Uncertainty principle for position and hamiltonian

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SUMMARY

The discussion centers on the uncertainty principle relating to position (delta x) and Hamiltonian (delta H) in quantum mechanics, specifically expressed as ΔxΔH ≥ (ħ*p)/2m. It concludes that in stationary states where momentum (p) equals zero, the uncertainty in energy (ΔH) can be precisely known, yet measuring position collapses the state, introducing uncertainty in momentum. The participants emphasize that while ΔH can be zero, the uncertainty in position is not guaranteed to be zero, highlighting the principle as a lower bound rather than an upper bound.

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  • Understanding of quantum mechanics principles
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  • Knowledge of Hamiltonian mechanics
  • Basic grasp of stationary states in quantum systems
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  • Explore Hamiltonian mechanics and its applications
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Quantum physicists, students of quantum mechanics, and researchers exploring the implications of the uncertainty principle in various quantum systems.

kehler
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I found the uncertainty between delta x (position) and delta H (Hamiltonian) to be greater or equal to (h_bar*<p>)/ 2m.
Does this mean for stationary states, where <p>=0, the uncertainty can be zero? ie we can precisely measure the position and energy?
 
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kehler said:
I found the uncertainty between delta x (position) and delta H (Hamiltonian) to be greater or equal to (h_bar*<p>)/ 2m.
Does this mean for stationary states, where <p>=0, the uncertainty can be zero? ie we can precisely measure the position and energy?

So, in a stationary state <delta H>=0. You thus already know the energy exactly. But, unless the position eigenstate is also a stationary state then when you measure the position you collapse to a different state, an give up info about the momentum. Thus, because most hamiltonians include a kinetic term, it would seem like you can not measure both exactly. But, for certain cooked up systems maybe you can.

You should probably explain more about the particular problem/system you have in mind. Give us a few equations as well that further explain your idea. This will help us come to a reasonable answer.
 
As Olgran said, in a stationary state <H> is definite. So \Delta H\Delta x=0 already. Your uncertainty in x is not guaranteed to be zero. Remember that the uncertainty principle is a lower bound on your uncertainty, not an upper bound. It does, however, mean you have some "hope" of finding a definite H and X state, but it's not guaranteed that you can just from looking at the Uncertainty Principle.
 

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