some zero point energy left over.  And that can bounce the electron around.  And then the question is, if it bounces a 2S electron around  or bounces a 2P electron around, what's the difference?  And if you actually follow through the calculation,  it's actually very, very beautiful work to follow.  And you can actually derive that the 2P should be lower.  And we can use the measured energy.  Once we have a series of these states,  we can use the measured energy, if we're careful,  to estimate the ionization potential,  also called the ionization energy.  And that's just an extrapolation  to what the energy would be at n equals infinity.  And that's very important to know what that number is.  That's one of the main things that's listed  in freshman chemistry books in the periodic table to try  to account for various trends.  If an electron is easy to ionize, and that means  that that element's much more likely to give an electron  up to another atom than an element  that has an electron that's extremely hard to ionize,  for example.  Let's try a practice problem then and have a look  at how we might tackle estimating the ionization energy.  Practice problem 18.  Consider the following unassigned,  that means we don't know n or l, emission lines  from a hydrogen arc, all of which terminate  in a common level, assign them  and estimate the ionization potential  or the ionization energy, same thing, of hydrogen.  While the hydrogen arc is just we put a big lightning strike,  a voltage through, and we excite the atoms  and then they emit light.  And if they terminate in a common level,  that means they're like the other diagram I drew.  We have arrows coming down to this, to this, to this,  and so forth, always to the same level.  And that means the change in energy has to do  with the upper level but not the lower level.  Of course, in actual fact, you just get all kinds of lines  and you have to be very clever to figure  out which ones go with which and which go together.  There are various tricks that were worked out to do that.  So here are the numbers.  They all have four or five digits in them, 1.8887  and so forth.  And let's just take these four numbers,  quoted in electron volts, and let's figure  out what the ionization energy should be.  The first thing is we have to figure  out what common level they terminate in.  And if we don't do that, then we aren't going to be able  to figure out anything.  And so the way we're going to do that is we first write this  formula that delta E is equal to the Rydberg constant times 1  over n squared common level minus 1 over n k squared  where nk is some higher number, higher up dropping down.  We don't know what the common level is, however.  And we don't know what the value of r is  because it wasn't given.  And on an exam, if you aren't given the value of r  and a problem like this, that means you're supposed  to figure it out without that.  So let's try to do that.  We don't know n common.  We don't know r. We have some values of delta E. And we have  to try to figure out if they fit this formula.  A very bad way is to just start plugging things in  and seeing what happens.  A very good way is to proceed systematically and figure  out if we can get a straight line plot.  Chemists love straight lines.  Before computers, straight lines were absolutely essential.  And the reason why is that the human eye can distinguish a  straight line from any curve.  But the human eye cannot distinguish curves  that are slightly different.  Even if they're statistically significantly different,  it's very hard to tell just by looking.  And before computers, people had to plot things  on paper often and look at them.  And in order to tell what's what,  if you could organize your equation so that it came  into a straight line plot, then you could tell.  You could tell if there was a systematic deviation,  some kind of slight curvature one way or the other way.  You could tell if there was scatter, noise, and so forth.  And all those are very important to quantify  when you're trying to estimate something.  We can cast our equation into a straight line form  by letting y equals to delta E,  letting x equal 1 over n squared,  and then letting the intercept b equals 1 over,  rather, r over n common.  And we can guess values of nk and see which look good.  So we have to assume a value of n common,  and then we guess values of nk all greater than that,  and we see what looks good.  Once the line is straight, then we can extract the slope,  y equals b plus mx, and m in this case should be minus r,  the Rydberg constant.  Once the line is straight, then we can get r.  Now the lowest common level could be n equals 1,  so in the absence of any other information,  we should assume that first and then work our way up.  If the common level is very high, really, really high,  this would be a disaster because we would take forever.  But we know that it can't be that high  because we see a couple of electron volts,  and so we know it has to be one of the levels further down.  If it's very high, then all the transitions would be very,  very small numbers because we'd already be most of the way  up to the ionization energy.  So let's try n common equals 1, and in that case,  we could have nk, the upper level, b2, 3, 4, and so forth.  If some transitions are missing for some reason,  we just didn't see them, they were faint, we made an error,  some other reason, then this kind  of exercise can get very frustrating  because the points don't appear to fit,  and if you have to start deleting values,  then it's very tedious, but in any kind of exam problem,  we'll never have something like that.  But in the lab, for various reasons, it can be  that some transitions are dark, something happens,  there's another pathway where the electron can go,  and it's very hard to see that, and sometimes it takes a lot  of patience to be able to work out what's going on.  So let's plot these.  I drew up a table, I have the delta E values for y,  and then I have the 1 over nk squared, and for 2, it's a quarter,  for 3, it's a ninth, and so forth as we go up.  And if we plot those, which I've done here, well, first  of all, we notice that the slope is negative, excuse me,  and that's, of course, exactly what we expect  because the slope should be negative r,  and r is a positive number, so that part's okay.  And let's plot them now, and here I've drawn a plot.  Now, there are four black dots on this plot on slide 370,  and if you just draw a line through the four black dots  by linear regression, you will get a pretty good fit.  In other words, there is a line that misses them all,  but is pretty good, and so you have to decide if that kind  of line is going to be good enough for what you're doing,  and in this case, it's not nearly good enough.  And what I've tried to do is emphasize that by first drawing a  line through the first two points on the left, the red line,  that has a certain slope, and keep in mind  that these points have very, very, very tiny errors.  So although I've put these black dots so that you can see them  on the slide, the points themselves are very,  very accurately determined.  We draw that red line, and then we compare that with the slope  from the last two points, the blue line.  We only have four points, so we have to be kind of careful,  and we plot that, and when you do it that way, what you see is  that the blue line has a much shallower negative slope  than the red line, that it is quite a bit different, and so there,  and if you took the center two points,  it would have a slope in between,  and what that means is that this is curved.  It's not n equals, n comma equals 1, and this is not good enough,  but you could easily assume it worked good enough if you're used  to other kinds of data that isn't this well determined.  In any case, we're going to discard that and say,  well, n comma equals 1 doesn't seem to work.  Let's go up to the next level.  So this is not close enough.  If we assume the next level up, n equals 2 for the common level,  and then we have 3, 4, 5, and 6 for nk squared,  then we get 1 ninth, 1 sixteenth, 1 twenty-fifth, 1 thirty-sixth.  We put those values in and we get this table  that I've shown on slide 371, and using this data,  we get the plot on the next slide.  And now the difference is it's dead on.  So I put one blue line and boy, it goes right  through all those points, perfectly through them, nice  and straight.  So there's quite a bit of difference between the two  when you view it this way.  And if you pick, I'll let you try it on your own.  If you pick the common level equals 3 as the terminating level  and do the same exercise and plot it, you will see  that it is significantly curved again.  But not by much.  It's curved, but you have to have a critical eye in order  to see it, and you have to understand how accurately  these plots should come out.  