Lecture 18. The Hydride Ion (Continued): Two-Electron Systems
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Transcript: English(auto-generated)
00:05
Hi, welcome back. Today, what we're going to discuss is the hydride ion. We're going to continue our calculation on that. Recall, we had just gotten to the point where we were looking at the electron repulsion term
00:22
as a perturbation. And we're going to look at some other two electron systems and we're going to be forced to work pretty hard today in order to try to reproduce any kind of properties that are known about these systems. And this is a very good test of the theory, of course,
00:41
because you should be able to reproduce things like the ionization energy of helium and so on if your theory is correct. And you can take the approximations far enough. Recall, here's what we had. We had an expression from first order, time independent perturbation theory.
01:01
And we were going to use it to compute the correction to the sum of the two hydrogen atom energies from each electron interacting independently with the positive nucleus. And so what we had is we had this sandwiched integral with psi star on the left and then 1 over R12
01:24
and then psi on the right. And the thing is the wave functions themselves are functions of the coordinates R1 and R2. And therefore, what I have to do is I have to be able to express 1 over R12, the distance between them, in terms of only R1, R2, and maybe some other coordinate
01:46
that I'm integrating over because I can't integrate a function if I don't know the functional dependence of the integrand. I can't just have some variable in an integral that's Y and I don't know how Y depends on X or if it does,
02:02
then I can't do the integral with respect to X. So with that in mind, let's take a look at this figure. What I've drawn here is an obtuse triangle with R1 oriented along Z. Remember, we were going to do that.
02:21
We were going to reorient the electrons every time we do the integral over the other variable so that the first one is along Z. And because it's spherically symmetric, that doesn't change the answer at all in terms of the energy. And what I've done is I've added a little length to R1
02:41
so that by the time it gets to R2, it's a right triangle. And I've, the little length I've added, I've called A. And the distance on the other side of the right triangle I've called B. And the angle between R1 and R2, which is bigger than 90 degrees
03:02
in this figure I've called theta, and that means that the other angle to the other side of the line is pi minus theta or 180 degrees minus theta. In order to get an expression then for the distance R12, which is between the two electrons,
03:23
there are two right triangles. There's one that involves the small figure, which is A squared plus B squared is R2 squared. And there's another one, which is a big triangle, which is R1 plus A quantity squared, always add them first
03:45
and then square them, plus B squared is equal to R12 squared. That's the big triangle. Expanding the second equation then and just writing it out,
04:00
we get R12 squared is equal to A squared plus 2AR1 plus R1 squared plus B squared. A squared plus B squared I can gather together. And by the other triangle, that's R2 squared. So R12 squared is equal to R2 squared, good, that's a variable R2, plus R1 squared, good,
04:25
that's a variable R1, plus 2 times A times R1, no good. What's A? I have to know what A is to be able to integrate over the thing. But luckily, A by that triangle is equal
04:44
to R1 times cosine of the angle that's nearest to A. And that angle is pi minus theta. And therefore, I could go back to Euler's identity and put an E to the I theta and figure it out. But I know, in fact, it just changes sign.
05:02
So that's minus R2 cosine theta. And that's A. So I substitute that value of A in. And I get what's called the law of the cosines, which you could go back to the Pythagoreans or Euclid and find that they were smart enough to figure this out as well. R12 squared is equal to R2 squared, plus R1 squared,
05:24
plus 2R1, R2 cosine theta. Theta is okay because remember, in spherical coordinates, theta being the angle between the two vectors, that's perfect. Because if one of them is along Z, then theta was the variable that I was integrating over for the other one.
05:41
So now we're set to go. And we can do the integral. In the case that theta is less than 90, so it's not an obtuse triangle, but it's an acute triangle, I'll let you draw the triangles. You draw them slightly differently.
06:00
But you come to exactly the same conclusion, namely that this formula is always valid. And of course, if theta is equal to 90 degrees, so that it's just a right triangle, then the cosine theta is zero. And that goes away very conveniently. And then we just have the Pythagorean theorem.
