CHEM 125a: Freshman Organic Chemistry I
|Transcript||Audio||Low Bandwidth Video||High Bandwidth Video|
Freshman Organic Chemistry I
CHEM 125a - Lecture 7 - Quantum Mechanical Kinetic Energy
Chapter 1. Limits of the Lewis Bonding Theory [00:00:00]
Professor Michael McBride: Okay, so we just glimpsed this at the end last time. This is a crystal structure of a complicated molecule that was performed by these same Swiss folks that we’ve talked about, and notice how very precise it is. The bond distances between atoms are reported to ±1/2000th of an angstrom. The bonds are like one and a half angstroms. So it’s like a part in a thousand, is the precision to which the positions of the atoms is known. Okay? But those positions are average positions, because the atoms are constantly in motion, vibrating. In fact, the typical vibration amplitude, which depends on — an atom that’s off on the end of something is more floppy than one that’s held in by a lot of bonds in various directions, in a cage sort of thing. But typically they’re vibrating by about 0.05 angstroms, which is twenty-five times as big as the precision to which the position of the average is known. Okay? So no molecule looks like that at an instant, the atoms are all displaced a little bit.
Now how big is that? Here, if you look at that yellow thing, when it shrinks down, that’s how big it is, that’s how big the vibration is. It’s very small. But these are very precise measurements. Right? Now why did they do so precise measurements? Did they really care to know bond distances to that accuracy? Maybe for some purposes they did, but that wasn’t the main reason they did the work very carefully. They did it carefully in order to get really precise positions for the average atom, so they could subtract spherical atoms and see the difference accurately. Okay? Because if you have the wrong position for the atom that you’re subtracting, you get nonsense. Okay, and what this is going to reveal is some pathologies of bonding, from the point of view of Lewis concept of shared electrons. Okay, so here’s a picture of this molecule. And remember, we had — rubofusarin, which we looked at last time, had the great virtue that it was planar. So you could cut a slice that went through all the atoms. This molecule’s definitely not planar, so you have to cut various slices to see different things. So first we’ll cut a slice that goes through those ten atoms. Okay? And here is the difference electron density. What does the difference density show? Somebody? Yes Alex?
Student: It’s the electron density minus the spherical —
Professor Michael McBride: It’s the total electron density minus the atoms; that is, how the electron density shifted when the molecule was formed from the atoms. Okay, and here we see just exactly what we expected to see, that the electrons shifted in between the carbon atoms, the benzene ring, and other pairs of carbon atoms as well. It also shows the C-H bonds, because in this case the hydrogen atoms were subtracted. We showed one last time where the hydrogen atoms weren’t subtracted. Okay, so this is — there’s nothing special here, everything looks the way you expect it to be; although it’s really beautiful, as you would expect from these guys who do such great work. Now we’ll do a different slice. This is sort of the plane of the screen which divides the molecule symmetrically down the middle, cuts through some bonds, cuts through some atoms and so on. So here’s the difference density map that appears on that slice.
Now the colored atoms, on the right, are the positions of the atoms through which the plane slices, but the atoms are subtracted out; so what you see is the bonding in that plane. So you see those bonds, because both ends of the bond are in the plane, so the bonds are in the plane, and you see just what you expect to see. But there are other things you see as well. You see the C-H bonds, although they don’t have nearly as much electron density as the C-C bonds did. Right? You also see that lump, which is the unshared pair on nitrogen. Right? But you see these two things, which are bonds, but they’re cross-sections of bonds, because this particular plane cuts through the middle of those bonds. Everybody see that? Okay, so again that’s nothing surprising. But here is something surprising. There’s another bond through which that plane cuts, which is the one on the right, through those three-membered rings, right? And what do you notice about that bond?
Student: It isn’t there.
Professor Michael McBride: It isn’t there. There isn’t any electron density for that bond. So it’s a missing bond. This is what we’ll refer to as pathological bonding, right? It’s not what Lewis would have expected, maybe; we don’t have Lewis to talk to, so we don’t know what he would’ve thought about this particular molecule. So here’s a third plane to slice, which goes through those three atoms, and here’s the picture of it. And again, that bond is missing, that we saw before. Previously we looked at a cross-section. Here we’re looking at a plane that contains the bond, and again there’s no there there. Okay? But there’s something else that’s funny about this slice. Do you see what? What’s funny about the bonds that you do see? Corey? Speak up so I can hear you.