If you do a linear regression on this  and you take the slope, then the slope M is equal  to minus 13.5984 electron volts.  That's R, and that's the ionization energy,  because if we put N equals 1 in the formula  and N equals infinity, the ionization energy is just R.  And therefore, the ionization energy for hydrogen,  according to this analysis with those four points,  should be 13.5984 electron volts.  If you go back to the NIST compendium of data,  what you'll find out is that we did pretty well.  That comes out very close to the exact value  of the ionization energy for hydrogen.  And doing that with just four points is pretty good,  because we didn't assign very many of the transitions.  It's especially important to note  that it's not enough to fit the data  if you don't do a statistical analysis of the fit.  If you just get a fit, it can look great,  and it can be completely wrong, and you have  to do a statistical analysis based on what you think the errors  in the data points might be and how well it fits.  If you don't bother to do that, and most  of the time people don't, they just do some fit,  it looks good, they assume it's right, and they quote an answer,  you can get really crossed up if you have systems  that are in fact very subtle, the curvature subtle,  and the data is very, very highly precise.  So in these cases, if it's not right on,  then your model isn't correct.  Keep that in mind when you're doing these kinds of problems.  Okay, now I want to talk about coupling schemes,  and this has to do with why there were two values of 2P  in the NIST database, there was one of them,  and then another one, and you think, well,  an electron can be in 2P, it can be in 2P,  what's the difference?  Is it 2PX2?  No, it's not, that's not the difference.  The difference is that the electron itself has a magnetic  moment, and we have to take that into account  when we have a nonzero value of L. Well, kind of,  in order to talk this through, we'll have to assume a kind  of a classical model of the atom, but this is just a heuristic  way to kind of look at what's going on.  We don't actually believe that something's orbiting around,  but anyway, if L is not equal to zero, we can think  that something is orbiting around in some sense.  If I'm an electron and I'm orbiting around the proton,  or the nucleus, if it's a bigger atom, and I go on the electron,  so I'm on the electron, boy, is that a wild ride,  what I see is I see the proton going around me,  because I'm the fixed guy, and the proton's whizzing  around me like that the other direction.  But the proton has a charge, and so what I see is I see a charge  whizzing around me in some kind of a pattern,  like a circle or something, and that to me  as the electron looks like a current loop.  It looks like an electromagnet, but as an electron,  I have an intrinsic bar magnet, the spin that we discovered  from the Stern-Gerlach experiment, and therefore,  my own bar magnet can either be aligned  with this magnetic field that I appear to feel,  or it can be the other way, and those two could have slightly  different energy because there's some magnetic energy.  We didn't take that into account in the Hamiltonian  when we wrote it down, but in fact, we know that there is,  and if we were more sophisticated,  we could take it into account.  And therefore, those two orientations there pretty much is  the resolution of why there's two P levels.  They're very close because this magnetic orientation  of the electron either being aligned or against this field  that the proton appears to induce is very tiny compared  to the electrostatic interactions with the potential.  And so, it's out in several digits out.  But as we'll see, we'll do a calculation probably  with sodium, I think this magnetic field  that this nucleus appears to generate at the electron as measured  by the difference in the energies is huge.  And it's bigger than almost any magnet we can make in the lab.  So, it has a very big effect because things are moving so fast  and it's close by, so it has a very large effect on the electron  compared to the kinds of magnetic fields  that we can generate in the lab.  So, quite a bit bigger than 1 Tesla.  1 Tesla would be the kind of magnetic field that if you went  in for an MRI in a whole body scanner and you had a state  of the art annular magnet, superconducting magnet,  you might have 1 Tesla in there.  And it's very much bigger, of course,  than any kind of normal magnet that you would ever find.  So, this kind of interaction between the spin magnetic moment  and the orbital motion is called spin orbit coupling  because it's a magnetic interaction  between the apparent magnetic field of the nucleus  and the intrinsic magnetic field of the spin.  Whenever there is more than one electron,  the electrons themselves can interact magnetically.  They're little bar magnets.  They can either align with each other or they can not align.  If it turns out, we'll see it when we get to it  that if they're in the same orbital, they have to be paired.  But if they aren't in the same orbital,  if there's an excited electron, there's another one here,  then they can have any kind of orientation, either for or against.  And if we have an open shell atom with more than one electron,  then these spins can add up and we first add up all  of the intrinsic bar magnets to give what's called the total spin  angular momentum, S. And if there's electrons with nonzero L,  then there's orbital angular momentum.  If we have more than one of those, we add all those up,  and I'll explain how in a second,  to give the total orbital angular momentum big L. And then L  and S, again magnetically, interact to give the total  electronic angular momentum for the atom,  which is given the symbol J, L and S, added together.  If you take it one stage further, sometimes the nucleus,  for example, proton, has a magnetic moment too.  And it's been one half like the electron.  And in that case, it can either be aligned or not aligned.  And we have to add its angular momentum.  It's magnetic interaction with J. And in that case,  we have to invent a new letter and that new letter is F. We'll  usually quit at J, but keep in mind  that nuclei do have magnetic properties as well.  Their spin is not always one half like the electron.  A deuteron has spin one.  Chlorine has spin three halves.  And if you look very closely at electronic transitions,  you actually have to be very careful to take  into account the properties of the nucleus  to explain what you see.  If you try to leave that out, then they're missing pieces  out of the puzzle and you don't appear  to get the right answer.  Here on slide 376 is an image adapted from Wikipedia  which shows how you can take these two cones,  a big cone for L. L is pointing somewhere.  And another cone for S. And you can combine them  to get a specific value of J. And then there are 2J plus 1  substates, magnetic substates of J,  which are the projections onto the Z axis  of the total angular momentum.  How do we add these things up to get big L?  Well, there's, you, in order to really do this correctly,  you have to do a little bit of work with the theory  of angular momentum.  But as I will explain as we go along, it makes sense  that if I've got two electrons, let's say,  and they have orbital angular momentum L1 and L2,  then I could add them up.  They could be parallel and that would be L1 plus L2.  So that would be the maximum if everything were co-linear,  everything were the wind blowing from behind.  And then they could be out of alignment.  And because things are quantized, then the next value is L1  plus L2 minus 1.  And the worst case is when they're completely misaligned.  But the total angular momentum L has to be a positive number  because it refers to L squared.  And so we terminate the series at the absolute value  of L1 minus L2.  We say absolute value because we don't know whether L1 is bigger  than L2 or less than L2.  And so we just put the absolute value there  to terminate the series.  It terminates at a positive number, which could be zero  because L1 and L2 could both be one, for example.  This series is called the Klebsch-Gordon series.  And the same algebra exactly applies  to the total spin angular momentum, big S. Big S can be S1  plus S2 or S1 plus S2 minus 1, blah, blah, blah,  down to the absolute value of S1 minus S2.  And we can see this in a vector picture keeping in mind  that the interaction between these guys is magnetic in nature.  And so it's the magnetic little bar magnets  that are doing the talking here.  And then they're adding up in certain ways  to give these overall angular momentum that we observe.  Let's, for an example, let's just take two electrons  and let's couple the two spin one-halves.  We've got the two bar magnets here.  Let's couple them together.  They're interacting.  How could they be?  Well, there are four possibilities.  They could be like this, like this, like that,  or like that because each electron can just be up or down.  