06:23
So the law of cosines is just a generalization of the Pythagorean theorem for the case where it's not exactly a right triangle. So now what we've done here in the bottom of slide 440 is,
06:40
we've put in 1 over R12. And because the first wave function that depends only on R1 has no dependence on R2, they're independent variables. I've factored it out, factored it out. And I have a shorthand notation here that I'm using.
07:02
And that's just to try to fit the equations onto the slide, basically. But I do a single integral, D vector R2. What that really means is I'm going to integrate over phi. I'm going to integrate over theta. And I'm going to integrate over the scalar R
07:21
from zero to infinity. I'm going to do all those things when it comes right down to it. But just to keep track as a placeholder, I've got that integral. It's going to be a triple integral. But I just write it as 1 to make the equation a little bit easier to see. But don't let that notation throw you off. We'll get to that.
07:41
Now, this still is not so easy because how do I do this integral here? I've got the square root of all this spinach in the denominator. And it doesn't necessarily suggest the answer right away. Well, let's put in the atomic wave function.
08:01
That's 1 over root pi, 1 over A naught to the 3 halves times E to the minus R over A naught. And it's the same whether it's R1 or R2. It's just the coordinate of the electron. But the wave function is the same, just has a different variable. And in atomic units, it's 1 over root pi times E
08:21
to the minus R. And therefore, I'll leave the first. Again, you see why because I can barely fit the equation on the slide. I am going to leave the first integration with respect to R1 as just a symbolic thing. And then the second integration with respect to R2,
08:41
I have the integral over theta of sine theta because remember that was part of the volume element that I needed. And then in the bottom, I have this 2R1, R2 cosine theta. Then I have the integral over phi. That doesn't bother me, integral over D phi 2.
09:01
I don't see any dependence on phi. Then I have the integral over R. And I have to remember to put in the R squared there. OK. So, if I make a substitution which you tend to do when you have trigonometric functions, if you have an algebraic function you can't do,
09:23
you tend to make a trigonometric substitution. And if you have a trigonometric function you can't do, you tend to make an algebraic substitution. Here, what I'm going to do is I'm going to let X, the variable X be cosine of theta 2. Then DX is equal to minus sine theta 2 D theta 2.
09:44
And therefore, the integral over D theta 2 of sine theta 2 over the radical is equal to minus the integral from 1 which is when cosine theta, when theta is equal to 0, cosine is 1 to minus 1 of DX over the square root
10:03
of R1 squared plus R2 squared plus minus 2R1 R2X. And then I can change the limits, make it from minus 1 to 1, get rid of the negative sign. And then that one, I can look up the antiderivative
10:20
because that one is a standard, very easy to do. And I'll let you verify it by taking this actual antiderivative I've given you here on the bottom of slide 442. And please differentiate it with respect to X and verify that you get the integrand that we started with. We get this, minus the square root
10:42
of R1 squared plus R2 squared minus 2R1 R2X divided by R1 R2. I think it's pretty easy to see where that came from once you start doing the derivative. If we put in the limits, then and do the subtraction,
11:02
we get the following, R1 plus R2 minus the absolute value of R1 minus R2 divided by R1 R2. And recall that the square root of X squared is the absolute value of X
11:21
because the square root is positive. Therefore, our integral over theta is equal to this. So it's already kind of interesting because you may not have encountered this before. The integral over theta of that sine theta over the square root is equal to 2 over R1
11:41
if R1 is bigger than R2. If R1 is bigger than R2, it's 2 over R1. But it's equal to 2 over R2 if R1 is less than R2. So it's equal to 2 over the bigger of the 2. And that means when I integrate over the other variable R1,
12:01
I have to be very careful here that I pick the right answer when I'm integrating the right variable. If I fix R2 and I integrate R1 up to R2, I should use one formula. And then when I integrate R1 the rest of the way to infinity, I should use the other formula.
