00:01
Hi there, so for this problem we are told for part a to normalize the wave function from problem 9 by adjusting the value of the multiplicative constant a so that the total probability of finding the associated particles somewhere in the region of the length a equals 1.
00:32
So in this, to normalize the wave function, we evaluate the following integrals.
00:39
So we will have that that is equal to 1, from minus a divided by 2 to positive a divided by 2.
00:48
And this is the conjugate of the wave function times the wave function integrated over x.
00:55
Now we know that the wave function is equal to zero upside this region.
01:06
Now, the wave function in the region is going to be equal to a sine of 2 times pi times x, and this divided by a, and this times the exponential of minus i times the energy times the time divided by h bar.
01:31
And then if we substitute this into the previous expression, we will have that 1 is equal to 2 times a square times the integral from 0 to a divided by 2 of the sign square of 2 times pi times x and this divided by a integrated over x so this is equal to a divided by pi times a square the integral from 0 to pi of the sign square of u of the differential in u.
02:12
So in here we obtained that this is equal to a divided by pi times a square times pi divided by two after we do the the integral...