I Free electron path without a nearby proton

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In the absence of a positive electric field from a proton, a free electron behaves differently than when bound in an atom. It retains quantized spin and charge, but its energy levels become continuous rather than discrete, as governed by Schrödinger's equation. The uncertainty principle still applies, affecting its position and momentum. Despite being unbound, quantum mechanics remains valid, allowing for phenomena like interference patterns in double-slit experiments. Thus, free electrons exhibit unique quantum behaviors distinct from those of electrons in atomic orbitals.
Tareq Naushad
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Quantum properties of an free electron outside atom
In absence of a positive electric field created by proton what type of behavior an electron shows? I am talking about a free electron like from a electron gun in deep space . Pauli's exclusion, orbitals, energy level etc. which are normal in an atom for electron will be absent for an electron out from an atom. So what type of quantum behavior it shows then ?
 
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It still has quantized spin and charge.
 
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thanks. what about its allowable energy level, uncertainty of position, momentum etc. because now it is not governed by the field of +ve proton
 
Tareq Naushad said:
what about its allowable energy level, uncertainty of position, momentum etc. because now it is not governed by the field of +ve proton
We're still solving Schrodinger's equation to find the wave function. Schrodinger's equation includes a term for the potential, which will be the Coulomb potential for an electron near a proton (whether bound or not) and zero for an isolated (and therefore obviously unbound) electron in vacuum.

The bound states will have discrete eigenvalues. The unbound ones will have a continuous spectrum (although there are some mathematical complications here that we can ignore for now) quantum mechanics still works: position/momentum uncertainty still works, Schrodinger's equation governs the evolution of the wave function, we can even in principle do a double-slit experiment with a beam of free electrons and produce an interference pattern.
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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