00:01
In this problem, we're going to be looking at the wave function, including its time dependence, of an infinite well.
00:08
So let's start at the basic point.
00:12
Shorlander's time dependent equation.
00:17
How do we solve something like that? what are your separation of variables? let me write this, and i'm going to talk about notation for a second.
00:26
Separation of variables, you have a function of x, one of the variables, and you have a separate function of the second variable.
00:32
In this case, t.
00:33
That's how you do it.
00:35
It'll allow these equations to decouple to give you two equations.
00:39
We'll see that in a minute.
00:40
But before we go on notation, notice this side has a bar underneath.
00:48
This side has no bar.
00:49
Later on, there's going to be one with bar on top and bottom.
00:54
So we just got to be mindful of what i'm writing.
00:56
I'll try to point it out to you as we do it.
00:59
Now, let's put this in.
01:01
Now, with this derivative, in terms of secondary perspective, x, the phi, there's no role in that.
01:09
Does not get affected by that.
01:10
So that's just kind of like a constant being brought out.
01:13
So i'll put that outside, minus h bar squared to m, d2 .5x, d x squared.
01:24
We don't need the partial anymore.
01:25
It's just a function of x.
01:27
Plus v, si of x, is equal to psi of x, ih bar, d, d, t, 5t, it's function of t.
01:45
So that's what we get.
01:47
I've dropped the x dependency on the potential, but i'm keeping the other ones to keep it clear what we're doing.
01:54
All right, let's divide by the psi and phi, the product of psi and phi.
02:03
So i get one over of x, nice h squared over 2m, g2, x, d2, x squared, v, si, v, si, y, is equal to ih bar over phi of t, d -fi d -t.
02:34
So that's what we have.
02:36
Now, there's a, this tells me something.
02:41
If this quantity on the left was a function of x, put it all together.
02:49
You did all these, if you had a particular wave function, you had a particular, sigh, and you calculated that it became out to be exponential x raised to the x power.
03:03
Could that be? cannot be.
03:06
Then that would give you an x dependence here.
03:09
But this is only dependent on t.
03:12
How can that be? how can you have x dependence on the left and just supposedly t dependence on the right? and likewise, if this was dependent on t, how can that be? this is an x.
03:34
So how can this how do these guys coexist the only way to do that is for that to be equal to a constant then there's no problem because if like i said if this had t dependence how you're going to solve for a si of x with a t dependence where to come from likewise if it's at an x dependence you're going to solve for si of t and you have an x dependence where'd that come from it's supposed to be only dependent on time it's supposed to be dependent on x so there's there's a clash.
04:09
There's a clash.
04:10
You get a result that has a t in it also, which is a contradiction.
04:17
So the only way this is going to work is if it's equal to a constant.
04:20
The left side, right side are equal to a constant.
04:23
Now, let me rewrite this left -hand equation.
04:31
I'm going to write it into operator form.
04:33
H -bar squared over 2m, d2, d -2, d -x squared, plus v, and let me not forget the x.
04:47
Here.
04:48
There we go.
04:50
Psi of x is equal to a constant times si of x.
04:58
Just multiply both sides by side.
05:01
Okay, now let's look at this.
05:03
And this should be, this is a c.
05:10
What does this look like? it looks like an eigenvalue equation.
05:16
Remember, this is the hamiltonian operator.
05:19
This is h.
05:20
So this is h.
05:24
Sigh of x equal to some constant times si.
05:28
Hmm.
05:30
H is the total energy operator for a particular system.
05:36
So what is its eigenvalue going to be called? the total energy.
05:42
E.
05:45
So we now know that that constant is equal to the energy of that state.
05:56
So it's the energy.
05:57
So we know that.
06:01
So from all this now, we get two equations.
06:14
First one, h bar squared, 2m, d2 .2 .5 .x.
06:23
X squared plus v, psi of x, e, si of x.
06:30
That's what i'm going to call equation one.
06:33
Second equation, d d t, si of t, is equal to minus i, e, si of t over hbri.
06:54
Now, this one's easy, just integrate.
06:57
You get a natural log, and then to get rid of that, you get an exponential.
