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)?


5 Answers 5


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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    $\begingroup$ 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 $\endgroup$ Commented Sep 7, 2011 at 2:39
  • $\begingroup$ 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 ? $\endgroup$ Commented Sep 7, 2011 at 2:40
  • $\begingroup$ 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. $\endgroup$
    – Ron Maimon
    Commented Sep 7, 2011 at 3:13
  • $\begingroup$ 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. $\endgroup$ Commented Sep 7, 2011 at 4:35
  • $\begingroup$ So both the likelihood of constructing a relativistic Bohmian interpertation and discovering that the schreodinger equation is nonlinear has vanished? $\endgroup$ Commented Sep 7, 2011 at 6:46

@Ron Maimon has given the canonical answer to this: the wavefunction is probabilities, and to preserve probabilities one must have a linear equation (indeed, also a norm-preserving evolution operator).

I offer another viewpoint, in the style of how Einstein thought about relativity, i.e. two postulates. The postulate is that it is not possible to solve NP-complete problems in polynomial time. Abraham and Lloyd showed that if quantum mechanics were non-linear at all, then this would be possible.

Aaronson has a nice paper, the start of which references a large literature on why quantum mechanics has to be the way it is.

  • $\begingroup$ Thanks yeah I discovered teh Aaronson paper right before you posted it and now feel that I have the answers to my questions. Thanks $\endgroup$ Commented Sep 8, 2011 at 3:28
  • $\begingroup$ @genneth: the postulate could be stated more broadly: it should be impossible to compute anything with a physical system exponentially faster than we can compute it with a Turing machine. Why not state it this way? Because Shor showed us that quantum mechanics can do that! So this principle is not very compelling. If quantum computers can factor, What is the philosophical objection to them solving NP complete problems too? $\endgroup$
    – Ron Maimon
    Commented Sep 11, 2011 at 15:42
  • $\begingroup$ @Ron Maimon: you answered it yourself: because factorisation of integers is not NP-complete. And it's precisely because of this that one states NP-complete and not just exponentially faster than a Turing machine. In addition, the fact remains that we don't have non-trivial lower bounds on factorisation. $\endgroup$
    – genneth
    Commented Sep 11, 2011 at 17:12

In addition to the classic Weinberg paper cited above, there's this shorter version, and then follow ups by Peres 1989 on how it violates the 2nd law, by Gisin on how it allows superluminal communications, and by Polchinski on how it would allow for an 'Everett' phone.

More recently, there's this mathematical argument against nonlinear QM by Kapustin.


Physicists like linearity because it is simpler. Sure you have heard the joke of the physicist and the spherical cow. Luckily for us the more interesting phenomena in Nature is nonlinear.

There are several reasons to consider nonlinear extensions of the Schrödinger equation, the most fundamental to me is that classical mechanics requires it!. Indeed using the Hamilton-Jacobi formulation we can cast classical mechanics in a wavelike formalism

$$i\hbar \frac{\partial \Psi}{\partial t} = \left(\frac{-\hbar^2 \nabla^2}{2m} + V - Q \right) \Psi$$

which is a nonlinear equation since $Q=Q(\Psi)$. Precisely its nonlinearity breaks the superposition principle and allows the description of classical systems. It is not strange that the quantum community working in decoherence and classicality use non-linear Schrödinger equations of generic form

$$i\hbar \frac{\partial \Psi}{\partial t} = \hat{G} \Psi$$

with nonlinear operator $\hat{G} = \hat{H} + \mathrm{nonlinear{-}corrections}$. It is worth to mention that the nonlinear terms can be chosen to preserve the norm of the state vector $||\Psi||^2$. I.e., the claim that linearity is obligatory to conserve probabilities is untrue.


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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    $\begingroup$ 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. $\endgroup$
    – Ron Maimon
    Commented Sep 7, 2011 at 2:09
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    $\begingroup$ 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 $\endgroup$ Commented Sep 7, 2011 at 2:44

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