Lecture 22. The Born-Oppenheimer Approximation and H2+
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. 
 Today in chemistry 131A we're going to move on from multi-electron atoms to the simplest molecule. We're going to talk first about the Born Oppenheimer approximation. This was a great idea that was introduced very shortly after the introduction of the Schrodinger equation itself. In fact, when you go back to this period in history, what you find out is that everybody was really working like crazy on this new field of quantum mechanics, much like if you come back to 2014 you may find a lot of biologists were working like crazy on stem cells now that they found a better way to produce them. And then after we talk about the Born Oppenheimer approximation, we're going to introduce the simplest molecule H2 plus and from there we'll go to H2. First, the Born Oppenheimer approximation. We're going to just look at the simplest molecules H2 plus and H2 and we're going to look at both of them in some detail to try to understand why it is that you can have a chemical bond. In fact, it's very interesting in the framework of classical mechanics and electrostatics to try to figure out if there could be such a thing as, for example, two positively charged particles and let's say one or two electrons. Let's say we have two so the whole thing is neutral and it's like H2 and we could propose that they're doing some classical thing. We could forget about the uncertainty principle. Let's have them going like this in some pattern and so forth and have all the electrostatic forces and try to see whether the molecule will hold together. And in fact, scientists at the time didn't really understand how molecules could hold together because it doesn't seem to work out very well. And we'll see exactly why when we come to the exchange integral. But if we're going to talk about molecules, even when we talked about helium, we had to use some approximations. We either had to make a good guess with the variational theory or try perturbation theory or something. And now that we've got two nuclei, two centers, even with one electron, we're going to guess that it's going to be extremely hard to solve things exactly. So the first idea is that we want to decouple the motion of the nuclei from the motion of the electrons. And we know that we can probably do that because the electrons are much less massive than the nuclei. And the electrons are therefore moving much more quickly. What we're trying to do then is say, let's say we've got an elephant and we've got a bunch of flies around it. And the elephant's moving slowly. And so the flies have plenty of time to readjust. And if the elephant rolls over slowly, the flies have plenty of time to find the best place where they want to alight on the elephant. And in much the same way, as the nuclear coordinates change, the electrons almost instantaneously, at least from the perspective of the nuclei readjust. And they readjust to the best possible position. And so that being the case, we can go one step further then. And we can just say supposing then that the, we just stop the motion of the nuclei. And we solve for the electrons. Then that we can do. And that's essentially what the Born-Appenheimer approximation is. First I want to talk about our notation here. Our equations are going to get more complicated. And that's going to happen quickly. So I'm going to use big M to denote the mass of the proton or the nucleus and little M to denote the mass of the electron. And our coordinate system is as on this slide 543 where you have two nuclei marked A and B. And in this case, I've put in both electrons, which I call one and two. This is H2 then. And this is at some instant. Some we're trying to set up how we're going to solve for the wave equation, for the Schrodinger equation. And we have all these repulsive and attractive terms between these charged particles. The Hamiltonian then, with all the terms in, has a kinetic energy for the first nucleus, A, a kinetic energy for the second nucleus, B, the inter-nuclear repulsion, which is a positive term pushing them apart. Then the kinetic energy for the first electron, one, the kinetic energy for the second electron, two. Then the attraction of electron one to the first nucleus to the second nucleus. The attraction of electron two to the first nucleus to the second nucleus. And then the repulsion, R12, which is the distance between the two electrons. And we are not going to be able to solve this exactly with all these terms for the total wave function, which includes all the nuclear and electronic coordinates. And in fact, even with the two electrons, we won't be able to solve it exactly. But like the H-atom, we can separate out the center of mass motion for sure. But we can't rigorously separate the relative nuclear motion. We have to make an argument about why it is that we're doing that. And Born and Oppenheimer wrote a classic paper that I've given you the reference to here if you'd like to look it up. Unfortunately, it's in German, and so you would have to look at the translation of it. I had to actually learn to read German when I got my degree in chemistry, which shows you how old I am. But it helps occasionally from time to time if you have to go back and read some of these papers. And their paper was on the quantum mechanics of molecules. And they have a very elegant derivation about why it is that you can ignore the nuclear motion. But I don't want to go through that because that would take at least this lecture and another lecture, and it would just take us a little bit too far a field to actually go through that. But we can certainly see that because the nuclei are moving slowly and because the electrons adjust, then if I just consider the nuclei here, and I solve the Schrodinger equation for the electrons, pretending the nuclei are frozen. And then I move the nuclei a little bit and resolve the Schrodinger equation. That's basically what the electrons are doing continuously as the nuclei move slowly from here to here, is they're just solving the Schrodinger equation again and again and again. And because they're readjusting so quickly, it appears to them as if the nuclei are basically frozen, and that means to us they're basically frozen as well. So if we write that the wave function of the molecule is, which depends on all the coordinates, is the factorization of product of the wave function of the electronic part and the wave function of the nuclear part, then the total energy is just the sum of the electronic energy and the nuclear energy. This is right back to particle in a box with the x, y directions, and this has been our strategy over and over and over, which shows you it's a very good strategy. Then what we find is that the electronic function depends instantaneously, so to speak, on the coordinates of the electrons, R1, R2, and it depends parametrically on this very slow variable, big R, which is just the distance between the two nuclei. And we can always use this approximation for molecules that have very high excited electronic states. So usually to excite an electron from any ground state to some higher state takes a lot of energy, and the slow motion of the nuclei cannot possibly supply that kind of energy. Where we have to be in trouble or where we have to be extremely careful is if we have radicals or open-shell atoms or something like that, where we have a lot of electronic states that are nearby, then what you can imagine happening is that the motion of the nuclei itself can somehow cause a transition of the electron to move out of the orbital that was originally and into a new one. And in that case, the motions are coupled, the whole thing's wobbling around, then we're in trouble and we can't use the Born-Oppenheimer approximation, so in that case it fails. But the upshot of our factorization is we can forget completely about the kinetic energy of the nuclei. We just clamp them down, we freeze them, and we solve for the electronic energy. And that's really an important simplification. Here I've written out the Hamiltonian then in atomic, in atomic units, the electronic Hamiltonian still has the kinetic energy of the electrons, the kinetic energy of the nuclei is gone. And I've added to it 1 over r, which is just the static, electrostatic repulsion of the two nuclei. We have all these instantaneous terms and then we have this static term, but we have to add that to the electronic energy because when the nuclei are very close that will be an overall unfavorable position. We don't care about the kinetic energy of the nuclei because we threw that away. And so what we want to do then is solve the Schrodinger equation for the electronic coordinates as a function of the internuclear separation, big r, and then see what that is. And in fact, if we do that, what we find is that when the nuclei are very close, it's very unfavorable. Partly it's just the internuclear repulsion, but partly it is that the electrons themselves are pretty much outside and so don't screen the positive charges. And as we move them apart, the potential comes down, and then there's a sweet spot where there's a minimum and then it comes back up again. And in fact, that's exactly the kind of shape that we argued was like a Morse potential or some potential like that when we were talking about vibrations and the harmonic oscillator. And now you can see that the nuclei themselves glide around on a potential energy surface that's constantly readjusting and we just draw it as if it's always there, but it's constantly readjusting as they move, it is there before they get to where they're going, so to speak. And we talked then about the nuclear coordinates moving around on a potential energy curve or a potential energy surface, which is the total electronic energy plus the internuclear repulsion. That's what a potential energy surface is how you should think of it. And the nuclei will respond to this. So if they're too close, they'll gradually pull apart. And as they pull apart, the electrons will jigger around and readjust. The simplest possible molecules H2 plus, I want to begin with that, I wrote down the Hamiltonian for H2, but that's a little bit intimidating for us right now, so I want to just do it with one electron. The second electron is now not present, so I don't need one and two, I can just use Dell. And there's only the attraction to the first nucleus, the attraction to the second nucleus. There's no 1 over R12, thank goodness, because that was quite a production when we did helium and hydride. And then there's 1 over R, which is no big deal, which is the internuclear repulsion. Here's our picture then on the bottom of slide 549, same coordinate system as before, except the second electron is missing. Well, much like we could solve the hydrogen atom, once we factored out the center of mass motion and assumed the proton was essentially fixed, by saying, well look, this problem has spherical symmetry. We should actually write the Schrodinger equation in spherical coordinates, and then what happened is, we knew we had the right coordinate system because the wave function factorized into three parts, and that made it easy to solve each part. This three-dimensional differential equation became three one-dimensional differential equations, and that is always the strategy if you can get it to work. Now in this case, spherical coordinates won't work because we have stretched the charge out in one direction. Well, if you take a circle and you stretch it in one dimension so that you have two centers, then you have an ellipse. And so not surprisingly, there's a coordinate system called elliptical coordinates, or confocal elliptical coordinates, where you take the sum of the distance between the electron and the nuclei divided by r, so that's a dimensionless thing, lambda, and then another one, which is the difference between them, rA minus rB divided by big r, call that mu. Then the third coordinate is still like it was in the hydrogen atom, it's that coordinate phi, and the solution for that because that's basically the same. That was the particle on a ring. We can still use that to solve. That decouples out, and so we don't have to worry about that one. The other two, we have to actually worry about what kind of differential equation that we get. You can think of, if you make a triangle between the two nuclei and the electron, you have a plane, three points to find a plane, and the position of that plane around this axis between the two nuclei is the azimuthal angle phi. And so this is just like a stretched hydrogen atom, really. Let's write the wave function again as a product. We're really beating a dead horse with this, but it's the only horse we've got. We're going to write it as a function of lambda, which I've called big lambda of lambda, a function of mu, and unfortunately capital mu looks like an m, but it's mu of mu, and then phi of phi. And those are just presupposing that we can write it like this, and I've given you a reference here at the bottom of page 550, slide 551, if you're interested. There's quite a long paper in the philosophical transactions of the Royal Society where they've gone through exactly how you separate these coordinate systems. It does separate, it does separate into a differential equation with respect to mu only, big m, and then another differential equation that only has big lambda. The problem for us is that these differential equations, even though they're one dimensional, if you look at them, they're very complicated, and they have a lot of moving parts, and we can't write down the solution to these kind of equations in any elementary way. There may be solutions in terms of repeated fractions, one plus one over x plus plus and so forth, and things like that, but that's not really too illuminating for us at this level. And so I don't want to go into writing, attempting to write down these very esoteric functions that might solve this because that's not going to move us that far forward. The azimuthal solutions though are pretty simple. That you're back to the e to the im phi divided by this square root of 2 pi, and m can be 0 plus or minus 1 plus or minus 2. And the reason why this one's so simple is that motion around this axis is that's still a good quantum number because we haven't monkey with the symmetry that way. When we pull the other thing apart, then the other stuff gets