So I've quoted that in a notation which you may see  in a more advanced course and these are called KETs.  But for our purposes, we don't need to know what they're called.  They just have the arrows in them.  They're either two up, up down, down up, or down down.  Those are the four possibilities.  And we have to see how these can give an overall spin, big S,  that's not going to be one-half like the either spin  of the electron because the Klebsch-Gordon series says,  well, big S could be S1 plus S2, that's one,  or it could be S1 plus S2 minus one, that's zero,  and that's the absolute value of S1 minus S2  since they're both spin one-half.  So we could only have big S be one or zero  when we couple the spins together.  So let's look at these then from a more pictorial view  in a vector diagram.  But here's a very nice picture on slide 379.  One side in blue is called the singlet.  We'll see why in a second.  The other side in red is called the triplet.  The singlet has one state.  The triplet has three states.  That part of it makes perfect sense.  And we can see I've written these kets next to each orientation.  For the triplet, they could be both parallel, both up.  They could have one up and one down,  but pointing the same direction like that, okay?  So this would be an S equals one, S equals zero,  and then they could both be pointing like this,  that's S equals minus one, M sub S, excuse me.  Those are the three values of the M quantum number  that would go with an S equals one state, M sub S,  big S equals plus one, zero, or minus one.  And then the singlet has just M sub S equals zero,  because S is zero, so that can be the only value of M sub S.  And for that one, we draw it slightly differently.  We draw it, instead of drawing it like this, we draw it like that.  So we've got one up and the other pointing the other way.  They add to zero, but they're kind of like the two guys  in the Star Trek time tunnel, they're just stuck battling each other  in this state, absolutely anti-symmetric.  Instead of calling one of them up, down,  and the other one down, up, you may notice  that I have this root two over two down,  up plus up, down for the triplet, and root two over two up,  down minus down, up for the singlet.  Why is that?  That we have to have them.  Well, we know of course that since the electron can go  through both slits, it's perfectly feasible for quantum systems  to decide that they want to exist in a superposition  of what we might think of as the most concrete things, namely,  well, it's either down, up or it's up, down.  Well, not in a quantum system, it could be 50% of that  and 50% of the other one together.  That could be the solution.  We see then we've got four states and they break apart  into three and one, and that's a common pattern  because the odd numbers add up to a perfect square.  And so if we have a certain angular momentum choice,  three here and three here, we get nine and we can write  that as a sum of one, three and five.  And that's basically what the Klebsch-Gordon series says.  It turns out that neither of the states up, down or down,  up will do and the reason why is because these guys either  in the triplet or the singlet or in other systems,  they have to have the same stripes.  They have to have the same characteristics with respect  to swapping the particles.  The particles are identical,  so which one we call one and two is unimportant.  But if we have two of them up and we swap them,  we get the same thing.  And if we have two of them down and we swap them,  we get the same thing.  But if we then say, well, I think the middle one then  for the triplet is up, down and we swap them,  we get a new guy, we get the other one,  which wasn't part of the mix and we can't have that.  We have to, when we swap them, we have to get the same thing  because it's part of the same series.  It has to have the same symmetry as the other ones.  Otherwise it doesn't work.  And therefore, we have to take a combination between up,  down and down, up such that when we swap them over,  we get the same thing as the M equals zero state of the triplet.  It's, unless you've seen how to do this, it's not,  not quite so easy to figure out how you might do it.  But let's explore this symmetry by artificially coloring the arrows,  red and blue.  That way we'll keep track of them.  When we swap them, we'll see what we get in terms  of whether we get the same thing or not.  At the top here on slide 381, I've shown a picture  with the coloring in.  We start with, we know we can't have one of them  because it switches to the other.  So we know we've got to have both of them  and we know we've got to have 50% of each.  Let's forget about the root 2 over 2.  That's the normalization constant just to keep the probability  of being in some state equal to 1.  And let's just take the combination of adding them.  Let's take up in red, down in blue,  plus down in red, up in blue.  And then let's swap the positions of the two arrows.  What we get then in the first one,  where we're up in red and down in blue, if we swap them,  we get down in blue, up in red.  Now in actual fact, the red and blue don't matter,  but it lets us keep track of it.  So we get down up.  The other one, which was down in red, up in blue,  we get up in blue, down in red.  And then if we just change the order, we find we get up  in blue, down in red, plus down in blue, up in red.  That's what we started with except the colors are swapped.  But the colors don't matter  because the electrons are identical particles.  We're just coloring them so that we can keep track  of what we're doing, otherwise we can't tell what we've swapped.  And the key is that we get plus one.  We get the same thing.  The eigenvalue for this swapping is plus one.  So that's kind of a parity value.  Plus one for swapping this way, plus one for this one,  plus one for swapping that way, they all go together.  All three of them are birds of a feather.  If one of them is plus, then the other one  to be orthogonal can't be plus as well.  And therefore it has to be minus.  And if we take that combination, which I've shown  in the second line here, and we take up down, minus down up,  and we swap them, we get down up, minus up down.  And that's equal to minus one times what we had before,  because of the fact that they're opposite.  And what that means then is that the singlet has different  symmetry than the triplet.  And so although they both have M equals zero states,  the states are different because one is symmetric  under exchange and the other is anti-symmetric  under exchange.  In more complex systems, you have to look at the symmetry  of each of the states and decide if it fits in or not.  And there are well-documented procedures  for doing exactly that if you have to.  So whether you're symmetric or anti-symmetric  under exchange is a very important property.  It's not just trivia.  Okay, suppose then we have these values.  We've seen we can take S2S1 half, spin one half particles.  We can get S equals one, S equals zero.  We can do the same thing with L, and we can follow  that through the same way.  Now we've got big L and big S. How do we know what values  of J will result?  Big L is something, could be two.  Big S is something, could be one or three halves,  depending how many particles we have.  And we want to figure out what values the total angular momentum  of the atom J can be.  Well, the answer is it's pretty much the same thing again  because angular momentum always adds  in exactly the same way.  Doesn't really matter what it is.  Well, if L and S happen to align, that's the maximum of value  of J, L plus S. And then if they don't,  then we decrement by one quantum, and we have L plus S minus one  and so forth.  And we go down to, again, the absolute value of L minus S  because we don't know whether L or S is bigger.  But we know that J referring to the square of the angular momentum  or the square root of the square of the angular momentum,  if you like, has to be a positive thing.  And so we can't have anything go negative.  And therefore, we get another Klebsch-Gordon series,  L plus S, L plus S minus one and so forth.  And therefore, what this means is that if you have a certain  configuration, an electronic configuration of an atom  that's open-shell that has several electrons,  depending on the details of how these magnetic moments are  interacting, and since these energies are small,  then if I have something like a blast, like an arc,  an electric arc or something like that,  they're all going to be there.  That's not like one of them is way high compared to the others.  If there's enough energy to populate the excited state,  there's probably enough energy to populate all the possibilities  randomly, statistically, just depending  on how many possibilities there are for each state.  