The spreading of a wave packet is very fast in quantum mechanics: for an electron, a gaussian wave packet spreads from one angström to 600km in one second! In his famous QM book, Sakurai mentions that there are numerical evidence that taking into account gravitation as $$-{\hbar^2\over 2m}\Delta\psi(\vec r) -{\cal G}m^2\int {|\psi(\vec r')|^2\over |\vec r-\vec r'|}d^3\vec r'\psi(\vec r)=i\hbar{\partial\psi\over\partial t}$$ stops the spreading at distances around 500nm. See for example https://arxiv.org/abs/1105.1921

Why should we take into account the gravitational interaction of the wave packet with itself but not the electromagnetic one? In the case of a free electron for example, the Coulomb repulsion of the wave packet by itself would accelerate the spreading.


1 Answer 1


At least in the linked paper they discuss using molecules (such as fluorofullerene C$_{60}$F$_{48}$) as the test particle. The molecules are uncharged.

  • $\begingroup$ The question seems valid. Why should gravitational interaction between different parts of the wavefunction be allowed if Coulomb interaction is not allowed? $\endgroup$
    – S. McGrew
    Commented Oct 16, 2018 at 13:53
  • $\begingroup$ Because the question seems to assume there is a net charge leading to Coulomb repulsion. $\endgroup$ Commented Oct 16, 2018 at 14:01
  • $\begingroup$ Perhaps @Christophe will clarify his question. $\endgroup$
    – S. McGrew
    Commented Oct 16, 2018 at 14:51
  • $\begingroup$ I had the case of the electron in mind (see the first sentence of the question). In contrast to molecules, it has an electric charge so why should we consider the gravitational self-attraction of the wave packet but not the Coulomb self-repulsion? $\endgroup$
    – Christophe
    Commented Oct 17, 2018 at 19:30

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