Student: They’re connected; they’re not totally separate.
Professor Michael McBride: What do you mean they’re connected separately?
Student: Usually you see separate electron densities, but they’re connected.
Professor Michael McBride: Somebody say it in different words. I think you got the idea but I’m not sure everybody understood it. John, do you have an idea?
Student: The top one seems to be more dense than the bottom one.
Professor Michael McBride: One, two, three, four; one, two, three four five. That’s true, it is a little more dense. That kind of thing could be experimental error, because even though this was done so precisely, you’re subtracting two numbers that are very large, so that any error you make in the experimental one, or in positioning things for the theoretical position of the atoms, any error you make will really be amplified in a map like this. But it’s true, you noticed. But there’s something I think more interesting about those — that. Yes John?
Student: The contour lines, they’re connected, the contour lines between the top and the bottom bonds are connected. So maybe the electrons, maybe — I don’t know if they’re connected.
Professor Michael McBride: Yes, they sort of overlap one another. But of course if they’re sort of close to one another, that doesn’t surprise you too much because as you go out and out and out, ultimately you’ll get rings that do meet, if you go far enough down. Yes Chris?
Student: The center of density on the bonds doesn’t intercept the lines connecting the atoms.
Professor Michael McBride: Ah. The bonds are not centered on the line that connects the nuclei. These bonds are bent. Okay? So again, pathological bonding; and in three weeks you’ll understand this, from first principles, but you’ve got to be patient. Okay, so Lewis pairs and octets provide a pretty good bookkeeping device for keeping track of valence, but they’re hopelessly crude when it comes to describing the actual electron distribution, which you can see experimentally here. There is electron sharing. There’s a distortion of the spheres of electron density that are the atoms, but it’s only about 5% as big as Lewis would’ve predicted, had he predicted that two electrons would be right between. Okay? And there are unshared pairs, as Lewis predicted. And again they’re less — but in this case they’re even less than 5% of what Lewis would’ve predicted. But you can see them.
Chapter 2. Introduction to Quantum Mechanics [00:08:35]
Now this raises the question, is there a better bond theory than Lewis theory, maybe even one that’s quantitative, that would give you numbers for these things, rather than just say there are pairs here and there. Right? And the answer, thank goodness, is yes, there is a great theory for this, and what it is, is chemical quantum mechanics. Now you can study quantum mechanics in this department, you can study quantum mechanics in physics, you can probably study it other places, right? And different people use the same quantum mechanics but apply it to different problems. Right? So what we’re going to discuss in this course is quantum mechanics as applied to bonding. So it’ll be somewhat different in its flavor — in fact, a lot different in its flavor — from what you do in physics, or even what you do in physical chemistry, because we’re more interested — we’re not so interested in getting numbers or solving mathematical problems, we’re interested in getting insight to what’s really involved in forming bonds. We want it to be rigorous but we don’t need it to be numerical. Okay? So it’ll be much more pictorial than numerical.
