00:01
In this problem, you're given two wave functions, or at least the first part of the problem, given two wave functions, psi -0, side 1, and you're told what their eigenvalue should be, and you're asked to show that.
00:15
So remember the eigenvalue equation, the h, the operator, times the wave function, is going to give me some number times the original wave function.
00:25
That's an aggregate equation.
00:26
So we're trying to prove lambda is one -half for si -0 and three -half for si -1.
00:33
And so we just have to construct the operator.
00:39
So let's look at, so this is part, this is part a.
00:44
So si, h of size zero, you have for the h, one -half p squared, and then you have plus one -half x squared.
00:59
That's what you have for your operator.
01:02
And the p is a differential operator.
01:05
It doesn't have any hs in it or anything like that.
01:09
So it's like they say, it's a dimensionless operator.
01:14
So we get minus 1 half, d2, dx squared.
01:21
That is minus i, d, d, x squared.
01:26
The minus i, comes from i squared is equal to minus 1, plus 1 half x squared.
01:34
E to minus x squared over 2.
01:39
So that's our first operation.
01:44
And so we're trying to get into a form like this.
01:48
So we'll just work on it and see what happens.
01:54
So we're going to get here minus one half, gdx.
02:01
Now the first derivative of e to minus x squared over 2 is minus 2x over 2, eat at minus x squared over two plus one half x squared edem minus x squared over two.
02:24
X is not a differential operator in this representation, just x.
02:31
Okay, so all we got to do is take the derivative again.
02:35
Now remember the, so this is really minus x.
02:39
The two is cancel out.
02:41
So let me write that.
02:43
So you see it.
02:43
Minus one half, d dx, minus x, eton minus x squared over two, plus one half x squared, ed in minus x squared over two.
03:05
Okay, so, minus one half.
03:13
Now, chain rule, when we have x times a function, e to the function of x.
03:19
So first we're going to differentiate the multiply, term and then the exponential again.
03:26
So minus eton minus x squared over 2.
03:32
Then we're going to have minus x minus 2x over 2 again because that's a differential of a minus x squared over 2, e .m minus x squared over 2 plus 1 1 half x squared e minus x squared over 2.
03:59
And now putting the minus 1 half all the way through, we could have taken care of the minus sign here.
04:09
Any way you want to do it.
04:12
It doesn't, one place or another.
04:15
It's fine.
04:16
Okay, so i'm going to get here, 1 .5, e to minus x squared over 2.
04:34
And then i'm going to get here, this would be plus minus minus one -half x -squared, either minus x squared, either minus x squared over two, plus one -half x -squared, either minus x squared over two.
04:56
Well, these two terms are exactly the same.
05:01
They're gone.
05:03
So this becomes one -half, either minus x squared, over 2, which is one half, psi not, the eigen value is one half, as we were expected to prove.
05:22
So there we have it.
05:25
Problem like this is just a matter of, just a matter of writing everything out.
05:31
It's really what you've got to do.
05:35
Now, for psi 1, which is going to be a little bit more because it's got an x in there.
05:44
So you can imagine that there's going to be a few more terms.
05:47
So, minus one half, d2, dx squared, plus one half x squared, x, e minus x squared over two.
06:06
So that's what we have.
06:09
Now let's start writing it.
06:11
Minus one half, d, d, x.
06:18
Okay, now we got to take the first derivative of this.
06:21
With the outside, it's going to leave me an minus x squared over two, plus x minus 2x over 2, e .m.
06:35
Minus x squared over 2.
06:42
So that's the first derivative, plus one half x cubed, e .m.
06:51
Minus x squared over 2.
07:01
Now we got, take the derivative again, and this derivative of the e minus x squared over 2, which is just a minus 2, x over 2 it's really a minus x but i'll write it out anyway aided minus x squared over 2 now here we also have a minus x squared either minus x squared so i'm going to get here minus 2x so this i'll just do it this way this is going to be minus x squared multiplying so it's going to be minus 2x edem minus x squared over two and then and then plus actually plus minus minus x squared minus 2x over 2 either minus x squared over 2 plus one half x cubed edin minus x squared over 2 so that's what we have do all these terms together.
08:24
The twos go away here, but i got a minus one half, so i'm going to get one half x, either minus x squared over two, and then here we get minus two over minus two, so i'll get plus x, even minus x squared over two.
08:46
Here, two is again go out, minus times minus is plus x cubed, but you've got a minus over two, minus one -half x -cubed, e to minus x squared over two, plus one -half x -c cubed, e -and -half x -cquared over two.
09:17
Okay, well, can we combine, can we get rid of anything, minus one -half x -cubed, e, plus one -fx, cubed gone.
09:30
So all we have left is one, one half plus one, three halves.
09:40
Three halves, x, an minus x squared over two, which is three halves, side one.
09:50
That gives me the eigenvalue value is three halves as we were asked to show.