And therefore, we need some way to keep track of these things  in order to figure out what kinds of emission lines we're going  to see in the spectrum and to assign it.  And the way we do that is with something called a term symbol.  The term symbol summarizes not only what state the atom is in,  but tells you what L, S, and J are.  And the term symbol has the structure 2S plus 1 on the left,  big L, which is like a letter like S, P, and so forth,  just like for the electron itself, little S, little P.  This is big L, the total angular momentum,  orbital angular momentum.  And then J as a right subscript to tell you which way S  and L are adding up to give the result  for the total angular momentum of the atom.  And in order to figure out what kinds  of transitions you're going to see,  you have to know what the term symbol is.  That's by far the best and most concise way to look at things.  And therefore, you have to know what these are.  2S plus 1 is called the multiplicity.  And if S is 0, it's called singlet.  Because there's one line, one state there, like we saw  with the S equals 0 for the two electrons.  S equals 1 is called triplet.  S equals 1 half, if we have that.  That's called doublet and so on.  We can have all the way up.  And let's try an example with this  then and see how we might use this symbol to figure out.  So let's take, here's practice problem 19.  Let's look at the sodium emission spectrum.  The ground state of sodium is got a 3S electron  outside a closed shell.  The first excited state is 3P because in the sodium electron,  3P and 3S are quite a bit far apart.  And so that turns out to be a transition  that has visible light.  It's not very tiny transition energy like radio waves.  So we can see that easily in the sodium spectrum.  The question is, if we have the electron in the 3P state  outside a closed shell, what terms can arise?  In other words, what are the possibilities  for these term symbols?  Which way can L, S and J add up?  Well, for the P state, little L is 1.  There's one electron and therefore big L is 1  because the Klebsch-Gordon series terminates.  Big L is little L in this case  because there's just one electron.  Why don't we include all the other electrons in sodium?  They're inside a closed shell.  And inside a closed shell,  the orbital angular momentum is zero  and the spin angular momentum is zero.  And so inside a closed shell, you just forget about it,  throw it away.  You don't want to include all those  or you'll spend the rest of your life playing around.  And there's one electron.  So S is 1 half and therefore the two possible values  of J are 1 plus a half and 1 minus a half, absolute value.  That's it.  Two possible values of J are 3 halves and 1 half  and therefore the two terms are 2S plus 1,  2 times a half plus 1 is 2, that's doublet P, 3 halves  and doublet P, 1 half.  Those are the two terms that can arise.  So as I said, closed shells ignore them.  They have a total angular momentum zero.  The two terms have close but not identical energy.  Both of them, it turns out, we'll see,  can emit a photon to the ground state, 3S.  And therefore we get two closely spaced lines  which are called the sodium D lines,  which are quite famous because they were studied very early.  In fact, even before the electron, it was figured  out that the electron existed as a particle by Thompson.  The year before, Zaman was looking at sodium  and putting on magnetic fields and looking  at how these lines changed and working out a ton  of information from these yellow lines,  which were very easy to excite, for example,  with sodium vapor with a flame.  And that's, of course, the sodium, the flame test  to see what kind of element it is.  You put a flame on and you look at the color  and you can tell if it's yellow, it's sodium.  These are also the characteristic yellow color  of the fog lamps you see down toward the beach in places  like San Diego or at least you used to.  They might not be so energy efficient now.  But the yellow color was thought to be much better  when you had fog because if you have white light,  sometimes what you see with very strong white light is you  just see the fog.  You don't see through the fog.  You instead light up the little droplets and you get a lot  of reflection from the fog so you don't see the pedestrian.  Whereas if you have the yellow color without the white,  which is why you use the sodium  because you have these two lines, then you just see the  pedestrian and then you put on the brakes.  The term symbol gives us a really quick way  to tell whether a transition is quote allowed or forbidden  according to the electric dipole selection rules.  In light atoms, L and S are still pretty good quantum numbers.  The reason why they aren't perfect quantum numbers is  that L refers to orbital angular momentum that came  from hydrogen. In hydrogen, everything is spherically  symmetric because there's that one proton and one electron.  But in other atoms, there are other electrons  and they muck things up.  And what that means is that L is not quite so good  because things are wobbling a little.  And so although, yeah, it's going around,  but there's a little bit of wobble and then there's S  in there and these magnetic interactions  with more than one particle.  And so they're 90% good but not 100% good anymore  and that means that you can get some forbidden transitions  and if you look closely, you'll see some weak things  that you can explain by that.  But one thing we know for sure if we're considering it's the  electric part of the light wave  that's moving the electron around.  The electric part of the light wave can't change the magnetic  bar magnet of the electron.  And therefore S, big S, can't change in the transition.  So delta big S has to be zero.  And because the photon has one unit of angular momentum,  delta big L has to be plus or minus one.  And putting those together, we can get the delta J is zero  or plus or minus one.  However, if J starts at zero, we can't terminate a J equals zero  because then it's impossible  to satisfy the delta L selection rule as well.  So in that case, we get shut down.  So those are the electric dipole selection rules  in terms of the term symbol.  Delta S is zero, delta L plus or minus one,  and delta J is zero plus or minus one.  We'll explore in the next lecture how we can use this notation  to analyze some of these spectra.
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Welcome back to Chemistry 131A. Today we're going to continue our exposition of atomic spectroscopy focusing on selection rules, coupling, and terms. Of course, atomic spectroscopy is the basic tool by which everybody learned that the atom was quantized and so it's of very central importance in this whole area of endeavor of quantum mechanics applied to small things. Various sets of transitions are observed. Here is energy level diagram for the hydrogen atom, not to scale because if you do it to scale, the levels crowd together too much at the top and it's very hard to see what's going on. There's a series called the Lyman series in which an electron jumps from the n equals 2 to n equals 1 or 3 to 1 or 4 to 1 and so forth. And then there's another series, the Ballmer series, that's in the visible range in which the electron jumps to the n equals 2 state and then there are other series and they have names according to who worked on them. The main thing to keep in mind here is that any change in the quantum number n in the energy is allowed in principle. The only thing that has to be satisfied is that there has to be conservation of energy and that's taken care of automatically because the photon is emitted with a quantum h nu which is exactly equal to the energy loss of the atom. So energy is conserved. In this diagram, the zero reference energy is at the top and all the energies quoted there are negative being more stable than a proton and an electron at rest at infinite separation. The selection rules are a little bit more complicated, however, and the first thing is that a photon has one unit of angular momentum, h bar. Angular momentum is conserved and so when the electron makes a transition there has to be a change by one unit of angular momentum. So that means that although we draw these things coming down to the one, in fact they're all coming from specific values of L. It's 2p to 1s, 3p to 1s, 4p to 1s and so forth and it's always p. We can't have 2s to 1s because then delta L is zero and that's not allowed. That doesn't conserve angular momentum and likewise we can't have 3d to 1s either because then delta L is equal to 2 and again that doesn't conserve angular momentum. Now if we have a different kind of transition or if more than one photon is emitted that's different because then, of course, that could be allowed because angular momentum could be conserved in a more complicated kind of process. Those kind of processes do happen all the time. They're usually just much slower than the single