00:01
So we're given a potential that looks like this because the fact that the potential is symmetric in x tells us that we should be able to talk about parity as a valid operator to kind of distinguish our states.
00:19
So let's look for states of definite parity.
00:27
So if p is 1, that means we have what's called even parity, which means that they are symmetric.
00:42
So that tells me i probably expect my psi to be some amplitude times a cosine of some wave vector or wave number times x, because this is symmetric in x okay and then because of the boundary conditions at our endpoints we have to have that k a has to be where the cosine is 0 is n plus 1 half times pi and this will work for any integer n all right so that psi psi is, we can solve that for k.
01:51
So we got cosine of n plus a half pi x over a.
02:06
And this will work for any n.
02:09
So to find the energy, we use the eigenvalue equation.
02:16
E psi is minus h bar squared over 2m second partial psi with respect to x.
02:27
This is a solution of that that was the whole point and we substitute that in we get like that and then we got h bar squared over 2m with a minus sign out in front but then the second derivative of this gives us a fact it gives us another minus sign and then it gives us this thing inside multiplying the x squared we get quantity squared.
03:14
Okay? and so we can solve, oh, there's an a, cosine the wave function.
03:22
Okay? so forget about the a cosine part.
03:25
We get e is h -bar squared over 2m.
03:30
The minus signs cancel out, and then we got n plus 1ā2 pi over a quantity squared.
03:47
We'll call that en.
03:48
N all right so these are for the even functions all right so now think about the odd ones so odd functions psi goes to minus x minus psi when x goes to minus x all right so that means our psi is going to be some amplitude times the sine of kx okay and then we know that k a has to be n pi where n not equal to zero is an integer because if it's equal to zero we don't have have anything in any interesting wave there okay so in this case k is n pi over a our white wave function is capital a our amplitude which is actually our normalization constant and we get n pi x over a okay then we got e e psi is equal to minus h bar squared over 2m times the second derivative of that, which is n pi over a quantity squared.
05:24
And we got a capital a sine of n pi x over a.
05:30
Just like we did before, there's another minus sign here so that our energy is h -bar squared over 2m times n -pi over a quantity squared.
05:47
All right...