What is the Momentum Probability for a Gaussian Wave-Packet?

[Campbe11]

Homework Statement

A free particle at time t=0 has the Gaussian wave-packet:

\Psi(x,t=0)=Ae^{-\tfrac{x^2}{2\sigma^2}}e^{ik_0x}

(a) What is A?
(b) What is the probability of measuring a momentum in the range between p
and p+dp?

Homework Equations

(a) \int^{\infty}_{-\infty}|\Psi(x,t)}|^2 dx=1

(b) \langle p\rangle=-ih\int^{\infty}_{-\infty}\left(\Psi^\ast\frac{\partial\Psi}{\partial x}\right)^2 dx

The Attempt at a Solution

(a) I think this is correct for A.
A=\frac{\sigma^2}{2\pi}

(b) This is where I'm having trouble. I tried evaluating this integral but it seems wrong:

\langle p\rangle=-ih\int^{p+dp}_{p}\left(\Psi^\ast\frac{\partial\Psi}{\partial x}\right)^2 dx

where
\Psi=\frac{\sigma^2}{2\pi}e^{-\tfrac{x^2}{2\sigma^2}}e^{ik_0x}

Please help I have a question similar to this on an exam this Monday!

 

[fzero]

\langle p\rangle=-ih\int^{\infty}_{-\infty}\left(\Psi^\ast\frac{\partial\Psi}{\partial x}\right) dx

is the most likely value of the momentum that would be found in a measurement. In order to determine the probability of a specific value of momentum, you need to find the wavefunction in momentum space, \psi(p). There's a simple expression for the probability of measuring a momentum in the range between p and p+dp in terms of \psi(p).

 

[Campbe11]

Thanks I think I'm on the right track now. So I've done the Fourier transform to get into momentum space and I'm left with this:

\psi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}\Psi(x,t=0)e^{\tfrac{-ipx}{\hbar}} dx

But (I think) that reduces to:

\psi(p)=\frac{\sigma^3}{2\sqrt{\hbar}}

Which doesn't make sense to me since I think we need to use

P=\int^{p+dp}_{p}|\psi(p)|^2dp

to get the result and if I plug what I got in I just get

\frac{\sigma^6dp}{4\hbar}


which is def wrong. What am I doing wrong? Plz help!

 

[arkajad]

Calculate carefully the Fourier transform. You have a Gaussian wave packet, and Gaussian wave packets have a peculiar property that they are Gaussian in momentum space as well.