 So we've found that if we want to make progress, finding out what the partition function is for a three-dimensional particle in a box, we can start by thinking about the partition function for a one-dimensional particle in a box, and then the 3D particle in a box can be composed by multiplying several of those together. So let's treat this smaller problem and ask ourselves, what is the partition function for a one-dimensional particle in a box? As always, partition functions are the sum of the Boltzmann factors. So for all the different states, n equals 1, 2, 3, 4, 5, and so on, all the way up to infinity, energy of the 1D particle in a box is h squared over 8m, n squared over a squared. That e to the minus that energy over kt is the Boltzmann factor. Sum all those up, and we have a partition function. So we're going to compute the value of that sum, but first let's ask ourselves, how many terms do we need to include in that sum? The sum goes out to infinity, but when n is large, the terms will become very small. So only some of the terms will contribute. And we can ask ourselves, how many of the terms are important? How many of the terms do we need to include in the sum? Equivalently, we could ask, how big is q? If q is a large number, we're going to use many of these terms. If it's a small number, we're going to use relatively few of these terms. And that's also equivalent to asking, how big are these energy levels compared to kt? Is this numerator in the exponent larger or smaller than the denominator by a lot or by a little? So those questions are all related. And we're essentially asking, remember what the particle in a box energy levels look like, there's an e1, e2, e3, e4, and so on. Those energy levels are more specifically the differences in energy between successive levels. How do those difference in energies compare to kt? That's a question we've addressed before. The very first time we started thinking about particle in a box problems using molecules as the particles and physical containers as the boxes they were confined in. And we found then that for, if we combine gas molecules like a nitrogen or oxygen molecule to a box of typical macroscopic size, we saw just by plugging into this expression that these energy levels were something like 10 to the minus 42nd joules apart from each other. And when compared to kt, in terms of order of magnitude, kt at room temperature is about 10 to the minus 21 joules. So there's many, many orders of magnitude difference, 21 orders of magnitude difference between the energy level spacings here and kt. So many energy levels can be populated for a system at a temperature of 298 Kelvin. So the answer to our question, how many of these terms are important? How large do we have to make n get before this energy is large compared to kt? And we don't have to include terms in the exponential anymore. Quite a few. A very large number of energy levels are going to be populated. So what that means is the partition function is going to be very large. The other thing that means is when we're solving this problem, we can at least approximately not think of this as a sum, an infinite sum from 1 to infinity, but an integral. So if I describe that graphically, the sum means that we should compute, so if this is n, we need the value for n equals 1, n equals 2, n equals 3, and so on. What we need to compute is e to the minus energy over kt. So when n is 1, this is a very small number that's compared to kt. So it's e to the minus very small number, and that's a value very close to 1. So let me, when n equals 2, same thing, the numerator is very small compared to the denominator. So it's also still very close to 1. Over the long run, these terms will look like e to the minus n squared because there's an n squared in the exponent. So it will look generally speaking like a Gaussian, e to the minus n squared, but the number of terms that we need to include, the number of these bars that make up this Gaussian that we need to calculate the individual areas of is very large because these energies are small compared to kt. Because many, many, many of these terms are occupied. It's not just 3 or 4 or a dozen, like I can draw in this picture, but it's many, many millions or 10 to the large number of states that we need to occupy. So it doesn't really matter if we add up the individual bars or if we compute the area under the curve because we've divided the curve up into so many small, narrow rectangles. So we're approximating this sum with an integral. And that turns out to be a very good approximation. So instead of summing overall values of n, we can integrate overall values of n running from 1 to infinity. And once we've noticed that there's a very large number of bars that contribute to this integral, it doesn't really matter whether I include one or a few extra or one or a few to a few of these bars because it's really more convenient for me to calculate this integral not from 1 to infinity, but from 0 to infinity. And we'll talk in just a second about why that's more convenient. But this integral we can do relatively easily. So I've converted the integral from going from 1 to infinity into 0 to infinity. And that's not correct. That's equivalent to throwing away one of the bars on the low end side of this integral. But because we've included so many of these terms, it's not a very big approximation for us to throw away one of them in order to make the integral easier to do. So what we're left with then is this Gaussian integral. And I'll tell you, we're going to run across a lot of these Gaussian integrals, probably the easiest one to remember. To memorize is if you have the integral of a Gaussian, e to the minus some constants times the variable squared, integrated everywhere. So if I have a Gaussian function, a bell curve, and I want to know the area under that Gaussian, that value is square root of pi over alpha, where alpha is this constant up in the exponent of the Gaussian. So that's something that you can get from an integral table. You can go look that up online. If you'd like to know how to actually do that integral, we'll have a separate video lecture on computing Gaussian integrals because they show up so often. But for now, we'll just say we know that the integral of a Gaussian is square root of pi over alpha. We only want to calculate the