Okay, so it came with the Schrödinger wave equation that was discovered in, or invented perhaps we should say, in — I don’t know whether — it’s hard to know whether to say discovered or invented; I think invented is probably better — in 1926. And here is Schrödinger the previous year, the sort of ninety-seven-pound weakling on the beach. Right? He’s this guy back here with the glasses on. Okay? He was actually a well-known physicist but he hadn’t done anything really earthshaking at all. He was at the University of Zurich. And Felix Bloch, who was a student then — two years before he had come as an undergraduate to the University of Zurich to study engineering, and after a year and a half he decided he would do physics, which was completely impractical and not to the taste of his parents. But anyhow, as an undergraduate he went to these colloquia that the Physics Department had, and he wrote fifty years later — see this was 1976, so it’s the 50th anniversary of the discovery of, or invention, of quantum mechanics. So he said:
And we write it now: HΨ=EΨ. He actually wrote it in different terms, but that’s his way, the Schrödinger equation. The reason the one we write is a little different from his is he included time as a variable in his, whereas we’re not interested in, for this purpose, in changes in time. We want to see how molecules are when they’re just sitting there. We’ll talk about time later. So within, what? Seven years, here’s Schrödinger looking a good deal sharper. Right? And where is he standing? He’s standing at the tram stop in Stockholm, where he’s going to pick up his Nobel Prize for this. Right? And he’s standing with Dirac, with whom he shared the Nobel Prize, and with Heisenberg, who got the Nobel Prize the previous year but hadn’t collected it yet, so he came at the same time. Okay? So the Schrödinger equation is HΨ=EΨ. H and E you’ve seen, but Ψ may be new to you. It’s a Greek letter. We can call it Sigh or P-sighi, some people call it Psee. Right? I’ll probably call it Psi. Okay. And it’s a wave function. Well what in the world is a wave function? Okay? So this is a stumbling block for people that come into the field, and it’s not just a stumbling block for you, it was a stumbling block for the greatest minds there were at the time.
So, for example, this is five years later in Leipzig, and it’s the research group of Werner Heisenberg, who’s sitting there in the front, the guy that — this was about the time he was being nominated or selected for the Nobel Prize. Right? So he’s there with his research group, and right behind him is seated Felix Bloch, who himself got the Nobel Prize for discovering NMR in 1952. So he’s quite a young guy here, and he’s with these other — there’s a guy who became famous at Oxford and another one who became the head of the Physics Department at MIT. Bloch was at Stanford. So these guys know they’re pretty hot stuff, so they’re looking right into the camera, to record themselves for posterity, as part of this distinguished group; except for Bloch. What’s he thinking about? [Laughter] What in the world is Ψ? Right? Now, in fact, in that same year, it was in January that Schrödinger announced the wave equation, and Ψ. Right?
And that summer these smart guys, who were hanging around Zurich at that time, theoretical physicists, the young guys went out on an excursion, on the lake of Zurich, and they made up doggerel rhymes for fun about different things that were going on, and the one that was made up by Bloch and Erich Hückel, whom we’ll talk about next semester, was about Ψ. “Gar Manches rechnet Erwin schon, Mit seiner Wellenfuction. Nur wissen möcht man gerne wohl, Was man sich dabei vorstell’n soll.” Which means: “Erwin with his Psi can do calculations, quite a few. We only wish that we could glean an inkling of what Psi could mean.” Right? You can do calculations with it, but what is it? — was the question. Okay? And it wasn’t just these young guys who were confused. Even Schrödinger was never comfortable with what Ψ really means. Now if we’re lucky, this’ll play this time. So this is a lecture by Schrödinger, “What is Matter”, from 1952.
[Short film clip is played]
Chapter 3. Understanding Psi as a Function of Position [00:16:36]
Professor Michael McBride: So twenty-six years later Schrödinger still didn’t really know what Ψ was. Okay? So don’t be depressed when it seems a little curious what Ψ might be. Okay? First we’ll — like Schrödinger and like these other guys — first we’ll learn how to find Ψ and use it, and then later we’ll learn what it means. Okay? So Ψ is a function, a wave function. What do you want to know, from what I’ve shown here? What is a function?
Student: A relationship.
Professor Michael McBride: Like sine is a function; what does that mean? Yes? I can’t hear very well.
Student: You put an input into a function and you get an output.
Professor Michael McBride: Yes, it’s like a little machine. You put a number in, or maybe several numbers, and a number comes out. Right? That’s what the function does; okay, you put in ninety degrees and sin says one. Okay? So what do you want to know about Ψ first?
Student: What does it do?