thrown off, the motion gets thrown off and it gets much more complicated, and so it gets much more complicated to describe it. So as I mentioned, we aren't going to get into lambda and mu, just to say that they can be figured out. There are numerical tables and you can compute energies and you can do everything with them. But we're going to actually get an approximate solution that's going to be a lot more insightful than either lambda or mu, although it won't be so quantitatively nice. The m equals 0 solutions are called s states by an out, they're called excuse me, sigma states by analogy with s states for an atom. And then states with one twist around this way are called pi states and plus or minus 2 are delta states. And much like we do with the px, py and pz orbitals, sometimes when we actually plot the solutions, we take linear combinations of them to get sine or cosine so that we don't have an imaginary part to the wave function. The other really important consideration for H2 is that the problem has symmetry. If I imagine the origin in the middle and a positive charge here and a positive charge here, then there's symmetry in the problem and that has to be reflected so if I swap the two nuclei or I invert all the coordinates, then nothing too drastic can happen because it's basically the same thing again. In particular, there's this operator I've called p hat that inverts the coordinates. It just takes all points x, y, z and puts them through the origin to minus x, minus y, minus z. And this operator, if you apply it to a wave function and then you apply the Hamiltonian and then you apply the Hamiltonian to the wave function and then this operator, what you find out is you get the same thing. Therefore, it's not like position and momentum where they didn't commute. This operator, the inversion operator of the coordinates, commutes with the Hamiltonian for H2 plus and that means that an eigenfunction of the Hamiltonian also has to be an eigenfunction of the inversion operator. Well, the inversion operator doesn't have any units so the eigenfunction of the inversion operator is just going to be a number. Question is what number is it? And the answer is let's apply the inversion operator twice. So we'll first take x, y, z and go to minus x, minus y, minus z. Great. Now we'll apply it again. We went back to where we started. Everything has to be the same then because all the coordinates are the same. And so if we do those two steps, we first put on the inversion operator onto an eigenfunction. We get a number c times the eigenfunction. That follows because it has to be an eigenfunction of the operator p. Can't turn the wave function into something else. And then if we apply it again, that's p squared, then we get c squared. But we get the same wave function back. And so what we conclude here is that c squared is 1 and that means that c is either plus 1 or minus 1. There's two possibilities there. Well, that means that the energy eigenfunctions, whatever they happen to be, are either even, if it's plus 1, or odd, if it's minus 1, under inversion of all the coordinates. And that's important because we can use that then to classify the solutions to say, well, you're in the even camp or you're in the odd camp. And historically, we use the symbol g for even and u for odd from the German grad and ungrad. Okay. Now, let's take, instead of writing down lambda and mu and writing all these very complicated formulas, and perhaps in a certain sense being none the wiser when we finish, let's try to guess a solution for the wave function for the electronic part of H2 plus. And then once we guess, let's use our hidden ace in the hole, the variational principle, to optimize our guess so that it's really good after we've proposed some kind of guess. Well, okay, let's explore this. There are a lot of kinds of functions I could guess. I could guess something where I have something in the exponent or I have something squared or I have a lot of things. And I could optimize those by the variational principle. But the problem is that might be a lot of work. You saw with the parameter zeta that we had before that we ended up with very complicated equations and then we had to really stare at them a long time to get what we wanted out. In this, make a contour plot even because we couldn't really see how to find the minima very easily. But in the special case that I fix the form of the function, I say my wave function is like a bowling ball plus a cube plus a pyramid or something. And I have percentages of each of these things. Of course, this is quantum mechanics. The percentages could be complex numbers. But I don't change the shape of the underlying function that I'm using. I just change the amount of it. In that case, there is a very elegant way to look at how this optimization works. And we get a series of equations that lead to something which we call the secular determinant, which I want to introduce now. This is not the slater determinant. This is something else. Let's then suppose that we have two fixed functions which I've called phi 1 and phi 2. And we have to pick them cleverly because if we pick them poorly, then even when we optimize them, we may get a terrible answer for the energy. So we need to have some physical insight, I would say, to pick these functions correctly. And some people have a lot more physical insight than others. But in any case, then we have an amount C1 of phi 1 and an amount C2 of phi 2. And all we're doing now is optimizing the amounts, the linear coefficients. We're adding up the functions. We are not multiplying them. We are not raising them to powers of each other or anything else. We're adding them up and we've got these linear coefficients. Okay. So we have to pick them correctly. Obviously, that's part of the physical intuition. We have to pick functions that go to 0 at infinity and other things like that so that they behave properly. Otherwise, if we optimize them, we may find out that the amount is 0 because we put in a function that can't be there. But for now, let's assume that the functions are fixed and known and they're real like the 1s orbital of hydrogen, for example. And let's also assume that the C1 and C2 are also real. So then we don't need the complex conjugate. And then what I've written here then is the expectation value of the energy which is just the integral of C1 phi 1 plus C2 phi 2. They would normally be star, but let's assume they're real. Then the Hamiltonian and then the same thing, C1 phi 1 plus C2 phi 2. And I can then multiply out. I get four terms. I get C1 squared times phi 1 H phi 1. And then I get the cross terms, two of those, C1 C2 plus C2 C1. And then I get C2 squared and then I have phi 2 H phi 2. And when I do these integrals of the Hamiltonian operating on a function and then integrate over the other function multiplying out on the left, that's just a number. The integral is just a number, whatever it is. It could be zero. It could be something. And I'm just going to write those numbers as the C1 squared. I'm going to call that one H sub 1 1. And for the 1 2, I'm going to call it that number, that integral H 1 2. And for the 2 1, I'm going to call it H 2 1. And for the 2 2, I'm going to call that integral H 2 2. And in general, these numbers H sub ij are called matrix elements. And that's because as you're going to see in a second, they are positions in a matrix. And this is a very beautiful systematic way to keep improving our guesses. That's why this linear combination of orbitals is a very, very powerful approach. Well, the Hamiltonians real and the Cs are real and the functions are real, so the matrix elements are symmetric. Of course, the Hamiltonians always are mission. And so if there's no imaginary part, the real part is symmetric. The imaginary part is anti-symmetric because H ij star is equal to H ji. But we don't have to worry about that. Well, we also, because we took these two functions, we mustn't assume that anything's normalized. We should actually make sure we can normalize things. And so the way you normalize things is you take the square of the wave function or you can consider the square of the wave function a slightly different way. Rather than putting the Hamiltonian in between the two sides, I'm just going to put the one operator. The one operator just takes the wave function and returns it. It does nothing to it. But by writing it that way and then expanding it out, I see that I get four terms again. C 1 squared, I get the integral of phi 1 squared. And then I get phi 1, phi 2, phi 2, phi 1. And I write those exactly the same way as I did with H. C 1 squared, S 1 1. And then I have S 1 2, S 2 1, and S 2 2. And S is called the overlap matrix because it's going to tell us how the different functions we picked overlap. Well, if the basis functions were orthonormal, then the overlap matrix would just be the identity matrix. If I had n functions and they were like different orbitals on a hydrogen atom, for example, on the same nucleus, and that were my basis, then because those are already orthonormal, I have nothing to do. But if I have atomic orbitals on different nuclei so that they're shifted and I have these exponentials, then they certainly are not orthonormal. They're normalized with respect to themselves. If I take this one and multiply by this one and square it, go over all space, it's one. And likewise with this one, but when I take the two, multiplying in between, it's positive everywhere, that can't be zero and therefore they are not orthogonal. So for the two basis functions, if they were orthogonal, this matrix S 1 1, S 1 2, S 2 1, which is the same as S 1 2 and S 2 2 would just amount to the unit matrix 1, then 0, then 0, then 1, but we usually can't, if we make a natural choice of functions, then we usually can't ensure that this happens. And we don't want to. What we want to do is just go ahead and compute the matrix S and then use that to figure out what's going on. So here's our formula then for the energy, keeping in mind that the wave function psi is not necessarily normalized. In fact, the exact overlap is going to depend on how far the nuclei are apart. We can write this energy functional, the energy as a function of the coefficient C 1 and the coefficient C 2, is the expectation value of the Hamiltonian on the top, divided by the square of the wave function on the bottom and the top, I got those four terms, but two of them are the same, so I end up with 2 C 1, C 2 times H 1 2, and on the bottom, I end up with those four terms in the S matrix, but two of them are the same because S 2 1 is equal to S 1 2, so I end up with three terms there. And rather than write that as a ratio the way I have, I'm going to multiply both sides by the denominator and write it as E times this spinach in S is equal to the Hamiltonian matrix elements. Those matrix elements are numbers that are fixed because the functions are fixed. That's the beauty of it. I specify the functions once, I do the integral once, and then they're done, and I don't play around with them anymore. If I had to keep doing the integrals over and over because I'm playing around with the functions themselves, it would be very, very time consuming, whether I do it on a computer or by hand. Well, the variational principle says if you've got the energy of a quantum system as a function of a parameter like C 1 and C 2, then if you minimize the energy with respect to the parameter, you're going to improve your guess of the wave function, and so that's exactly what we're going to do because now we have a formula here for the energy. What I've done here then is I've just taken the derivative of both sides of this equation with respect to C 1. So we have a product, so it's the derivative of the first, that's the first term, times the second, which is all the S matrix elements, plus the first, which is just the energy, times the derivative of the second, the term with C 2 goes away because this is a partial derivative with respect to C 1. And on the Hamiltonian side, I end up with 2C 1H 1 1 plus 2C 2H 1 2. And interestingly enough, there's a lot of 2's in this thing. At the minimum or maximum, but it's going to be a minimum, in fact, the derivative should be 0, therefore we can say that the E times the S part is, in fact, equal to the other part because the term with D E, D C 1, is 0. And that's nice because now we have E left over. And if we write that out, we get what I have on the bottom of slide 560, which is that C 1 times H 1 1 minus E S 1 1 plus C 2 times H 1 2 minus E S 1 2 is equal to 0. If we do the derivative with respect to C 2, you get a similar equation. And these two equations then, in two unknowns, can be written in matrix notation as a matrix that now has elements H 1 1 minus E S 1 1 and so forth in each of the entries. It's always H minus E S times the column vector C 1 C 2 is equal to 0. This is a system of linear equations. And one solution to this system of linear equations, the simplest one actually, is just say, okay, no matter what the matrix is, let's let C 1 and C 2 equals to 0, both of them, then anything times 0 plus 0 times anything is 0. That works and the other one works. And so the whole thing comes to 0. Well, that one's no good because if C 1 and C 2 are both 0, then unfortunately our wave function, which is C 1 times the first function plus C 2 times the second, is also 0 and that's not going to help us find a good wave function. So that trivial solution is not the one we want. In some sense then what we want is instead of the column vector being 0, we want the matrix somehow to be 0. But we can't control the matrix to be 0 per se because it has these fixed numbers in it. What we can do is we can make the determinant of the matrix equal to 0. And in order to see why this works, you really need a course in linear algebra, which is why it's very important to take a course in linear algebra. So you can get to the bottom of stuff like this once and for all. And I've written then the straight lines like I did for the Slater determinant, the determinant of this array of numbers is equal to 0. That's the necessary and sufficient condition that we get a solution that we want. Well, this determinant is in fact called the secular determinant. The secular determinant then is just the matrix element of the Hamiltonian, take the function, put H on it, take the other function, start if you need to, if it's