00:01
Hello, and in this question here, we'll be looking at how to normalize the wave function and calculate the probability of the particle in some interval x.
00:12
So the wave function we've been given is equal to b times a squared minus x squared, sorry, b times a squared minus x squared, when x is less than a and greater than minus a.
00:27
The wave function is equal to zero in all other cases.
00:34
So we've been asked to find using normalization this real number b in terms of this real number a.
00:44
Okay? well, what does the normalization condition mean? well, this means that the probability of finding the particle in a region from minus infinity to plus infinity is equal.
00:59
To 1.
01:00
Well, how do we write this? well, the probability of finding the particle in a region from minus infinity to plus infinity is given by this integral here, where we're taking the complex conjugate of the wave function, multiplying it by the conjugate, and integrating it from minus infinity to plus infinity.
01:23
And this must equal 1, because we know the particle must be located somewhere between plus infinity and minus infinity.
01:31
So if we were to compute this integral, we'll be able to find b in terms of a.
01:38
So to do this, i'm first going to break up the limits of integration in this integral.
01:45
So the left -hand side is equal to the integral of the wave of the...
01:52
So one thing just to note before we start is that in this case, size squared is equal to the complex conjugate of si times si.
02:02
But we're told in the question that both a and b are real so si the complex conjugate of si is equal to si so this means we're just integrating the wave function to be squared so this here is equal to the wave function to be squared from minus infinity to plus infinity now rewriting over different limits of integration so we're gonna say we start from minus infinity and go to minus a the integral of si square dx and now we're going to integrate it from minus a to plus a and we're integrating si squared dx and finally we integrate from a to plus infinity of si squared dx now as you can see here we we've integrated this this wave function to be squared in three separate places and the reason why we've done this is because the wave function in the region from minus infinity to minus a is equal to zero.
03:14
Similarly, the wave function from plus a to plus infinity is equal to zero.
03:21
So an integrate, the integral of zero is equal to zero.
03:25
So the only bit of this integral on the left hand side that will contribute is the integral from minus a to plus a.
03:36
So we just need to evaluate the integral from minus a to plus a of si squared d x well this is equal to squaring outside it's going to be b squared times a squared minus x squared all to be squared d x expanding this square x and moving the constant b outside the integral it's going to be equal to b squared times the integral from minus a to plus a of a to the power of four plus x to the power of four minus two a squared x squared d x now when we're integrating something all we do is we increase the power by one and divide by the new power so this gives us b squared multiplied by a to the power of four times x plus x to the power of 5 divided by 5 minus 2 to the power minus 2 times a squared times x to the power of 3 divided by 3 and we evaluate this at a to from minus a to a so filling in from a to minus a to minus a we get b squared is equal to a to the power of 5 plus sorry so it would be 2a to the power of 5 because there one one 1 a to the power of 5 will come from the plus a contribution, and the second a to the power 5 will come from the minus a.
05:16
Then the next term is twice, a to the power of 5 over 5, and the final term is minus 4 over 3, a to the power of 5...