integral from 0 to infinity. So exactly half of this integral. That explains why it's easier to take the integral from 0 to infinity instead of 1 to infinity because 0 to infinity is exactly the positive half of the integral. So the result here is square root of pi over alpha, where alpha is everything that doesn't include the variable, the n squared. So square root of pi over h squared divided by 8m a kt. And because I'm only calculating the positive half of the integral, this integral is one half of the value that would be for the infinite integral. So that's our result. That's the partition function for a 1D particle in a box. We can simplify it quite a bit. So let's get the fraction out of the denominator. So I can rewrite this as 1 half square root. There's a pi in the numerator. There's an h squared in the denominator. The 8m a kt needs to come up to the numerator. So I've got 8m a kt. I've got a 1 half outside the square root and an 8 inside the square root. So if I pull the 1 half inside the square root, it becomes 1 quarter. So the 8 turns into a 2. So we see that, oh, and I've made a mistake. There's an a squared. This should have been an a squared, a squared, and I left that out entirely. So the a squared is inside the square root. So I can pull that out of the square root and just call that a. So the result we've got here is that the 1D particle in a box partition function looks like the square root of a bunch of constants. 2 times pi times the mass of our particle times kt divided by Planck's constant squared. Take the square root of all that and then multiply by the box length in the x direction for this one dimensional particle confined to a box length a in the x direction. That quantity, the square root of 2 pi m kt over h squared, we're going to give a name to that quantity. If you think about what the units on this quantity must be, so this quantity right here, we know that a partition function is not allowed to have any units. It must be unitless. Because remember, that's just a sum of a bunch of Boltzmann factors. The Boltzmann factors are e to the something. Those are all unitless themselves. So when the partition function is unitless, this quantity that I'm going to give a name to multiplies a distance, a box length, has units of length. So this quantity in the square root, after taking the square root, it has to have units of one over length. So the quantity we're actually going to define has a slightly cumbersome name. It's called the thermal de Broglie wavelength. We usually use the variable lambda, capital lambda, to describe that wavelength. That quantity is exactly the reciprocal of what we have done here. And we take the reciprocal because this quantity has units of length, which is easy to think about physically. It represents the wavelength of the particle in some sense. So I've taken one over the square root and it's going to have units of length. What that means is after making this definition, we can say that the one-dimensional particle in a box, after I've renamed this constant, that square root down there looks like a one over lambda. And I multiply that by the box length. So the important thing to know about the thermal de Broglie wavelength, it's similar to the de Broglie wavelength we talked about when we defined the basic properties of quantum mechanics. But it's the thermal de Broglie wavelength, meaning the de Broglie wavelength that's important for the particle when it's at a temperature T, when it has the energy in the vicinity of kT. And what's important about the value of this quantity is under circumstances where the de Broglie wavelength itself is, let's say, much less than the box length. So this ratio of the thermal de Broglie wavelength and the box length is the important comparison to make when lambda is much smaller than the box length than this means the one-dimensional particle in a box partition function is going to be a very large number. And remember what that means physically is that many, many, many of these energy levels are going to be occupied. On the other hand, if the opposite were true, that's not our focus right now, but if the opposite were true, if lambda were bigger than a, which is not in this case, then the opposite would be true. The one-dimensional partition function will either be close to 1, or depending on how the zero of energy is defined in this case, the zero of energy is below state E1. We can just say that the partition function is going to be a small number. But again, at the moment, we're interested in this case when we can find gas molecules to a box. Their thermal de Broglie wavelength is very small compared to the box length. That's equivalent to saying the energy spacing is small compared to kT. Also equivalent to saying the partition function is greater than 1. Those are all different ways of saying the same thing. And under those conditions, that means the approximation we made in converting the sum to an integral because there's so many different slices that we've converted that we're using in the sum. That's a good approximation. And we say that that approximation that we've made is the approximation of doing this computation in what's called the classical limit. Even though we started out using quantum mechanics, these are the quantum mechanical energies for a one-dimensional particle in a box. By the time we've written this equation, essentially what we've said is, even though the real energy levels are discretized, only certain energy levels are allowed, in thinking about an integral instead of a sum, we said, well, whatever, any energy level is allowed. Instead of only caring about integer values of n, I'm allowing any real value of n to be used. And when I calculate the partition function under those conditions, it's like saying quantum mechanics isn't terribly important. We can do this calculation pretending the system is classical and there's no discretization or quantization of the energy levels. So we've succeeded in computing the one-dimensional particle in a box partition function, whether we write it with the derogatory wavelength or whether we write it with the more explicit formula here. And remember, our next step is going to be combine those 1D particle in a box partition functions and learn what we can about the three-dimensional particle in a box partition function.