Professor Michael McBride: What’s it a function of? What are the things you have to put in, in order to get a number out? Okay? So it’s different from the name. The wave functions have names. That’s not what they’re a function of. Right? You can have sine, sine2, cosine. Those are different functions, but they can be functions of the same thing, an angle. Right? So we’re interested in what’s it a function of; not what the function is but what’s it a function of? So you can have different Ψs. They have names and quantum numbers give them their names. For example, you can have n, l, and m. You’ve seen those before, n, l, m, to name wave functions. Those are just their names. It’s not what they’re a function of. Or you can have 1s or 3dxy, or σ, or π, or π*. Those are all names of functions. Right? But they’re not what it’s a function of. What it’s a function of is the position of a particle, or a set of particles. It’s a function of position, and it’s also a function of time, and sometimes of spin; some particles have spin and it could be a function of that too. But you’ll be happy to know that for purposes of this course we’re not so interested in time and spin. So for our purposes it’s just a function of position. So if you have N particles, how many positions do you have to specify to know where they all are? How many numbers do you need? You need x, y, z for every particle. Right? So you need 3N arguments for Ψ. So Ψ is a function that when you tell it where all these positions are, it gives you a number. Now curiously enough, the number can be positive, it can be zero, it can be negative, it can even be complex, right, although we won’t talk about cases where it’s complex. The physicists will tell you about those, or physical chemists. Okay? And sometimes it can be as many as 4N+1 arguments. How could it be 4N+1?
Professor Michael McBride: Because if each particle also had a spin, then it would be x, y, z and spin; that’d be four. And if time is also included, it’s plus one. Okay, so how are we going to go through this? First we’ll try — this is unfamiliar territory, I bet, to every one of you. Okay? So first we’re going to talk about just one particle and one dimension, so the function is fairly simple. Okay? And then we’ll go on to three dimensions, but still one particle, the electron in an atom; so a one-electron atom, but now three dimensions, so it’s more complicated. Then we’ll go on to atoms that have several electrons. So you have now more than three variables, because you have at least two electrons; that would be six variables that you have to put into the function to get a number out. Then we’ll go into molecules — that is, more than one atom — and what bonding is. And then finally we get to the payoff for organic chemistry, which is talking about what makes a group a functional group and what does it mean to be functional, what makes it reactive? That’s where we’re heading. But first we have to understand what quantum mechanics is.
So here’s the Schrödinger equation, ΗΨ=EΨ, and we’re talking about the time-independent Schrödinger equation, so time is not a variable, and that means what we’re talking about is stationary states. We don’t mean that the atoms aren’t moving, but just that they’re in a cloud and we’re going to find how is the cloud distributed. If a molecule reacts, the electrons shift their clouds and so on; it changes. We’re not interested in reaction now, we’re just interested in understanding the cloud that’s sitting there, not changing in time. Okay, now the right part of the equation is E times Ψ, right? And E will turn out to be the energy of the system; maybe you won’t be surprised at that. So that’s quite simple. What’s the left? It looks like H times Ψ. If that were true, what could you do to simplify things? Knock out Ψ. But HΨ is not H times Ψ. H is sort of recipe for doing something with Ψ; we’ll see that right away. So you can’t just cancel out the Ψ, unfortunately. Okay, so ΗΨ=EΨ. Oops sorry, what did I do, there we go. Now we can divide, you can divide the right by Ψ, and since it was E times Ψ, the Ψ goes away. But when you divide on the left, you don’t cancel the Ψs, because the top doesn’t mean multiplication.
Now I already told you the right side of this equation is the total energy. So when you see a system, what does the total energy consist of? Potential energy and kinetic energy. So somehow this part on the left, ΗΨ/Ψ, must be kinetic energy plus potential energy. That recipe, H, must somehow tell you how to work with Ψ in order to get something which, divided by Ψ, gives kinetic energy plus potential energy. So there are two parts of it. There’s the part that’s potential energy, of the recipe, and there’s the part that’s kinetic energy. Now, the potential energy part is in fact easy because it’s given to you. Right? What’s Ψ a function of?
Professor Michael McBride: Position of the particles. Now if you know the charges of the particles, and their positions, and know Couloumb’s Law, then you know the potential energy, if Couloumb’s Law is right. Is everybody with me on that? If you know there’s a unit positive charge here, a unit negative charge here, another unit positive charge here and a unit negative charge here, or something like that, you — it might be complicated, you might have to write an Excel program or something to do it — but you could calculate the distances and the charges and so on, and what the energy is, due to that. So that part is really given to you, once you know what system you’re dealing with, the recipe for finding the potential energy. So that part of HΨ/Ψ is no problem at all. But hold your breath on kinetic energy. Sam?
