Allowed momentum values for a plane wave

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Homework Help Overview

The discussion revolves around the allowed momentum values for a particle described by a specific wavefunction in quantum mechanics. The context involves understanding the implications of the wavefunction's form on momentum probabilities, particularly in a scenario without a potential well.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to determine the allowed momentum values based on the wavefunction provided, questioning why it is not an eigenfunction of momentum despite having a single k value. Some participants discuss the implications of momentum being a vector operator and the significance of sign in momentum values.

Discussion Status

The discussion is exploring the relationship between the wavefunction and momentum probabilities, with some participants suggesting specific probabilities for different momentum values. There is an ongoing examination of the assumptions regarding the wavefunction's characteristics and its implications for momentum measurements.

Contextual Notes

Participants are considering the infinite uncertainty in position and its effect on momentum, as well as the implications of the wavefunction being a superposition of terms with different momentum components.

alec_grunn
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Hi all,
This is from a past exam paper: At t=0 the state of a particle is described by the wavefunction

Code:
 $$ \Psi (x,0) =A(iexp(ikx)+2exp(-ikx)) $$

This is between positive and negative infinity - not in a potential well.

What values of momentum are allowed, and with what probability in each case?

Relevant Equations:
## \hat p = -i \hbar \frac{\partial}{\partial x} ##

My attempted solution:
Since there's only one k value present I was thinking there is one momentum value: ##p = \hbar k## with 100% chance of measuring this. And the fact that the uncertainty in position is infinite means that its momentum is sharp. But if this is the case, then why isn't it an eigenfunction of momentum?

Please help,
Cheers
 
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Momentum is a vector operator - the sign matters.
 
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Likes   Reactions: alec_grunn
Ok so that means 20% chance of +p and 80% chance of -p?
 
alec_grunn said:
Ok so that means 20% chance of +p and 80% chance of -p?
Yes.
 

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