12:21
And if I'm not careful about how I do that, then I get the wrong answer. Now, you can imagine if you have a lot of electrons and you have a lot of integrals that keep doing this kind of thing that they depend on who's where, it can get mighty tricky to figure out how
12:41
to keep track of the right way to do things. But so this might be your first introduction to this kind of straightforward problem that seems to give a conditional antiderivative that depends on what's going on. The integral over phi 2 gives 2 pi.
13:00
Big deal. Nothing to do there. The integral over theta 1, there's no theta 1 left. Theta is gone. It was theta 2. There's no theta 1 left. And phi 1, that just gives 4 pi like it always does because the 4 pi is the angular extent of the spherical coordinates.
13:23
And so there's nothing else there to do except the radial part. And we have to break the radial part into two integrals. We have to first break it into 1 over r1 times the integral of dr2 r2 squared e
13:40
to the minus 2r2 from 0 to r1. And then the rest of the way from r1 to infinity it's just dr2 r2 because it was 1 over r2. So one of them went away, e to the minus 2r2. And both of these, so we have one that's r2 e
14:04
to the minus 2r2. We've got the other one r2 squared e to the minus 2r2. Boy, do you get good at doing these. In fact, you get so good you just know them by heart if you start doing this kind of thing often because you don't want to waste time to flip through pages.
14:22
It's like knowing somebody's phone number. If you call them up a lot, you just know it. And you will know this stuff if you work through these things and you will suddenly seem very smart to people who don't do these kind of calculations or very nerdy perhaps. Anyway, you can do these integrals by parts.
14:41
And here's what we get. The integral of r squared e to the minus 2r2 is equal to minus e to the minus 2r2 over 4 times the polynomial in r2. 1 plus 2r2 plus 2r2 squared. And the other one has one less term and is
15:01
in the middle of slide 445. Again, if you worry about whether these are correct or you just want to reassure yourself, take the derivative. And whenever you're by yourself and you don't have software and you're proposing an antiderivative, the antiderivative is always a bit harder.
15:22
It's like dividing in a way. You have to kind of see if it's going to work. The derivative is more like multiplication. You can just do it. And so you can always go backwards and figure out. Now let's put in the limits. And we can finish off the integral over r2.
15:41
And we put in these limits, 0 to r1 on the first integral. We get the expression at the top of slide 446. And r1 to infinity on the second integral. Well, at infinity, the exponential vanishes. So we just get 0. And then we have a term that I've shown.
16:01
And if you tidy everything up, you get the following. You get that this radial integral is equal to 1 fourth times 1 over r1 minus e to the minus 2 over r1 times 1 over r1 plus 1. A little bit messy but not too bad.
16:25
The integral over r1 can also be done now by parts. So you've got the integral over r2. Which was conditional. Now you've done that. The integral over theta 1 and phi 1 I said was 4 pi.
16:41
The integral over r1. Now you've got this function of r1. You've got to do it by parts again. And that integral turns out to just be 5 over 128. And that's the way these things sometimes work. And that usually means you've done it right
17:01
if it works like that. But what? Well, we get a factor of 16 because we got the two 4 pis. But the pi squared goes away because of the 1 over the square root of pi in the each 1s wave function. I got two of them but I got them on both sides.
17:22
So I got pi squared. That goes away conveniently. And so we get that the energy correction by first order perturbation theory. E1 is equal to 5 eighths because of the factor of 16.
17:43
The first reaction when you do this calculation is to curse and say 5 eighths of what? 5 eighths. Well, it's 5 eighths of a hard tree because we were working
18:02
in atomic units. And we know that that's the unit of energy in atomic units. So one way is to close your eyes and just say, look, I didn't do anything wrong. I set up these units. I know this is energy. It's got to be a hard tree.