Student: Didn’t we just throw away an equation? There was an adjusted Couloumb’s Law equation.
Professor Michael McBride: Yes, that was wrong. That was three years earlier, remember? 1923 Thomson proposed that. But it was wrong. This is what was right.
Student: How did they prove it wrong?
Professor Michael McBride: How did what?
Student: Did they prove it wrong or just —
Professor Michael McBride: Yes, they proved this right, that Couloumb’s Law held, because it agreed with a whole lot of spectroscopic evidence that had been collected about atomic spectra, and then everything else that’s tried with it works too. So we believe it now. So how do you handle kinetic energy? Well that’s an old one, you did that already in high school, right? Forget kinetic energy, here it is. It’s some constant, which will get the units right, depending on what energy units you want, times the sum over all the particles of the kinetic energy of each particle. So if you know the kinetic energy of this particle, kinetic energy of this particle, this particle, this particle; you add them all up and you get the total kinetic energy. No problem there, right? Now what is the kinetic energy that you’re summing up over each particle? It’s ½mv2. Has everybody seen that before? Okay, so that’s the sum of classical kinetic energy over all the particles of interest in the problem, and the constant is just some number you put in to get the right units for your energy, depending on whether you use feet per second or meters per year or whatever, for the velocity.
Okay, but it turned out that although this was fine for our great-grandparents, it’s not right when you start dealing with tiny things. Right? Here’s what kinetic energy really is. It’s a constant. This is the thing that gets it in the right units: (h2)/8(π2) times a sum over all the particles — it’s looking promising, right? — of one over the mass — not the mass mv2, but one over the mass of each particle — and here’s where we get it — [Students react] — times second derivatives of a wave function. That’s weird. I mean, at least it has twos in it, like v2, right? [Laughter] That’s something. And in fact it’s not completely coincidental that it has twos in it. There was an analogy that was being followed that allowed them to formulate this. And you divide it by the number Ψ. So that’s a pretty complicated thing. So if we want to get our heads around it, we’d better simplify it. And oh also there’s a minus sign; it’s minus, the constant is negative that you use. Okay, now let’s simplify it by using just one particle, so we don’t have to sum over a bunch of particles, and we’ll use just one dimension, x; forget y and z. Okay? So now we see something simpler. So it’s a negative constant times one over the mass of the particle, times the second derivative of the function, the wave function, divided by Ψ. That’s kinetic energy really, not ½mv2. Okay? Or here it is, written just a little differently. So there’s a constant, C, over the mass, right? And then we have the important part, is the second derivative. Does everybody know that the second derivative is a curvature of a function. Right? What’s the first derivative?
Professor Michael McBride: Slope, and the second derivative is how curved it is. It can be curving down, that’s negative curvature; or curving up, that’s positive curvature. So it can be positive or negative; it can be zero if the line is straight. Okay. So note that the kinetic energy involves the shape of Ψ, how curved it is, not just what the value of Ψ is; although it involves that too. Maybe it’s not too early to point out something interesting about this. So suppose that’s the kinetic energy. What would happen if you multiplied Ψ by two? Obviously the denominator would get twice as large, if you made Ψ twice as large. What would happen to the curvature? What happens to the slope? Suppose you have a function and you make it twice as big and look at the slope at a particular point? How does the slope change if you’ve stretched the paper on which you drew it?
Student: It’s sharper.
Professor Michael McBride: The slope will double, right, if you double the scale. How about the curvature, the second derivative? Does it go up by four times? No it doesn’t go up by four times, it goes up by twice. So what would happen to the kinetic energy there if we doubled the size of Ψ every place?
Student: Stay the same.
Professor Michael McBride: It would stay the same. The kinetic energy doesn’t depend on how you scale Ψ, it only depends on its shape, how curved it is. Everybody see the idea? Curvature divided by the value. Okay, now solving a quantum problem. So if you’re in a course and you’re studying quantum mechanics, you get problems to solve. A problem means you’re given something, you have to find something. You’re given a set of particles, like a certain nuclei of given mass and charge, and a certain number of electrons; that’s what you’re given. Okay, the masses of the particles and the potential law. When you’re given the charge, and you know Couloumb’s Law, then you know how to calculate the potential energy; remember that’s part of it. Okay? So that part’s easy, okay? Now what do you need to find if you have a problem to solve? Oh, for example, you can have one particle in one dimension; so it could be one atomic mass unit is the weight of the particle, and Hooke’s Law could be the potential, right? It doesn’t have to be realistic, it could be Hooke’s Law, it could be a particle held by a spring, to find a Ψ. You want to find the shape of this function, which is a function of what?
