Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I heard that there have been some propositions about describing the collapse of the wave-function by adding non-linear terms, but I couldn't anything in any any textbooks or even articles (probably those propositions never reached a good level of consistency). However, I'd like to read about it. Could someone send me a reference?

share|improve this question

4 Answers 4

up vote 2 down vote accepted

The Ghirardi-Rimini-Weber Model is such a theory. See for instance http://arxiv.org/abs/quant-ph/0406094.

share|improve this answer

As far as I know, nonlinearities aren't compatible with Lorentz invariance. The overall probability renormalization factor also needs to be rescaled globally, although that might not be a problem if rejecting a probabilistic ontology.

share|improve this answer
Can you explain to us what is the probabilistic ontology and what is the issue of global rescaling? –  stupidity Jun 29 '12 at 19:37
Well, based on Weinberg proposition for non linear QM, the non linear observables need to be of order one in $\psi$ and $\psi^*$ so non linear operators in the Hamiltonian will be suppresed by factors $n = \int d^3x \;\psi(x)\psi^*(x)$ which is a non-local quantity. –  toot Jul 13 '12 at 17:24

Collapse of the State Vector

Phys. Rev. A 85, 062116 (2012)


share|improve this answer

Roger Penrose advanced the notion that gravity causes wave function collapse, giving handwavy arguments involving the Schrodinger-Newton equation (one particular flavor of the nonlinear Schrodinger equation).

The references I'm aware of:

  1. Roger Penrose, "On Gravity's Role in Quantum State Reduction", General Relativity and Gravitation 28 5 (1996) 581-600. DOI:10.1007/BF02105068
  2. Roger Penrose, "Quantum computation, entanglement and state reduction", Phil. Trans. R. Soc. Lond. A 356 no. 1743 (1998) 1927-1939. DOI:10.1098/rsta.1998.0256
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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