 Alright, so now we're going to talk a little bit about units and develop something called the atomic system of units. So what I've reminded us of on the board here is the Schrodinger equation for a hydrogen atom. So I've not written out all the DDR, DD theta, DD phi terms, I've just written all them as del squared, but even so this is a relatively complicated equation only in the sense that the del squared, the kinetic energy term is prefixed by a collection of constants. The potential energy term itself has a whole bunch of constants and a 1 over r. When we solved that Schrodinger equation to determine the allowable energies for an electron and a hydrogen atom, we found the energy was z squared over n squared times a whole bunch more constants. And even in this expression, this a naught, that constant is itself a large collection of constants as well. So constants are littered all throughout the equations we use in quantum mechanics when we deal with hydrogen atoms. And we're very used to the fact that we can work in certain systems of units and we choose the values of these constants in different units depending on what's convenient to us at the time. For example, if we're using r, the gas constant, sometimes we might use it in units of joules per mole kelvin, 8.314 joules per mole kelvin if that's going to be convenient, or if we're working in liters and atmospheres we might prefer to use it as 0.08206 liter atmospheres per mole kelvin. Or in other problems, we might not call it the gas constant, we might call it Planck's constant, I'm sorry not Planck's constant, we might call it Boltzmann's constant, and use 1.38 times 10 to the minus 23rd joules per kelvin. The point is we can use a different value of that constant depending on what units we want to work in, and when we convert those units to each other they're all the same value just expressed in different units. The same thing is true for atomic problems, there's a set of units that's not the SI units that is going to turn out to be much more convenient for us, and those units we'll call atomic units. For every type of quantity, for example, mass in SI units, the unit of mass is a kilogram. In atomic units we're going to let the mass, the fundamental mass unit in atomic units be whatever the mass of an electron is. We know the value of that expressed in SI units, the mass of an electron is 9.1 times 10 to the minus 31 kilograms. So that value is one electron mass, which is the quantity that we use as the unit of mass in atomic units just as the kilogram is the fundamental quantity we use in SI units. For units like length and time and charge, I'll compare those values in SI, the unit of length is a meter, the unit of time is a second, the unit of charge is a coulomb. In atomic units our fundamental unit of length is going to be the Bohr radius, and as we've seen that value is 0.529 angstroms, roughly half an angstrom, if I write it in SI units that's 5.29 times 10 to the minus 11th meters. Rather than using seconds as our unit of time, we're going to use something, time is not all that frequently used so it doesn't actually have its own variable in the atomic system of units so we just say an atomic unit of time. That atomic unit of time, I'll have to look up because that one's not terribly common. The one atomic unit of time is 2.4 times 10 to the minus 17 seconds and the unit of charge that we're going to use is the charge of an electron which is 1.6 times 10 to the minus 19th coulombs. So all we've done is defined a different base unit for each of these quantities and the whole purpose of using a system of units like sticking it with SI for an entire problem is because it makes your life easier when you get to composite units. When you're working in SI, for example, if you want a quantity in energy, you don't have to work out every time that a kilogram meter squared per second squared is going to work out to be a joule. You know that a joule is defined as whatever the unit of mass is times two units of length divided by two units of time. So the same thing is going to be true in atomic units and we can work that as an example. Once I've told you what the atomic units are for some of these quantities we can figure out what the atomic unit would be for other units like the energy. So like we've just mentioned, energy is a mass times a length squared divided by time squared. That statement is true in any system of units regardless of whether it's SI units, kilogram meter squared per second squared. The atomic unit of energy turns out to be an electron mass times a Bohr radius squared divided by the atomic unit of time squared. But if we want to get a value for that, if we want to see what that equals in SI units, we can say that the atomic unit of energy is going to be the atomic unit of mass, nine point one times 10 to the minus 31 kilograms, atomic unit of distance squared, divide all that by the atomic unit of time, each of