# Could the Schrödinger equation be nonlinear?

Is there any specific reasons why so few consider the possibility that there might be something underlying the Schrödinger equation which is nonlinear? For instance, can't quantum gravity (QG) be nonlinear like general relativity (GR)?

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–  Qmechanic Oct 16 '11 at 15:33

I would advise the following reference. In this thesis, we study a variant of the Schrodinger equation Nonlinear Schrodinger equation nonlinear spatially inhomogeneous nonlinearity. The thesis is divided into 4 parts: Following an introductory chapter on the nonlinear Schrodinger equation, we divide the study of nonlinear Schrodinger equation inhomogeneous in 3 blocks: The first results are given on the existence and stability of solutions of this equation. This uses various techniques such as variational approximation techniques, dynamical systems, eigenvalue problem ...

In the second block, once proven the existence of solutions, we proceed to calculate analytical solutions of this equation using different analytical methods, such as the method of Lie symmetries, transformacions of similarity, and so on.

In the third and last block, discusses some physical applications of this equation for Bose Einstein condensates and nonlinear optics.

http://dialnet.unirioja.es/servlet/tesis?codigo=18730

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ah indeed. I forgot. the reference is in Spanish –  jormansandoval Sep 6 '11 at 23:57
Hehe thanks, yeah I don't speak or read Spanish, so that'll be a problem. However can you tell me your personal opinion or what you conclude in that paper? –  SchroedingersGhost Sep 7 '11 at 0:01
I will to try find you a better paper. –  jormansandoval Sep 7 '11 at 0:56
@SchroedingersGhost First of all, NLSE is a generalisation of Schroedinger eq only on mathematical ground, not physical -- it is a theory claiming that a gas of many bosons can be described by a single 3D creature analogous to weave function driven by an eq that is mathematically equivalent to SE of one particle + this nonlinear term. And it is not even a ab initio theory -- it contains empirical parameter. –  mbq Sep 7 '11 at 11:53

Although it's not a very satisfying (or informative) answer, nonlinear equations are a pain in the butt to solve so we prefer to avoid them whenever possible. It makes sense that the first equation(s) developed to describe quantum systems would be linear, simply because they're the simplest.

That being said, there's no reason that the "true" theory underlying QM would have to be linear. In fact, for exactly the reason you pointed out (i.e. that general relativity is nonlinear), it's commonly believed that we will need some kind of nonlinear theory to properly explain the universe at its most basic level.

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What? No! The waveunction is exactly linear for much the same reasons that probability distribution functions are exactly linear. It is extraordinarily difficult, if not impossible, to deform QM with a nonlinearity. –  Ron Maimon Sep 7 '11 at 2:09
David Zaslavsky, thanks for the answer. Could you elaborate a little on the last part of your post? Or perhaps you will do so in a answer to Ron Maimon and Ill just watch and see if I get my answers from your disagreement –  SchroedingersGhost Sep 7 '11 at 2:44

A classical paper on this is Weinberg's Testing Quantum Mechanics.

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Thanks, I didn't see your post right away. I'll take a look at the paper now –  SchroedingersGhost Sep 7 '11 at 2:54

There are nonlinear versions of the Schrodinger equation that are completely irrelevant to your question. These are like the Gross-Pitaevski equation, they are nonlinear classical field equations that describe the flow of a self-interacting superfluid or BEC. These equations have nothing to do with the evolution of probability amplitudes, and I will not consider them further.

### Probability theory is exactly linear

To understand why the concept of a nonlinear equation for probability amplitudes is not reasonable, and most likely completely impossible, consider first classical probability. Suppose I have a classical equation of motion of the form

$${dx\over dt} = V(x)$$

where the vector field V describes the future behavior as a flow on phase space, coordinatized by x. Now I can ask what is the evolution of a probability distribution $\rho(x)$, if I have incomplete knowledge of the initial position.

