# Wavefunction questions

1. Nov 18, 2013

1. The problem statement, all variables and given/known data
Ok the problem is this bad boy:
ψ(x) = \begin{cases}
A & \text{for $|x| < d$} \\
0 & \text{for $|x|>d$} \\
\end{cases}

2. Relevant equations

(a)Find a value of A which makes the wavefunction normalized.
(b)Find the momentum wavefunction ψ'(p).
(c)what is the relationship between d, the half width of the position probability distribution and the half-width of the momentum probability distribution. is this relationship consistent with the Heisenberg uncertainty principle?
(d)If the momentum of the particle is measured, what is the probability of it being observed moving to the right?

3. The attempt at a solution
(a) Well to normalize it I tried to break up the peicewise, such that I would get two integrals, one from neg infinity to d the other d to infinity. the one from d to infinity drops out because of the zero. so I am left with an integral that diverges? I dont know where I went wrong.

(b)no idea how to do this.. please start me off.

(c) Perhapes once I get the momentum wave function this will be become apparent? if not can someone start me off?

(d) Do I just apply the uncertainty principle? ΔxΔp>=hbar/2 ??

thanks for all the help!!!

2. Nov 18, 2013

### Dick

Just exactly how did you get an integral that diverges for a)? ψ(x) is also zero for x<-d. I'm assuming this is a 1 dimensional problem. Show your work, ok?

3. Nov 19, 2013

∫(from -∞ to d) A^2dx = 1

This integral has no answer and no way to solve for A?

4. Nov 19, 2013

### Dick

The problem says ψ(x)=0 if |x|>d. That means if x<(-d) then ψ(x)=0. You only integrate A^2 from -d to +d. Do you know what |x| means?

5. Nov 19, 2013

yes aboslute value.

So A = i/sqrt(2d) ?

6. Nov 19, 2013

### Dick

I would have said A=1/sqrt(2d), not i/sqrt(2d), but that works as well. But why did you put the i in?

7. Nov 19, 2013

dont know... haha. YA I got the answer now thank you!!

so part b says Find the momentum wavefunction ψ'(p).

How can i do this?

8. Nov 19, 2013

### Dick

The momentum wavefunction is related to position wavefunction by a Fourier transform. You might have to do some research on this if you didn't know that.