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

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Homework Help Overview

The discussion revolves around a Gaussian wave-packet for a free particle and the calculation of momentum probability. The original poster presents a wave function and seeks to determine the normalization constant and the probability of measuring momentum within a specified range.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the normalization constant A and the probability of momentum measurement using integrals. Some participants suggest the need to transform the wave function into momentum space to determine the probability distribution. Others question the results of the Fourier transform and the subsequent calculations.

Discussion Status

Participants are actively exploring the problem, with some providing guidance on the properties of Gaussian wave packets in momentum space. The original poster expresses confusion regarding their calculations and seeks further clarification.

Contextual Notes

The original poster mentions an upcoming exam, indicating a time constraint and the need for clarity on similar problems. There is also a reference to specific integrals and expressions that may require careful evaluation.

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Homework Statement



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

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

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

Homework Equations



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

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

The Attempt at a Solution



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

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

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

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

Please help I have a question similar to this on an exam this Monday!
 
Last edited:
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[tex]\langle p\rangle=-ih\int^{\infty}_{-\infty}\left(\Psi^\ast\frac{\partial\Psi}{\partial x}\right) dx[/tex]

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, [tex]\psi(p)[/tex]. There's a simple expression for the probability of measuring a momentum in the range between p and p+dp in terms of [tex]\psi(p)[/tex].
 
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:

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

But (I think) that reduces to:

[tex]\psi(p)=\frac{\sigma^3}{2\sqrt{\hbar}}[/tex]

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

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

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

[tex]\frac{\sigma^6dp}{4\hbar}[/tex]


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

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