complex, integrate it, get a number. That number is fixed, you shove them into a matrix, do the same thing with the overlap where there's no Hamiltonian, shove those numbers in, put E in there, and then you will get an equation in E and following the rules for taking a determinant. Remember it's a crisscross rule for a 2 by 2. So I have H minus ES, H minus ES. Well, H11 and H22 are the same in this problem and H12 is the same. And so we're going to get a quadratic equation in the energy and we're going to get two roots and take the smaller one to be our best estimate of the energy. As usual, when you solve mathematical equations, mathematicians like to give you all possible answers, not necessarily the one you're looking for. And so because they don't know which one you're looking for among other things. And so we have to look through and then after we get all the energies out of solving this secular determinant, we have to then figure out which one's the best. When we've got three basis functions, we end up with a cubic because we end up with a 3 by 3 determinant, which means E, every term has E times E times E. So we've got a cubic equation to solve, which can be very messy. And of course it's got other powers too because there's more than one term. They don't all have E or we could just take the cube root. And with a quartic, it's very messy indeed. And so in actual fact, we don't ever try to just write down a formula for the energy except if it's a quadratic equation. And usually if we're lucky, we don't even have to write down the formula for a quadratic equation. Either we can just look at it. But you can always calculate them numerically and there are standard packages to root polynomials. Once you set up the secular determinant and whack it out, you end up with a big polynomial in the energy and you want to solve for the values of E that are real that make that thing equal to zero. And there are standard methods to root polynomials, packages that you can get and I think you can see why having something like that is like having a hammer or a screwdriver. It's incredibly useful if you want to pound in a nail or screw in a screw and if you don't have that tool, you're kind of stuck. And there are lots of course mathematical tools like that available to us these days. Okay, let's go back to H2 with this secular determinant and let's do this one then as practice problem 28. Let's guess, since we've got no better ideas, two 1S orbitals. Let's take one of them centered on proton A and the other one centered on proton B. Then the question is, we're going to take C1 of this one and C2 of that one. What's the form for the two energies from solving the secular determinant? That's the question. Well, okay. We know that the 1S orbitals are normalized, so we don't have to worry about that. If we integrate them over all space, the answer is one. And unfortunately, they're not orthogonal because once they're off nuclei, they aren't neither one nor zero. They're something and we have to figure out what the something is. And I've called that something just S because it's the only thing that's not going to be one in the overlap matrix. It's definitely not equal to zero, that much we can see because they're both positive exponentials. It's got to integrate to something between zero and one. And the problem has symmetry, so luckily we don't have to do all the matrix elements over and over. We can just do first the matrix element I'll call HAa. That's got 1SA, the Hamiltonian 1SA. We'll work out what it is later, but to solve this, we don't need to actually work it out. We just need to assume it has some value. And that's exactly the same as HBB because we got the same orbital. We got the same nuclear charge. Everything's the same. And then there's this one HAb, which is mixed term. And that again has symmetry, whether I call this one A or this one B or vice versa, they have to be the same. So that one's equal to HBA. So I only have to do two of the four possible matrix elements and then I can just fill in the other ones. Keep in mind here that although we're writing two functions, 1SA, 1SB, these integrals are one three-dimensional integral because I've got one orbital that's the sum of these two guys. I'm writing them out as four terms because that's the way it pans out. But this is different than when we took helium where we had a product of this times this. That's different because then you've got all the variables and we had six-dimensional integrals. We had to integrate over R1 and then integrate over R2 as well. These integrals just have R, one R. And I have to figure out how to express the orbitals 1SA and 1SB in terms of this variable R and maybe some other angular variables like theta and it won't surprise you that the law of cosines is going to come in again. But keep in mind that this is one three-dimensional integral and nothing like what we were doing with helium even though it might look like it. It's easy to get mixed up. Well, here's our secular determinant then putting in all the values that are the same. We have HAA minus E and then the next one off diagonal HAB minus E times S where S is just the one that isn't zero. And then same thing in the 2, 1 entry. And then same thing on the bottom. And so this times this minus this times this just gives us HAA minus E quantity squared minus HAB minus ES quantity squared equals to zero. Well, I could expand this out and solve it as a quadratic equation. And that would be quite messy to do because rather than just having AX squared plus BX plus C, A, B and C would be a very messy thing and then I'd have to try to simplify it and that would be a very slow way to go. But luckily I can factorize this by looking at it because whenever I see this squared minus that squared, I remember that that's something plus something times something minus something and therefore I can think, well, if I say A squared minus B squared is A plus B times A minus B then I can write that out with A and B. My terms there, HAA minus E plus HAB minus ES and then the other way around equals to zero. I've got two things equal to zero. The only way that can be true is if either this one's zero or that one's zero. And so that means we get two energy roots. The first one I've called E plus, you get HAA minus E plus HAB minus ES equals to zero. You solve that for E and you call it E plus and you get HAA plus HAB divided by one plus S. The other one is just the other part and I get HAA minus HAB divided by one minus S and I've called that E minus. To quantify things though, what we're going to have to do, so that's the end of the problem. But to quantify things, we need to know what these numbers are and in particular, we need to know what these numbers are as a function of big R, the separation between the two nuclei. If we can work that out and then we can track E plus and E minus pretending we're an electron riding along on this aircraft carrier of this molecule, we can pretend and we can see where we're going to have the sweet spot of energy or if we have one. What would be good is to, we tried to predict the ionization potential of helium, we tried to predict whether hydride would be stable or not, we saw sometimes it could be tough. We want to see if we take something simple like this with just a quadratic equation here in two roots and we pick the most stable one, does it predict that H2 plus, which does in fact exist, does exist and if it does, what value of big R is the most stable value and how does that compare with the actual value, which has been measured of course spectroscopically. And then next time what we'll see then is we can probably take this and take this orbital that we solved with this and we can probably pop another electron in there and we can argue well the electron-electron repulsion is important but it's not all important and so we can just kind of adjust our theory a little bit and then we may be able to predict things about the properties of a molecule like H2. So next time we'll take a bit of time and we'll actually go through how to calculate with these 1S orbitals, the matrix elements, how to do these integrals. They aren't so easy to do. They aren't so hard but they aren't trivial either and it's probably good one time to go through them and see how to do them and then you'll really appreciate when the computer is doing it for you how much labor saving it is. If you set up a quantum chemistry calculation, sometimes the program will tell you how much percent overlap you've got in terms of setting up the calculation and now I think you can see basically what it's talking about is talking about this matrix S and the fact that if you don't set up an orthonormal basis then it's having to do a lot more calculation but sometimes it's very hard to see what kind of orthonormal basis I'm going to use when I've got things at different positions in space and I just can't visualize what I'm going to pick then I'm going to tend to pick atomic orbitals that are centered on the atoms and in that case I certainly can't expect them to all be orthogonal to each other. So next time we'll do those integrals, we'll work through them systematically and we'll see if we can predict whether H2 plus is stable or not and if so how stable is it?
 Today in chemistry 131A we're going to move on  from multi-electron atoms to the simplest molecule.  We're going to talk first about the Born Oppenheimer  approximation.  This was a great idea that was introduced very shortly  after the introduction of the Schrodinger equation itself.  In fact, when you go back to this period in history,  what you find out is that everybody was really working  like crazy on this new field of quantum mechanics,  much like if you come back to 2014 you may find a lot  of biologists were working like crazy on stem cells now  that they found a better way to produce them.  And then after we talk  about the Born Oppenheimer approximation,  we're going to introduce the simplest molecule H2 plus  and from there we'll go to H2.  First, the Born Oppenheimer approximation.  We're going to just look at the simplest molecules H2 plus  and H2 and we're going to look at both of them  in some detail to try to understand why it is  that you can have a chemical bond.  In fact, it's very interesting in the framework  of classical mechanics and electrostatics to try  to figure out if there could be such a thing as, for example,  two positively charged particles and let's say one  or two electrons.  Let's say we have two so the whole thing is neutral  and it's like H2 and we could propose  that they're doing some classical thing.  We could forget about the uncertainty principle.  Let's have them going like this in some pattern and so forth  and have all the electrostatic forces and try  to see whether the molecule will hold together.  And in fact, scientists at the time didn't really understand  how molecules could hold together  because it doesn't seem to work out very well.  And we'll see exactly why when we come  to the exchange integral.  But if we're going to talk about molecules,  even when we talked about helium,  we had to use some approximations.  We either had to make a good guess with the variational theory  or try perturbation theory or something.  And now that we've got two nuclei, two centers,  even with one electron, we're going to guess  that it's going to be extremely hard  to solve things exactly.  So the first idea is that we want to decouple the motion  of the nuclei from the motion of the electrons.  And we know that we can probably do that  because the electrons are much less massive than the nuclei.  And the electrons are therefore moving much more quickly.  What we're trying to do then is say, let's say we've got an  elephant and we've got a bunch of flies around it.  And the elephant's moving slowly.  And so the flies have plenty of time to readjust.  And if the elephant rolls over slowly, the flies have plenty  of time to find the best place where they want  to alight on the elephant.  And in much the same way, as the nuclear coordinates change,  the electrons almost instantaneously, at least from the  perspective of the nuclei readjust.  And they readjust to the best possible position.  And so that being the case, we can go one step further then.  And we can just say supposing then that the,  we just stop the motion of the nuclei.  And we solve for the electrons.  Then that we can do.  And that's essentially what the Born-Appenheimer approximation is.  First I want to talk about our notation here.  Our equations are going to get more complicated.  And that's going to happen quickly.  So I'm going to use big M to denote the mass of the proton  or the nucleus and little M to denote the mass of the electron.  And our coordinate system is as on this slide 543  where you have two nuclei marked A and B. And in this case,  I've put in both electrons, which I call one and two.  This is H2 then.  And this is at some instant.  Some we're trying to set up how we're going to solve  for the wave equation, for the Schrodinger equation.  And we have all these repulsive and attractive terms  between these charged particles.  The Hamiltonian then, with all the terms in, has a kinetic energy  for the first nucleus, A, a kinetic energy  for the second nucleus, B, the inter-nuclear repulsion,  which is a positive term pushing them apart.  Then the kinetic energy for the first electron, one,  the kinetic energy for the second electron, two.  Then the attraction of electron one  to the first nucleus to the second nucleus.  The attraction of electron two to the first nucleus  to the second nucleus.  And then the repulsion, R12, which is the distance  between the two electrons.  And we are not going to be able to solve this exactly  with all these terms for the total wave function,  which includes all the nuclear and electronic coordinates.  And in fact, even with the two electrons, we won't be able  to solve it exactly.  But like the H-atom, we can separate out the center  of mass motion for sure.  But we can't rigorously separate the relative nuclear motion.  We have to make an argument about why it is  that we're doing that.  And Born and Oppenheimer wrote a classic paper  that I've given you the reference to here  if you'd like to look it up.  Unfortunately, it's in German, and so you would have  to look at the translation of it.  I had to actually learn to read German when I got my degree  in chemistry, which shows you how old I am.  