Suppose two atoms are falling away from each other with velocity $v$ each. Suppose they have evenly distributed momentum among their electrons. Suppose one atom swaps an electron with the other atom. Suppose then over a short period of time the electrons and the nucleus redistribute their momentums equally by various forces. Then the momentum of the first atom should be reduced by the $m_e v$. Then I think the change in momentum at any one moment should be $\mathrm{d} p = \mathrm{d} P \, m_e v$ where $\mathrm{d} P$ is the chance of swapping an electron at any instant.

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    $\begingroup$ Do you know what a Feynman diagram is? $\endgroup$
    – user108787
    Jul 27, 2016 at 17:53
  • $\begingroup$ You can't describe this in Classical terms. $\endgroup$
    – Gert
    Jul 27, 2016 at 17:54
  • $\begingroup$ @Gert I suspected that. How would you describe it nonclassically? $\endgroup$ Jul 27, 2016 at 17:56
  • $\begingroup$ @StevenStewart-Gallus: you'll need to start studying QM, to understand what momentum of an electron in an atom actually is. $\endgroup$
    – Gert
    Jul 27, 2016 at 17:58
  • $\begingroup$ In molecules, like $H_2$, electrons are 'shared' by two atoms all the time. A molecular orbital containing 2 electrons 'binds' the two H nuclei into an $H_2$ molecule. Simply put. $\endgroup$
    – Gert
    Jul 27, 2016 at 18:00

1 Answer 1


Your question is posed in classical language, yet it involves quantum systems (many-body systems in fact) that cannot be treated with classical physics. The best that can be said in this framework is that the probability of such a swap is greater than zero. In fact, quantum many-body systems are theoretically formulated in such a way to allow such swapping and there is a name in many-body physics for this phenomenon: it is called the exchange interaction. When one calculates the binding energy of a quantum many-body system, part of that energy is associated with this exchange interaction and this allows one to quantify its importance in a given circumstance.

The history of this phenomenon is quite interesting. Shortly after Schrodinger introduced his equation and demonstrated that it was in almost exact agreement with the hydrogen atom spectrum, physicists investigated the many-electron version of the Schrodinger equation. There was no problem in writing down the many-body equation, but it was immediately obvious that unlike the hydrogen atom there was no exact mathematical procedure for solving it to get the many-body wave functions and their eigenvalues. Approximations had to be introduced before any progress could be made.

The British physicist Hartree suggested that a product of single electron wave functions (similar to those found for the hydrogen atom) might be appropriate and proceeded to develop numerical techniques for solving the coupled set of equations that resulted from his approximation scheme. The Pauli exclusion principle had already been introduced by that time so Hartree made sure that no two of his single electron wave functions had the same set of quantum numbers. He proceeded to calculate his orbital wave functions and found that they could represent the many-body spectra of light atoms to a reasonable approximation.

The American physicist Slater (who taught my first quantum mechanics course) noted that if a determinant of these single electron wave functions were used instead of the Hartree product form then the Pauli exclusion principle was automatically satisfied because the determinant would yield zero if two identical wave functions were selected. The Russian physicist Fork then showed that the numerical Hartree procedure could be extended for the Slater determinant form to yield the original Hartree coupled set of equations plus an extra term that he called the exchange term. The full procedure is now known as the Hartree-Fock procedure and it is behind most of the theoretical calculations performed for quantum many-body systems in atomic, molecular, solid-state, and nuclear physics. These calculations are very numerically challenging so most of the progress was delayed until the development of modern computers.

  • $\begingroup$ But how does the exchange interaction lead to repulsion between atoms? The way I phrased it suggests that atoms should be attracted to each other. $\endgroup$ Jul 27, 2016 at 21:11
  • $\begingroup$ @StevenStewart-Gallus The exchange interaction works both ways, basically the photon can transfer both positive (repulsion) or negative momentum (attraction). Don't ask me how, I dunno. $\endgroup$
    – user108787
    Jul 27, 2016 at 21:42
  • $\begingroup$ @StevenStewart-Galls Between atoms the interaction gets complicated as count_to_10 suggests. The most obvious interaction is repulsion since the electron clouds around each atom should repell. The polarization, however changes this and allows attraction. Exchange is another beast altogether and what sign it takes depends on the circumstance. Note that the exchange interaction contributes to a single atom's binding as well (not just between atoms) $\endgroup$ Jul 27, 2016 at 22:07
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    $\begingroup$ The Hartree approximation includes a self interaction that is cancelled by exchange. Since self-interaction of electron is repulsive, part of the exchange term is attractive. $\endgroup$ Jul 27, 2016 at 22:15

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