10:04
There we have it.
10:05
So we showed the three halves.
10:09
So it's just a matter of dealing with the operator and writing out all the terms and doing whatever, however you want to write it out, the algebra, just a matter being careful.
10:31
So that was part a just to show those eigenvalues.
10:37
They're eigen functions and they have specific eigenvalues.
10:43
If you don't, if you don't meet the criteria h.
10:47
Si is equal to some number times si, then it's not an eigen function of that operator.
10:55
All, nothing wrong with that, but that's not what we were told.
11:00
Okay, part b, they're giving us a psi 2 that they want us to find alpha so that it is orthogonal.
11:10
That has a zero inner product with si naught.
11:25
Now this is the normal inner product in quantum mechanics with the psi star side idea.
11:32
The first argument is the star.
11:36
Second argument is not.
11:41
Here, it's all real so it doesn't really matter but that is the form we have.
11:52
Okay.
11:56
Now putting it in the form of side 2 that they want us to use 1 plus alpha x squared, e minus x squared.
12:07
So that is si 2.
12:11
Like i said, no complex aspect of this.
12:15
So we just write it down.
12:18
And then we have si not, which is just e minus x squared over 2, dx, which is minus infinity to infinity, 1 plus alpha x squared.
12:36
We combined these, this is an x squared over 2.
12:41
X squared over 2, both are.
12:44
So we combine that, that just gives us a minus x squared dx.
12:51
So that's what we have.
12:52
That's the property of the exponentials.
12:55
And this is equal, they want an orthogonal, so zero inner product.
13:01
So that's what we're looking for.
13:03
So how do we proceed? certainly, at this point, you could say, well, i don't know how to deal with any minus x squared, and you'd be correct.
13:13
It's not standard, cannot find it through standard approaches.
13:20
So what do you do? you know, do you go into a table, vinegar goes? certainly that works.
13:26
Do you go online, symbolic, some type of symbolic program online or program that you have access to, certainly, that would work.
13:36
But let me show you how actually you can actually find, you can, you can a way, trick your way to getting the e to minus x squared, and from that, the x squared to e minus x squared.
13:51
Let me show you.
13:53
Third digression.
13:58
Let me define i zero.
14:00
And you might say, what in the world is this zero? well, that's going to be the power of x, that would be multiplying to the e, with one times.
14:09
To eat and minus x squared, that power is zero.
14:11
So i is zero.
14:13
But this is not a function.
14:14
We're not worrying about this.
14:16
This is a definite integral.
14:17
So this would be a number.
14:19
But a number could involve some parameter.
14:24
So i -0 -beta as infinity to infinity.
14:36
Just because we have a definite integral, that means we get rid of the x aspect of things.
14:46
It does not mean that that is true.
14:48
A number just means it cannot involve x.
14:51
You don't have any functional form anymore.
14:54
Ead minus beta x squared d x.
15:00
So that is the form.
15:02
Now i've just leading to this, i2 beta would be x squared, eating minus x squared over 2.
15:08
I'll show you from all this, you'll be able to get that also.
15:14
So keep that in mind.
15:16
You might say, well, what did you have to for us? you said we couldn't do that with any, you know, traditional approach is what are we supposed to do? well, let me take you through the trickery that i was talking about.
15:28
That's squared.
15:38
So that's going to be i squared.
15:40
I not squared to be precise.
15:42
I's infinity.
15:43
And again, remember, it's just some function of, it's just some function of beta.
15:51
So here's, here's that once and here's that twice.
16:06
Notice, got to have different, you've got to have different dx and 1, d, y, and others.
16:12
It's just two different d somethings.
16:14
It can't be the same.
16:17
It's like multiplied squirre in two series together.
16:21
You can't use the same index.
16:23
It doesn't work.
16:24
You've got to have two different indices that run from, say, from n equals zero to a thousand, and then m equals zero to a thousand.
16:34
You've got to have two different indices.
16:36
And the same way here.
16:38
It's got to be done twice.
16:40
Anybody say, what did that do for me? it looked even worse.
16:42
Well, let's take it one step at a time.
16:58
So let's combine.
17:01
X squared plus y squared.
17:07
Dx, d, y, d, y.
17:09
All i've done is just combined, that's all.
17:14
I've done anything else.
17:16
But now, here's the trick.
17:20
X squared plus y squared.
17:22
Seems like i'd be a circle.
17:27
Well, that's sweet.
17:28
To polar coordinates.
17:32
So now we're going to have row and we're going to have phi.
17:37
Zero to infinity.
17:38
It's a radius, right? zero to infinity, zero to pi, e minus beta, row squared, row, de row, d, phi.
17:56
That's our area element in terms of polar coordinates.
18:00
And the phi integration is trivial.
18:02
2 pi, 0 to infinity, eden minus beta, row squared, row, d row...