But it helps occasionally from time to time if you have  to go back and read some of these papers.  And their paper was on the quantum mechanics of molecules.  And they have a very elegant derivation about why it is  that you can ignore the nuclear motion.  But I don't want to go through that because that would take  at least this lecture and another lecture,  and it would just take us a little bit too far a field  to actually go through that.  But we can certainly see that because the nuclei are moving  slowly and because the electrons adjust,  then if I just consider the nuclei here,  and I solve the Schrodinger equation for the electrons,  pretending the nuclei are frozen.  And then I move the nuclei a little bit  and resolve the Schrodinger equation.  That's basically what the electrons are doing continuously  as the nuclei move slowly from here to here,  is they're just solving the Schrodinger equation again  and again and again.  And because they're readjusting so quickly, it appears  to them as if the nuclei are basically frozen,  and that means to us they're basically frozen as well.  So if we write that the wave function of the molecule is,  which depends on all the coordinates, is the factorization  of product of the wave function of the electronic part  and the wave function of the nuclear part,  then the total energy is just the sum  of the electronic energy and the nuclear energy.  This is right back to particle in a box  with the x, y directions, and this has been our strategy over  and over and over, which shows you it's a very good strategy.  Then what we find is that the electronic function depends  instantaneously, so to speak, on the coordinates  of the electrons, R1, R2, and it depends parametrically  on this very slow variable, big R, which is just the distance  between the two nuclei.  And we can always use this approximation for molecules  that have very high excited electronic states.  So usually to excite an electron from any ground state  to some higher state takes a lot of energy,  and the slow motion of the nuclei cannot possibly supply  that kind of energy.  Where we have to be in trouble or where we have  to be extremely careful is if we have radicals  or open-shell atoms or something like that,  where we have a lot of electronic states that are nearby,  then what you can imagine happening is that the motion  of the nuclei itself can somehow cause a transition  of the electron to move out of the orbital that was originally  and into a new one.  And in that case, the motions are coupled,  the whole thing's wobbling around, then we're in trouble  and we can't use the Born-Oppenheimer approximation,  so in that case it fails.  But the upshot of our factorization is we can forget  completely about the kinetic energy of the nuclei.  We just clamp them down, we freeze them,  and we solve for the electronic energy.  And that's really an important simplification.  Here I've written out the Hamiltonian then in atomic,  in atomic units, the electronic Hamiltonian still has the  kinetic energy of the electrons, the kinetic energy  of the nuclei is gone.  And I've added to it 1 over r, which is just the static,  electrostatic repulsion of the two nuclei.  We have all these instantaneous terms  and then we have this static term, but we have to add  that to the electronic energy because when the nuclei are very  close that will be an overall unfavorable position.  We don't care about the kinetic energy of the nuclei  because we threw that away.  And so what we want to do then is solve the Schrodinger equation  for the electronic coordinates as a function  of the internuclear separation, big r, and then see what that is.  And in fact, if we do that, what we find is  that when the nuclei are very close, it's very unfavorable.  Partly it's just the internuclear repulsion,  but partly it is that the electrons themselves are pretty  much outside and so don't screen the positive charges.  And as we move them apart, the potential comes down,  and then there's a sweet spot where there's a minimum  and then it comes back up again.  And in fact, that's exactly the kind of shape  that we argued was like a Morse potential  or some potential like that when we were talking  about vibrations and the harmonic oscillator.  And now you can see that the nuclei themselves glide around  on a potential energy surface that's constantly readjusting  and we just draw it as if it's always there,  but it's constantly readjusting as they move,  it is there before they get to where they're going, so to speak.  And we talked then about the nuclear coordinates moving  around on a potential energy curve  or a potential energy surface,  which is the total electronic energy  plus the internuclear repulsion.  That's what a potential energy surface is how you should think  of it.  And the nuclei will respond to this.  So if they're too close, they'll gradually pull apart.  And as they pull apart, the electrons will jigger around  and readjust.  The simplest possible molecules H2 plus, I want to begin  with that, I wrote down the Hamiltonian for H2,  but that's a little bit intimidating for us right now,  so I want to just do it with one electron.  The second electron is now not present,  so I don't need one and two, I can just use Dell.  And there's only the attraction to the first nucleus,  the attraction to the second nucleus.  There's no 1 over R12, thank goodness,  because that was quite a production  when we did helium and hydride.  And then there's 1 over R, which is no big deal,  which is the internuclear repulsion.  Here's our picture then on the bottom of slide 549,  same coordinate system as before,  except the second electron is missing.  Well, much like we could solve the hydrogen atom,  once we factored out the center of mass motion  and assumed the proton was essentially fixed,  by saying, well look, this problem has spherical symmetry.  We should actually write the Schrodinger equation  in spherical coordinates, and then what happened is,  we knew we had the right coordinate system  because the wave function factorized into three parts,  and that made it easy to solve each part.  This three-dimensional differential equation became three  one-dimensional differential equations,  and that is always the strategy if you can get it to work.  Now in this case, spherical coordinates won't work  because we have stretched the charge out in one direction.  Well, if you take a circle and you stretch it in one dimension  so that you have two centers, then you have an ellipse.  And so not surprisingly, there's a coordinate system  called elliptical coordinates,  or confocal elliptical coordinates,  where you take the sum of the distance between the electron  and the nuclei divided by r, so that's a dimensionless thing,  lambda, and then another one, which is the difference  between them, rA minus rB divided by big r, call that mu.  Then the third coordinate is still like it was  in the hydrogen atom, it's that coordinate phi, and the solution  for that because that's basically the same.  That was the particle on a ring.  We can still use that to solve.  That decouples out, and so we don't have to worry about that one.  The other two, we have to actually worry  about what kind of differential equation that we get.  You can think of, if you make a triangle between the two nuclei  and the electron, you have a plane, three points to find a plane,  and the position of that plane around this axis  between the two nuclei is the azimuthal angle phi.  And so this is just like a stretched hydrogen atom, really.  Let's write the wave function again as a product.  We're really beating a dead horse with this,  but it's the only horse we've got.  We're going to write it as a function of lambda,  which I've called big lambda of lambda, a function of mu,  and unfortunately capital mu looks like an m, but it's mu  of mu, and then phi of phi.  And those are just presupposing that we can write it  like this, and I've given you a reference here at the bottom  of page 550, slide 551, if you're interested.  There's quite a long paper in the philosophical transactions  of the Royal Society where they've gone  through exactly how you separate these coordinate systems.  It does separate, it does separate  into a differential equation with respect to mu only,  big m, and then another differential equation  that only has big lambda.  The problem for us is that these differential equations,  even though they're one dimensional,  if you look at them, they're very complicated,  and they have a lot of moving parts, and we can't write  down the solution to these kind of equations  in any elementary way.  There may be solutions in terms of repeated fractions,  one plus one over x plus plus and so forth,  and things like that, but that's not really too illuminating  for us at this level.  And so I don't want to go into writing, attempting to write  down these very esoteric functions that might solve this  because that's not going to move us that far forward.  The azimuthal solutions though are pretty simple.  That you're back to the e to the im phi divided by this square  root of 2 pi, and m can be 0 plus or minus 1 plus or minus 2.  And the reason why this one's so simple is that motion  around this axis is that's still a good quantum number  because we haven't monkey with the symmetry that way.  When we pull the other thing apart,  then the other stuff gets thrown off, the motion gets thrown  off and it gets much more complicated,  and so it gets much more complicated to describe it.  So as I mentioned, we aren't going to get into lambda and mu,  just to say that they can be figured out.  There are numerical tables and you can compute energies  and you can do everything with them.  But we're going to actually get an approximate solution  that's going to be a lot more insightful  than either lambda or mu,  although it won't be so quantitatively nice.  The m equals 0 solutions are called s states by an out,  they're called excuse me, sigma states by analogy  with s states for an atom.  And then states with one twist around this way are called pi  states and plus or minus 2 are delta states.  And much like we do with the px, py and pz orbitals,  sometimes when we actually plot the solutions,  we take linear combinations of them to get sine or cosine  so that we don't have an imaginary part to the wave function.  The other really important consideration  for H2 is that the problem has symmetry.  If I imagine the origin in the middle and a positive charge here  and a positive charge here, then there's symmetry in the problem  and that has to be reflected so if I swap the two nuclei  or I invert all the coordinates, then nothing too drastic  can happen because it's basically the same thing again.  In particular, there's this operator I've called p hat  that inverts the coordinates.  It just takes all points x, y, z and puts them through the origin  to minus x, minus y, minus z. And this operator,  if you apply it to a wave function and then you apply  the Hamiltonian and then you apply the Hamiltonian  to the wave function and then this operator,  what you find out is you get the same thing.  Therefore, it's not like position and momentum  where they didn't commute.  This operator, the inversion operator of the coordinates,  commutes with the Hamiltonian for H2 plus and that means  that an eigenfunction of the Hamiltonian also has  to be an eigenfunction of the inversion operator.  Well, the inversion operator doesn't have any units  so the eigenfunction of the inversion operator is just  going to be a number.  Question is what number is it?  And the answer is let's apply the inversion operator twice.  So we'll first take x, y, z and go to minus x, minus y,  minus z. Great.  Now we'll apply it again.  We went back to where we started.  Everything has to be the same then  because all the coordinates are the same.  And so if we do those two steps, we first put  on the inversion operator onto an eigenfunction.  We get a number c times the eigenfunction.  That follows because it has to be an eigenfunction  of the operator p. Can't turn the wave function  into something else.  And then if we apply it again, that's p squared,  then we get c squared.  But we get the same wave function back.  And so what we conclude here is that c squared is 1  and that means that c is either plus 1 or minus 1.  There's two possibilities there.  Well, that means that the energy eigenfunctions,  whatever they happen to be, are either even, if it's plus 1,  or odd, if it's minus 1, under inversion  of all the coordinates.  And that's important because we can use that then  to classify the solutions to say, well,  you're in the even camp or you're in the odd camp.  And historically, we use the symbol g for even and u  for odd from the German grad and ungrad.  Okay. Now, let's take, instead of writing down lambda  and mu and writing all these very complicated formulas,  and perhaps in a certain sense being none the wiser  when we finish, let's try to guess a solution  for the wave function for the electronic part of H2 plus.  And then once we guess, let's use our hidden ace  in the hole, the variational principle,  to optimize our guess so that it's really good  after we've proposed some kind of guess.  Well, okay, let's explore this.  There are a lot of kinds of functions I could guess.  I could guess something where I have something in the exponent  or I have something squared or I have a lot of things.  And I could optimize those by the variational principle.  But the problem is that might be a lot of work.  You saw with the parameter zeta that we had before  that we ended up with very complicated equations  and then we had to really stare at them a long time  to get what we wanted out.  In this, make a contour plot even  because we couldn't really see how  to find the minima very easily.  But in the special case that I fix the form of the function,  I say my wave function is like a bowling ball plus a cube  plus a pyramid or something.  And I have percentages of each of these things.  Of course, this is quantum mechanics.  The percentages could be complex numbers.  But I don't change the shape  of the underlying function that I'm using.  I just change the amount of it.  In that case, there is a very elegant way to look  at how this optimization works.  