Student: Positions of the particles.
Professor Michael McBride: Positions of the particles, and if you’re higher, further on than we are, time as well; maybe spin even. But that function has to be such that HΨ/Ψ is the total energy, and the total energy is the same, no matter where the particle is, right, because the potential energy and the kinetic energy cancel out. It’s like a ball rolling back and forth. The total energy is constant but it goes back and forth between potential and kinetic energy, right? Same thing here. No matter where the particles are, you have to get the same energy. So Ψ has to be such that when you calculate the kinetic energy for it, changes in that kinetic energy, in different positions, exactly compensate for the changes in potential energy. When you’ve got that, then you’ve got a correct Ψ; maybe. It’s also important that Ψ remain finite, that it not go to infinity. And if you’re a real mathematician, it has to be single valued; you can’t have two values for the same positions. It has to be continuous, you can’t get a sudden break in Ψ. And Ψ2 has to be integrable; you have to be able to tell how much area is under Ψ2, and you’ll see why shortly. But basically what you need is to find a Ψ such that the changes in kinetic energy compensate changes in potential energy.
Chapter 4. Understanding Negative Kinetic Energy and Finding Potential Energy [00:33:24]
Now what’s coming? Let’s just rehearse what we did before. So first there’ll be one particle in one dimension; then it’ll be one-electron atoms, so one particle in three dimensions; then it will be many electrons and the idea of what orbitals are; and then it’ll be molecules and bonds; and finally functional groups and reactivity. Okay, but you’ll be happy to hear that by a week from Friday we’ll only get through one-electron atoms. So don’t worry about the rest of the stuff now. But do read the parts on the webpage that have to do with what’s going to be on the exam. Okay, so normally you’re given a problem, the mass and the charges — that is, that potential energy as a function of position — and you need to find Ψ. But at first we’re going to try it a different way. We’re going to play Jeopardy and we’re going to start with the answer and find out what the question was. Okay? So suppose that Ψ is the sine of x; this is one particle in one dimension, the position of the particle, and the function of Ψ is sine. If you know Ψ, what can you figure out? We’ve just been talking about it. What can you use Ψ to find?
Student: Kinetic energy.
Professor Michael McBride: Kinetic energy. How do you find it? So we can get the kinetic energy, which is minus a constant over the mass times the curvature of Ψ divided by Ψ at any given position. Right? And once we know how the kinetic energy varies with position, then we know how the potential energy varies with position, because it’s just the opposite, in order that the sum be constant. Right? So once we know the kinetic energy, then we know what the potential energy was, which was what the problem was at the beginning. Okay? So suppose our answer is sin(x). What is the curvature of sin(x), the second derivative?
[Students speak over one another]
Professor Michael McBride: It’s -sin(x). Okay, so what is the kinetic energy?
Professor Michael McBride: C/m. Does it depend on where you are, on the value of x? No, it’s always C/m. So what was the potential energy? How did the potential energy vary with position?
Student: It doesn’t.
Professor Michael McBride: The potential energy doesn’t vary with position. So sin(x) is a solution for what? A particle that’s not being influenced by anything else; so its potential energy doesn’t change with the position, it’s a particle in free space. Okay? So the potential energy is independent of x. Constant potential energy, it’s a particle in free space. Now, suppose we take a different one, sin(ax). How does sin(ax) look different from sin(x)? Suppose it’s sin(2x). Here’s sin(x). How does sin(2x) look? Right? It’s shorter wavelength. Okay? Now so we need to figure out — so it’s a shortened wave, if a > 1. Okay, now what’s the curvature? Russell?
Student: It’s -a2times sin(ax).