photon electric dipole allowed transition. If we have transitions to 2p in the Bombers series, for example, they could be from 3s or 3d. If we have transitions to the 2s state then they have to be again from 3p, 4p and so on. And in general, delta L is plus or minus 1 for electric dipole allowed transitions and these are generally the very strongest transitions that you see in the spectrum. Not always, however, if the atom is very heavy like mercury as we'll see in the next lecture, we could have a strong transition that doesn't seem to follow this rule. There are weaker transitions, magnetic dipole transitions and electric quadrupole transitions. We won't touch on those much but you should keep in mind that those can and do occur and therefore you mustn't just assume that this is an ironclad rule. When we use the terms allowed and forbidden in spectroscopy, what we're talking about really is strong and weak. Allowed is the light is green, forbidden is it's red but some people run red lights. Here's a compendium from the NIST database just to show you how accurate and fastidious these spectroscopists have been in cataloging these energy levels. This one has a different energy zero. In this case, energies are quoted from the 1S level of the hydrogen atom being zero and then up from there. But the amazing thing about these numbers is that they have six, seven, eight, and nine digits of accuracy. So you can see that they had to take a lot of precautions when they were doing these. And if you look closely, there are multiple entries. There's 2P and then there's 2S and there's just 2 and then there's 2P again and so forth. And we'll see in a second what these things mean. But you have to keep track of individual substates in these systems and oftentimes you put on a magnetic field to be able to tell which state is which and how they move around to keep track of them. And they go all the way up to 6H. So in the periodic table, we don't go beyond F. But if we go to excited states of atoms, it's easy to go very high and to have very high values of the angular momentum L. If you look closely, you find out that the first 2P state that's quoted is ever so slightly lower in energy than 2S, which is quite a surprise because in other elements, of course, 2S is lower than 2P. That's why lithium is 2S1 and not 2P1, for example. And we don't fill up the 2P until we get over to boron. But in hydrogen, it's kind of an anomaly, one of those 2P states is lower than the 2S. And that was extremely interesting to theoreticians how that could possibly be. And in studying that in great detail, they worked out important clues about actually the residual electromagnetic field that permeates our entire universe like the zero point energy of a harmonic oscillator. You can think of the electromagnetic field as some kind of electromagnetic oscillation. And it turns out that even in a vacuum at absolute zero, in this universe, apparently, we can't have nothing. There's still some zero point energy left over. And that can bounce the electron around. And then the question is, if it bounces a 2S electron around or bounces a 2P electron around, what's the difference? And if you actually follow through the calculation, it's actually very, very beautiful work to follow. And you can actually derive that the 2P should be lower. And we can use the measured energy. Once we have a series of these states, we can use the measured energy, if we're careful, to estimate the ionization potential, also called the ionization energy. And that's just an extrapolation to what the energy would be at n equals infinity. And that's very important to know what that number is. That's one of the main things that's listed in freshman chemistry books in the periodic table to try to account for various trends. If an electron is easy to ionize, and that means that that element's much more likely to give an electron up to another atom than an element that has an electron that's extremely hard to ionize, for example. Let's try a practice problem then and have a look at how we might tackle estimating the ionization energy. Practice problem 18. Consider the following unassigned, that means we don't know n or l, emission lines from a hydrogen arc, all of which terminate in a common level, assign them and estimate the ionization potential or the ionization energy, same thing, of hydrogen. While the hydrogen arc is just we put a big lightning strike, a voltage through, and we excite the atoms and then they emit light. And if they terminate in a common level, that means they're like the other diagram I drew. We have arrows coming down to this, to this, to this, and so forth, always to the same level. And that means the change in energy has to do with the upper level but not the lower level. Of course, in actual fact, you just get all kinds of lines and you have to be very clever to figure out which ones go with which and which go together. There are various tricks that were worked out to do that. So here are the numbers. They all have four or five digits in them, 1.8887 and so forth. And let's just take these four numbers, quoted in electron volts, and let's figure out what the ionization energy should be. The first thing is we have to figure out what common level they terminate in. And if we don't do that, then we aren't going to be able to figure out anything. And so the way we're going to do that is we first write this formula that delta E is equal to the Rydberg constant times 1 over n squared common level minus 1 over n k squared where nk is some higher number, higher up dropping down. We don't know what the common level is, however. And we don't know what the value of r is because it wasn't given. And on an exam, if you aren't given the value of r and a problem like this, that means you're supposed to figure it out without that. So let's try to do that. We don't know n common. We don't know r. We have some values of delta E. And we have to try to figure out if they fit this formula. A very bad way is to just start plugging things in and seeing what happens. A very good way is to proceed systematically and figure out if we can get a straight line plot. Chemists love straight lines. Before computers, straight lines were absolutely essential. And the reason why is that the human eye can distinguish a straight line from any curve. But the human eye cannot distinguish curves that are slightly different. Even if they're statistically significantly different, it's very hard to tell just by looking. And before computers, people had to plot things on paper often and look at them. And in order to tell what's what, if you could organize your equation so that it came into a straight line plot, then you could tell. You could tell if there was a systematic deviation, some kind of slight curvature one way or the other way. You could tell if there was scatter, noise, and so forth. And all those are very important to quantify when you're trying to estimate something. We can cast our equation into a straight line form by letting y equals to delta E, letting x equal 1 over n squared, and then letting the intercept b equals 1 over, rather, r over n common. And we can guess values of nk and see which look good. So we have to assume a value of n common, and then we guess values of nk all greater than that, and we see what looks good. Once the line is straight, then we can extract the slope, y equals b plus mx, and m in this case should be minus r, the Rydberg constant. Once the line is straight, then we can get r. Now the lowest common level could be n equals 1, so in the absence of any other information, we should assume that first and then work our way up. If the common level is very high, really, really high, this would be a disaster because we would take forever. But we know that it can't be that high because we see a couple of electron volts, and so we know it has to be one of the levels further down. If it's very high, then all the transitions would be very, very small numbers because we'd already be most of the way up to the ionization energy. So let's try n common equals 1, and in that case, we could have nk, the upper level, b2, 3, 4, and so forth. If some transitions are missing for some reason, we just didn't see them, they were faint, we made an error, some other reason, then this kind of exercise can get very frustrating because the points don't appear to fit, and if you have to start deleting values, then it's very tedious, but in any kind of exam problem, we'll never have something like that. But in the lab, for various reasons, it can be that some transitions are dark, something happens, there's another pathway where the electron can go, and it's very hard to see that, and sometimes it takes a lot of patience to be able to work out what's going on. So let's plot these. I drew up a table, I have the delta E values for y, and then I have the 1 over nk squared, and for 2, it's a quarter, for 3, it's a ninth, and so forth as we go