The evolution equation is determined by considering the probability of ending in a little box surrounding x'. This probability is the sum of all possible paths that lead to x' times the probability of being at the beginning of the path. This sum gives the probability equation:

$${\partial \rho\over \partial t} = V(x) \cdot {\partial \rho \over \partial x} - \rho(x)\nabla\cdot V$$

The point is that this equation is exactly linear, for fudamental reasons. It is impossible to even conceive of a nonlinear term in the evolution equation of a probability distribution, because the very definition of probability is lack of information, as represented by a linear space.

Note that classical probability distributions are defined on the entire phase space, so they are enormous dimensional linear equations which completely include the nonlinear dynamics if you restrict to delta-function sharp probability distributions on x. The only difference with quantum mechanics is that there are no delta-function sharp distributions in the presence of non-commuting observables on all observables. Otherwise the two types of descriptions are similar

### Quantum mechanics mixes amplitudes and probabilities

If you have a quantum mechanical system, the wavefunction mixes with classical probability in a nontrivial way. If you consider a quantum system of two entangled spin 1/2 particles in a spin singlet, the projection of the wavefunction onto one of the two particles is a density matrix which is a classical probability.

This is extremely important to preserve, because the probabilities are nonlocally correlated, so if there were any way to extract the far-away component of the spin wavefunction, you would be almost certainly be able to use this to signal faster than light, because you can collapse the wavefunction where you are, and the far-away density matrix would then not have a probability interpretation.

These types of nonlinear theories are so difficult to conceive, that Weinberg suggested in the 1960s that quantum mechanics has absolutely no deformation of any kind which is consistent with no-signalling. Although this conjecture is not proved, to my knowledge, it is certainly plausible, and there are no nonlinear deformations which could serve as counterexamples (the link to this paper has just been posted as I write by Oda).

It is wrong to think that there is any nonlinear deformation of the Schrodinger amplitude equation. Such modifications do not exist, and almost certainly cannot exist. If the world obeyed such an equation with a tiny nonlinearity, different Everett branches would become interacting, and we would be able to see the ghosts of our other selves, and other nonsense. It would rule out any form of hidden-variable interpretation of the wavefunction, and it would almost certainly lead to violations of no-signalling.

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Thanks for your answer and towards the end you touch upon exactly why I am asking. There is currenlty no Everettian interpretation that makes sense, it doesn't give us Born Rule, it doesn't give us a ontological structure and when someone tries they run into problems with relativity etc. Hidden variables as in deBroglie Bohm can indeed derive Born Rule, but there you have the nonrelativistic fact of the interpretation... People like Tim Palmer is working on "deeper" underlying theories as described here: physorg.com/news169725980.html Continued in next post –  SchroedingersGhost Sep 7 '11 at 2:39
Gerard 't Hooft is also working on something deeper and more fundamental. And it seems this is the only way to restore determinism. So why couldn't it be nonlinear at a deeper level ? –  SchroedingersGhost Sep 7 '11 at 2:40
To fix Bohm for relativity, you can just consider a bosonic field as the variable which is doing the Bohmian motion. I haven't thought about this for Fermionic fields, but I am sure it is doable. So it is not correct to say Bohm is nonrelativistic. I have read t'Hooft's stuff on quantum mechanics, and I have never been able to understand it (not for lack of trying). The key problem I have is that it is still amplitudes. A question focused on that would be good. If you use an t'Hooft style theory, the wavefunction (or density matrix) should be a derived quantity, obeying a linear equation. –  Ron Maimon Sep 7 '11 at 3:13
I believe the standard Bohmian attitude towards fermionic fields is just to say that only the boson fields are "beables" - and that this is fine since the Higgs boson will still tell you where all the matter is. But Bohmian field theory's main problem with relativity is that it isn't covariant. As Bell pointed out in "Beables for quantum field theory", there is a preferred frame but it is experimentally undetectable, as in electrodynamics before Einstein. –  Mitchell Porter Sep 7 '11 at 4:35
So both the likelihood of constructing a relativistic Bohmian interpertation and discovering that the schreodinger equation is nonlinear has vanished? –  SchroedingersGhost Sep 7 '11 at 6:46