00:01
So in this problem, we're looking at the schroeder equation for a three -dimensional cube.
00:06
It's a three -dimensional analog of the infinite square well potential.
00:12
So the problem doesn't explicitly say it, but the potential inside the cube is zero, and outside its length l, the potential energy is infinite, so the weight function is zero at l or greater distance.
00:33
From the origin.
00:40
So it wants us to find the energy in this particular form where it's h bar instead of h.
00:49
So hold on a sec.
01:06
So what we want to do is note that we're given this wave function a times sine kx, sine ky, sine kz, where the ks are different.
01:18
What this is really saying is that these side functions are one is dependent on x one is dependent on y one is dependent on z there's no cross dependence so they're separate from each other so the key of this problem is that you can decompose that wave function is a product of one -dimensional wave functions and for notational simplicity i'm just going to call those big x y and z so those aren't variables x, y, and c, those are functions x, y, and z.
02:04
So if you write it that way, you can rewrite the schrodinger equation.
02:12
It's a h bar squared 2m.
02:19
So the point of this is that the schrodinger equation was partial derivatives here.
02:27
Because we're doing a separation of variables where things depend on only one dimension.
02:39
Means we're turning the three -dimensional partial derivative partial differential equation into three ordinary differential equations, which is essentially just doing the infinite square well, the three infinite square well problems simultaneously.
03:20
So these functions are effectively constant because they don't, y and z here don't have any x dependent, so they're effectively constant.
03:41
So this is supposed to be times size, so that is xyz.
03:47
So if you were to then divide x, y, z through on both sides, you get three ordinary differential equations of this form.
04:31
Then if we notice that the potential u is zero inside the cube and infinite outside, where the whole thing goes the whole way of function goes to zero if you is infinite then we can just since this is zero we can write that as equal to minus e um so first thing i want to point out is that since these x y and z are sign functions uh we do two derivatives on them what you're really doing so you do the first, you do the river the first time, you pull out a kx, so the constant factor inside the sign function.
05:54
But then, so then it turns into a cosine kx.
05:57
Then the second derivative turns the cosine into a minus side kx and brings out another kx.
06:05
So that becomes this...