up. And if we plot those, which I've done here, well, first of all, we notice that the slope is negative, excuse me, and that's, of course, exactly what we expect because the slope should be negative r, and r is a positive number, so that part's okay. And let's plot them now, and here I've drawn a plot. Now, there are four black dots on this plot on slide 370, and if you just draw a line through the four black dots by linear regression, you will get a pretty good fit. In other words, there is a line that misses them all, but is pretty good, and so you have to decide if that kind of line is going to be good enough for what you're doing, and in this case, it's not nearly good enough. And what I've tried to do is emphasize that by first drawing a line through the first two points on the left, the red line, that has a certain slope, and keep in mind that these points have very, very, very tiny errors. So although I've put these black dots so that you can see them on the slide, the points themselves are very, very accurately determined. We draw that red line, and then we compare that with the slope from the last two points, the blue line. We only have four points, so we have to be kind of careful, and we plot that, and when you do it that way, what you see is that the blue line has a much shallower negative slope than the red line, that it is quite a bit different, and so there, and if you took the center two points, it would have a slope in between, and what that means is that this is curved. It's not n equals, n comma equals 1, and this is not good enough, but you could easily assume it worked good enough if you're used to other kinds of data that isn't this well determined. In any case, we're going to discard that and say, well, n comma equals 1 doesn't seem to work. Let's go up to the next level. So this is not close enough. If we assume the next level up, n equals 2 for the common level, and then we have 3, 4, 5, and 6 for nk squared, then we get 1 ninth, 1 sixteenth, 1 twenty-fifth, 1 thirty-sixth. We put those values in and we get this table that I've shown on slide 371, and using this data, we get the plot on the next slide. And now the difference is it's dead on. So I put one blue line and boy, it goes right through all those points, perfectly through them, nice and straight. So there's quite a bit of difference between the two when you view it this way. And if you pick, I'll let you try it on your own. If you pick the common level equals 3 as the terminating level and do the same exercise and plot it, you will see that it is significantly curved again. But not by much. It's curved, but you have to have a critical eye in order to see it, and you have to understand how accurately these plots should come out. If you do a linear regression on this and you take the slope, then the slope M is equal to minus 13.5984 electron volts. That's R, and that's the ionization energy, because if we put N equals 1 in the formula and N equals infinity, the ionization energy is just R. And therefore, the ionization energy for hydrogen, according to this analysis with those four points, should be 13.5984 electron volts. If you go back to the NIST compendium of data, what you'll find out is that we did pretty well. That comes out very close to the exact value of the ionization energy for hydrogen. And doing that with just four points is pretty good, because we didn't assign very many of the transitions. It's especially important to note that it's not enough to fit the data if you don't do a statistical analysis of the fit. If you just get a fit, it can look great, and it can be completely wrong, and you have to do a statistical analysis based on what you think the errors in the data points might be and how well it fits. If you don't bother to do that, and most of the time people don't, they just do some fit, it looks good, they assume it's right, and they quote an answer, you can get really crossed up if you have systems that are in fact very subtle, the curvature subtle, and the data is very, very highly precise. So in these cases, if it's not right on, then your model isn't correct. Keep that in mind when you're doing these kinds of problems. Okay, now I want to talk about coupling schemes, and this has to do with why there were two values of 2P in the NIST database, there was one of them, and then another one, and you think, well, an electron can be in 2P, it can be in 2P, what's the difference? Is it 2PX2? No, it's not, that's not the difference. The difference is that the electron itself has a magnetic moment, and we have to take that into account when we have a nonzero value of L. Well, kind of, in order to talk this through, we'll have to assume a kind of a classical model of the atom, but this is just a heuristic way to kind of look at what's going on. We don't actually believe that something's orbiting around, but anyway, if L is not equal to zero, we can think that something is orbiting around in some sense. If I'm an electron and I'm orbiting around the proton, or the nucleus, if it's a bigger atom, and I go on the electron, so I'm on the electron, boy, is that a wild ride, what I see is I see the proton going around me, because I'm the fixed guy, and the proton's whizzing around me like that the other direction. But the proton has a charge, and so what I see is I see a charge whizzing around me in some kind of a pattern, like a circle or something, and that to me as the electron looks like a current loop. It looks like an electromagnet, but as an electron, I have an intrinsic bar magnet, the spin that we discovered from the Stern-Gerlach experiment, and therefore, my own bar magnet can either be aligned with this magnetic field that I appear to feel, or it can be the other way, and those two could have slightly different energy because there's some magnetic energy. We didn't take that into account in the Hamiltonian when we wrote it down, but in fact, we know that there is, and if we were more sophisticated, we could take it into account. And therefore, those two orientations there pretty much is the resolution of why there's two P levels. They're very close because this magnetic orientation of the electron either being aligned or against this field that the proton appears to induce is very tiny compared to the electrostatic interactions with the potential. And so, it's out in several digits out. But as we'll see, we'll do a calculation probably with sodium, I think this magnetic field that this nucleus appears to generate at the electron as measured by the difference in the energies is huge. And it's bigger than almost any magnet we can make in the lab. So, it has a very big effect because things are moving so fast and it's close by, so it has a very large effect on the electron compared to the kinds of magnetic fields that we can generate in the lab. So, quite a bit bigger than 1 Tesla. 1 Tesla would be the kind of magnetic field that if you went in for an MRI in a whole body scanner and you had a state of the art annular magnet, superconducting magnet, you might have 1 Tesla in there. And it's very much bigger, of course, than any kind of normal magnet that you would ever find. So, this kind of interaction between the spin magnetic moment and the orbital motion is called spin orbit coupling because it's a magnetic interaction between the apparent magnetic field of the nucleus and the intrinsic magnetic field of the spin. Whenever there is more than one electron, the electrons themselves can interact magnetically. They're little bar magnets. They can either align with each other or they can not align. If it turns out, we'll see it when we get to it that if they're in the same orbital, they have to be paired. But if they aren't in the same orbital, if there's an excited electron, there's another one here, then they can have any kind of orientation, either for or against. And if we have an open shell atom with more than one electron, then these spins can add up and we first add up all of the intrinsic bar magnets to give what's called the total spin angular momentum, S. And if there's electrons with nonzero L, then there's orbital angular momentum. If we have more than one of those, we add all those up, and I'll explain how in a second, to give the total orbital angular momentum big L. And then L and S, again magnetically, interact to give the total electronic angular momentum for the atom, which is given the symbol J, L and S, added together. If you take it one stage further, sometimes the nucleus, for example, proton, has a magnetic moment too. And it's been one half like the electron. And in that case, it can either be aligned or not aligned. And we have to add its angular momentum. It's magnetic interaction with J. And in that case, we have to invent a new letter and that new letter is F. We'll usually quit at J, but keep in mind that nuclei do have magnetic properties as well. Their spin is not always one half like the electron. A deuteron has spin