2
$\begingroup$

I tried to figure it out myself. If you take the integro-differential form of the equation, a complex square of the time-dependent wavefunction appears.. It seems to me that this means the equation cannot be time symmetric but I'm not sure about this reasoning. The same goes for PT-symmetry, since the Laplacian is the same even under a P reversal.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for posting this. I had never heard about this equation: en.wikipedia.org/wiki/Schr%C3%B6dinger%E2%80%93Newton_equation . Maybe I'm dumb, but I don't see why the time-reversal symmetry would be in doubt. Under time-reversal or complex conjugation, doesn't the gravitational term have the same behavior as the kinetic and $V$ terms, maintaining the same symmetry? $\endgroup$
    – user4552
    May 9, 2018 at 4:41
  • $\begingroup$ Yeah, that's partially why I'm confused. The reason I'm concerned that it might not be is because the equation involves taking the magnitude of the time-dependent wavefunction, so the derivation of the Schroedinger equation's time symmetry doesn't quite work...at least I don't see how it would work. The other reason I'm confused is that in the S-N equation, it's possible for a wavefunction to "collapse" to a set of basis states but I see no mention of the time-reverse being possible. $\endgroup$ May 9, 2018 at 6:40

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.