Exploring the De Broglie Wave and Momentum in a Trapped Particle's Excited State

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SUMMARY

The discussion focuses on the de Broglie wave and momentum of a particle in a trapped state, specifically the first excited state. It establishes that for a particle confined in a region of length L, the de Broglie wave manifests as a pure sine wave from 0 to 2π. The momentum can be calculated using the de Broglie relation as 2h/L, while the energy quantization relation gives p=nh/2L, resulting in p=h/L for the second state (n=2). The conversation highlights that particle-in-a-box eigenstates do not correspond to specific momentum values due to their non-eigenstate nature with respect to the momentum operator.

PREREQUISITES
  • Understanding of de Broglie wavelength and its implications
  • Familiarity with quantum mechanics concepts such as eigenstates
  • Knowledge of the particle-in-a-box model
  • Basic grasp of momentum and energy quantization relations
NEXT STEPS
  • Study the implications of the de Broglie wavelength in quantum mechanics
  • Explore the particle-in-a-box model and its eigenstates
  • Learn about the momentum operator in quantum mechanics
  • Investigate energy quantization in different quantum systems
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, wave-particle duality, and the behavior of particles in confined systems.

Apashanka
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For a particle trapped in a region of length L the de broglie wave for the 1st excited state is a pure sine wave from 0 to 2pi
for which the particle momentum can be calculated as 2h/L from de broglie relation
Whereas from energy quantisation relation p=nh/2L where n is the state integer,for n=2 ,p=h/L
Why does so happen?
 
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Apashanka said:
Whereas from energy quantisation relation p=nh/2L where n is the state integer,for n=2 ,p=h/L
Particle-in-a-box eigenstates are not eigenstates of the momentum operator, therefore you can't say that it corresponds to a given value of momentum.
 

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