Wavefunction collapse and measurement

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SUMMARY

The discussion centers on the concept of wavefunction collapse in quantum mechanics, particularly regarding stationary states and measurements. When a measurement of energy is made, the wavefunction collapses into an energy eigenstate, which remains stable unless perturbed. In contrast, position measurements lead to wavefunction evolution post-measurement due to the non-commutation of position operator x with the Hamiltonian H. The key takeaway is that if an observable's operator A commutes with H, the system remains in the eigenstate of A; otherwise, it evolves into a combination of eigenstates over time.

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  • Understanding of quantum mechanics principles, particularly wavefunction behavior
  • Familiarity with operators and observables in quantum systems
  • Knowledge of Hamiltonian mechanics and time evolution in quantum physics
  • Concept of eigenstates and eigenvalues in quantum theory
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  • Study the implications of non-commuting operators in quantum mechanics
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  • Explore the concept of eigenstates and their significance in measurement theory
  • Investigate specific examples of measurements that lead to wavefunction evolution
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Students and professionals in quantum mechanics, physicists exploring measurement theory, and anyone interested in the foundational aspects of wavefunction behavior and evolution.

wduff
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So, rookie question, I know, but I'm having a little trouble with the idea of wavefunction collapse as it pertains to stationary states:

If a measurement of energy collapses a wavefunction into an energy eigenstate, it stays there forever (unless perturbed). But my impression is that although position measurements collapse a wavefunction, the wavefunction will begin to evolve rather quickly after the measurement.

Does this have something to do with the fact that x doesn't commute with H? What are some other measurements that will yield an evolving wavefunction?

Say I have some operator associated with an observable that has two values: 'happy' and 'sad.' What else do I need to know about this situation to predict whether a measurement of 'happy' will leave the particle in that state for all time?

Thanks
 
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Yes, it has to do with whether or not the operator A corresponding to the measurement you are doing commutes with the Hamiltonian H. H governs the time evolution of the wave function. A measurement of A will collapse the system into an eigenstate of A. If A commutes with H, the system will remain in that eigenstate of A as it evolves in time as determined by H. If A and H don't commute, then it will be in an eigenstate of A immediately after the measurement, but will then begin evolving into some linear combination of all the eigenstates of A as time passes.
 
Okay, makes sense. Thanks for your time phyzguy, glad to hear I wasn't too far from home there!
 

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