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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.