one. Chlorine has spin three halves. And if you look very closely at electronic transitions, you actually have to be very careful to take into account the properties of the nucleus to explain what you see. If you try to leave that out, then they're missing pieces out of the puzzle and you don't appear to get the right answer. Here on slide 376 is an image adapted from Wikipedia which shows how you can take these two cones, a big cone for L. L is pointing somewhere. And another cone for S. And you can combine them to get a specific value of J. And then there are 2J plus 1 substates, magnetic substates of J, which are the projections onto the Z axis of the total angular momentum. How do we add these things up to get big L? Well, there's, you, in order to really do this correctly, you have to do a little bit of work with the theory of angular momentum. But as I will explain as we go along, it makes sense that if I've got two electrons, let's say, and they have orbital angular momentum L1 and L2, then I could add them up. They could be parallel and that would be L1 plus L2. So that would be the maximum if everything were co-linear, everything were the wind blowing from behind. And then they could be out of alignment. And because things are quantized, then the next value is L1 plus L2 minus 1. And the worst case is when they're completely misaligned. But the total angular momentum L has to be a positive number because it refers to L squared. And so we terminate the series at the absolute value of L1 minus L2. We say absolute value because we don't know whether L1 is bigger than L2 or less than L2. And so we just put the absolute value there to terminate the series. It terminates at a positive number, which could be zero because L1 and L2 could both be one, for example. This series is called the Klebsch-Gordon series. And the same algebra exactly applies to the total spin angular momentum, big S. Big S can be S1 plus S2 or S1 plus S2 minus 1, blah, blah, blah, down to the absolute value of S1 minus S2. And we can see this in a vector picture keeping in mind that the interaction between these guys is magnetic in nature. And so it's the magnetic little bar magnets that are doing the talking here. And then they're adding up in certain ways to give these overall angular momentum that we observe. Let's, for an example, let's just take two electrons and let's couple the two spin one-halves. We've got the two bar magnets here. Let's couple them together. They're interacting. How could they be? Well, there are four possibilities. They could be like this, like this, like that, or like that because each electron can just be up or down. So I've quoted that in a notation which you may see in a more advanced course and these are called KETs. But for our purposes, we don't need to know what they're called. They just have the arrows in them. They're either two up, up down, down up, or down down. Those are the four possibilities. And we have to see how these can give an overall spin, big S, that's not going to be one-half like the either spin of the electron because the Klebsch-Gordon series says, well, big S could be S1 plus S2, that's one, or it could be S1 plus S2 minus one, that's zero, and that's the absolute value of S1 minus S2 since they're both spin one-half. So we could only have big S be one or zero when we couple the spins together. So let's look at these then from a more pictorial view in a vector diagram. But here's a very nice picture on slide 379. One side in blue is called the singlet. We'll see why in a second. The other side in red is called the triplet. The singlet has one state. The triplet has three states. That part of it makes perfect sense. And we can see I've written these kets next to each orientation. For the triplet, they could be both parallel, both up. They could have one up and one down, but pointing the same direction like that, okay? So this would be an S equals one, S equals zero, and then they could both be pointing like this, that's S equals minus one, M sub S, excuse me. Those are the three values of the M quantum number that would go with an S equals one state, M sub S, big S equals plus one, zero, or minus one. And then the singlet has just M sub S equals zero, because S is zero, so that can be the only value of M sub S. And for that one, we draw it slightly differently. We draw it, instead of drawing it like this, we draw it like that. So we've got one up and the other pointing the other way. They add to zero, but they're kind of like the two guys in the Star Trek time tunnel, they're just stuck battling each other in this state, absolutely anti-symmetric. Instead of calling one of them up, down, and the other one down, up, you may notice that I have this root two over two down, up plus up, down for the triplet, and root two over two up, down minus down, up for the singlet. Why is that? That we have to have them. Well, we know of course that since the electron can go through both slits, it's perfectly feasible for quantum systems to decide that they want to exist in a superposition of what we might think of as the most concrete things, namely, well, it's either down, up or it's up, down. Well, not in a quantum system, it could be 50% of that and 50% of the other one together. That could be the solution. We see then we've got four states and they break apart into three and one, and that's a common pattern because the odd numbers add up to a perfect square. And so if we have a certain angular momentum choice, three here and three here, we get nine and we can write that as a sum of one, three and five. And that's basically what the Klebsch-Gordon series says. It turns out that neither of the states up, down or down, up will do and the reason why is because these guys either in the triplet or the singlet or in other systems, they have to have the same stripes. They have to have the same characteristics with respect to swapping the particles. The particles are identical, so which one we call one and two is unimportant. But if we have two of them up and we swap them, we get the same thing. And if we have two of them down and we swap them, we get the same thing. But if we then say, well, I think the middle one then for the triplet is up, down and we swap them, we get a new guy, we get the other one, which wasn't part of the mix and we can't have that. We have to, when we swap them, we have to get the same thing because it's part of the same series. It has to have the same symmetry as the other ones. Otherwise it doesn't work. And therefore, we have to take a combination between up, down and down, up such that when we swap them over, we get the same thing as the M equals zero state of the triplet. It's, unless you've seen how to do this, it's not, not quite so easy to figure out how you might do it. But let's explore this symmetry by artificially coloring the arrows, red and blue. That way we'll keep track of them. When we swap them, we'll see what we get in terms of whether we get the same thing or not. At the top here on slide 381, I've shown a picture with the coloring in. We start with, we know we can't have one of them because it switches to the other. So we know we've got to have both of them and we know we've got to have 50% of each. Let's forget about the root 2 over 2. That's the normalization constant just to keep the probability of being in some state equal to 1. And let's just take the combination of adding them. Let's take up in red, down in blue, plus down in red, up in blue. And then let's swap the positions of the two arrows. What we get then in the first one, where we're up in red and down in blue, if we swap them, we get down in blue, up in red. Now in actual fact, the red and blue don't matter, but it lets us keep track of it. So we get down up. The other one, which was down in red, up in blue, we get up in blue, down in red. And then if we just change the order, we find we get up in blue, down in red, plus down in blue, up in red. That's what we started with except the colors are swapped. But the colors don't matter because the electrons are identical particles. We're just coloring them so that we can keep track of what we're doing, otherwise we can't tell what we've swapped. And the key is that we get plus one. We get the same thing. The eigenvalue for this swapping is plus one. So that's kind of a parity value. Plus one for swapping this way, plus one for this one, plus one for swapping that way, they all go together. All three of them are birds of a feather. If one of them is plus, then the other one to be orthogonal can't be plus as well. And therefore it has to be minus. And if we take that combination, which I've shown in the second line here, and we take up down, minus down up, and we swap them, we get down up, minus up down. And that's equal to minus one times what we had before, because of the fact that they're opposite. And what that means then is that the singlet has different symmetry than the triplet. And so although