00:01
In this prime, you're given a wave function that is different, x greater and equal to zero.
00:09
We have this, a x, a .m.
00:12
Minus a, little a x, and zero for negative x values, x less than zero.
00:17
And the first thing we want is the normalization.
00:20
Find a.
00:21
So we use normalization.
00:23
Remember, what is normalization? that basically says, you're asking the question, is this particle somewhere? the answer is yes, with certainty.
00:33
So the probability is one.
00:35
So you add up all the probabilities of it being everywhere else, anywhere in the universe, every single x, from minus infinity to plus infinity, and that's got to give you one.
00:49
So in general, this would be what you write for the normalization.
00:57
But we have a restricted, our wave function for minus infinity to zero is zero.
01:02
So this can be rewritten as zero to infinity.
01:06
And i'll put in the complex, the modulus squared of a.
01:10
This could be a complex number.
01:13
So it might as well be consistent.
01:16
X squared, a .m.
01:18
2ax, dx.
01:20
So that's what we have.
01:24
And this turns into modules of a squared, 0 to infinity, x squared, ed .m.
01:34
2ax, dx.
01:35
So that's what we got to work on.
01:37
The a squared being a constant, we just bring it out.
01:39
Now, how do we proceed? integration by parts.
01:44
U is equal to x squared.
01:46
D u is 2x -tx -d -v, eid minus 2 -a -x -t -x, v is e -2a -x over minus 2a2a.
02:05
Remember the formula for integration by parts, u -d -v is equal to uv minus v -d -u.
02:14
That's the integration by parts.
02:17
So we get from that, 1 is equal to modules of a squared.
02:22
We're going to get from this, uv, minus x squared over 2a, e to minus 2ax, 0 to infinity, minus the integral 0 to infinity, e to minus 2ax or minus 2a, 2a, 2a, 2x, 2a, 2x, 2x, 2x, 2x, 2x, 2x, so that's what we get.
02:50
Now, certainly, you see when x is zero, this term is gone.
02:56
This is one, that's zero, so that term is gone.
02:58
When it's infinity, you might say, wait a second, i got infinity here.
03:01
I'm going to have infinity on the bottom.
03:03
But if you were to look at this limit as x goes to infinity, when you use lopatiles rule, you're going to end up getting in zero.
03:14
Because this never goes away when you take derivatives of the denominator.
03:17
It's always going to be an exponential down there.
03:19
This is going to go away.
03:21
So this term is gone both of it.
03:24
So we are left with modules a squared over a, zero to infinity, x, e to minus 2ax, dx.
03:40
Integration by parts again, u is equal to x, du is equal to dx, dv is equal to e minus 2ax, dx, same exact.
03:53
E minus 2ax or minus 2a.
03:56
Exactly the same as what we just did for that one.
04:00
So i get 1 is equal to modulus of a squared over a, that's the lead part.
04:09
Then we get x minus x over 2a at a minus 2a so this is again going to be gone minus the integral 0 to infinity em minus 2a x over minus 2ax over minus 2 and i should mention in case you didn't see it, why the two went away because of the two in the du.
04:35
That's why there's no two out here.
04:38
E to minus and then minus 2a here, dx.
04:43
So, what do we get from this? modulus of a squared over 2a squared, integral, 0 to infinity, a minus 2ax, dx.
05:01
So we've reduced it to something simple.
05:04
So 1 is equal modules of a squared or 2a squared.
05:11
And we're going to get from this, e to minus 2ax.
05:16
We've done this over now the third time, minus 2a, 0 to infinity.
05:23
So i'm going to get minus modules of a squared over 4a cubed, e to minus infinity, minus infinity, minus 1.
05:37
E minus infinity, as we've talked about, that's zero, one over e to infinity, it's zero.
05:44
So that's gone.
05:46
So we're left with modules of a squared over 4a cubed.
05:52
And i'm going to choose a to be real.
05:55
So a is equal to 2a3 afts.
06:01
So that is the normalization constant.
06:03
So our wave function now becomes 2a3 -half x, e .m minus a x, and 0, x gradin equals 0, x less than 0.
06:27
Notice, when x is 0, both wave functions, both regions are matching up 0 and 0.
06:35
So we have no issues.
06:37
We have no issues of continuity at that boundary at x equals 0.
06:44
This isn't a.
06:50
Okay.
06:57
Question two.
06:59
We are looking for what vx is and what e is.
07:03
We're told we are in an eigenstate of the hematonian.
07:07
Remember, that means h -si is equal to e -si.
07:13
E is the energy eigenvalue.
07:14
So we're in a stationary state, an eigenstate of h.
07:18
And h, p squared, 2m, psi -x, plus v, x.
07:27
Psi x, e, si, si, x.
07:34
So, remember, e's a constant.
07:37
I can value.
07:39
The operator for p, or px, to be precise, is i .h bar, d, d, x.
07:45
And i'm sure you've seen this form.
07:47
And i'm going to write in a second, h -bar squared over 2m, d2, dx squared, psi -x, plus vx.
07:58
This is the time -independent scha -lander's equation.
08:03
This is the form you get when you have a stationary state.
08:06
That means you're going to stay with the energy value of e.
08:23
Okay, so that's what we've got to work on now.
08:26
Really, the only thing we've got to work on is derivatives.
08:28
So let's take it one step at a time.
08:31
Decide the x and you can just see it up there.
08:36
So i'm going to leave the two a to three halves outside.
08:39
It doesn't, it just gets in my way if i do anything else with it.
08:43
And so i'm going to have to take the derivative first of the x.
08:50
So i get e to minus ax.
08:52
Then i take derivative of e to minus ax...