# Fourier Transfrom and expectation value of momemtum operator

1. Feb 17, 2013

### black_hole

1. The problem statement, all variables and given/known data
Using <$\hat{p}$n> = ∫dxψ*(x)($\hat{p}$)nψ(x) and $\hat{p}$ = -ihbar∂x and the definition of the fourier transform

show that <$\hat{p}$> = ∫dk|$\tilde{ψ}$(k)|2hbar*k

2. The attempt at a solution

Let n = 1 and substitute the expression for the momentum operator. Transform the wavefunction and its conjugate. Take out all constants.

$\hat{p}$ = -ihbar/2pi∫dx∫dkeikx$\tilde{ψ}$*(k)∂xdkeikx$\tilde{ψ}$

Here I'm stuck. I tired applying the ∂x operator onto the eikx next to it. That cancels the negative sign, the i, and brings down a k. I thought I could change the the transform of the wavefunction and its conjugate into its norm squared and then I'd be left with ∫dxe2ikx but that integral does not give me 2pi.

have a made a mistake?

Last edited by a moderator: Feb 18, 2013
2. Feb 18, 2013

### clamtrox

You need to be more careful. The two Fourier transformations you do don't have the same index: you integrate over two different k's.

3. Feb 18, 2013

### black_hole

Um, ok. You're probably right. But I'm not sure I see what to do ...

4. Feb 18, 2013

### clamtrox

You will need an identity concerning Dirac delta functions: $$\delta(k+k') = \frac{1}{2\pi} \int e^{ix(k+k')} dx$$

5. Feb 18, 2013

### vela

Staff Emeritus
What clamtrox is saying is that you should start off with
$$\langle \hat{p} \rangle = -\frac{i\hbar}{2\pi}\int dx \left[\int dk\,\tilde{\psi}(k)e^{ikx}\right]^* \frac{\partial}{\partial x} \left[\int dk'\,\tilde{\psi}(k')e^{ik'x}\right].$$ Do what you tried the first time, but this time you'll get something that looks like the identity clamtrox provided above, which will allow you to perform one of the integrals, leaving you with the result you want.

6. Feb 18, 2013

### black_hole

Ok Thanks guys! That makes more sense