18:22
If that doesn't reassure you, you can go back and put in all the constants now that you've done everything. Let them ride along, very messy, messy stuff. And you can see that it is 5 eighths of a hard tree. All the constants just ride along. Now, the two electrons, each one interacting
18:41
with the nucleus is minus half a hard tree because that's what hard tree is twice the ionization of hydrogen atom approximately. So that's minus a half, minus a half plus 5 eighths. And therefore, I've added the total energy
19:00
of the hydride anion now is the energy of the first electron, minus a half, plus the energy of the second electron, minus a half, plus 5 eighths, minus 3 eighths. Perfect, you might think. It's negative. So that means that H minus is stable compared to a proton
19:26
and an electron and another electron at rest at infinity. Well, unfortunately, that is a pretty low bar to have to meet
19:41
because that's not the question. The question really is, is hydride stable compared to a hydrogen atom which we know is stable and an electron at infinity? And that's a real sour ending because the energy
20:01
of the hydride is minus 3 eighths of a hard tree and the energy of a hydrogen atom plus an electron is minus a half. And therefore, what we're predicting is if we have a hydrogen atom and we have an electron and we bring it up that the energy becomes more unstable.
20:24
In other words, it just kicks it back out. It ionizes it back out, kicks it out and the repulsion force wins. And that would mean that hydride wouldn't exist. And if hydride didn't exist, we wouldn't have a name for it.
20:40
Well, maybe I shouldn't say that. We have names for plenty of things that don't exist except in our heads. But hydride is a real thing that we can see. It has a very big radius. The radius of the hydride ion is bigger than the fluoride anion. So it's very puffed up but it does exist.
21:04
And unfortunately, what this means is it could mean two things. It could mean that quantum mechanics is a crock and it doesn't work and this is the proof. Or it could mean that perturbation theory to first order is not good enough
21:20
to give us the correct answer. And in fact, in this case, it's the second thing. That means that we've got to somehow work harder in order to figure out a better wave function or a better way to calculate the energy that we know is more accurate than what we've done here.
21:41
After all that work, then we can't reproduce the simple fact that H minus exists. It has a well-known radius. It has a positive ionization energy. Well, we should have anticipated that in retrospect and here's why. Our perturbation has exactly the same size
22:05
as the two attractive parts. Recall that the idea behind perturbation theory was that you had a large problem that was simple that you had solved. And then you were adding kind of a small complicated part
22:21
that what you couldn't solve exactly but because it was small, you could expand it and you could close in on the answer as a power series and you could decide when to stop. But if we look at this, then suppose we propose there's a parameter X that indicates the size of the various energies.
22:44
And the ratio of X is sort of the ratio of the repulsion energy to the attraction energy for any particular electron. Then if we're trying to get a solution in X and X is near to be 1, then X doesn't get small very quickly.
23:02
And so just assuming that E to the X, for example, is 1 plus X if X is not small is a very bad approximation especially if X is near 1, you're comparing, you know, 2 and it's way off.
23:21
And if X in the perturbation series happens to be bigger than unity, then what that means is as you calculate successive corrections, they might get more and more violent. So you might say, well, first the energy should go down like this and then it should go up like that and it should go down like this.
23:42
And you might have a sum of terms that just doesn't add up to anything, just diverges and furthermore becomes very inaccurate. After you do all these tons of work, you still get garbage out at the end. So we shouldn't be so surprised that this is not so easy to do.
24:05
And whenever you look at a problem like this, it's a good idea to try to estimate how big these energies are likely to be and get an idea whether perturbation theory may or may not work. Here, in retrospect, we didn't expect it to work well
24:22
but of course if you go through the calculation, you do all those integrals, you do all that work and it doesn't work, then you really scratch your head and you remember that a long time. How can we improve them? What's wrong? Well, using the hydrogen 1s orbitals is not a good idea
24:43
because the hydrogen 1s orbitals have their maximum probability of finding the electron in a shell at the Bohr radius. But we know experimentally the hydride ion has a much bigger radius than the hydrogen atom.
