Measuring momentum in a square well

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SUMMARY

The discussion centers on the complexities of measuring momentum in both infinite and finite square wells within quantum mechanics. The participant expresses confusion regarding the implications of measurement, specifically how measuring an observable leads to an eigenstate of the corresponding operator. The participant highlights that for an infinite square well, this results in a wavefunction extending beyond boundaries, while for a finite square well, it implies the wavefunction becomes an eigenstate of the momentum operator, suggesting free particle behavior. Clarification is sought on the momentum operator specific to square wells.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wavefunctions and eigenstates.
  • Familiarity with the concept of observables and their corresponding operators in quantum mechanics.
  • Knowledge of the differences between infinite and finite square wells.
  • Basic grasp of momentum operators in quantum systems.
NEXT STEPS
  • Research the momentum operator specific to infinite and finite square wells.
  • Study the implications of measurement in quantum mechanics, focusing on eigenstates and observables.
  • Explore the mathematical formulation of wavefunctions in square wells.
  • Examine case studies or examples of momentum measurement in quantum systems.
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Students and professionals in quantum mechanics, physicists exploring wavefunction behavior, and anyone interested in the implications of measurement in quantum systems.

thoughtgaze
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Hi

I'm a little confused when it comes to measuring the momentum in a infinite or finite square well. Is this even possible? One of the postulates of quantum mechanics states that measuring an observable will leave it as an eigenstate of that observables corresponding operator just after measurement. Fine and dandy but for an infinite square well this would result in the wavefunction existing beyond the boundaries. For a finite square well this would also result in the wavefunction being an eigenstate of the momentum operator and therefore a free particle. Can someone explain where I am going wrong/what I'm missing. Thanks.
 
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George Jones said:
I think this is a bit subtle. What is the momentum operator for the square well? See

http://arxiv.org/abs/quant-ph/0103153.

This looks great George, thanks. Reading it now.
 

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