00:01
All right, so we have the first two wave functions of the quantum harmonic oscillator, and they have the form, so m omega over pi h bar, 4 through to that.
00:14
And the function here is going to be e to the minus m omega x squared over 2h bar.
00:23
And then our next wave function is similar, but it's not as pretty.
00:31
So this should be equal to square root of two times, let's see, n cubed, times omega squared over pi hbar cubed.
00:48
And this is the fourth root of that as well, times x times the same exponential we encountered in the first one.
00:57
Okay, so if it took me that long just to write out the equations, you know this is going to be, you know, quite a problem.
01:02
So now we want to ensure that both of these are normalized.
01:06
So, of course, the normalization and condition is that the integral from negative infinity to infinity of psi star times si, respect to x is, you know, equal to one.
01:20
So for the first wave function, si -zero, when we integrate this from negative infinity to infinity, let's move, so let's do si.
01:34
Now, si is real.
01:35
So basically we don't need to worry about any complex conjugates.
01:38
We're basically just integrating this term.
01:41
And so this is going to be equal to m omega over pi h bar to the one -half when we square it, times the integral from, sorry, negative infinity to infinity of e to the minus m -omega -x squared over h -bar.
02:01
So not h -bar times 2, but just h -bar by itself.
02:06
And this is a familiar gaussian integral and the solution of that so pi so here's our initial constant the solution of this integral here is going to be the square root of pi divided by the coefficient of x squared so what that comes out to be is pi h bar over the coefficient of you know pi times h bar over m omega and you can see that's exactly what we have here so this doesn't be equal one so that checks out and now let's do this for the next wave function.
02:40
So once again, it's real.
02:41
So we don't have to worry about complex conjugates and things like that.
02:45
So we're going to be integrating psi squared with respect to x.
02:49
And so we're going to get a two times, what is it, m cubed, omega cubed over pi h bar cubed to the one half.
03:03
Let me make sure i wrote that down the first time.
03:04
I wrote this as an omega squared.
03:06
This should have been an omega cubed.
03:08
I think.
03:12
Let me double check from the notes.
03:14
Anyway, so it's this times the integral of the function, which is x squared times e to the minus m omega x squared over h bar.
03:30
And we're integrating with respect to x.
03:32
Now there's a little bit of trick we can use, which is to note that x squared, or sorry, x squared times e to the minus m omega x squared over h bar is the same thing as taking the partial derivative of this term with respect to the mass if we multiply by the right constants to cancel this so if we multiply by h over omega then when we take this partial derivative we're just or sorry there should be a negative sign as well we'll just produce x square so what i'm going to do is write this integral then that we were just doing as two times m cubed omega cubed over pi h bar cubed to the one half times these other constants i have so h bar over omega and now we're going to take the partial derivative with respect to m of this integral which is now just a regular galsene integral we don't have an x squared term inside the integran which makes things a lot easier because we are we've already computed this integral from the previous part.
04:42
So this is two, let's see, two times this constant over pi h bar cubed, square root of that times h bar divided by omega.
04:58
And we're taking the partial derivative of this integral, which evaluates to the square root, remember of pi over the coefficient of x squared.
05:06
So it is just pi h bar over m omega square root.
05:11
This is what we're taking the derivative of.
05:14
So the only part we care about in this derivative is the actual term with the m inside.
05:18
So i can bring all the other constants out front.
05:22
So it's going to take a second, but this is m cubed, omega, cubed, over pi, hbar cubed, so that's the first term.
05:31
And then we have h bar over omega times pi, h bar, omega, square root.
05:41
So we can absorb these two terms into the square root by the way by just cubing them inside the square root sorry this is never running his line it's times that and then times the partial derivative the remaining part with respect to m well that's just going to give us a negative one -half m to the negative three halves and so if i finally combine all this the one half the negative one -half cancels with the negative two at the start you've got m cubed omega cubed over pi h -bar cubed uh this is to the one -half i forgot to write the one half on the previous part and then times pi h bar cubed this is m to the negative three half so if we bring it inside of a square root it becomes an m cubed and lo and behold these are just reciprocals of each other so this once again equals one okay so that is only part a we have much more laborious calculations to go next we want to calculate the expectation value of x in each of these two states.
06:44
Now, admittedly, this is actually a little bit easier, even though it might not seem that way at first.
06:48
So the expectation of x in the 0th state is going to be m omega over pi hbar to the one -fourth times the integral of negative infinity to infinity of x times e to the minus m -omega -x squared over h -bar.
07:13
And this is integrating with respect to x.
07:15
Now, the reason this is easy is because this x is an odd function.
07:21
This next term is even.
07:23
And hopefully you know that whenever you integrate an odd function, the product of this is odd.
07:28
Whenever you integrate an odd function over a symmetric interval, you always get zero.
07:32
So we actually don't really have to do much more in the interval other than that.
07:37
Likewise, when we do this for the other state, so this becomes 2 over m cubed omega cubed over pi hbar cubed to the one half sorry this should be one half not one fourth but not that it matters because it's still zero times the integral of negative infinity to infinity now it's of x squared times this so m um or sorry x cubed because when we square side we get an x squared term inside the integral when we then multiply x, get an x cubed.
08:14
So we should have something like this.
08:16
And by exactly the same argument, this whole function is odd.
08:21
And so it's integrated from negative infinity to infinity.
08:23
So it's just going to be zero again.
08:26
All right.
08:27
Next up, we want to evaluate the expectation of the momentum operator in each of these states.
08:31
This is a little bit harder.
08:33
So this is going to be negative ih bar times m omega pi over h bar to the one -half comes the integral i'm sorry from negative infinity to infinity and here this is si times the momentum operator we've already factored out the negative ih bar in front of the integral but um this will be negative m omega x squared over two h bar times the derivative of the next term respect to x okay so we continue along then we take the derivative of the right -hand term it's going to reproduce the exponential but we're also going to have a new term of m omega x over h bar and this is integrated with respect to x so notice i changed this exponential no longer has the two in the denominator because i've combined the exponentials after i took the derivative so that's why in my have looked like i skipped a step, but i didn't.
09:52
So anyway, this becomes like i, h -bar, times a bunch of stuff, m -omega over pi h -bar.
10:04
In fact, this would be like m -qued, omega -qued, over h -bar -qued.
10:09
This is to the one -half.
10:11
This is now the integral from negative infinity to infinity of just x times m -o -m -squared x over h -bar.
10:20
And conveniently this is also zero because once again odd function over a symmetric interval.
10:27
Another way you can kind of see this is like this exponential term is like the probability distribution for a normal distribution.
10:35
And if you know anything about normal distributions, you know that they look like negative x minus mu over two sigma or mu is the average value.
10:45
So if mu is zero, then you just get an x squared over two sigma, which is kind of like the same kind of term we have.
10:50
Here, which once again indicates, like, if there's no average value to this, then this integral is going to be zero.
10:59
Maybe that helps me.
11:00
That doesn't.
11:00
But then let's do the same thing for the other wave function.
11:05
So this is going to be negative ih bar times two times m cubed, omega cubed, pi h bar cubed to the one -half times this integral, and this integral is a little bit worse because we have linear and exponential terms in the weight function...