07:01
So, si of t, is equal to e, minus i, e, t over h bar.
07:08
There is the time dependent aspect of this.
07:14
Now, you might say what happened to the constant? isn't there supposed to be a constant? well, you got two choices.
07:24
You know that in the end, this is all got to be combined with the psi of x, this guy, and that has been normalized.
07:35
Now, when you have products of complex, complex, quantities put together.
07:44
It doesn't even have, it doesn't have to be complex, but it's true also with their complex.
07:49
When you look at the modulus squared of that multiplication, it is the module square of the first quantity, which would be psi x, times the modulus square to the second quantity, which is phi t.
08:02
Well, so in a way, you're individually normalizing each, but this is already normalized.
08:09
It's modulus is one.
08:11
Its norm is one.
08:14
So if we were to go through it, if we were to say, well, let me put a normalization constant here and do that.
08:21
We'd find it's one.
08:23
It's trivially one because it's already normalized.
08:26
It's already normalized.
08:28
If that bothers you, you then think, if you want to think if there's still a constant here, group that constant in with the constant that comes from the first differential equation here.
08:38
And in the end, you'll normal.
08:40
Normalize that.
08:42
We'll find what that needs to be to have the whole wave function normalized.
08:46
It all leads to the same answer.
08:48
Whatever you're comfortable with, either combine the two constants together or think of two individual normalizations.
08:56
This already normalized, so it's got a constant one in front of it.
09:01
Whatever works for you.
09:04
Okay.
09:04
So we know the time dependency.
09:06
We'll be using this a little later.
09:08
Not just yet.
09:09
Got a little bit of work to do before that time.
09:11
Now here is our well plus infinity everywhere but in the interior, which is from zero to l.
09:25
So everywhere, zero.
09:27
I mean everywhere, infinity except for in the interior.
09:32
All right.
09:37
Solution of one, if, well, let me say it this way.
09:48
4.
09:51
Vx equals zero.
09:54
0, 0, less than x, less than l, infinity, otherwise.
10:04
So that means there's going to be no chance of ever finding it anywhere from minus infinity to 0 and l to infinity.
10:13
It's infinite, infinite potential.
10:18
That means, unlike the finite cases, that's different.
10:22
You'll see that if you've not already seen it.
10:24
This is a, let me fix this, this is a implied sign.
10:37
With the infinity here, potential being infinite, you cannot be ever found at the wall or to its left.
10:50
Likewise, from l to infinity, cannot find it there.
10:57
Think about what would go on.
10:59
You're having effectively an infinite force arise when it reaches the wall, sending it back the other way.
11:07
You cannot penetrate an infinite force.
11:10
With these infinities, you cannot do there.
11:13
Now, like i said, it'll be different when you deal with finite wells, when you talk about tunneling, when you talk about transmission and reflection coefficients, that's different.
11:22
They are finite situations.
11:24
This is not finite.
11:26
This is infinite.
11:28
Okay.
11:30
We'll come back to the boundary conditions in a couple of minutes.
11:33
So let's write the equation out.
11:36
We're going to solve.
11:37
H bar squared over 2m, d2, i'm going to drop the x now.
11:44
I'm going to, well, i mean, let me, i probably should carry it.
11:49
Continue it, seeing that there's going to be so many different symbols.
11:53
And i can rewrite this.
12:06
All i'm doing is multiply, i'm just multiplying by minus 2m over h bar squared.
12:12
2m .e over h bar squared, sine of x, and i will define.
12:19
A constant k here it's not an independent quantity it's defined it's equal to the square root k is equal to square root of 2 m e over h bar square it's a definition of k uh si of x so that's the equation we want to solve well we can solutions through that are either a sum of complex exponentials or sign and cosine either way will lead to the same the sine and cosine will be a little bit faster.
12:45
A little less work to be done.
12:48
You'll get to the same answer.
12:51
A of x.
12:52
I'm not going to use a or b.
12:54
A is used.
12:55
If you look at the problem, description, it's got a in there already.
12:58
So i don't want to do that because i'm going to use c and d for my two constants.
13:03
C -sign kx plus d cosine kx.
