00:01
In this problem, you have a linear superposition of the ground and first excited state of a harmonic oscillator, and they have equal chance, equal probability, which would actually be in both cases, would be a squared, would be the probability of both.
00:23
Now, the first thing it wants is to find a based on that, the way we constructed.
00:28
So we normalize.
00:29
1 is equal to minus infinity to infinity, si star x t, si xx, and this should not be a t yet, this is just zero.
00:47
Later on we'll get to the later times, but this is just zero right now.
00:54
Okay, and let me fix this, so it looks a little better here, there we go.
01:05
And this becomes, you're going to have a star a, is the modulus of a squared, so that can just be factored out in front.
01:16
Behind it's infinity.
01:18
Infinity.
01:20
It's just a matter of multiplying the complex conjugate times the original.
01:26
So, si star, zero.
01:29
Now when i write now, from now on, when anything's got a zero or one on the side, that's referring to just the x dependency.
01:38
The time dependency, when it comes in, will be written explicitly.
01:41
So it's just a spatial part of the state.
01:46
So, si -not star, si -not -d -x, minus infinity to infinity, psi -star, psi -1 -d -x, plus minus infinity to infinity, psi -1 -star, psi -0 -d -d -x, plus minus infinity, to infinity, psi 1 star xx now the eigenstates the harmonic oscillator are orthormal.
02:29
Normal meaning they have unit norm and ortho stand for orthogonal to each other.
02:36
So remember this is an inner product.
02:42
Si 0 star this inner go will be 1.
02:51
Si 0 star, side 1, this inner star, side 1.
02:54
This integral will be zero.
02:58
So this is going.
03:01
Likewise, this is zero, and this will be one.
03:09
So, when all is said and done, we get two times the module is a squared, and i'm going to choose a to be real.
03:20
Don't have to get fancy.
03:25
A is equal to one over the square root of two.
03:29
So our current status now, x0, 1 over the square root of 2, psi 0 plus side 1.
03:41
So there's our normalized wave function that we're in.
03:49
So that's part a, part b.
03:54
Now we've got to get xt for some later time.
04:03
And that, 1 over a square root of 2, is really, if you have a supposition at t equals 0, that superposition is not changed any later time.
04:12
It just now has a time dependence to it.
04:17
Size 0, if you're solving shoulders equation for size 0 state as a spatial and time dependence, and so on you get size 0, e minus i, e0, t, h bar, plus side 1, e minus i, e1t over h bar.
04:43
So that is the time dependence.
04:46
So you're really just appending the exponential with the appropriate energy connected.
05:02
All right.
05:04
Now, we can clean this up a little bit more for a harmonic oscillator, e .n plus one half, h bar omega, applies e0, h bar omega over two.
05:20
E1 is equal to three, h bar omega over two.
05:29
So our current status now, psi xt, 1 over square root 2, si 0, e.
05:43
Minus i, omega t divided by 2, plus psi 1, e .mise i 3, mega t divided by 2.
05:57
So there is our, those the h bars go away, because e has the h bar and it's the h bars go away.
06:03
We just pick up the two on the bottom and the omega on the top.
06:09
Okay, now we are going to need the probability density.
06:18
1 over square root 2, si 0 star e .i, omega, t divided by 2, plus psi 1 star, e to i, 3, omega, t divided by 2.
06:44
So i have to complex conjugate all of the terms.
06:49
And then we got one over a square over two again.
06:52
And now the original sign -a -e -e -m -i omega -t over 2 plus i -1 e -m minus i 3 omega -t over 2 and now it's just a matter of expanding everything multiplying everything through so i'm going to get a one -half out in front now when you do term -by -term size 0 stars size 0 e i omega -t over 2 e minus omega -t so this should become size 0 star, size 0.
07:37
Now i get size 0 star with side 1, but notice the omega terms do not cancel out.
07:44
So though the original states were stationary states, this linear super position is not a stationary state.
07:54
The probability density does depend on time.
07:59
Size 0 star, side 1, e minus i, omega -t you got minus three plus one so that's minus two divided by two is minus one then we're to the second term side one star size zero e now you get three minus one divided by two so two divided by two is one so this becomes i mega t and then we got the psi 1 with the 3i and minus 3i are giving you a unit norm out of that.
08:44
So, si 1 star, psi 1.
