# Question: Minimum uncertainty in ground state electrons

## Main Question or Discussion Point

Hey all, I'm a student curently studying in a Singaporean Junior College (American 12th grade equivalent). I was curious and just thought of the following:

Suppose a system with a electron and a proton nucleus, a hydrogen atom.

From electrostatic force and circular motion equations-

F = e²/(4πr²ɛ) , F = mv²/r
p = e/2 *√(m/(πrɛ))

Suppose the electron has a momentum classically determined to be 0, hence its position is right in the position the proton is.
p = 0 , r = 0
However, due to HUP, there is an uncertainty involved for both momentum and position of the electron.

Therefore,
p = Δp , r = Δx,

Δp = e/2 *√(m/(πΔxɛ))
ΔpΔx = e/2 *√(mΔx/(πɛ))
From HUP,
e/2 *√(mΔx/(πɛ)) ≥ ħ/2
mΔx/(πɛ) ≥ ħ²/e²
Δx ≥ ħ²πɛ/me²

And there is a precise, minimum uncertainty in position (and also momentum) for an electron in ground state.

Assuming my calculations and theory to be correct (if it is not, please tell me why), what happens when we confine an electron (by an infinitely strong potential barrier perhaps?) in such a way that its position uncertainty is less than ħ²πɛ/me². What would happen then? Maybe a wavefunction collapse?

Any help and input would be really appreciated here. I would prefer if you not use complex mathematical ideas or notation as I do not even have a university education. I do have little understanding of calculus, vectors, and differential equations, but bra-ket, matrices, PDEs are totally out of my league currently.

Thanks.

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Also from the de broglie relation and HUP,

Δp ≥ ħ/2Δx , Δp = 2πħ/λ
2πħ/λ ≥ ħ/(2Δx)
Δx ≥ λ/4π

is there any relation between ħ²πɛ/me² and λ/4π?
Perhaps ħ²πɛ/me² = λ/4π for electrons in ground states?

Demystifier