which just has the values that I've listed over here. And if we use a calculator and plug numbers in and figure out what that equals, we'll find out that that equals 4.36 times 10 to the minus 18 joules. So that's actually a number we've seen before. That is, turns out to be the same as if I collected some collection of these constants and that is exactly the value that we have called in the past one Hartree. So the reason we defined a particular collection of constants to be a Hartree is because that Hartree is the atomic unit of energy. If we do another example like this, let's ask ourselves about the value of Planck's constant. So this constant right there in atomic units. So we can do this, we can treat this one just like a conversion factor. If I know the value of Planck's constant, 6.626 times 10 to the minus 34 joule seconds in SI units, if I want to convert units, first I need to convert the unit of joules. So I've just discovered that one atomic unit of energy is equal to 4.36 times 10 to the minus 18 units of energy in the SI system. So that many joules is equal to one atomic unit of energy. And then if I also convert the time unit, 2.42 times 10 to the minus 17 seconds is one atomic unit of time. So if I take Planck's constant divided by the size of a Hartree divided by the size of an atomic unit of time, what I find to the two, three sig figs I've been using here is I find that that equals 6.28, which looks suspiciously like twice the value of pi. It turns out even if I were to work this out to five or six or eight sig figs, I'm always going to get twice the value of pi to that many sig figs. And that's no accident because it turns out that atomic units are defined in a way to make sure that h comes out equal to the value of 2 pi. So h is essentially defined to be equal to 2 pi in atomic units. So essentially what we've done in atomic units, in SI units what we've said is anything that weighs a kilogram is one SI unit of mass. Anything that has a length of one meter is one SI unit of length in the SI system. We've chosen a different standard in the atomic units system. We've defined the quantity so that in atomic units the mass of an electron is one atomic unit. The radius of a hydrogen atom, the most likely distance at which we find the electron in a hydrogen atom is defined to be one in the atomic system of units. Likewise, the charge on an electron is one in atomic units. And we haven't talked about this one, but it's also true that we choose one over 4 pi epsilon not the, this shows up here in the Coulomb potential energy term of the Schrodinger equation. We've defined the electrostatic constant so that one over 4 pi epsilon not comes out to be one when we're working in atomic units. So what that means by making that convenient set of definitions is that rather than writing the energy as this collection of constants multiplied by one over n squared turns out that since h is equal to 2 pi, h squared is 4 pi squared so that almost cancels this 8 pi squared. m is one, a not is one, all we're left with when we're done is that the energy is, since it almost cancels but not quite, I still have a 2 left over is minus a half z squared over n squared. That looks similar to an equation we've written before. Before I've written minus half z squared over n squared times a collection of constants that equals a heart tree, but the value of a heart tree in atomic units, that's just one atomic unit of energy. So, in atomic units, the energy is just minus a half z squared over n squared period. Likewise, we can rewrite the Schrodinger equation. So that's the last thing we need to put on the board. The Schrodinger equation, if I rewrite this expression with every time I have something like a 1 over 4 pi epsilon naught, if I rewrite it in atomic units, 1 over 4 pi epsilon naught has the value of 1 in atomic units, then h squared again becomes 4 pi squared, so the minus h squared over 8 pi squared just becomes minus a half. m, mass of an electron has the value of 1 in atomic units, so I am left with just del squared of the wave function for the Coulomb term. z is still z, but e has the value of 1, 1 over 4 pi epsilon naught has the value of 1, I still have an r, so the Coulomb term just looks like z over r, and the energy, because we haven't solved the Schrodinger equation yet for an arbitrary problem, we don't know what that value is, it's still a variable that has some value. So the Schrodinger equation, rather than being littered with all these constants, when we rewrite it in this atomic system of units is quite simple, just minus 1 half del squared wave function minus z over r wave function must equal e times wave function. So that's going to come in handy for us when we extend beyond the hydrogen atom and start talking about more complicated systems where the Schrodinger's equation gets even more complicated, we're going to want to simplify it as much as we can by working in atomic units.