they both have M equals zero states, the states are different because one is symmetric under exchange and the other is anti-symmetric under exchange. In more complex systems, you have to look at the symmetry of each of the states and decide if it fits in or not. And there are well-documented procedures for doing exactly that if you have to. So whether you're symmetric or anti-symmetric under exchange is a very important property. It's not just trivia. Okay, suppose then we have these values. We've seen we can take S2S1 half, spin one half particles. We can get S equals one, S equals zero. We can do the same thing with L, and we can follow that through the same way. Now we've got big L and big S. How do we know what values of J will result? Big L is something, could be two. Big S is something, could be one or three halves, depending how many particles we have. And we want to figure out what values the total angular momentum of the atom J can be. Well, the answer is it's pretty much the same thing again because angular momentum always adds in exactly the same way. Doesn't really matter what it is. Well, if L and S happen to align, that's the maximum of value of J, L plus S. And then if they don't, then we decrement by one quantum, and we have L plus S minus one and so forth. And we go down to, again, the absolute value of L minus S because we don't know whether L or S is bigger. But we know that J referring to the square of the angular momentum or the square root of the square of the angular momentum, if you like, has to be a positive thing. And so we can't have anything go negative. And therefore, we get another Klebsch-Gordon series, L plus S, L plus S minus one and so forth. And therefore, what this means is that if you have a certain configuration, an electronic configuration of an atom that's open-shell that has several electrons, depending on the details of how these magnetic moments are interacting, and since these energies are small, then if I have something like a blast, like an arc, an electric arc or something like that, they're all going to be there. That's not like one of them is way high compared to the others. If there's enough energy to populate the excited state, there's probably enough energy to populate all the possibilities randomly, statistically, just depending on how many possibilities there are for each state. And therefore, we need some way to keep track of these things in order to figure out what kinds of emission lines we're going to see in the spectrum and to assign it. And the way we do that is with something called a term symbol. The term symbol summarizes not only what state the atom is in, but tells you what L, S, and J are. And the term symbol has the structure 2S plus 1 on the left, big L, which is like a letter like S, P, and so forth, just like for the electron itself, little S, little P. This is big L, the total angular momentum, orbital angular momentum. And then J as a right subscript to tell you which way S and L are adding up to give the result for the total angular momentum of the atom. And in order to figure out what kinds of transitions you're going to see, you have to know what the term symbol is. That's by far the best and most concise way to look at things. And therefore, you have to know what these are. 2S plus 1 is called the multiplicity. And if S is 0, it's called singlet. Because there's one line, one state there, like we saw with the S equals 0 for the two electrons. S equals 1 is called triplet. S equals 1 half, if we have that. That's called doublet and so on. We can have all the way up. And let's try an example with this then and see how we might use this symbol to figure out. So let's take, here's practice problem 19. Let's look at the sodium emission spectrum. The ground state of sodium is got a 3S electron outside a closed shell. The first excited state is 3P because in the sodium electron, 3P and 3S are quite a bit far apart. And so that turns out to be a transition that has visible light. It's not very tiny transition energy like radio waves. So we can see that easily in the sodium spectrum. The question is, if we have the electron in the 3P state outside a closed shell, what terms can arise? In other words, what are the possibilities for these term symbols? Which way can L, S and J add up? Well, for the P state, little L is 1. There's one electron and therefore big L is 1 because the Klebsch-Gordon series terminates. Big L is little L in this case because there's just one electron. Why don't we include all the other electrons in sodium? They're inside a closed shell. And inside a closed shell, the orbital angular momentum is zero and the spin angular momentum is zero. And so inside a closed shell, you just forget about it, throw it away. You don't want to include all those or you'll spend the rest of your life playing around. And there's one electron. So S is 1 half and therefore the two possible values of J are 1 plus a half and 1 minus a half, absolute value. That's it. Two possible values of J are 3 halves and 1 half and therefore the two terms are 2S plus 1, 2 times a half plus 1 is 2, that's doublet P, 3 halves and doublet P, 1 half. Those are the two terms that can arise. So as I said, closed shells ignore them. They have a total angular momentum zero. The two terms have close but not identical energy. Both of them, it turns out, we'll see, can emit a photon to the ground state, 3S. And therefore we get two closely spaced lines which are called the sodium D lines, which are quite famous because they were studied very early. In fact, even before the electron, it was figured out that the electron existed as a particle by Thompson. The year before, Zaman was looking at sodium and putting on magnetic fields and looking at how these lines changed and working out a ton of information from these yellow lines, which were very easy to excite, for example, with sodium vapor with a flame. And that's, of course, the sodium, the flame test to see what kind of element it is. You put a flame on and you look at the color and you can tell if it's yellow, it's sodium. These are also the characteristic yellow color of the fog lamps you see down toward the beach in places like San Diego or at least you used to. They might not be so energy efficient now. But the yellow color was thought to be much better when you had fog because if you have white light, sometimes what you see with very strong white light is you just see the fog. You don't see through the fog. You instead light up the little droplets and you get a lot of reflection from the fog so you don't see the pedestrian. Whereas if you have the yellow color without the white, which is why you use the sodium because you have these two lines, then you just see the pedestrian and then you put on the brakes. The term symbol gives us a really quick way to tell whether a transition is quote allowed or forbidden according to the electric dipole selection rules. In light atoms, L and S are still pretty good quantum numbers. The reason why they aren't perfect quantum numbers is that L refers to orbital angular momentum that came from hydrogen. In hydrogen, everything is spherically symmetric because there's that one proton and one electron. But in other atoms, there are other electrons and they muck things up. And what that means is that L is not quite so good because things are wobbling a little. And so although, yeah, it's going around, but there's a little bit of wobble and then there's S in there and these magnetic interactions with more than one particle. And so they're 90% good but not 100% good anymore and that means that you can get some forbidden transitions and if you look closely, you'll see some weak things that you can explain by that. But one thing we know for sure if we're considering it's the electric part of the light wave that's moving the electron around. The electric part of the light wave can't change the magnetic bar magnet of the electron. And therefore S, big S, can't change in the transition. So delta big S has to be zero. And because the photon has one unit of angular momentum, delta big L has to be plus or minus one. And putting those together, we can get the delta J is zero or plus or minus one. However, if J starts at zero, we can't terminate a J equals zero because then it's impossible to satisfy the delta L selection rule as well. So in that case, we get shut down. So those are the electric dipole selection rules in terms of the term symbol. Delta S is zero, delta L plus or minus one, and delta J is zero plus or minus one. We'll explore in the next lecture how we can use this notation to analyze some of these spectra.
in this whole area of endeavor of quantum quoted from the 1S level of the hydrogen atom atoms and then they emit light. And if they it, and you have to understand how accurately the curvature subtle, and the data is very, very compared to the electrostatic interactions with electronic configuration of an atom that's