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.

 

[paranormal]

Hello,

Thank you for your question. Let me try to provide a response to the content you have provided.

(a) A is the normalization constant, which ensures that the wave function is properly normalized to 1. Therefore, we can calculate A using the normalization condition:

\int^{\infty}_{-\infty}|\Psi(x,t=0)|^2 dx=1

Substituting the given wave function, we get:

A^2 \int^{\infty}_{-\infty}e^{-\tfrac{x^2}{\sigma^2}}dx=1

This integral can be evaluated using the Gaussian integral formula, which gives:

A^2\sqrt{\pi}\sigma=1

Therefore, the value of A is:

A=\frac{1}{\sqrt{\sqrt{\pi}\sigma}}

(b) The probability of measuring a momentum in the range between p and p+dp is given by the expression:

P=\int^{p+dp}_{p}|\Psi(p,t)|^2 dp

where \Psi(p,t) is the momentum representation of the wave function, given by:

\Psi(p,t)=\int^{\infty}_{-\infty}\Psi(x,t)e^{-ipx/\hbar}dx

Substituting the given wave function, we get:

\Psi(p,t)=\frac{A}{\sqrt{2\pi}}\int^{\infty}_{-\infty}e^{-\tfrac{(x-ip\sigma^2/\hbar)^2}{2\sigma^2}}dx

This integral can be evaluated using the Gaussian integral formula, which gives:

\Psi(p,t)=Ae^{-\tfrac{p^2\sigma^2}{2\hbar^2}}

Therefore, the probability of measuring a momentum in the range between p and p+dp is:

P=\int^{p+dp}_{p}|\Psi(p,t)|^2 dp=A^2\int^{p+dp}_{p}e^{-\tfrac{p^2\sigma^2}{\hbar^2}}dp

Substituting the value of A from part (a), we get:

P=\frac{\sigma^2}{\pi}\int^{p+dp}_{p}e^{-\tfrac{p^2\sigma^2}{\hbar^2}}dp

This integral can be evaluated using