25:02
And therefore, we would be much better because of the electron-electron repulsion among other things that puffs it up. We would expect that it might be too tight a squeeze to just sit there with the hydrogen wave functions
25:20
and just calculate a correction. Now, we could go further in perturbation theory and calculate the correction to the hydrogen wave function based on perturbation theory and what we would find is that it would get bigger when we corrected it. But that would be a lot of integrals to do, not a few,
25:41
a lot, and it would take a long, long time. Even if we had software helping us out along the way, and we still have to keep track of things very accurately and add them all up. So I want to try a slightly different approach. We would use perturbation theory
26:01
to correct the wave function. But why don't we correct the wave function with physical intuition? Why don't we introduce an artificial parameter into the orbitals that controls their radius? Before, we had the wave function is e to the minus r1, for example.
26:22
Now, let's put something else in there. So let's put an e to the minus zeta r1. And now zeta is something we can control. It's a dial. We can dial it in and out and we can compute the energy of the hydride anion as a function of this parameter,
26:43
zeta, and then we can find the optimum. It won't necessarily guarantee success, but you learn a lot by failing in this kind of endeavor. It's very important never to look at the answer before you've really tried like crazy.
27:01
Because if you do, you miss everything about it. It's like somebody giving the answer to crosswords or anything else, telling you in a Sudoku to put a 3 there. It completely ruins it in a way. And you never really learn it. All right. So let's keep in mind that our evaluation of the energy,
27:23
although we did it by perturbation theory there, it's in fact exact because we had the hydrogen atom energy, that's exact, minus 1 half Hartree. And we did the integral over R12 exactly.
27:40
We didn't make any approximations there. So that's the exact energy for that wave function. In order to get a better estimate, we have to correct the wave function. And we're just going to correct the wave function now. And hopefully, if we just tweak the wave function slightly but don't change it in any essential way,
28:02
the math won't get too difficult. We can still use all our other results and do our integrals and so forth and then see what we get. So we'll recycle most of our work. I'll refer back to those integrals if you want to look back at how to do them. And just slightly adjust the most probable radius.
28:23
And what we expect physically is that it should get more stable if we make it a little bit bigger. Now, if we make it too big, it'll be very bad because if it's very far apart, then the energy is zero because it's like the proton and two electrons
28:40
at rest at infinity. So as I said, rather than doing perturbation theory to the next order, which would be interesting but it'll take a long time, we're just going to guess a way to adjust the wave function. And then we're going to try this variational approach. We have energy that we've calculated that depends
29:01
on a parameter and then we know if the energy goes lower that it's a closer approximation to the truth. So we minimize the energy with respect to the parameter. So we introduce a parameter that I'm calling zeta which looks like a Greek squiggle. A lot of people don't even know what the letter is
29:22
but now you're part of the club. There's two of them that are squiggles. There's zeta and there's xi. And don't get them mixed up. Zeta has less squiggles and so let's put in our new normalized guess. Here it is at the bottom of slide 454. The wave function which is a function of r1 and r2,
29:45
xi of r1, r2 is equal to zeta cubed divided by pi times e to the minus zeta r1 times e to the minus zeta r2. And when zeta is 1, we get exactly the same thing
30:00
that we had before. And the question is, is zeta equal to 1, the best guess? Well, we can't do worse by introducing this. It may not improve but we can't do worse because we can always pick zeta is equal to 1 and we got what we got before. Now, we have to calculate the expectation value of the energy
30:23
with this wave function. And that means we have to do the two hydrogen-like terms, each electron interacting with the proton. And then we have to do the darn repulsion integral again. Oh, here we have our three Hamiltonians, h1, h2,
30:44
in atomic units and h12, I'll call, which is the interaction term that mucks it up, prevents the energy from just being the sum. And we have three integrals to do but thank goodness two of them are identical just with electron 1 and electron 2.
31:02
There's no point in doing that one over. And here's what we get, the energy which is equal to the double integral, again, I've used the shorthand here of integral d vector r1, d vector r2, of psi h psi,
31:21
excuse me, psi star h psi. But in this case, the wave function is real so it does not matter whether we have the complex conjugate or not, and that breaks into three terms, the integral with h1 sandwiched in between, the integral with h2 sandwiched in between, whatever they are, they're the same,
31:41
no difference between 1 and 2, and then the repulsive integral with h12 which is just 1 over r12. Sandwiched in between, which we saw how to do. And therefore, it's not going to be too bad because the only real difference is this thing zeta
32:02
in the exponent, and that doesn't change in any fundamental way whether we can do the integral or not. It's not like we suddenly switched it to r squared in the exponent or something that might make it quite difficult or change or r cubed and make it difficult to do the integral.
