00:01
This problem asks us to solve the canonical partition function, solve for the canonical partition function, where we have the classical particle in a box scenario in three dimensions.
00:15
So this n squared here, or this k squared, writing out our energy, our n squared is just an x squared plus n y squared plus nz squared.
00:27
So the definition of our many particle partition function is just the sum overall energies of e to the negative beta of our hamiltonian, where our hamiltonian is a many particle hamiltonian.
00:55
And beta is defined as just one over kv times the temperature.
01:04
However, in this case, we can simplify this somewhat by noticing that each of these particles is going to have the same energy.
01:14
And so our multi -particle hamiltonian is just going to be basically a sum of these different possible energies for the different particles.
01:28
So we're going to have two pi squared, hbar squared over going to be.
01:39
Going back up, ml squared for ml squared times n1 squared plus 2, pi squared, each bar squared over ml squared, n2 squared, plus blah, blah, blah.
02:08
And this is basically where this is talking about particle number.
02:13
So then putting this into our many particle partition function, we can see that z of n equals the sum over nx1, y1, and z1, as well as all these other sums over all the different particle numbers.
02:55
So i'm just going to write down the last one here, which would be the nth particle, and of e to the negative beta.
03:09
And then we can actually, as well, more than just taking the beta, we can also take all of our multiplicative factors out in front as well as those don't change between the different particles.
03:28
And it's just going to be n1 squared plus n2 squared dot dot, dot, dot, plus.
03:37
And sub n.
03:40
And we can see that since all of these sums are going to be the same for every particle, then we can write out our many particle partition function instead in terms of just the single particle partition function.
04:03
So if we just had one particle, we'd have nx1, nx2, or nx1, and y1, and nz1, of e to the negative beta 2 pi 2x4 over n l squared of n1 squared.
04:29
So this would just be our single particle partition function.
04:32
And if we look and compare that to this many particle partition function, we can see that since this is going to be the same for every particle, and since e to the x plus y just equals e to the x times z to the y, we can say that our z sub -n is just going to equal a, first off, a normalization term where we just include that so that we aren't double counting particles.
05:08
So basically, as soon as a particle is accounted for, we need to add this one over n factorial.
05:14
And then we have our z1.
05:20
And then we just take n copies of that z1, and that's our many particle partition function.
05:30
So what we can do now is if we now say in the large limit, we're going to say that these sums over n sub x, y, or z are actually going to go to integrals over x, y, or z, then what we can say is if we have, our single particle projection function, remember being nx and z of e to the negative beta, 2 pi hbar squared over 2m l squared n squared.
06:32
Then what we can say is that instead of a sum, we can rewrite this as a triple integral.
06:43
And at this point, what we can do is notice that our integral is going to actually be spherically symmetric.
07:01
Since the only thing that we're integrating over is our end values and all of these are being squared, it would actually be good at this point.
07:17
And i just realized i made a typo here.
07:20
Copying that.
07:21
Whoops.
07:24
It's okay.
07:24
We'll just keep going by that.
07:28
If we look at this, this is spherically symmetric.
07:42
And so instead of doing an integral over each one of the cartesian coordinates, what we can do is instead convert to spherical.
08:01
So in that case, our z1 is going to equal an integral from 0 to 2 pi for our phi, integral from 0 to pi, and integral from 0 to infinity over the same expression, h -fr squared, over m -l -squared.
08:32
And we can say n -squared now.
08:36
And now we need our volume element in spherical volume space, which is n -squared, sine theta, d -n, d -theta, d -fi.
08:47
Since there's no angular dependence in our expression, we can just integrate over both theta and phi for a solid angle of 4 pi.
09:01
And we're left with negative beta times 2 pi h bar squared over m squared, n squared times n squared d.
09:23
But luckily, we have a general form for integrals like this, which say that negative infinity, infinity, x squared, e to the negative ax squared, dx equals one half square root of pi over a cubed.
09:46
So if we then come over here, and we define negative beta, 2 pi hbar squared over m l squared.
10:06
We define that.
10:07
And actually we want to keep the negative outside of this definition so that we can match exactly the form of the expression.
10:16
If we define this as a, then we can compare these two and see that they're exactly the same except for we have a bounds issue.
10:28
So this one goes from negative infinity to infinity.
10:32
This one really goes from zero to infinity.
10:35
But again, we can use the fact that our integral is even, meaning that because n squared is the same for plus or minus n, we can say that instead of an integral from negative infinity to infinity, that's the same in this case because our function is even as taking two times the integral from zero to infinity...