Position of a proton in Quantum mechanics.

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SUMMARY

The discussion focuses on calculating the maximum accuracy of a proton's position in quantum mechanics without altering its non-relativistic kinetic energy by more than 1 keV. The Heisenberg uncertainty principle, expressed as ΔxΔp = ħ/2, is central to this calculation. By determining the momentum p using the kinetic energy formula E(kinetic) = 1/2mv², the relationship p = √(2mE) is established. The conclusion emphasizes that the uncertainty in position Δx can be derived from Δp using the equation Δp ≤ √(mΔE), leading to Δx ≥ ħ/(2√(mΔE)).

PREREQUISITES
  • Understanding of Heisenberg's uncertainty principle
  • Knowledge of non-relativistic kinetic energy equations
  • Familiarity with momentum calculations in quantum mechanics
  • Basic grasp of quantum mechanics terminology
NEXT STEPS
  • Study the implications of Heisenberg's uncertainty principle in quantum mechanics
  • Learn about non-relativistic kinetic energy and its applications
  • Explore advanced momentum calculations in quantum systems
  • Investigate the relationship between energy and position uncertainty in quantum particles
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Students of quantum mechanics, physicists focusing on particle behavior, and anyone interested in the principles governing position and momentum in quantum systems.

Skullmonkee
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Homework Statement


Find the maximum accuracy that can be found to a proton's position without changing it's (not-realativistic) kinetic energy by more that 1 keV

I think this involves Heisenberg's uncertainty principle [tex]\Delta x\Delta p=hbar/2[/tex] but I am not sure at all.

Now to find the momentum p of a non-realativistic kinetic energy we can use [tex]E(kinetic) = 1/2mv^2 = p^2/2m[/tex]
which gives [tex]p = \sqrt{}2mE[/tex]
however this is as far as i got.
 
Last edited:
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Usually, professors expect you to realize that [tex]p\ge\Delta p[/tex]. Use this and see whether the answer matches!
 
That is, use that
[tex]p\Delta p=m\Delta E[/tex]
[tex]\Rightarrow \Delta p\le\sqrt{m\Delta E}[/tex]
[tex]\Rightarrow \Delta x\ge \frac{\hbar}{2\sqrt{m\Delta E}}[/tex]
 

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