I'm trying to build a quantum memory system that uses the superposition principle to model specific phenomenon I am trying to predict.

Is the universe linear? The superposition principle would apply in all cases if the universe is linear, which would make sense as the quantum theory is the most accurate theory so far.

If the universe is linear, why is it? Why not semi or non linear?

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    $\begingroup$ General Relativity isn't linear ... $\endgroup$ Aug 2, 2012 at 17:37
  • $\begingroup$ Related: physics.stackexchange.com/q/16262/2451 $\endgroup$
    – Qmechanic
    Aug 2, 2012 at 19:14
  • $\begingroup$ @JohnRennie I'm a bit surprised to see that general relativity can be used to reason about QM, because I heard that we still have no theory which satisfactorily unifies QM and GE. I think the wavefunction collapse during a measurement is the place where QM becomes nonlinear. And the fact that the wavefunction is normally normed makes the remaining linearity pretty useless anyway. $\endgroup$ Aug 2, 2012 at 22:13
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    $\begingroup$ @ThomasKlimpel: I'm not implying that we need to include GR to understand QM. However the question wasn't "is QM linear", it was "is the universe linear". You can't understand the universe using just QM. $\endgroup$ Aug 3, 2012 at 5:43
  • $\begingroup$ What about the Standard Model? $\endgroup$
    – jim
    May 16, 2016 at 17:23

1 Answer 1


This question is very hard to answer at a fundamental level, because quantum mechanics seems to be exact so far, yet one cannot be sure in the scientific sense without confirmation that nontrivial quantum computation is possible. If this is so, then one would have to renounce any classical descriptions, at least within the bounds of scientific reason, and it is likely that quantum mechanics is the final word. If this is not so, then of course, all bets are off.

One should say at first that any system can be converted to an exactly linear system by making probability distributions. Suppose you have a nonlinear equation, for example:

$$ {dx\over dt} = {1\over x^2 + 1} $$

Now you have a nonlinear relation between $x(0)$ and $x(t)$. Suppose you say "I don't know $x(0)$, but I have an idea that it is described by a probability distribution $\rho(x_0)$". Then you can say, knowing how x is changing, how the probability distribution $\rho$ is changing. The equation for $\rho$ is always linear, even if the underlying dynamics for x is nonlinear.

The reason is simple--- if you have a probability distribution which is a linear combination of two others:

$$ \rho(x_0) = \alpha \rho_1(x_0) + (1-\alpha)\rho_2(x_0) $$

you can interpret this distribution by the following process: flip a coin which has probability $\alpha$ of landing heads, then pick $x_0$ according to $\rho_1$ if you got heads, and according to $\rho_2$ if you got tails. Then evolve $x_0$ according to the nonlinear equation.

You can also pick an $x_0$ from $\rho_1$ and pick and $x_0'$ from $\rho_2$ and evolve both of these, and then choose one of the outcomes using the same coin. Clearly (meaning it is equivalent to the axioms of probability) flipping the coin at the beginning to find $x_0$ is the same as evolving a pick from $\rho_1$ and $\rho_2$ without knowing which is which, and flipping the coin at the end. The reason is that we know there is a secret $x_0$ underneath, and the probability of the $x_0$ being what it is doesn't care when we find out the answer as to which it was, so long as we find out the answer eventually.

This tells you that the time evolution operator for probability distributions obeys:

$$ T\rho = \alpha T \rho_1 + (1-\alpha)T \rho_2 $$

This is the fundamental linearity of probability theory.

Similarly, for GR, and you consider probability distributions on metrics, the equation for the probability distribution is always linear. This doesn't make solving the equations any easier, because the linear space is so much more enormous--- the difficulty of solving for the evolution of probability distributions on classical systems, is by Monte-Carlo roughly the same as solving the classical equations many times.

The thing about quantum mechanics is that it is like probability, in that the equation of motion is always linear. It is also like probability in that it is formulated over the space of all configurations, so the number of real numbers you use grows exponentially with the size of the system.

But Quantum mechanics is not like probability in that it has interference. This means that you can't use the ignorance justification for the perfect linearity--- you can't say "the reason the wavefunction evolves perfectly linearly is because it is representing the ignorance of coin-tosses at the beginning", because the only reasonable calculus for ignorance is probability.

It also means that it is not possible to use Monte-Carlo in general to simulate quantum mechanics with multiple samples (although it is amazing to what extent you can do quantum monte-carlo, you need to do an impossible analytic continuation to turn it into general real space results). This is almost a theorem, since Monte Carlo can't have exponential speed-up for classical computation. So the existence of Shor factoring and other exponential speed-ups means you can't simulate a quantum computer efficiently with a stochastic classical one.

But quantum mechanics is still mixed up with probability, because the probabilities are the square of the wavefunction values. In general, the diagonal elements of the density matrix are probabilities when you measure, and need to obey exact classical linearity of probability theory when you have multiple measurements.

This restriction means that you have a very hard time imagining quantum mechanics having a slight nonlinearity, since it is difficult or impossible to make sure that the probability theory included in the classical limit is going to be exactly linear (in the limit of classical measurements) if the underlying quantum dynamics is not exactly linear. There are attempts from time to time to collapse the wavefunction using nonlinearity, but these are generally misguided. If you do a nonlinear interaction on a probability like entity like the wavefunction, you get interaction between different possible worlds, or Everett branches, not a sensible collapse to one of the possible worlds.

So yes, with scientific standards of knowledge, if quantum mechanics is exact, it is linear. If quantum mechanics fails, it is likely because it is a stochastic description of some underlying variables, but what this could be is very difficult to imagine.

  • $\begingroup$ About your last paragraph, isn't that what Bohmian mechanics does? $\endgroup$
    – MBN
    Aug 3, 2012 at 7:45
  • $\begingroup$ @MBN: In Bohmian mechanics, there are wavefunctions plus particles, this is a computational description which is marginally larger than quantum mechanics itself. So I don't consider it a real option, since you might as well just consider the wavefunction and do Everett, if you are philosophically able. A real alternative would reduce the computation to 10^80 bits from 10^10^80. $\endgroup$
    – Ron Maimon
    Aug 3, 2012 at 17:45
  • $\begingroup$ I fully agree that the partial differential equations related to QM are linear. However, the wavefunction collapse is part of the old Copenhagen interpretation of QM, and such a collapse is not linear. Also the interference phenomenon is quadratic instead of linear. I guess the Everett interpretation allows you to decouple the underlying linear partial differential equation from the nonlinear phenomenons, but this probably comes at the price to blow up the size of the phase space. $\endgroup$ Aug 3, 2012 at 23:51
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    $\begingroup$ @mike4ty4: A quantum computer which factors an arbitrary 10,000 digit number is sufficient to rule out all classical 10^140 bit computers even with extra miracles (like order 10^sqrt(n) factoring realized easily in nature), so that's the condition I choose for saying "quantum mechanics is experimentally proved to be exponentially large". That's enough for me to say that you've proven quantum mechanics is exact for all intents and purposes, since exponentially large alternatives are not particularly interesting to me personally. $\endgroup$
    – Ron Maimon
    Nov 12, 2014 at 2:30
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    $\begingroup$ Or, say, let's not. $\endgroup$
    – Ron Maimon
    Nov 12, 2014 at 2:34

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