Professor Michael McBride: It’s -a2 times sin(ax). Right? The a comes out, that constant, each time you take a derivative. So now what does the kinetic energy look like? It’s a2 times the same thing. Okay? So again, the potential energy is constant. Right? It doesn’t change with position. But what is different? It has higher kinetic energy if it’s a shorter wavelength. And notice that the kinetic energy is proportional to one over the wavelength squared, right?; a2; a shortens the wave, it’s proportional to a2, one over the wavelength squared. Okay. Now let’s take another function, exponential, so ex. What’s the second derivative of ex? Pardon me?
Professor Michael McBride: ex. What’s the 18th derivative of ex?
Professor Michael McBride: Okay, good. So it’s ex. So what’s this situation, what’s the kinetic energy?
Professor Michael McBride: -C/m. Negative kinetic energy. Your great-grandparents didn’t get that. You can have kinetic energy that’s less than zero. What does that mean? It means the total energy is lower than the potential energy. Pause a minute just to let that sink in. The total energy is lower than the potential energy. The difference is negative. Okay? So the kinetic energy, if that’s the difference, between potential and total, is negative. You never get that for ½m(v2). Yes?
Student: Does this violate that Ψ has to remain finite?
Professor Michael McBride: Does it violate what?
Student: That Ψ has to remain finite?
Professor Michael McBride: No. You’ll see in a second. Okay, so anyhow the constant potential energy is greater than the total energy for that. Now, how about if it were minus exponential, e-x? Now what would it be? It would be the same deal again, it would still be -C/m, and again it would be a constant potential energy greater than the total energy. This is not just a mathematical curiosity, it actually happens for every atom in you, or in me. Every atom has the electrons spend some of their time in regions where they have negative kinetic energy. It’s not just something weird that never happens. And it happens at large distance from the nuclei where 1/r — that’s Couloumb’s Law — where it stops changing very much. When you get far enough, 1/r gets really tiny and it’s essentially zero, it doesn’t change anymore. Right? Then you have this situation in any real atom. So let’s look at getting the potential energy from the shape of Ψ via the kinetic energy. Okay, so here’s a map of Ψ, or a plot of Ψ, it could be positive, negative, zero — as a function of the one-dimension x, wherever the particle is. Okay? Now let’s suppose that that is our wave function, sometimes positive, sometimes zero, sometimes negative. Okay? And let’s look at different positions and see what the kinetic energy is, and then we’ll be able to figure out, since the total will be constant, what the potential energy is. Okay? So we’ll try to find out what was the potential energy that gave this as a solution? This is again the Jeopardy approach. Okay? Okay, so the curvature minus — remember it’s a negative constant — minus the curvature over the amplitude could be positive — that’s going to be the kinetic energy; it could be positive, it could be zero, it could be negative, or it could be that we can’t tell by looking at the graph. So let’s look at different positions on the graph and see what it says. First look at that position. What is the kinetic energy there? Positive, negative, zero? Ryan, why don’t you help me out?
Professor Michael McBride: Well no, you can help me out. [Laughter] Look, so what do you need to know? You need to know — here’s the complicated thing you have to figure out. What is minus the curvature divided by the amplitude at this point? Is it positive, negative or zero? So what’s the curvature at that point? Is it curving up or down at that point? No idea. Anybody got an idea? Keith? Kevin?
Student: It looks like a saddle point so it’s probably zero.
Professor Michael McBride: It’s not a saddle point. What do you call it?
Students: Inflection points.
Professor Michael McBride: A saddle point’s for three dimensions. In this it’s what? Inflection point. It’s flat there. It’s curving one way on one side, the other way on the other side. So it’s got zero curvature there; okay, zero curvature. Now Ryan, can you tell me anything about that, if the curvature is zero?
Professor Michael McBride: Ah ha.
Professor Michael McBride: Not bad. So that one we’ll color grey for zero. The kinetic energy at that point is zero, if that’s the wave function. Now let’s take another point. Who’s going to help me with this one? How about the curvature at this point right here?
[Students speak over one another]
Professor Michael McBride: It’s actually — I choose a point that’s not curved.
Professor Michael McBride: It’s straight right there. I assure you that’s true. So I bet Ryan can help me again on that one. How about it?
