Does the non-linear Schrodinger equation satisfy quantum mechanics rules?

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?

• I'm fairly certain that's not the nonlinear Schrodinger equation. en.wikipedia.org/wiki/Nonlinear_Schr%C3%B6dinger_equation – probably_someone Nov 29 '18 at 16:15
• What happened to the term with wavefunction partially double differentiated with respect to position in that equation? – exp ikx Nov 29 '18 at 16:24
• It's a one dimensional version hence no space terms. 1 dimensional = time only – zooby Nov 29 '18 at 16:33
• 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. – By Symmetry Nov 29 '18 at 16:39
• @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." – probably_someone Nov 29 '18 at 16:45

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.