Momentum values in quantum mechanics

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SUMMARY

The ground state wavefunction in an infinite potential well is described by ψ=A*sin(πx/a). This wavefunction does not have a definite momentum, as it does not satisfy the condition \hat{p}ψ=pψ for a constant p. The momentum operator \hat{p} applied to this wavefunction yields a result that indicates the absence of a definite momentum value. The allowable momentum values correspond to the eigenfunctions of the momentum operator, which need to be expressed in relation to the given wavefunction.

PREREQUISITES
  • Understanding of wavefunctions in quantum mechanics
  • Familiarity with the momentum operator \hat{p} in quantum mechanics
  • Knowledge of eigenfunctions and eigenvalues in quantum mechanics
  • Basic calculus for differentiation of functions
NEXT STEPS
  • Study the properties of eigenfunctions of the momentum operator
  • Learn how to express wavefunctions in terms of momentum eigenstates
  • Explore the implications of non-definite momentum in quantum mechanics
  • Review the concept of potential wells and their impact on wavefunctions
USEFUL FOR

Students of quantum mechanics, physicists analyzing wavefunctions, and anyone interested in the implications of momentum in quantum states.

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Homework Statement



In an infinite potential hole, the ground state is described by the wavefunction ψ=A*sin([itex]\frac{πx}{a}[/itex]). Does the ground state have a definite momentum? If not, then what are the values of momentum in ground state?

Homework Equations


Wavefunction ψ=A*sin([itex]\frac{πx}{a}[/itex])
Momentum operator [itex]\hat{p}[/itex]=-iħ[itex]\frac{d}{dx}[/itex]


The Attempt at a Solution



Momentum has a definite value when [itex]\hat{p}[/itex]ψ=pψ.
In this case [itex]\frac{dψ}{dx}[/itex]=[itex]\frac{π}{a}[/itex]A*cos([itex]\frac{πx}{a}[/itex])
and [itex]\hat{p}[/itex]ψ=-[itex]\frac{iħπA}{a}[/itex]*cos([itex]\frac{πx}{a}[/itex])
So this is the point where I am stuck, what does this result show me?
 
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You're almost there! You're correct that the momentum has a definite value when [itex]\hat{p}\psi = p\psi[/itex] for some constant [itex]p[/itex]. Is that the case here? If not, you will have to conclude that this state has no definite momentum.

As for the second half, we know that the allowable momentum values are the eigenfunctions of the momentum operator. Do you know what these are? Can you express the given wavefunction in terms of these eigenfunctions?

HTH
 
The first part of your reply was just what I was expecting, so thank you for that. I have given all the information concerning the problem, so I don't know if I can express it in terms of eigenfunctions, in case I haven't missed anything. Can you help me formulate an answer, what are the values of momentum in ground state? Sorry if you didn't understand something, I am finding it hard to "talk physics" in English.
 

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