00:01
In this problem we are given a particle that is represented by this following normalized wave function.
00:10
So we have si of x equal to, so it is piecewise continuous or piecewise defined to be more general.
00:21
Some constant times 1 minus x squared or a squared for x between minus a minus a, and a and it is zero otherwise.
00:42
With that we have four parts.
00:45
We are going to compute first this normalization constant capital a, then we will compute the expectation value or the average of x squared and also the average of p squared, the momentum squared.
01:03
Finally we will show that these operators, x and p satisfy the heisenberg's uncertainty relation.
01:14
Okay, let's get started with the first part, namely the normalization.
01:19
So the general rule says we should have one.
01:25
Okay, let's cite more clearly.
01:27
When we integrate our wave function everywhere in the, between the infinities for the entire space.
01:40
So since we have a one -dimensional space here, we should integrate our wave function from minus infinity to infinity.
01:49
But our function is defined piecewise, so it is, it is non -zero only in a certain range.
01:58
So we can split this integral into three parts from minus infinity to minus a, dx, si squared, plus from minus a to a, d x side squared plus from a to infinity d x side squared okay i'm going to write this only once to show the reduction of the integral limit integration limits for the rest of the problem okay so in this first first range from minus infinity to minus a our way function is zero in the last piece we have also a vanishing wave function so beyond this value x equal to a our wave function vanishes again we have only this non -zero piece between minus a and a so we have from minus a to a dx okay let's take the score of this wave function a squared one minus x squared or lowercase a squared squared here the usual assumption is that we assume this normalization constant is a real number.
03:32
So that's why we ignore these absolute signs most of the time because it is convenient.
03:41
Okay.
03:43
Now let us bring this constant outside the integral and also let us observe that this integrant is even or symmetric about the origin, namely about the point x .e.
03:57
Equal to 0.
03:58
So i can put this factor 2 and divide or consider only half of the integral.
04:07
So from 0 to a dx, 1 minus x squared over a squared squared squared.
04:17
So let's expand this integrant from 0 to a dx.
04:21
We have 1 plus x to the power 4 divide by a to the power 4 minus 2 x squared over a squared.
04:33
So let's do this integral turn by turn.
04:37
We get x plus x to the power 5 over 5 a to the power 4 minus 2 x cube over 3 a squared from 0 to a.
04:54
The lower limit gives 0, so we can consider the upper limit only.
05:00
So we have 2a squared.
05:03
Notice that we will always end up with a single power of a.
05:07
A here, a here, a here.
05:11
To be consistent with the units, that's important.
05:14
So i can directly write this overall constant as a.
05:19
Now i have a numerical part, 1 plus, 1 5th, minus 2 thirds.
05:28
So if we carry out this calculation, you will obtain 16 a over 15 times.
05:36
Xx4 so this should be equal to 1 and we have a equal to 15 over 16 a under the square root so this is our normalization factor now let's do part b here we are going to compute the expectation value or the average of the operator x squared so the rule is if we are given a function of x and if you are representing if you are explicitly using some wave function for our particle then we just have this integral over the appropriate domain for this particle we have the wave function squared times whatever that function is it can be x squared or some other crazy function of x it doesn't matter but we will see that when we consider the momentum things will be a bit different.
06:44
Okay, so let's attack this problem now.
06:48
We have from minus a to a, dx, wayfunction squared, capital a squared times one minus x squared over a squared squared squared, square times x squared.
07:06
Let's bring this normalization factor outside and expand the intake, integrate but before that okay let's also do the same trick we have this integrant symmetric about the origin so if you let x to minus x the integrant we will be the same so i can consider only half of the integration by including a factor of two outside so from zero to a i have dx one minus x squared or a squared squared x squared using or keeping or having a having zero as the lower limit simplifies our calculations eras so that's why we like to emphasize this symmetry about the origin okay so we have two a squared from zero to a d x okay let's expand this integrant we have one plus x to power four over a to the power 4 minus 2 x squared over a squared times x squared 2 capital a squared from 0 to a d x squared plus x to the power 6 divide by 8 to power 4 minus 2 x to the power 4 divide by a squared let's do this integration turn by turn we get x cube over 3 plus x to the power 6 plus x to the power 7 over 7 a to the power 4 minus 2 x to the power 5 over 5 a squared from 0 to a the lower limit gives 0 again so we can consider the upper limit only we have 2 capital a squared and notice that we can factor out some a's in this parentheses here we will always have an a -cube turn so a cube here, 7 minus 4 a cube here, 5 minus 2 a cube here.
09:33
Let's do that.
09:35
And the rest will be just some numbers.
09:38
We have 1 3rd plus 1 over 7 minus 2 over 5.
09:47
Okay, so let's also expand the a square, the capital a squared expression.
09:56
We have 2 times 15 over 16.
10:01
A times this a squared term times the rest of the numbers it goes like 8 over 105 so if we simplify this expression we obtain 1 over 7 times a squared so this is the expectation value of x squared now let's do the expectation value okay part c not b now let's so the expectation value of the momentum squared operator, p squared.
10:45
Now we have to make a sandwich using our wave function.
10:51
Again from, again in this available region or the relevant legion, the region relevant to our wave function, namely from minus a to a, we have the conjugate of the wave function times this operator, we know that momentum corresponds to a differential operator in quantum mechanics.
11:19
It is not just a usual function of x...