And we get a series of equations that lead to something  which we call the secular determinant,  which I want to introduce now.  This is not the slater determinant.  This is something else.  Let's then suppose that we have two fixed functions  which I've called phi 1 and phi 2.  And we have to pick them cleverly because if we pick them poorly,  then even when we optimize them,  we may get a terrible answer for the energy.  So we need to have some physical insight, I would say,  to pick these functions correctly.  And some people have a lot more physical insight than others.  But in any case, then we have an amount C1 of phi 1  and an amount C2 of phi 2.  And all we're doing now is optimizing the amounts,  the linear coefficients.  We're adding up the functions.  We are not multiplying them.  We are not raising them to powers of each other or anything else.  We're adding them up and we've got these linear coefficients.  Okay. So we have to pick them correctly.  Obviously, that's part of the physical intuition.  We have to pick functions that go to 0 at infinity  and other things like that so that they behave properly.  Otherwise, if we optimize them, we may find out that the amount is 0  because we put in a function that can't be there.  But for now, let's assume that the functions are fixed and known  and they're real like the 1s orbital of hydrogen, for example.  And let's also assume that the C1 and C2 are also real.  So then we don't need the complex conjugate.  And then what I've written here then is the expectation value  of the energy which is just the integral of C1 phi 1 plus C2 phi 2.  They would normally be star, but let's assume they're real.  Then the Hamiltonian and then the same thing,  C1 phi 1 plus C2 phi 2.  And I can then multiply out.  I get four terms.  I get C1 squared times phi 1 H phi 1.  And then I get the cross terms, two of those, C1 C2 plus C2 C1.  And then I get C2 squared and then I have phi 2 H phi 2.  And when I do these integrals of the Hamiltonian operating  on a function and then integrate over the other function  multiplying out on the left, that's just a number.  The integral is just a number, whatever it is.  It could be zero.  It could be something.  And I'm just going to write those numbers as the C1 squared.  I'm going to call that one H sub 1 1.  And for the 1 2, I'm going to call it that number,  that integral H 1 2.  And for the 2 1, I'm going to call it H 2 1.  And for the 2 2, I'm going to call that integral H 2 2.  And in general, these numbers H sub ij are called matrix elements.  And that's because as you're going to see in a second,  they are positions in a matrix.  And this is a very beautiful systematic way  to keep improving our guesses.  That's why this linear combination  of orbitals is a very, very powerful approach.  Well, the Hamiltonians real and the Cs are real  and the functions are real, so the matrix elements are symmetric.  Of course, the Hamiltonians always are mission.  And so if there's no imaginary part, the real part is symmetric.  The imaginary part is anti-symmetric  because H ij star is equal to H ji.  But we don't have to worry about that.  Well, we also, because we took these two functions,  we mustn't assume that anything's normalized.  We should actually make sure we can normalize things.  And so the way you normalize things is you take the square  of the wave function or you can consider the square  of the wave function a slightly different way.  Rather than putting the Hamiltonian in between the two sides,  I'm just going to put the one operator.  The one operator just takes the wave function and returns it.  It does nothing to it.  But by writing it that way and then expanding it out,  I see that I get four terms again.  C 1 squared, I get the integral of phi 1 squared.  And then I get phi 1, phi 2, phi 2, phi 1.  And I write those exactly the same way as I did with H. C 1  squared, S 1 1.  And then I have S 1 2, S 2 1, and S 2 2.  And S is called the overlap matrix because it's going  to tell us how the different functions we picked overlap.  Well, if the basis functions were orthonormal,  then the overlap matrix would just be the identity matrix.  If I had n functions and they were like different orbitals  on a hydrogen atom, for example, on the same nucleus,  and that were my basis, then because those are already  orthonormal, I have nothing to do.  But if I have atomic orbitals on different nuclei  so that they're shifted and I have these exponentials,  then they certainly are not orthonormal.  They're normalized with respect to themselves.  If I take this one and multiply by this one  and square it, go over all space, it's one.  And likewise with this one, but when I take the two,  multiplying in between, it's positive everywhere,  that can't be zero and therefore they are not orthogonal.  So for the two basis functions, if they were orthogonal,  this matrix S 1 1, S 1 2, S 2 1, which is the same as S 1 2  and S 2 2 would just amount to the unit matrix 1, then 0,  then 0, then 1, but we usually can't,  if we make a natural choice of functions,  then we usually can't ensure that this happens.  And we don't want to.  What we want to do is just go ahead  and compute the matrix S and then use that  to figure out what's going on.  So here's our formula then for the energy, keeping in mind  that the wave function psi is not necessarily normalized.  In fact, the exact overlap is going to depend  on how far the nuclei are apart.  We can write this energy functional, the energy as a  function of the coefficient C 1 and the coefficient C 2,  is the expectation value of the Hamiltonian on the top,  divided by the square of the wave function on the bottom  and the top, I got those four terms, but two of them are the  same, so I end up with 2 C 1, C 2 times H 1 2,  and on the bottom, I end up with those four terms  in the S matrix, but two of them are the same  because S 2 1 is equal to S 1 2, so I end up with three terms  there.  And rather than write that as a ratio the way I have,  I'm going to multiply both sides by the denominator  and write it as E times this spinach in S is equal  to the Hamiltonian matrix elements.  Those matrix elements are numbers that are fixed  because the functions are fixed.  That's the beauty of it.  I specify the functions once, I do the integral once,  and then they're done, and I don't play around with them anymore.  If I had to keep doing the integrals over and over  because I'm playing around with the functions themselves,  it would be very, very time consuming, whether I do it  on a computer or by hand.  Well, the variational principle says if you've got the energy  of a quantum system as a function of a parameter  like C 1 and C 2, then if you minimize the energy with respect  to the parameter, you're going to improve your guess  of the wave function, and so that's exactly what we're going  to do because now we have a formula here for the energy.  What I've done here then is I've just taken the derivative  of both sides of this equation with respect to C 1.  So we have a product, so it's the derivative of the first,  that's the first term, times the second,  which is all the S matrix elements, plus the first,  which is just the energy, times the derivative  of the second, the term with C 2 goes away  because this is a partial derivative with respect to C 1.  And on the Hamiltonian side, I end up with 2C 1H 1 1 plus 2C 2H 1 2.  And interestingly enough, there's a lot of 2's in this thing.  At the minimum or maximum, but it's going to be a minimum,  in fact, the derivative should be 0, therefore we can say  that the E times the S part is, in fact, equal to the other part  because the term with D E, D C 1, is 0.  And that's nice because now we have E left over.  And if we write that out, we get what I have on the bottom  of slide 560, which is that C 1 times H 1 1 minus E S 1 1  plus C 2 times H 1 2 minus E S 1 2 is equal to 0.  If we do the derivative with respect to C 2,  you get a similar equation.  And these two equations then, in two unknowns, can be written  in matrix notation as a matrix that now has elements H 1 1 minus E S 1 1  and so forth in each of the entries.  It's always H minus E S times the column vector C 1 C 2 is equal to 0.  This is a system of linear equations.  And one solution to this system of linear equations,  the simplest one actually, is just say, okay,  no matter what the matrix is, let's let C 1 and C 2 equals to 0,  both of them, then anything times 0 plus 0 times anything is 0.  That works and the other one works.  And so the whole thing comes to 0.  Well, that one's no good because if C 1 and C 2 are both 0,  then unfortunately our wave function, which is C 1 times the first function  plus C 2 times the second, is also 0 and that's not going  to help us find a good wave function.  So that trivial solution is not the one we want.  In some sense then what we want is instead  of the column vector being 0, we want the matrix somehow to be 0.  But we can't control the matrix to be 0 per se  because it has these fixed numbers in it.  What we can do is we can make the determinant  of the matrix equal to 0.  And in order to see why this works, you really need a course  in linear algebra, which is why it's very important  to take a course in linear algebra.  So you can get to the bottom of stuff like this once and for all.  And I've written then the straight lines like I did  for the Slater determinant, the determinant of this array  of numbers is equal to 0.  That's the necessary and sufficient condition  that we get a solution that we want.  Well, this determinant is in fact called the secular determinant.  The secular determinant then is just the matrix element  of the Hamiltonian, take the function, put H on it,  take the other function, start if you need to, if it's complex,  integrate it, get a number.  That number is fixed, you shove them into a matrix,  do the same thing with the overlap  where there's no Hamiltonian, shove those numbers in,  put E in there, and then you will get an equation in E  and following the rules for taking a determinant.  Remember it's a crisscross rule for a 2 by 2.  So I have H minus ES, H minus ES.  Well, H11 and H22 are the same in this problem  and H12 is the same.  And so we're going to get a quadratic equation in the energy  and we're going to get two roots and take the smaller one  to be our best estimate of the energy.  As usual, when you solve mathematical equations,  mathematicians like to give you all possible answers,  not necessarily the one you're looking for.  And so because they don't know which one you're looking  for among other things.  And so we have to look through and then after we get all the  energies out of solving this secular determinant,  we have to then figure out which one's the best.  When we've got three basis functions, we end up with a cubic  because we end up with a 3 by 3 determinant,  which means E, every term has E times E times E.  So we've got a cubic equation to solve,  which can be very messy.  And of course it's got other powers too  because there's more than one term.  They don't all have E or we could just take the cube root.  And with a quartic, it's very messy indeed.  And so in actual fact, we don't ever try to just write  down a formula for the energy except  if it's a quadratic equation.  And usually if we're lucky, we don't even have to write  down the formula for a quadratic equation.  Either we can just look at it.  But you can always calculate them numerically  and there are standard packages to root polynomials.  Once you set up the secular determinant and whack it out,  you end up with a big polynomial in the energy and you want  to solve for the values of E that are real  that make that thing equal to zero.  And there are standard methods to root polynomials,  packages that you can get and I think you can see why having  something like that is like having a hammer or a screwdriver.  It's incredibly useful if you want to pound in a nail  or screw in a screw and if you don't have that tool,  you're kind of stuck.  And there are lots of course mathematical tools  like that available to us these days.  Okay, let's go back to H2 with this secular determinant  and let's do this one then as practice problem 28.  Let's guess, since we've got no better ideas,  two 1S orbitals.  Let's take one of them centered on proton A  and the other one centered on proton B. Then the question is,  we're going to take C1 of this one and C2 of that one.  What's the form for the two energies  from solving the secular determinant?  That's the question.  Well, okay.  We know that the 1S orbitals are normalized,  so we don't have to worry about that.  If we integrate them over all space, the answer is one.  And unfortunately, they're not orthogonal  because once they're off nuclei, they aren't neither one nor zero.  They're something and we have to figure out what the something is.  And I've called that something just S  because it's the only thing that's not going to be one  in the overlap matrix.  It's definitely not equal to zero, that much we can see  because they're both positive exponentials.  It's got to integrate to something between zero and one.  And the problem has symmetry,  so luckily we don't have to do all the matrix elements over  and over.  We can just do first the matrix element I'll call HAa.  That's got 1SA, the Hamiltonian 1SA.  We'll work out what it is later, but to solve this,  we don't need to actually work it out.  We just need to assume it has some value.  And that's exactly the same as HBB  because we got the same orbital.  We got the same nuclear charge.  Everything's the same.  And then there's this one HAb, which is mixed term.  And that again has symmetry, whether I call this one A  or this one B or vice versa, they have to be the same.  So that one's equal to HBA.  So I only have to do two of the four possible matrix elements  and then I can just fill in the other ones.  Keep in mind here that although we're writing two functions,  1SA, 1SB, these integrals are one three-dimensional integral  because I've got one orbital that's the sum of these two guys.  I'm writing them out as four terms  because that's the way it pans out.  But this is different than when we took helium  where we had a product of this times this.  That's different because then you've got all the variables  and we had six-dimensional integrals.  We had to integrate over R1 and then integrate over R2 as well.  