32:23
The first two terms are the same, and the angular parts integrate to 4 pi in each case because there's no angular dependence in the wave function, and therefore, all we have to do to do either of the first two terms is do the angular,
32:41
excuse me, the radial part. And I've written it out in full here for the variable r1 at the bottom of slide 457. We have r1 squared. That's what comes in from the volume element. We've got zeta cubed over pi. Then we have the first wave function.
33:00
Then we've got the Hamiltonian. Then we've got the second wave function, and the 4 pi comes out on the angular variables. So let's then take a closer look at how to do this. Well, we've got the kinetic energy part.
33:22
Now, you might say, well, why don't you just kind of divine? This is so similar to the 1s orbital of hydrogen. Why don't you just divine where to put zeta into the answer? And the answer is I don't trust myself to be able to get that right. So I'm going to go back and put
33:41
in what the kinetic energy is, and I've written it in atomic units here. It's minus 1 half del squared, and I know how to write that out in spherical polar coordinates. And the wave function has no dependence on phi or theta. And therefore, the part that I end up with is the second derivative with respect
34:01
to r1 squared minus 1 over r1 times the first derivative with respect to r1. And I have to keep in mind that I have the factor of minus 1 half, which I've included there. Therefore, here's what we've got.
34:21
Now, this is pretty messy because I'm integrating with respect to r1. There's the wave function, and then there's this thing that's taking a lot of derivatives with respect to r1. And then there's the potential energy, which is minus 1 over r1 that I've put in there.
34:40
So we've got all those three things in there. And then they're operating on the wave function. So I've got to go step by step. I've got to take the derivatives, write down everything, put them there, write them there, and then I'm going to have to integrate by parts. Anything that doesn't depend on r1 can be pulled out in front.
35:02
So I've pulled out the two wave, the two exponentials that depend on r2, and now I've got this mess to do with r1, but the derivatives are pretty easy. All they do is bring down a zeta each time and change the sign depending on how many derivatives I took. So I end up with this little thing here
35:21
to integrate r1 squared e to the minus zeta r1 times minus zeta squared over 2 plus 1 over r1 times zeta minus 1. And then there's another e to the minus zeta r1. And the integrals are standard, by which I mean while you just look them up,
35:40
r to the n, e to the minus r. And if you go ahead and do the integral and do it out, you get a result that looks pretty nice. It's 1 over 8 zeta minus 1 over 4 zeta squared. And that's then our result.
36:01
If you integrate over r2, then the leading constants out front, including the leading constants out front, excuse me, and you do the whole thing then. You get the following result for the single electron energy. The expectation value of just the Hamiltonian H1 is zeta
36:25
over 2 times zeta minus 2. And when we put zeta equals 1, we get minus a half. And so I wouldn't, for, I wouldn't have been able to figure out that it had this functional form
36:42
without a very quiet room, without actually doing these integrals. And that's why I don't just put in something like, well, it's minus zeta over 2, because that's not correct. It might be that. That also gives a half when zeta is 1, but it might not be.
37:01
And in fact, it isn't. It's, and there's good reasons why it has to have a linear quadratic term. The electron repulsion integral is the same thing. And I'm going to let you do that, because that one we did in great detail with the sine theta
37:21
and the conditional 2 over R1. And if you want to do that, and you go through and you do it with a big piece of paper and a quiet room, you will get 5 zeta divided by 8. That's the answer for the electron-electron repulsion.
37:42
And that'll be quite a lot of paper to hand in when you do it as one of our problems. Now, when, whenever you get something, you should check whether it makes sense. Sometimes things don't seem to make sense, like the electron going through both slits.
