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We know that an electron behaves like a point charge, and the probability density of its position is given by the Born rule. Now suppose that we shoot an electron toward an atom, so the electron gets scattered due to the Coulomb forces the nucleus and the electrons of the atom push on it.

How is the direction of this coulomb force determined? If the electrons are acting as point charges, and not smeared charges inside the orbitals, the coulomb force vector on the shot electron should have a random but certain direction changing by time. Is this true?

Is this how the scattering of the electrons work? How is the probability density for the deflection angle of the shot electron calculated?

Alex L
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1 Answers1

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You are mixing incommensurable notions in this question. In quantum mechanics, where there is a "probability density", there is no such thing as a "Coulomb force", there is just the quantum interaction described by the Coulomb potential.

The Coulomb force only emerges from the quantum scattering process in the non-relativistic classical limit, see this answer of mine.

The quantum mechanical scattering process has certain probabilities for certain directions of the outgoing momentum of the electron after scattering, usually expressed as a differential cross section with respect to a solid angle. There is no "force vector" in this quantum mechanical scattering - quantum mechanics simply offers no description of the scattering process in such terms, see also this answer of mine.

In fact, in quantum field theory the "point-like" characteristic of the electron really refers to the form of its differential cross sections for scattering at high energies, see e.g. this question and its linked questions or this question, and not to it acting as a "point charge" in any classical sense.

ACuriousMind
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