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As I understand it, the linearity of quantum mechanics is considered to be an inviolable principle - e.g., this paper - because (among other things) causality would be violated or and/or superluminal communication would be possible if quantum mechanics were nonlinear. However, even in the simple case of multiple electrons, the path (or evolution of the wavefunction) of an individual electron would influence the paths of the electrons, which would in turn influence the path of the individual electron. This looks a lot like self-interaction which would imply nonlinearity.

I think it is possible to represent the multiple-electron system by a set of coupled quasi-linear equations; but I also think that an equation in terms of the wavefunction of the whole multiple-electron system would be nonlinear. {By "quasi-linear" I mean linear in some of the unknowns, in such a way that the other unknowns resemble constants.) This SE answer may be relevant, but seems not to directly address my problem: Does separability into coupled quasi-linear equations somehow meet the overall linearity requirement?

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    $\begingroup$ The standard model is non-linear. Where did these speculations about the linearity of quantum mechanics come from? $\endgroup$ Commented Jan 11, 2020 at 10:30
  • $\begingroup$ It may depend on representation. For example position representation may be nonlinear while momentum ( Fourier transform) may be linear and vice versa. For more complicated systems, various quasiparticle ( aka Bogoliubov transformations) may occur, for which dynamics may be linear ( or even decoupled) while for physical particles it may be quite complicated. Etc. So it is hard in advance even tell what do you mean by "linear". Basic Schroedinger equation is linear $\endgroup$
    – kakaz
    Commented Jan 11, 2020 at 12:41

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Depends on what you mean by "linear" - it is certainly not linear in the sense that the full state can be composed of a product of many single-particle wave functions + (anti)symmetrization. For an interacting theory, the many-body wave function contains many correlations that do not allow for the decomposition into separable independent single-particle wave functions.

It is linear in the sense that if $|\psi(t)\rangle, |\varphi(t)\rangle$ are two solutions of the Schrödinger equation then $|\psi(t)\rangle + |\varphi(t)\rangle$ will also be a solution. This is relevant also for a many-body wave function $\langle \vec{r}_1,\ldots \vec{r}_N | \psi(t) \rangle = \psi(\vec{r}_1,\ldots,\vec{r}_N, t)$. It doesn't mean that there is a way to write $\psi(\vec{r}_1,\ldots,\vec{r}_N, t) = \prod_j \psi_j(\vec{r}_j)$ with each of these functions a solution.

The non-linear Schrödinger equation, for example one that involves $|\psi|^2$, is an approximation. Usually (in the contexts that I am familiar with) a mean-field approximation where the effects of the other particles are taken as an external potential, and then solved self-consistently.

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  • $\begingroup$ That is helpful. I think you're saying that the wave function for the whole system is always linear, although the wave functions for the individual particles may not be linear. Can you provide a simple example, or a reference? $\endgroup$
    – S. McGrew
    Commented Jan 14, 2020 at 15:02
  • $\begingroup$ I'm not sure what exactly you want an example for, but consider the simple Hamiltonian for two $1/2$-spins $H=B \vec{s}_1 \cdot \vec{s}_2$. For $B>0$ the ground state is the singlet $(|\uparrow, \downarrow\rangle - |\downarrow,\uparrow\rangle)/\sqrt{2}$, and it has the time evolution with a phase which is the energy times the time. It cannot be decomposed into $|\psi\rangle = |s_1\rangle \otimes |s_2\rangle$ as they are correlated. Yet we can linearly add to it another eigenstate, with its own time evolution, and the combined state will satisfy the TDSE. Is that what you had in mind? $\endgroup$
    – user245141
    Commented Jan 14, 2020 at 16:41
  • $\begingroup$ Yes, that's what I had in mind. $\endgroup$
    – S. McGrew
    Commented Jan 14, 2020 at 17:52

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