00:02
Okay, so we have a square well with an infinite wall at x equals 0, and a wall of height u at x equals l.
00:14
So for the case, e is less than u, we need to obtain solutions to the schrodinger equation inside the well and in the region beyond the well.
00:25
That satisfied the appropriate boundary conditions at x equals 0 and x equals infinity.
00:30
So we also need to match the boundary conditions at x equals l to find the equation for the allowed energies.
00:41
Okay, so let's start at the beginning.
00:45
Inside the well, the particle is free.
00:48
So in portion a, the particle is free, and the wave function is trigometric.
00:54
So you'll get that inside the well, si of x is equal to a sine of kx plus b cosine of kx.
01:17
And this is 4x between 0 and l.
01:24
So at x equals 0, there is an infinite wall.
01:28
This requires that si at zero needs to equal to zero.
01:39
Therefore, b is equal to zero.
01:45
Okay, outside of the well, so in portion 2, x equals l, u of x equals to u, and the schrodinger equation has exponential solutions.
02:00
So the exponential solutions, so this is for portion 2, your wave function is equal to c .e.
02:15
To the minus alpha x plus d e to the positive alpha x.
02:25
And this is for x greater than l.
02:30
Okay, and we know that alpha is equal to 2m u minus e over hbar squared the square root of this.
02:48
Okay, so now we need to bind the wave function at x equals infinity.
02:54
So to do that as x goes to infinity, d will go to 0.
03:03
So d needs to be 0.
03:08
And at x equals l, you need it to be continuous.
03:16
So this is where we will start to link the two equations.
03:21
So we need continuity of si and d -sci dx...