Electron in a hydrogen atom (uncertainty principle)

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SUMMARY

The discussion focuses on calculating the uncertainty product (Δx)(Δp) for an electron in a hydrogen atom in the state ψ(100). Participants confirm that the uncertainty principle dictates that (Δx)(Δp) ≥ ħ/2, where ħ is the reduced Planck's constant. The integrals for Δx and Δp are defined as Δx = ∫[ψ(x - ⟨x⟩)² ψ] dv and Δp = ∫[ψ(p - ⟨p⟩)² ψ] dv. Clarification is sought regarding whether p refers to the momentum component px.

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Homework Statement



For an electron in a hydrogen atom: in state psi(100):

find (del-x)(del-p) where

del x = integral of :[psi (x-<x>)2 psi]dv
and
del p = integral of: [psi(p-<p>)2 psi]dv



Homework Equations





The Attempt at a Solution


i attempted to multiply out both equations but i am not sure if this is the right way to go,
assume the answer is something like (delx) (del p)>= h/2 ? but i am not sure how to get there?
 
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I suppose you could just do the integrals. BTW is p supposed to be px?
 
Last edited:

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