Fourier Transfrom and expectation value of momemtum operator

[black_hole]

Homework Statement

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∫dke^ikx\tilde{ψ}*(k)∂_xdke^ikx\tilde{ψ}

Here I'm stuck. I tired applying the ∂_x operator onto the e^ikx 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 ∫dxe^2ikx but that integral does not give me 2pi.

have a made a mistake?

 

[clamtrox]

Here I'm stuck. I tired applying the ∂_x operator onto the e^ikx 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 ∫dxe^2ikx but that integral does not give me 2pi.

have a made a mistake?

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.

 

[black_hole]

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

 

[clamtrox]

You will need an identity concerning Dirac delta functions: \delta(k+k&#039;) = \frac{1}{2\pi} \int e^{ix(k+k&#039;)} dx

 

[vela]

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

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.

 

[black_hole]

Ok Thanks guys! That makes more sense