These integrals just have R, one R. And I have to figure  out how to express the orbitals 1SA and 1SB in terms  of this variable R and maybe some other angular variables  like theta and it won't surprise you that the law  of cosines is going to come in again.  But keep in mind that this is one three-dimensional integral  and nothing like what we were doing with helium even  though it might look like it.  It's easy to get mixed up.  Well, here's our secular determinant then putting  in all the values that are the same.  We have HAA minus E and then the next one off diagonal HAB  minus E times S where S is just the one that isn't zero.  And then same thing in the 2, 1 entry.  And then same thing on the bottom.  And so this times this minus this times this just gives us HAA  minus E quantity squared minus HAB minus ES quantity squared  equals to zero.  Well, I could expand this out and solve it  as a quadratic equation.  And that would be quite messy to do because rather  than just having AX squared plus BX plus C, A, B and C would be  a very messy thing and then I'd have to try to simplify it  and that would be a very slow way to go.  But luckily I can factorize this by looking at it  because whenever I see this squared minus that squared,  I remember that that's something plus something times something  minus something and therefore I can think, well,  if I say A squared minus B squared is A plus B times A  minus B then I can write that out with A and B. My terms there,  HAA minus E plus HAB minus ES and then the other way  around equals to zero.  I've got two things equal to zero.  The only way that can be true is if either this one's zero  or that one's zero.  And so that means we get two energy roots.  The first one I've called E plus, you get HAA minus E  plus HAB minus ES equals to zero.  You solve that for E and you call it E plus and you get HAA  plus HAB divided by one plus S. The other one is just the  other part and I get HAA minus HAB divided by one minus S  and I've called that E minus.  To quantify things though, what we're going to have to do,  so that's the end of the problem.  But to quantify things, we need to know what these numbers are  and in particular, we need to know what these numbers are  as a function of big R, the separation between the two nuclei.  If we can work that out and then we can track E plus  and E minus pretending we're an electron riding along  on this aircraft carrier of this molecule, we can pretend  and we can see where we're going to have the sweet spot  of energy or if we have one.  What would be good is to, we tried to predict the ionization  potential of helium, we tried to predict whether hydride would  be stable or not, we saw sometimes it could be tough.  We want to see if we take something simple like this  with just a quadratic equation here in two roots  and we pick the most stable one, does it predict that H2 plus,  which does in fact exist, does exist and if it does,  what value of big R is the most stable value  and how does that compare with the actual value,  which has been measured of course spectroscopically.  And then next time what we'll see then is we can probably take  this and take this orbital that we solved with this  and we can probably pop another electron in there  and we can argue well the electron-electron repulsion is  important but it's not all important and so we can just  kind of adjust our theory a little bit and then we may be  able to predict things about the properties  of a molecule like H2.  So next time we'll take a bit of time and we'll actually go  through how to calculate with these 1S orbitals,  the matrix elements, how to do these integrals.  They aren't so easy to do.  They aren't so hard but they aren't trivial either  and it's probably good one time to go through them  and see how to do them and then you'll really appreciate  when the computer is doing it for you how much labor saving it is.  If you set up a quantum chemistry calculation,  sometimes the program will tell you how much percent overlap  you've got in terms of setting up the calculation  and now I think you can see basically what it's talking  about is talking about this matrix S and the fact  that if you don't set up an orthonormal basis then it's  having to do a lot more calculation but sometimes it's  very hard to see what kind of orthonormal basis I'm going  to use when I've got things at different positions in space  and I just can't visualize what I'm going to pick then I'm going  to tend to pick atomic orbitals that are centered on the atoms  and in that case I certainly can't expect them  to all be orthogonal to each other.  So next time we'll do those integrals, we'll work  through them systematically and we'll see  if we can predict whether H2 plus is stable or not  and if so how stable is it?
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Today in chemistry 131A we're going to move on from multi-electron atoms to the simplest molecule. We're going to talk first about the Born Oppenheimer approximation. This was a great idea that was introduced very shortly after the introduction of the Schrodinger equation itself. In fact, when you go back to this period in history, what you find out is that everybody was really working like crazy on this new field of quantum mechanics, much like if you come back to 2014 you may find a lot of biologists were working like crazy on stem cells now that they found a better way to produce them. And then after we talk about the Born Oppenheimer approximation, we're going to introduce the simplest molecule H2 plus and from there we'll go to H2. First, the Born Oppenheimer approximation. We're going to just look at the simplest molecules H2 plus and H2 and we're going to look at both of them in some detail to try to understand why it is that you can have a chemical bond. In fact, it's very interesting in the framework of classical mechanics and electrostatics to try to figure out if there could be such a thing as, for example, two positively charged particles and let's say one or two electrons. Let's say we have two so the whole thing is neutral and it's like H2 and we could propose that they're doing some classical thing. We could forget about the uncertainty principle. Let's have them going like this in some pattern and so forth and have all the electrostatic forces and try to see whether the molecule will hold together. And in fact, scientists at the time didn't really understand how molecules could hold together because it doesn't seem to work out very well. And we'll see exactly why when we come to the exchange integral. But if we're going to talk about molecules, even when we talked about helium, we had to use some approximations. We either had to make a good guess with the variational theory or try perturbation theory or something. And now that we've got two nuclei, two centers, even with one electron, we're going to guess that it's going to be extremely hard to solve things exactly. So the first idea is that we want to decouple the motion of the nuclei from the motion of the electrons. And we know that we can probably do that because the electrons are much less massive than the nuclei. And the electrons are therefore moving much more quickly. What we're trying to do then is say, let's say we've got an elephant and we've got a bunch of flies around it. And the elephant's moving slowly. And so the flies have plenty of time to readjust. And if the elephant rolls over slowly, the flies have plenty of time to find the best place where they want to alight on the elephant. And in much the same way, as the nuclear coordinates change, the electrons almost instantaneously, at least from the perspective of the nuclei readjust. And they readjust to the best possible position. And so that being the case, we can go one step further then. And we can just say supposing then that the, we just stop the motion of the nuclei. And we solve for the electrons. Then that we can do. And that's essentially what the Born-Appenheimer approximation is. First I want to talk about our notation here. Our equations are going to get more complicated. And that's going to happen quickly. So I'm going to use big M to denote the mass of the proton or the nucleus and little M to denote the mass of the electron. And our coordinate system is as on this slide 543 where you have two nuclei marked A and B. And in this case, I've put in both electrons, which I call one and two. This is H2 then. And this is at some instant. Some we're trying to set up how we're going to solve for the wave equation, for the Schrodinger equation. And we have all these repulsive and attractive terms between these charged particles. The Hamiltonian then, with all the terms in, has a kinetic energy for the first nucleus, A, a kinetic energy for the second nucleus, B, the inter-nuclear repulsion, which is a positive term pushing them apart. Then the kinetic energy for the first electron, one, the kinetic energy for the second electron, two. Then the attraction of electron one to the first nucleus to the second nucleus. The attraction of electron two to the first nucleus to the second nucleus. And then the repulsion, R12, which is the distance between the two electrons. And we are not going to be able to solve this exactly with all these terms for the total wave function, which includes all the nuclear and electronic coordinates. And in fact, even with the two electrons, we won't be able to solve it exactly. But like the H-atom, we can separate out the center of mass motion for sure. But we can't rigorously separate the relative nuclear motion. We have to make an argument about why it is that we're doing that. And Born and Oppenheimer wrote a classic paper that I've given you the reference to here if you'd like to look it up. Unfortunately, it's in German, and so you would have to look at the translation of it. I had to actually learn to read German when I got my degree in chemistry, which shows you how old I am. But it helps occasionally from time to time if you have to go back and read some of these papers. And their paper was on the quantum mechanics of molecules. And they have a very elegant derivation about why it is that you can ignore the nuclear motion. But I don't want to go through that because that would take at least this lecture and another lecture, and it would just take us a little bit too far a field to actually go through that. But we can certainly see that because the nuclei are moving slowly and because the electrons adjust, then if I just consider the nuclei here, and I solve the Schrodinger equation for the electrons, pretending the nuclei are frozen. And then I move the nuclei a little bit and resolve the Schrodinger equation. That's basically what the electrons are doing continuously as the nuclei move slowly from here to here, is they're just solving the Schrodinger equation again and again and again. And because they're readjusting so quickly, it appears to them as if the nuclei are basically frozen, and that means to us they're basically frozen as well. So if we write that the wave function of the molecule is, which depends on all the coordinates, is the factorization of product of the wave function of the electronic part and the wave function of the nuclear part, then the total energy is just the sum of the electronic energy and the nuclear energy. This is right back to particle in a box with the x, y directions, and this has been our strategy over and over and over, which shows you it's a very good strategy. Then what we find is that the electronic function depends instantaneously, so to speak, on the coordinates of the electrons, R1, R2, and it depends parametrically on this very slow variable, big R, which is just the distance between the two nuclei. And we can always use this approximation for molecules that have very high excited electronic states. So usually to excite an electron from any ground state to some higher state takes a lot of energy, and the slow motion of the nuclei cannot possibly supply that kind of energy. Where we have to be in trouble or where we have to be extremely careful is if we have radicals or open-shell atoms or something like that, where we have a lot of electronic states that are nearby, then what you can imagine happening is that the motion of the nuclei itself can somehow cause a transition of the electron to move out of the orbital that was originally and into a new one. And in that case, the motions are coupled, the whole thing's wobbling around, then we're in trouble and we can't use the Born-Oppenheimer approximation, so in that case it fails. But the upshot of our factorization is we can forget completely about the kinetic energy of the nuclei. We just clamp them down, we freeze them, and we solve for the electronic energy. And that's really an important simplification. Here I've written out the Hamiltonian then in atomic, in atomic units, the electronic Hamiltonian still has the kinetic energy of the electrons, the kinetic energy of the nuclei is gone. And I've added to it 1 over r, which is just the static, electrostatic repulsion of the two nuclei. We have all these instantaneous terms and then we have this static term, but we have to add that to the electronic energy because when the nuclei are very close that will be an overall unfavorable position. We don't care about the kinetic energy of the nuclei because we threw that away. And so what we want to do then is solve the Schrodinger equation for the electronic coordinates as a function of the internuclear separation, big r, and then see what that is. And in fact, if we do that, what we find is that when the nuclei are very close, it's very unfavorable. Partly it's just the internuclear repulsion, but partly it is that the electrons themselves are pretty much outside and so don't screen the positive charges. And as we move them apart, the potential comes down, and then there's a sweet spot where there's a minimum and then it comes back up again. And in fact, that's exactly the kind of shape that we argued was like a Morse potential or some potential like that when we were talking about vibrations and the harmonic oscillator. And now you can see that the nuclei themselves glide around on a potential energy surface that's constantly readjusting and we just draw it as if it's always there, but it's constantly readjusting as they move, it is there before they get to where they're going, so to speak. And we talked then about the nuclear coordinates moving