Momentum eigenfunctions in an infinite well

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SUMMARY

The discussion centers on the properties of momentum eigenfunctions in the context of quantum mechanics, specifically within an infinite potential well. It is established that the wavefunctions derived from the Schrödinger equation, sin(nπx/L), are not momentum eigenfunctions, leading to the conclusion that momentum cannot be directly measured in this scenario. The Hamiltonian operator p²/(2m) commutes with the momentum operator, but the latter is not self-adjoint in the infinite well, resulting in the absence of momentum observables. The conversation also touches on the implications of momentum in both classical and relativistic frameworks.

PREREQUISITES
  • Understanding of the Schrödinger equation and its solutions
  • Familiarity with quantum mechanics concepts such as eigenfunctions and operators
  • Knowledge of Hamiltonian mechanics, specifically p²/(2m)
  • Basic principles of self-adjoint operators in quantum mechanics
NEXT STEPS
  • Study the concept of self-adjoint operators in quantum mechanics
  • Explore the implications of momentum eigenstates in finite potential wells
  • Learn about the Fourier transform method in quantum mechanics
  • Investigate the relationship between momentum and energy in quantum systems
USEFUL FOR

This discussion is beneficial for quantum mechanics students, physicists exploring the foundations of quantum theory, and anyone interested in the mathematical underpinnings of momentum in quantum systems.

  • #31
dyn said:
Are you telling me that in QM , CM and SR there is no limit on how big momentum can be ?
Yes, that's what they have told you.

dyn said:
Surely v not being able to reach c places a limit on momentum in SR
Instead of waving your hands, do the math. If you have questions about the math, you can start a new thread in the appropriate forum. This thread is closed as the question you asked in your OP has been answered.
 
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  • #32
dyn said:
Are you telling me that in QM , CM and SR there is no limit on how big momentum can be ? Surely v not being able to reach c places a limit on momentum in SR
It does not.
The momentum of a particle moving at speed ##v## is ##p=mv/\sqrt{1-(v^2/c^2)}##.

That value can be made as large as we please by choosing a value of ##v## sufficiently close to but still less than ##c##.
 
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