13:09
So that's my, that's my, that's my side of x.
13:13
Now the boundary conditions, what you're doing is matching up.
13:17
And this is a zero here.
13:20
You're matching up the wave function to the left, which is zero, with what it has to be from the right, from the interior.
13:33
So, si of zero, equal to zero, and si of l, is equal to zero.
13:44
We're matching up.
13:46
Can't have a discontinuity.
13:50
Otherwise, you'd have two different probability densities at that point.
13:52
Got to have a single probability density.
13:55
Cannot have two different probability densities at a point.
14:01
Okay, so first, side of zero equals zero.
14:08
Sign of zero is zero, so this term doesn't even appear.
14:11
It does not make c zero.
14:13
It just means the term doesn't appear.
14:15
So this becomes d cosine of zero, but that's one.
14:20
So that means d is equal to zero.
14:27
So at this point in time, we have a wave function.
14:33
C, sine, kx.
14:40
Now, let's do the other boundary condition.
14:43
Psi of l is equal to zero, is equal to c, sine kl.
14:50
Now, do we want to make c zero? then we got no wave function.
14:56
There's really no particle that, because you've got no probability of being anywhere.
15:03
So that's like having no particle.
15:04
So c cannot be zero.
15:07
So we've got to find that the, you got to set the condition so at the interior, it will be zero at the wall.
15:17
So that means kl n pi, and n will be one, two, three, and so on.
15:30
Why not zero? because if i made n zero, what do i get here? for any x, psi zero.
15:43
It's back to the thing.
15:44
What's that? represents nothing there's no it cannot be cannot represent a particle it's got be somewhere that be somewhere so no n equals zero so k n equals n pi over l now i can find the energy eigenvalues now for each of the states to m e or h bar squared notice i put a subsequent and indicating now that there is is a set of k values that i can have, not just one.
16:26
And likewise, i'm going to do the same thing for e in a second.
16:32
So e sub n squared, h bar squared, pi squared, 2m, l squared.
16:43
So that is the energy.
16:45
When n is one, we've got a certain energy.
16:47
When it is two.
16:48
And so on.
16:48
We have a collection of of states, bound states.
16:53
These are bound states.
16:55
Okay, so at this point in time, psi n of x is equal to c, sine n pi x over l.
17:08
And you might say, well, i didn't, wouldn't we supposed to get c as a specific value? why is it still there? no, you guys do have one last freedom, the freedom to normalize.
17:18
So if you set everything up, where's the freedom to normalize? you don't have it.
17:23
So c could not have been found.
17:28
Okay, so now, normalization.
17:33
One is equal.
17:33
Now, what is the one to present? you have 10 ,000 measurements of a particle.
17:38
That says that with certainty it's going to be found between zero and l.
17:42
Well, technically between minus infinity to infinity, but there's not going to be outside of the zero to l interval.
17:52
The probability is zero in those other regions.
17:55
So let's not even worry about it.
17:56
So it exists.
17:58
It's going to be found.
17:59
So 10 ,000 measurements, it's going to be found 10 ,000 times between zero and l.
18:04
Don deal, 100 % certainty.
18:06
So that's what the one represents.
18:08
You're 100 % certain.
18:09
You're going to find it.
18:09
It exists.
18:11
Now, but in reality, when you take those 10 ,000 measurements, you're going to find some time, you know, 25 times i found it at, at, let's just say l is, you know, it's obviously large for a well, but i mean, let's not worry about that.
18:28
Let's say l is 3 meters.
18:32
At x equals 1 meter, i found it 600 times.
18:35
At l is 1 .5 meters, i found it 900 times.
18:39
And you look at all the points.
18:43
But isn't that got to add, that's got to add up to 10 ,000 or 10 ,000 times? but it just tells you where you got them.
18:50
But still got to add up to 10 ,000 times.
18:52
You were between 0 and l.
18:54
You didn't find anything.
18:55
You didn't find any measurements outside of that.
18:58
So when you add up 900 and 600 and 5 ,000 and whatever, that comes out to be 10 ,000 out of 10 ,000 between zero and whatever i said, three meters, whatever i said, as the well...