around on a potential energy curve or a potential energy surface, which is the total electronic energy plus the internuclear repulsion. That's what a potential energy surface is how you should think of it. And the nuclei will respond to this. So if they're too close, they'll gradually pull apart. And as they pull apart, the electrons will jigger around and readjust. The simplest possible molecules H2 plus, I want to begin with that, I wrote down the Hamiltonian for H2, but that's a little bit intimidating for us right now, so I want to just do it with one electron. The second electron is now not present, so I don't need one and two, I can just use Dell. And there's only the attraction to the first nucleus, the attraction to the second nucleus. There's no 1 over R12, thank goodness, because that was quite a production when we did helium and hydride. And then there's 1 over R, which is no big deal, which is the internuclear repulsion. Here's our picture then on the bottom of slide 549, same coordinate system as before, except the second electron is missing. Well, much like we could solve the hydrogen atom, once we factored out the center of mass motion and assumed the proton was essentially fixed, by saying, well look, this problem has spherical symmetry. We should actually write the Schrodinger equation in spherical coordinates, and then what happened is, we knew we had the right coordinate system because the wave function factorized into three parts, and that made it easy to solve each part. This three-dimensional differential equation became three one-dimensional differential equations, and that is always the strategy if you can get it to work. Now in this case, spherical coordinates won't work because we have stretched the charge out in one direction. Well, if you take a circle and you stretch it in one dimension so that you have two centers, then you have an ellipse. And so not surprisingly, there's a coordinate system called elliptical coordinates, or confocal elliptical coordinates, where you take the sum of the distance between the electron and the nuclei divided by r, so that's a dimensionless thing, lambda, and then another one, which is the difference between them, rA minus rB divided by big r, call that mu. Then the third coordinate is still like it was in the hydrogen atom, it's that coordinate phi, and the solution for that because that's basically the same. That was the particle on a ring. We can still use that to solve. That decouples out, and so we don't have to worry about that one. The other two, we have to actually worry about what kind of differential equation that we get. You can think of, if you make a triangle between the two nuclei and the electron, you have a plane, three points to find a plane, and the position of that plane around this axis between the two nuclei is the azimuthal angle phi. And so this is just like a stretched hydrogen atom, really. Let's write the wave function again as a product. We're really beating a dead horse with this, but it's the only horse we've got. We're going to write it as a function of lambda, which I've called big lambda of lambda, a function of mu, and unfortunately capital mu looks like an m, but it's mu of mu, and then phi of phi. And those are just presupposing that we can write it like this, and I've given you a reference here at the bottom of page 550, slide 551, if you're interested. There's quite a long paper in the philosophical transactions of the Royal Society where they've gone through exactly how you separate these coordinate systems. It does separate, it does separate into a differential equation with respect to mu only, big m, and then another differential equation that only has big lambda. The problem for us is that these differential equations, even though they're one dimensional, if you look at them, they're very complicated, and they have a lot of moving parts, and we can't write down the solution to these kind of equations in any elementary way. There may be solutions in terms of repeated fractions, one plus one over x plus plus and so forth, and things like that, but that's not really too illuminating for us at this level. And so I don't want to go into writing, attempting to write down these very esoteric functions that might solve this because that's not going to move us that far forward. The azimuthal solutions though are pretty simple. That you're back to the e to the im phi divided by this square root of 2 pi, and m can be 0 plus or minus 1 plus or minus 2. And the reason why this one's so simple is that motion around this axis is that's still a good quantum number because we haven't monkey with the symmetry that way. When we pull the other thing apart, then the other stuff gets thrown off, the motion gets thrown off and it gets much more complicated, and so it gets much more complicated to describe it. So as I mentioned, we aren't going to get into lambda and mu, just to say that they can be figured out. There are numerical tables and you can compute energies and you can do everything with them. But we're going to actually get an approximate solution that's going to be a lot more insightful than either lambda or mu, although it won't be so quantitatively nice. The m equals 0 solutions are called s states by an out, they're called excuse me, sigma states by analogy with s states for an atom. And then states with one twist around this way are called pi states and plus or minus 2 are delta states. And much like we do with the px, py and pz orbitals, sometimes when we actually plot the solutions, we take linear combinations of them to get sine or cosine so that we don't have an imaginary part to the wave function. The other really important consideration for H2 is that the problem has symmetry. If I imagine the origin in the middle and a positive charge here and a positive charge here, then there's symmetry in the problem and that has to be reflected so if I swap the two nuclei or I invert all the coordinates, then nothing too drastic can happen because it's basically the same thing again. In particular, there's this operator I've called p hat that inverts the coordinates. It just takes all points x, y, z and puts them through the origin to minus x, minus y, minus z. And this operator, if you apply it to a wave function and then you apply the Hamiltonian and then you apply the Hamiltonian to the wave function and then this operator, what you find out is you get the same thing. Therefore, it's not like position and momentum where they didn't commute. This operator, the inversion operator of the coordinates, commutes with the Hamiltonian for H2 plus and that means that an eigenfunction of the Hamiltonian also has to be an eigenfunction of the inversion operator. Well, the inversion operator doesn't have any units so the eigenfunction of the inversion operator is just going to be a number. Question is what number is it? And the answer is let's apply the inversion operator twice. So we'll first take x, y, z and go to minus x, minus y, minus z. Great. Now we'll apply it again. We went back to where we started. Everything has to be the same then because all the coordinates are the same. And so if we do those two steps, we first put on the inversion operator onto an eigenfunction. We get a number c times the eigenfunction. That follows because it has to be an eigenfunction of the operator p. Can't turn the wave function into something else. And then if we apply it again, that's p squared, then we get c squared. But we get the same wave function back. And so what we conclude here is that c squared is 1 and that means that c is either plus 1 or minus 1. There's two possibilities there. Well, that means that the energy eigenfunctions, whatever they happen to be, are either even, if it's plus 1, or odd, if it's minus 1, under inversion of all the coordinates. And that's important because we can use that then to classify the solutions to say, well, you're in the even camp or you're in the odd camp. And historically, we use the symbol g for even and u for odd from the German grad and ungrad. Okay. Now, let's take, instead of writing down lambda and mu and writing all these very complicated formulas, and perhaps in a certain sense being none the wiser when we finish, let's try to guess a solution for the wave function for the electronic part of H2 plus. And then once we guess, let's use our hidden ace in the hole, the variational principle, to optimize our guess so that it's really good after we've proposed some kind of guess. Well, okay, let's explore this. There are a lot of kinds of functions I could guess. I could guess something where I have something in the exponent or I have something squared or I have a lot of things. And I could optimize those by the variational principle. But the problem is that might be a lot of work. You saw with the parameter zeta that we had before that we ended up with very complicated equations and then we had to really stare at them a long time to get what we wanted out. In this, make a contour plot even because we couldn't really see how to find the minima very easily. But in the special case that I fix the form of the function, I say my wave function is like a bowling ball plus a cube plus a pyramid or something. And I have percentages of each of these things. Of course, this is quantum mechanics. The percentages could be complex numbers. But I don't change the shape of the underlying function that I'm using. I just change the amount of it. In that case, there is a very elegant way to look at how this optimization works. And we get a series of equations that lead to something which we call the secular determinant, which I want to introduce now. This is not the slater determinant. This is something else. Let's then suppose that we have two fixed functions which I've called phi 1 and phi 2. And we have to pick them cleverly because if we pick them poorly, then even when we optimize them, we may get a terrible answer for the energy. So we need to have some physical insight, I would say, to pick these functions correctly. And some people have a lot more physical insight than others. But in any case, then we have an amount C1 of phi 1 and an amount C2 of phi 2. And all we're doing now is optimizing the amounts, the linear coefficients. We're adding up the functions. We are not multiplying them. We are not raising them to powers of each other or anything else. We're adding them up and we've got these linear coefficients. Okay. So we have to pick them correctly. Obviously, that's part of the physical intuition. We have to pick functions that go to 0 at infinity and other things like that so that they behave properly. Otherwise, if we optimize them, we may find out that the amount is 0 because we put in a function that can't be there. But for now, let's assume that the functions are fixed and known and they're real like the 1s orbital of hydrogen, for example. And let's also assume that the C1 and C2 are also real. So then we don't need the complex conjugate. And then what I've written here then is the expectation value of the energy which is just the integral of C1 phi 1 plus C2 phi 2. They would normally be star, but let's assume they're real. Then the Hamiltonian and then the same thing, C1 phi 1 plus C2 phi 2. And I can then multiply out. I get four terms. I get C1 squared times phi 1 H phi 1. And then I get the cross terms, two of those, C1 C2 plus C2 C1. And then I get C2 squared and then I have phi 2 H phi 2. And when I do these integrals of the Hamiltonian operating on a function and then integrate over the other function multiplying out on the left, that's just a number. The integral is just a number, whatever it is. It could be zero. It could be something. And I'm just going to write those numbers as the C1 squared. I'm going to call that one H sub 1 1. And for the 1 2, I'm going to call it that number, that integral H 1 2. And for the 2 1, I'm going to call it H 2 1. And for the 2 2, I'm going to call that integral H 2 2. And in general, these numbers H sub ij are called matrix elements. And that's because as you're going to see in a second, they are positions in a matrix. And this is a very beautiful systematic way to keep improving our guesses. That's why this linear combination of orbitals is a very, very powerful approach. Well, the Hamiltonians real and the Cs are real and the functions are real, so the matrix elements are symmetric. Of course, the Hamiltonians always are mission. And so if there's no imaginary part, the real part is symmetric. The imaginary part is anti-symmetric because H ij star is equal to H ji. But we don't have to worry about that. Well, we also, because we took these two functions, we mustn't assume that anything's normalized. We should actually make sure we can normalize things. And so the way you normalize things is you take the square of the wave function or you can consider the square of the wave function a slightly different way. Rather than putting the Hamiltonian in between the two sides, I'm just going to put the one operator. The one operator just takes the wave function and returns it. It does nothing to it. But by writing it that way and then expanding it out, I see that I get four terms again. C 1 squared, I get the integral of phi 1 squared. And then I get phi 1, phi 2, phi 2, phi 1. And I write those exactly the same way as I did with H. C 1 squared, S 1 1. And then I have S 1 2, S 2 1, and S 2 2. And S is called the overlap matrix because it's going to tell us how the different functions we picked overlap. Well, if the basis functions were orthonormal, then the overlap matrix would just be the identity matrix. If I had n functions and they were like different orbitals on a hydrogen atom, for example, on the same nucleus, and that were my basis, then because those are already orthonormal, I have nothing to do. But if I have atomic orbitals on different nuclei so that they're shifted and I have these exponentials, then they certainly are not orthonormal. They're normalized with respect to themselves. If I take this one and multiply by this one and square it, go over all space, it's one. And likewise with this one, but when I take the two, multiplying in between, it's positive everywhere, that can't be zero and therefore they are not orthogonal. So for the two basis functions, if they were orthogonal, this matrix S 1 1, S 1 2, S 2 1, which is the same as S 1 2 and S 2 2 would just amount to the unit matrix 1, then 0, then 