A quantum particle which is almost at rest but whose position is rando

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SUMMARY

The discussion centers on the behavior of quantum particles with low momentum and uncertain position, specifically using the wave function ##\psi(x) = e^{-x^2/L^2}##. Participants emphasize the calculation of key quantities such as ##\langle x \rangle##, ##\langle p \rangle##, ##\Delta x##, and ##\Delta p## to understand the implications of the uncertainty principle. The analysis reveals that a large value of ##L## results in high position uncertainty and low momentum uncertainty, illustrating the inherent trade-off dictated by quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles
  • Familiarity with wave functions and their properties
  • Knowledge of the uncertainty principle in quantum physics
  • Ability to perform calculations involving expectation values
NEXT STEPS
  • Calculate expectation values for the wave function ##\psi(x) = e^{-x^2/L^2}##
  • Explore the time evolution of quantum wave functions
  • Study the implications of the uncertainty principle in various quantum systems
  • Investigate the general wave function ##\psi(x) = e^{-x^2/L^2}e^{i k x}## and its effects on momentum and position uncertainty
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, wave functions, and the uncertainty principle will benefit from this discussion.

rajesh_d
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I had posted this question here : http://physics.stackexchange.com/q/69003/540

I guess its appropriate to post links in here as question.

This question is really puzzling me and any suugestion/comments are much appreciated and welcome.
 
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Yes, you can have a particle with very small momentum and a very uncertain position. A simple case in which you can work everything out analytically is the wave function

##\psi(x) = e^{-x^2/L^2}##

If you haven't done this before you should calculate various quantities for this wave function like ##\langle x \rangle##, ##\langle p \rangle##, ##\Delta x = \sqrt{\langle x^2 \rangle - {\langle x \rangle}^2}##, and ##\Delta p = \sqrt{\langle p^2 \rangle - {\langle p \rangle}^2}##. It's also enlightening to calculate the time evolution of this wave function, which isn't too hard. If you do end up doing this, make sure to do it again with the more general wave function ##\psi(x) = e^{-x^2/L^2}e^{i k x}##

You'll find that the uncertainty in position is of order ##L##. The expectation of momentum is 0, with uncertainty of order ##\hbar/L##. Taking ##L## large gives a high position uncertainty and a small momentum uncertainty.

This is not a problem with quantum mechanics; it's a straightforward consequence of it. In fact, the intuitive content of the uncertainty principle can basically be summarized as: "If you want a particle to have a certain momentum (such as zero) with high precision, then its position must be very uncertain."
 
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