00:01
So we start with the wave function for a particle in a box, and we'll assume that it is a sign dependence, basically a sign function.
00:11
And we're going to calculate the quantities that are needed for the uncertainties in the uncertainty principle, namely the mean or expectation values of x, x, x squared, p, and p squared.
00:25
And so in doing so, this will allow us to get the uncertainties.
00:31
And we'll see what happens when n approaches zero, basically when n is equal to zero.
00:39
So we'll start with x, with its expectation value, and we'll see what happens basically in getting a contradiction for the uncertainty principle.
00:52
So the expectation value is obtained by putting the operator x in the integral times, in this case the square of the wave function, and we obtain the following integral, which you can obtain just by looking at a table of integrals.
01:10
It's kind of a complicated integral.
01:12
You can also do this integral with integration by parts.
01:16
But this is the integral we obtain on the interval from zero to l.
01:29
And i simplify this further.
01:32
In the next line, i'm going to go ahead and scroll down.
01:51
And we'll basically just keep going and we'll get the final result.
01:55
And we see that for the expectation value of x, it's really just just l over 2, which is what we would expect for a particle on a box of length l.
02:07
So on average, a particle is halfway along.
02:11
It's somewhere in that interval, on average, halfway.
02:15
So now we'll go on and get the expectation value for x squared.
02:21
So we just substitute in x squared inside the integral and multiply by the square of the wave function and integrate from 0 to l.
02:28
And we get again another complicated integral, but we can use a table of integrals or use integration by parts, and we get the following.
02:47
So notice in this integral that a lot of it will go away.
02:52
You have the sign multiplying this quantity in parentheses involving x squared times l and l cubed, but because of the sine function, it's going to be zero at x equal to zero as well as x equal to l.
03:08
And you'll know that because you have a 2 pi times in inside of the sign function.
03:15
So automatically those terms will vanish because they multiply by that sign function.
03:26
And again, all of that is evaluated at 0 and l.
03:30
But this greatly simplifies.
03:32
And we have the following that comes about.
03:48
And this is the final result we get...