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Thinking about the 0+1 dimensional (time-only) non-linear Schrodinger equation:

$$i\frac{\partial}{\partial t} \psi(t) =\kappa |\psi(t)|^2 \psi(t).$$

Treating $\psi$ as a wave function instead of a field does this satisfy the rules of quantum mechanics? Because it seems like probability would be conserved. But it is non-linear.

Yes you could treat $\psi$ as a quantum field. Then you would have a wave function $\Psi[\psi]$. But if you treated $\psi$ as a wave function by itself, would this work? (It would be a modification of quantum mechancs). If not why not?

The rules of quantum mechanics as I can make out satisfy two things:

  1. Probabilities always add up to 1.

  2. States that are orthogonal stay orthogonal.

Are these rules obeyed?

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    $\begingroup$ I'm fairly certain that's not the nonlinear Schrodinger equation. en.wikipedia.org/wiki/Nonlinear_Schr%C3%B6dinger_equation $\endgroup$ – probably_someone Nov 29 '18 at 16:15
  • $\begingroup$ What happened to the term with wavefunction partially double differentiated with respect to position in that equation? $\endgroup$ – exp ikx Nov 29 '18 at 16:24
  • $\begingroup$ It's a one dimensional version hence no space terms. 1 dimensional = time only $\endgroup$ – zooby Nov 29 '18 at 16:33
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    $\begingroup$ If you have not spatial variation (and you don't seem to have any discrete indices either) then to conserve probabilities you must have $|\psi| = 1$, but in that case you have a stratforward linear equation and $\psi(t) = \exp(-\imath \kappa t)$. Essentially you only seem to have 1 state, so the only allowed time evolution is trivial. $\endgroup$ – By Symmetry Nov 29 '18 at 16:39
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    $\begingroup$ @zooby That's not what "1-dimensional" typically means. One might call this "0+1-dimensional" (0 spatial dimensions, 1 time dimension) whereas most of the time "1-dimensional" motion is "1+1-dimensional." $\endgroup$ – probably_someone Nov 29 '18 at 16:45
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In QM, the states are elements of a projective Hilbert space. Equivalently, $\left|\psi\right>$ and $c \left|\psi\right>$ refer to the same state for any $c \in \mathbb{C} \setminus \{0\}$. Your equation doesn’t preserve this symmetry, so it doesn’t satisfy the axioms of QM.

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