0, then 1, but we usually can't, if we make a natural choice of functions, then we usually can't ensure that this happens. And we don't want to. What we want to do is just go ahead and compute the matrix S and then use that to figure out what's going on. So here's our formula then for the energy, keeping in mind that the wave function psi is not necessarily normalized. In fact, the exact overlap is going to depend on how far the nuclei are apart. We can write this energy functional, the energy as a function of the coefficient C 1 and the coefficient C 2, is the expectation value of the Hamiltonian on the top, divided by the square of the wave function on the bottom and the top, I got those four terms, but two of them are the same, so I end up with 2 C 1, C 2 times H 1 2, and on the bottom, I end up with those four terms in the S matrix, but two of them are the same because S 2 1 is equal to S 1 2, so I end up with three terms there. And rather than write that as a ratio the way I have, I'm going to multiply both sides by the denominator and write it as E times this spinach in S is equal to the Hamiltonian matrix elements. Those matrix elements are numbers that are fixed because the functions are fixed. That's the beauty of it. I specify the functions once, I do the integral once, and then they're done, and I don't play around with them anymore. If I had to keep doing the integrals over and over because I'm playing around with the functions themselves, it would be very, very time consuming, whether I do it on a computer or by hand. Well, the variational principle says if you've got the energy of a quantum system as a function of a parameter like C 1 and C 2, then if you minimize the energy with respect to the parameter, you're going to improve your guess of the wave function, and so that's exactly what we're going to do because now we have a formula here for the energy. What I've done here then is I've just taken the derivative of both sides of this equation with respect to C 1. So we have a product, so it's the derivative of the first, that's the first term, times the second, which is all the S matrix elements, plus the first, which is just the energy, times the derivative of the second, the term with C 2 goes away because this is a partial derivative with respect to C 1. And on the Hamiltonian side, I end up with 2C 1H 1 1 plus 2C 2H 1 2. And interestingly enough, there's a lot of 2's in this thing. At the minimum or maximum, but it's going to be a minimum, in fact, the derivative should be 0, therefore we can say that the E times the S part is, in fact, equal to the other part because the term with D E, D C 1, is 0. And that's nice because now we have E left over. And if we write that out, we get what I have on the bottom of slide 560, which is that C 1 times H 1 1 minus E S 1 1 plus C 2 times H 1 2 minus E S 1 2 is equal to 0. If we do the derivative with respect to C 2, you get a similar equation. And these two equations then, in two unknowns, can be written in matrix notation as a matrix that now has elements H 1 1 minus E S 1 1 and so forth in each of the entries. It's always H minus E S times the column vector C 1 C 2 is equal to 0. This is a system of linear equations. And one solution to this system of linear equations, the simplest one actually, is just say, okay, no matter what the matrix is, let's let C 1 and C 2 equals to 0, both of them, then anything times 0 plus 0 times anything is 0. That works and the other one works. And so the whole thing comes to 0. Well, that one's no good because if C 1 and C 2 are both 0, then unfortunately our wave function, which is C 1 times the first function plus C 2 times the second, is also 0 and that's not going to help us find a good wave function. So that trivial solution is not the one we want. In some sense then what we want is instead of the column vector being 0, we want the matrix somehow to be 0. But we can't control the matrix to be 0 per se because it has these fixed numbers in it. What we can do is we can make the determinant of the matrix equal to 0. And in order to see why this works, you really need a course in linear algebra, which is why it's very important to take a course in linear algebra. So you can get to the bottom of stuff like this once and for all. And I've written then the straight lines like I did for the Slater determinant, the determinant of this array of numbers is equal to 0. That's the necessary and sufficient condition that we get a solution that we want. Well, this determinant is in fact called the secular determinant. The secular determinant then is just the matrix element of the Hamiltonian, take the function, put H on it, take the other function, start if you need to, if it's complex, integrate it, get a number. That number is fixed, you shove them into a matrix, do the same thing with the overlap where there's no Hamiltonian, shove those numbers in, put E in there, and then you will get an equation in E and following the rules for taking a determinant. Remember it's a crisscross rule for a 2 by 2. So I have H minus ES, H minus ES. Well, H11 and H22 are the same in this problem and H12 is the same. And so we're going to get a quadratic equation in the energy and we're going to get two roots and take the smaller one to be our best estimate of the energy. As usual, when you solve mathematical equations, mathematicians like to give you all possible answers, not necessarily the one you're looking for. And so because they don't know which one you're looking for among other things. And so we have to look through and then after we get all the energies out of solving this secular determinant, we have to then figure out which one's the best. When we've got three basis functions, we end up with a cubic because we end up with a 3 by 3 determinant, which means E, every term has E times E times E. So we've got a cubic equation to solve, which can be very messy. And of course it's got other powers too because there's more than one term. They don't all have E or we could just take the cube root. And with a quartic, it's very messy indeed. And so in actual fact, we don't ever try to just write down a formula for the energy except if it's a quadratic equation. And usually if we're lucky, we don't even have to write down the formula for a quadratic equation. Either we can just look at it. But you can always calculate them numerically and there are standard packages to root polynomials. Once you set up the secular determinant and whack it out, you end up with a big polynomial in the energy and you want to solve for the values of E that are real that make that thing equal to zero. And there are standard methods to root polynomials, packages that you can get and I think you can see why having something like that is like having a hammer or a screwdriver. It's incredibly useful if you want to pound in a nail or screw in a screw and if you don't have that tool, you're kind of stuck. And there are lots of course mathematical tools like that available to us these days. Okay, let's go back to H2 with this secular determinant and let's do this one then as practice problem 28. Let's guess, since we've got no better ideas, two 1S orbitals. Let's take one of them centered on proton A and the other one centered on proton B. Then the question is, we're going to take C1 of this one and C2 of that one. What's the form for the two energies from solving the secular determinant? That's the question. Well, okay. We know that the 1S orbitals are normalized, so we don't have to worry about that. If we integrate them over all space, the answer is one. And unfortunately, they're not orthogonal because once they're off nuclei, they aren't neither one nor zero. They're something and we have to figure out what the something is. And I've called that something just S because it's the only thing that's not going to be one in the overlap matrix. It's definitely not equal to zero, that much we can see because they're both positive exponentials. It's got to integrate to something between zero and one. And the problem has symmetry, so luckily we don't have to do all the matrix elements over and over. We can just do first the matrix element I'll call HAa. That's got 1SA, the Hamiltonian 1SA. We'll work out what it is later, but to solve this, we don't need to actually work it out. We just need to assume it has some value. And that's exactly the same as HBB because we got the same orbital. We got the same nuclear charge. Everything's the same. And then there's this one HAb, which is mixed term. And that again has symmetry, whether I call this one A or this one B or vice versa, they have to be the same. So that one's equal to HBA. So I only have to do two of the four possible matrix elements and then I can just fill in the other ones. Keep in mind here that although we're writing two functions, 1SA, 1SB, these integrals are one three-dimensional integral because I've got one orbital that's the sum of these two guys. I'm writing them out as four terms because that's the way it pans out. But this is different than when we took helium where we had a product of this times this. That's different because then you've got all the variables and we had six-dimensional integrals. We had to integrate over R1 and then integrate over R2 as well. These integrals just have R, one R. And I have to figure out how to express the orbitals 1SA and 1SB in terms of this variable R and maybe some other angular variables like theta and it won't surprise you that the law of cosines is going to come in again. But keep in mind that this is one three-dimensional integral and nothing like what we were doing with helium even though it might look like it. It's easy to get mixed up. Well, here's our secular determinant then putting in all the values that are the same. We have HAA minus E and then the next one off diagonal HAB minus E times S where S is just the one that isn't zero. And then same thing in the 2, 1 entry. And then same thing on the bottom. And so this times this minus this times this just gives us HAA minus E quantity squared minus HAB minus ES quantity squared equals to zero. Well, I could expand this out and solve it as a quadratic equation. And that would be quite messy to do because rather than just having AX squared plus BX plus C, A, B and C would be a very messy thing and then I'd have to try to simplify it and that would be a very slow way to go. But luckily I can factorize this by looking at it because whenever I see this squared minus that squared, I remember that that's something plus something times something minus something and therefore I can think, well, if I say A squared minus B squared is A plus B times A minus B then I can write that out with A and B. My terms there, HAA minus E plus HAB minus ES and then the other way around equals to zero. I've got two things equal to zero. The only way that can be true is if either this one's zero or that one's zero. And so that means we get two energy roots. The first one I've called E plus, you get HAA minus E plus HAB minus ES equals to zero. You solve that for E and you call it E plus and you get HAA plus HAB divided by one plus S. The other one is just the other part and I get HAA minus HAB divided by one minus S and I've called that E minus. To quantify things though, what we're going to have to do, so that's the end of the problem. But to quantify things, we need to know what these numbers are and in particular, we need to know what these numbers are as a function of big R, the separation between the two nuclei. If we can work that out and then we can track E plus and E minus pretending we're an electron riding along on this aircraft carrier of this molecule, we can pretend and we can see where we're going to have the sweet spot of energy or if we have one. What would be good is to, we tried to predict the ionization potential of helium, we tried to predict whether hydride would be stable or not, we saw sometimes it could be tough. We want to see if we take something simple like this with just a quadratic equation here in two roots and we pick the most stable one, does it predict that H2 plus, which does in fact exist, does exist and if it does, what value of big R is the most stable value and how does that compare with the actual value, which has been measured of course spectroscopically. And then next time what we'll see then is we can probably take this and take this orbital that we solved with this and we can probably pop another electron in there and we can argue well the electron-electron repulsion is important but it's not all important and so we can just kind of adjust our theory a little bit and then we may be able to predict things about the properties of a molecule like H2. So next time we'll take a bit of time and we'll actually go through how to calculate with these 1S orbitals, the matrix elements, how to do these integrals. They aren't so easy to do. They aren't so hard but they aren't trivial either and it's probably good one time to go through them and see how to do them and then you'll really appreciate when the computer is doing it for you how much labor saving it is. If you set up a quantum chemistry calculation, sometimes the program will tell you how much percent overlap you've got in terms of setting up the calculation and now I think you can see basically what it's talking about is talking about this matrix S and the fact that if you don't set up an orthonormal basis then it's having to do a lot more calculation but sometimes it's very hard to see what kind of orthonormal basis I'm going to use when I've got things at different positions in space and I just can't visualize what I'm going to pick then I'm going to tend to pick atomic orbitals that are centered on the atoms and in that case I certainly can't expect them to all be orthogonal to each other. So next time we'll do those integrals, we'll work through them systematically and we'll see if we can predict whether H2 plus is stable or not and if so how stable is it?
from multi-electron atoms to the simplest introduction of the Schrodinger equation itself. framework of classical mechanics and time didn't really understand how molecules could charged particles. The Hamiltonian then, with all electronic energy and the nuclear energy. This is solve the hydrogen atom, once we factored out the Hamiltonian also has to be an eigenfunction of course, this is quantum mechanics. The it is. It could be zero. It could be something.