Professor Michael McBride: Ah ha. So we’ll make that one grey too. Now I’ll go to someone else. How about there? What’s the curvature at that point do you think? Shai?
Student: It looks straight, zero curvature.
Professor Michael McBride: It looks straight, zero curvature. So does that mean that this value is zero?
Student: Not necessarily, because the amplitude —
Professor Michael McBride: Ah, the amplitude is zero there too. So really you can’t be sure. Right? So that one we’re going to have to leave questionable, that’s a question mark. How about out here? Not curved. So what’s the kinetic energy? Josh?
Professor Michael McBride: Questionable, right? Because the amplitude is zero again; zero in the numerator, also zero in the denominator; we really don’t know. Okay, how about here? Tyler, what do you say? Is it curved there?
Professor Michael McBride: Curving up or down?
Professor Michael McBride: So negative, the curvature is negative. The value of Ψ?
Professor Michael McBride: Positive. The energy, kinetic energy?
Professor Michael McBride: Positive. Okay, so we can make that one green. Okay, here’s another one. Who’s going to help me here? Kate?
Professor Michael McBride: Okay, so how about the curvature; curving up, curving down?
Student: It’s curving down, that’s negative.
Professor Michael McBride: Yes. Amplitude?
Student: Zero. So it should be green.
Professor Michael McBride: Ah, green again. Okay. How about here? Ah, now how about the curvature? Seth?
Student: I don’t know.
Professor Michael McBride: Which way is it curving at this point here?
Student: Curving up.
Professor Michael McBride: Curving up. So the curvature is —
Professor Michael McBride: Positive.
Student: The amplitude is negative, so it’s positive.
Professor Michael McBride: Yeah. So what color would we make it? Green again. Okay, so if you’re — you can have — be curving down or curving up and still be positive; curving down if you’re above the baseline, curving up if you’re below the baseline. Right? So as long as you’re curving toward the baseline, towards Ψ=0, the kinetic energy is positive. How about here? Zack? Which way is it curving? Curving up or curving down?
Student: It should be curving up.
Professor Michael McBride: Curving up, curvature is positive. The value?
Professor Michael McBride: Positive.
Student: I guess it’ll be negative.
Professor Michael McBride: So it’s negative kinetic energy there. Make that one whatever that pinkish color is. Okay? Here’s another one, how about there? Alex? Which way is it curving at the new place? Here?
Student: Curving down.
Professor Michael McBride: Curving down, negative curvature.
Student: Negative amplitude.
Professor Michael McBride: Negative amplitude.
Student: Negative kinetic energy.
Professor Michael McBride: Negative kinetic energy; pink again. Is that enough? Oh, there’s one more, here, the one right here. Okay?
Professor Michael McBride: Pardon me?
Professor Michael McBride: Negative, because it’s — how did you do it so quick? We didn’t have to go through curvature.
Student: It goes away from the line.
Professor Michael McBride: Because it’s curving away from the baseline, negative. Okay, pink. Okay, curving away from Ψ=0 means that the kinetic energy is negative. So now we know at all these positions whether the kinetic energy is positive, negative or zero, although there are a few that we aren’t certain about. Right? So here’s the potential energy that will do that. If you have this line for the total energy, right? Then here and here you have zero. Right? Also, incidentally, here and here, you have zero kinetic energy. With me? Okay. So no curvature, right? At these green places, the total energy is higher than the potential energy. So the kinetic energy is positive. Okay? At these places, the potential energy is higher than the total energy. So the kinetic energy is negative and the thing is curving away from the baseline. Right? And now we know something about this point. If the potential energy is a continuous kind of thing, then, although we couldn’t tell by looking at the wave function, it’s curving away from the baseline, but very slightly, right? It’s negative kinetic energy there, and also on the right here is negative kinetic energy. And here we know, just by continuity, that at this point it must’ve been positive kinetic energy, even though we couldn’t tell it by looking at the curve. There must be an inflection point when you go through zero, otherwise you’d get a discontinuity in the potential energy. Okay, so that one was green. Okay, now I have to stop.
[end of transcript]Back to Top
|mp3||mov [100MB]||mov [500MB]|