08:49
Okay, there's what we have for probability density.
08:55
That's one of the things you were asked, these last two things here.
09:00
Let me circle them for you.
09:04
Those were the, this is what was asked.
09:09
In part b, i think i put in the, this is really what was asked for.
09:18
There we have that.
09:20
Okay.
09:22
All right.
09:24
C wants the expectation value for x.
09:29
And that's minus infinity to infinity.
09:33
Sigh star xt.
09:37
X.
09:37
X.
09:38
X.
09:38
Sigh t.
09:45
X.
09:46
That's the definition of the mean expectation value, the mean, average.
09:53
But x is not a differential operator.
09:57
So the order does not.
09:58
Matter.
09:58
So this can be rewritten as minus infinity to infinity x.
10:04
And then the probability density, si x t, the modulus of that squared.
10:11
So we have that.
10:12
That's why we, that's why they asked you to write it.
10:15
Okay.
10:16
Now, could i write out every conceivable thing here and, you know, all gory detail? yes, it's fine.
10:27
But let's look at some things.
10:30
We're going to have four interregnable things.
10:31
We're going to have four integrations x times size 0 star size 0 and then obviously each of these four terms so let me just look at those separately and see what i can get as infinity infinity x now why is it zero this is an odd function.
11:00
X is odd.
11:01
So if f of x equal to x is what is, you know, odd is what? f of minus x minus f of x, even f minus x is equal to f of x.
11:31
So if you have over symmetric limits and minus infinity is symmetric, if you have an odd function for the integrand, unless you have a dx in there, you have an odd function for the integrand, you have equal amounts of, negative area and you have positive area under the curve.
11:49
So zero, get zero.
11:52
So this term here will not contribute at all.
12:03
And you might say, well, how do you know that size zero? if this is odd, this, i should really say, this is odd, this is even.
12:16
If i had two odds, then i get even out of it.
12:18
This is even.
12:21
I'll show you the form of it to make it explicit.
12:30
So i'll be showing you size 0 and side 1.
12:33
But actually for a harmonic oscillator, any state it's going to be, well, what is this? this is the expectation value.
12:50
If you were in size 0, expectation value, if you're in that state, it's zero.
12:56
Expectation value of x in any state, any eigenstate, i should say, is zero for the harmonic oscillator.
13:12
This again, odd, even.
13:19
So that will be zero.
13:20
So we've taken four big integrations and reduced them to two.
13:26
Now let me write out the expressions.
13:29
Size zero, m omega, pi h bar, one quarter, e minus m, mega x squared over 2 h bar you can see it already put it minus x in here x squared it's even si 1 now these i could have written in more general forms using the hermite polynomials that's uh that's don't need to go i mean that that um fancy here i'm just going to use the results so i'm just writing it down.
14:37
Those are not complex exponentials.
14:40
This is actual, this is e, there's no i in this, so remember that.
14:46
Okay, there we go.
14:49
There's sie 0 and sye 1.
14:56
Now we have to put all this together.
15:00
So we're going to be doing the sie 0 star, side 1, that integration.
15:19
All right.
15:19
Now, when you put all this in, actually i could let me, clean this up one.
15:28
Let me go back and clean this one up just a little bit more.
15:33
1 .0.
15:34
Pi, one quarter, m mega, h bar, 3 quarters, square root 2 x, e minus m mega x squared 2h bar.
15:57
This is just 2 over, square root of 2 with square root of 2 and so on.
16:01
So i've just cleaned it up a little bit.
16:04
All right, all right, where i would.
16:06
Here.
16:07
Okay.
16:08
Now, putting these together, putting these together, notice this term and this term, other than the pie, got a pie of one quarter, pie of one quarter, and then we've got a square a two, so i'm going to get here, square root of two over a pie.
16:28
That takes care of the pie and this square root of two.
16:30
Now we got the m omega over h bar, one quarter, m omega over h bar three quarters.
16:35
Well, add them up.
16:37
One quarter plus three quarters is 4 over 4.
16:46
1m omega over h bar.
16:50
It's infinity to infinity.
16:53
Now we get an x from the size 0, side 1.
17:00
This has no x.
17:01
This has an x, but we've got an x from the operator x to do the expectation value.
17:08
So i'm going to get an x.
17:09
X squared, e...