38:00
But in that case, you keep doing the experiment over and over. In this case, you should expect the calculation to tell you that the thing got bigger. And when we look at this formula, 5 zeta over 8, if zeta's reduced, that means since, if zeta's reduced,
38:22
the thing gets larger, then the repulsion is lower. And that makes sense. That's exactly what we expect to happen. The total expectation value of the energy then as a function of zeta is the sum of these three terms.
38:42
And finally, we get this formula. When everything comes out in the wash, we add everything up and we do the algebra very carefully and don't make any errors. We get zeta squared minus 11 zeta over 8.
39:01
That is the zeta-dependent energy. And when zeta is equal to 1, we get our prior value of minus 3 eighths of a Hartree. Now, however, we have this energy as a function of zeta. We have the variational principle that says
39:22
when the energy goes lower, you did better. And this kind of function with a quadratic and a linear term with different signs clearly has a minimum. And so we can optimize this by the variational principle
39:40
and get a much better estimate of the energy of the hydride anion. Let's do this then as a practice problem. This is practice problem 24. Let's optimize the value of zeta to minimize the energy. Is the hydride ion predicted to be stable?
40:05
Here's the answer. The simple problem in calculus, what do we do? If there's a minimum, that means that's the lowest it was. The slope is 0 there. And so while saying the function has a minimum is kind of a vague statement in a way saying
40:22
that the derivative is equal to 0 is something you can actually work with. It's an actual equation that gives you a solution. And so that's, of course, what you translate it to mean. And therefore, we take the derivative of that function and we get, so we have zeta squared minus 11 eighths zeta.
40:45
And if we take the derivative, we get 2 zeta minus 11 eighths. And if we set that equal to 0, then zeta should be 11 sixteenths. Now, again, we say, does that make sense? And the answer is yes because 16 sixteenths was 1.
41:04
And what we did is we puffed it out. We let it go out quite a bit more. And that lowered the energy because of the repulsion term going away faster than the attraction terms did. And now the question is, is the hydride ion stable?
41:25
By which I mean, is the energy of the darn hydride ion less than half a heart tree, which is the energy of a hydrogen atom and an electron just hanging out in the wind?
41:40
Well, let's put in the optimum value of zeta. The energy minimum then as a function of zeta is 11 over 16 squared minus 11 over 8 times 11 over 16. And that becomes minus 121 over 256.
42:06
So there's no joy in Mudville here because to be stable, we would have to get an energy lower than minus 128 over 256.
42:21
We're at minus 121 or much better than the three-eighths before that we had. That was pretty bad. This is better for sure. And that shows that we're closing in on the correct way of looking at the problem. But unfortunately, it's not good enough.
42:43
It looks like hydride then is a tough nut to crack. And indeed it is. We're going to crack it but we're going to have to go into the kitchen and we're going to have to get some tools out. And, you know, some nuts are Brazil nuts and other nuts are peanuts. You can open them with your fingers.
43:01
This is definitely a macadamia or a Brazil nut and it's going to be tough to crack. It's not us. We didn't do anything wrong. We've taken a perfectly sensible approach. It's just that this is a tricky system. And it's interesting that it's so simple and yet it's so hard
43:21
at the same time to get the right answer. But for some reassurance that it's not the quantum mechanics, that it's not completely wrong or something crazy, let's try helium. Why would helium be better? Well, helium has a plus 2 charge for the nucleus.
43:41
So that's going to attract the electrons much more strongly than the wimpy single charge of the proton. And that might mean that it's much easier to get it to be stable and actually get a reasonable. And it also teaches you why you love atomic units.
44:03
So here I've written the Hamiltonian in atomic units for helium, the kinetic energy is exactly the same, minus a half del squared. And then the only thing that changes is we've got minus 2 over r1, minus 2 over r2, and then plus 1 over r12.
44:21
So the repulsion integral is the same. That's done. And the only other thing is the other things have a 2. And so that's easy to do. So that we can scale. The kinetic and potential energies are going to be both doubled for each electron. And the repulsion is going to be increased as well
44:43
because the orbitals are going to get pulled in toward the nucleus. And what we get when we work it out is the, instead of getting minus a half, minus a half, we get minus 2 squared for the zeroth order energy.
