00:01
In this problem, we have a wave function, a normalized wave function.
00:04
There's a superposition of three eigenstates of a one -dimensional infinite well.
00:11
Those eigenstates are for n -equals 3, 7, and 5.
00:14
All the problems get the probability of finding yourself getting, being in an n -equals -5 state.
00:22
I've changed, i've taken away the numbers.
00:25
All i'm writing here, i don't want to have to be writing down the numbers continually, so i've reduced a and b for the coefficients.
00:34
Okay.
00:35
Now, as i mentioned, normalized size and normalized wave function, which means when we say that that's the case, it's basically saying that, yes, for certain, you're going to get your, you're going to be in n equals three, equals seven, and equals five, one of those states.
00:55
That's what it's basically telling us.
00:59
So let's write that out.
01:01
1 is equal to si star, si -d -x.
01:08
And i won't put any limits of integration, it's not important.
01:12
The one -dimensional well could be from zero to l.
01:16
It could be from minus l over 2 to l over 2.
01:19
Not important.
01:23
We're going to be using the orthormal condition between the eigenstates.
01:31
Psi -m si -m star.
01:35
Psi n dx is equal to chronicle delta mn.
01:43
So when m is and n are equal, this is one.
01:45
When they're not equal, it's zero.
01:48
So that gives us the orthogonality and the unit norm when it's the same.
01:59
Okay.
02:01
So now let's write this all out.
02:03
First, let's take the complex conjugate of this.
02:06
And i'm going to assume, at least for the writing, we can tell that a is not, it's just a real number, so really a star and a are the same, certainly.
02:15
B though is not, it is a complex, but i'm going to write it out as if everything was complex, just for completing the sake.
02:27
So what i'm doing is complex conjugating every piece.
02:35
So there's si star.
02:37
Now i just write over psi.
02:48
Okay, that's what we have.
02:50
Now we just expand it out of this.
02:52
You would, if you were given x plus, 3 plus 6 times, you know, times some other product of sum of multiple terms.
03:03
Just distribute through.
03:05
Okay.
03:07
Now, a star, a, that's the modulus of a square.
03:16
Okay, and this would be si -3 star, psi -3, tx, plus a -star b, psi 3 star size 7 you can already tell by this condition this is gone remember a and b are just constants so the integration is just on the size they're not part of it okay let's see a star oh i lost a star on the 5 okay a star g si 3 star psi 5 dx so we like i said we already know so, gone, gone.
04:09
Now we'll do the second term, the b star term.
04:12
So we got b star, a, size 7 star, side 3, dx, plus modulus of b square, size 7 star, size 7, dx.
04:33
And if you didn't want to keep writing all the integral signs and the dx, you certainly could have just expanded this all out under here and put big parentheses.
04:43
It's fine.
04:44
Whatever.
04:45
I'm doing it just so that you can see the separate integrations.
04:50
It's just a matter of choice.
04:52
Okay, what do we have? b star, g, si 7 star, sci 5, dx.
05:03
So again, this is gone.
05:05
This is gone.
05:06
So notice we're just getting the, we're just getting the, in a way, diagonal terms.
05:11
The way i'm writing this.
05:13
And now the last term.
05:14
Our last set of terms with the g star, si 5 star, g star a, sci 5 star, psi 3 plus g star b, sci 5 star, psi 7.
05:34
Oh, and i lost the dx here, didn't i? let me put this in, and i will just move this over a little, give myself a little room.
05:48
There we go.
05:52
There we go...