00:02
In this problem, you have a collection of subprout.
00:05
So first is to find the normalization constant for this wave function.
00:09
Notice, x does not have units of meters.
00:11
It's got units of radiance.
00:12
There's normally as a factor multiplying x that has radiance per meter as its units.
00:18
So we use normalization condition.
00:20
1 is equal to 0 to pi, psi star, x0, psi x0.
00:29
That's the complex conjugate dx.
00:31
Notice the normalization condition would normally be over from minus infinity to infinity, but the wave function, we're assuming here, is zero elsewhere.
00:44
So this becomes a squared, sine squared x, dx, from zero to pi.
00:54
So this becomes a squared, and we can use a trigonimic identity to get the answer here, which is one half, zero to pi, one minus cosine, two.
01:08
X dx so that's the trigonim identity and now we have a squared over two x minus sine 2x don't forget to divide by the factor multiplying x in the cosine so one -half zero to pi so this becomes a squared over two pi minus zero minus one -half sine two pi minus two pi minus sine two pi minus sign of zero which are both zero.
01:48
So this just becomes pi.
01:50
A squared pi over two, so a equals square root of two over pi.
02:00
So that's the normalization constant in that case.
02:03
Remember, normalization, basically this one represents the fact you ask the question.
02:07
Is that particle inside? and if it exists, the answer is yes, with certainty.
02:15
So there's where the probability of one comes in.
02:19
100%.
02:19
That's where that represents.
02:21
Okay, question two.
02:24
Once a node of side 20, then equals 20 for an infinite well, is an odd function around the origin.
02:37
And the answer is it is.
02:39
And the way you can see this, n is equal to 20 implies n minus 1 is equal to 19 nodes.
02:48
Now that's between the walls, not counting the walls.
02:52
And this is 20 anti -nodes.
02:57
Maximum points.
02:59
These are zeros.
03:00
These are maximums.
03:01
This being odd, it tells me that si 20 is odd function.
03:14
So is an odd function.
03:17
Now let me show you, let me show you how this looks for, let's look at psi 4, something simple and quick for me to draw.
03:37
So, n minus 1 nodes, not coming the walls.
03:41
One, two, three.
03:45
There's our three nodes.
03:47
One, two, four, here's my center line.
03:53
Remember, classic, what does it mean to be an odd function? sign of minus x, sine being an odd function is sine of x.
04:01
That defines it.
04:03
You know, or really, if you don't want to use sign, f minus x, minus f.
04:15
So here is f of x.
04:18
F of minus x.
04:20
So, so this is odd.
04:23
So notice, even power, leading to an odd number of nodes that defines.
04:29
It gives you that it's going to be odd.
04:33
Question three, wants to know what's the first excited state, towards the n.
04:41
That will be n equal two.
04:45
N equals one is the ground state.
04:48
Not n equals zero.
04:49
N equals zero is not the ground state, n equals one, four.
04:54
He's looking for the boundary conditions for an infinite well.
04:59
Minus a over 2, a over 2.
05:03
These are rigid walls, plus infinity, plus infinity.
05:07
We've got three realms, 1, 2, and 3.
05:11
Psi 1 must be equal to 0.
05:14
Si 3 must also be equal to 0.
05:17
If it is not, you then have non -zero probability densities, which means you have a probability of beam between x and x was dx somewhere out here, and that's impossible physically.
05:28
So, continuity, boundary conditions, with the really continuity conditions.
05:33
Psi 1 minus a over 2, is equal to psi 2.
05:38
You have to be single valued at the wall.
05:42
You can't have two different probability densities, which are two different probabilities of being between x and x plus dx.
05:49
These would be zero in this case for an infinite well.
05:53
But this part is true no matter of infinite well or not.
05:57
Finite well.
05:58
Likewise on the right, side 3, a over 2, i 2, a over 2 is equal to 0.
06:07
Again, this part is true no matter finite or infinite, but this is only for the infinite.
06:13
If you had a finite one, you also pick up continuity of the first derivative at the walls.
06:18
But that's not our problem here.
06:22
Okay, that was question four.
06:24
Question five, stationary states, what really, in terms of being formal, probability density, is independent of time.
06:51
And from that, from that, everything follows.
06:54
Expectation values for operators that are not dependent on time are constant.
06:59
That's what leads to the energy being constant.
07:03
But let me go, i'll talk about one thing at a time.
07:07
So here's si of xt.
07:09
If you're in a, if you were in a stationary state, you can factor your wave function into this form.
07:17
Si of x, just a dependent on x minus i .e .t over h bar.
07:24
So if you can put it in that form, you have a stationary state.
07:28
This leads to this.
07:30
So if you consider this as the definition of it, this is the form that gives you that satisfies that definition.
07:39
Now let me show you how that is.
07:41
Sigh xt squared, si star x t, psi, which is si star x, e to i, because you've got a complex conjugate of this becomes minus, i becomes minus i, which gives you an i plus i.
08:03
E, i t over h bar, psi x, e minus i, et t over h bar.
08:11
Well, e i times e minus i, that's one.
08:16
So it's just become si star x, psi x, which is the modulus of psi x.
08:28
So there's no time dependence.
08:32
It is the same at all times.
08:36
Now from that, from that, from this idea, if some operator is time independent, applies the expectation value is time, independent.
09:05
That's the key.
09:08
That's really, i mean, everything's coming from, this is experiments, right? expectation values are you get that from experiments in terms of that...