45:05
And instead of just getting 5 eighths, we get 5 eighths times 2. So therefore, the total energy of the helium atom, if you follow through the exact same math as what we did with the hydride anion, then you just have this 2 instead of a 1 there everywhere through.
45:21
You can do it very quickly if you just keep track of it. You get minus 11 fourths. That means that the heat, the atom is stable already by perturbation theory compared to a helium ion and an electron at infinity at rest.
45:42
And the experimental energy that's listed in the literature in Hartree's is minus 2.9033. And recall, I said if you quote things in Hartree's, you never have to re-quote them because you don't depend on what the units are.
46:02
It just depends on what the calculation is. It doesn't depend on the fundamental constants. And our calculated value, minus 11 fourths, is minus 2.75 Hartree's. Here's the real good helium experimentally down here.
46:23
Here's the best we could do before we, just by perturbation theory. So at least we got that it was stable. It wasn't unstable, but it's quite a bit of a difference between the correct answer.
46:42
Five percent error in these kinds of things is way, way, way, way, way off unfortunately. That's like you didn't even bomb the right city. You didn't even bomb the right country. You're on a different planet in terms of your target.
47:02
And therefore, we can't do that. But what, why don't we just do the same thing as what we did with hydride, with helium? Since it's already better, at least it's predicted to be stable, let's puff the orbitals up a little bit with zeta and take the minimum and follow
47:20
through the same math again with the 2 instead of the 1. And we get a similar expression to be what we had before, but now E of zeta is zeta squared minus 27 over 8 zeta. And the same technique, obviously, you get 27
47:42
over 16 for zeta being the optimum value. And then if you put that in to the minimum, what you get is about, you get minus 27 over 16 squared. That's what, exactly what you get.
48:01
And what that approximately becomes is minus 2.84766 Hartree. That is looking pretty good because the real one is minus 2.9, but the real question is, well, how good are our results
48:21
in terms of something that a chemist might be interested in? Could we calculate base with this kind of approach? Could we calculate something accurately enough that somebody would really invest money in some scheme based on that calculation? And the answer is no. The energy of the helium ion is minus 2 Hartree
48:45
because it's just like a hydrogen atom so that with a different charge. So that is no. And our ionization energy then, if we're minus 2.846, our ionization energy, the energy to boot one electron off is 0.84766.
49:06
The true value is 0.9037. To boot an electron off helium. When we convert that from Hartree to kilojoules per mole, which are more familiar units to a chemist, we find that we're in error
49:23
by about 147 kilojoules per mole. That is similar to the value of many chemical bonds. Many chemical bonds have a value that could be a weak bond. But that's a lot of energy as far as a chemist is concerned.
49:45
That's not a tiny error. That's a huge error. And so we're nowhere close. Even doing this variational calculation, inserting this parameter zeta, doing all this work,
50:03
doing all these integrals, taking the derivative, setting it to 0, optimizing it, putting it in, hold your breath, calculate it. And we're still nowhere close to what we would call chemical accuracy. I think you can see why this kind of field
50:22
in atomic physics and quantum chemistry gets called computational chemistry very quickly. Because you may have to introduce functions that aren't so very easy to integrate, but happen to be very close to the true wave function. And maybe part of our trouble is we're introducing
50:43
exponentials because we know how to integrate them. We have a closed form for the antiderivative. But maybe once we start getting more than one electron in there, they aren't so close to the correct result. Next time then what we're going to look at is can we,
51:01
with a little bit more physical insight, so we said, well, it puffed out. But could we do better than that? Could we somehow change the wave function in such a way that it's not so bad to integrate, but that it's much, much better in terms of results?
51:21
And that will be what I call hydride tri number 3. We're either going to strike out or we're going to at least get a single. And hopefully we can figure out that hydride is, in fact, stable, so we'll pick it up there next time. Thanks for watching.