What is the physical reason for Schrodinger equation to be linear? Though in physics many interactions or dynamics are found non linear.


2 Answers 2


It should be understood that physics - at least in it's current form - does not provide answers to "Why these laws?" questions. It can only describe an emergent law from a deeper and more fundamental one. Quantum theory is so far the most fundamental framework we have, so there is no more fundamental "reason" to describe it's structure aside from finding links between various properties of the theory.

The linearity of the Schrödinger equation is a consequence of the more general superposition principle. This principle states that causes add up linearly towards effects and it is postulated.

But what lead us to this postulate? Experimental observations - wave effects such as interference, and certain experiments with spin/polarization of particles. See e.g. the double-slit experiment for interference and Malus's law for polarized light - even though you pass a beam of perfectly polarized photons through a polarizer at a different angle, they can be "linearly decomposed" into photons of different polarizations and a part of them passes through. I.e. the photons that pass through will be polarized in accord with the orientation of the polarizer and this process can be fully understood just through the linearity of quantum-mechanical states.

However, the postulation of linearity was only a consequence of knowing mainly linear wave equations. These effects are conceivable in a theory with slight non-linearities and, indeed, this has been proposed. This article briefly reviews the proposals and their experimental tests that yielded that the proposed nonlinearities are beyond detection scope.

The linked article also provides a "proof" of linearity of evolution of quantum mechanics under some reasonable assumptions. But I would understand it more as a proof of a deeper connection between the usual structure of operators and linear state spaces with the general linearity of quantum mechanical evolution. I.e., the article shows that we would have to switch to a different framework, without states $|\psi\rangle$, linear hermitian operators and their usual interpretation, to include non-linearity in quantum mechanics.

So the conclusion is - it seems that linearity of the quantum-mechanical evolution (aka Schrödinger equation) is a vital part of the structure of the theory. Nevertheless, we can never justify the linearity completely, the main reason for it is "it just works". But that does not preclude the possibility of a paradigm shift including the introduction of non-linearity.

  • $\begingroup$ Could you please fix the reference to the paper you mention? That's the last link in the answer. It seems nobody followed it in all those years, since it goes to the wiki page on polarizer (the same as the link before). $\endgroup$ Sep 17, 2018 at 14:21

It is better seen in the Heisenberg representation. Physical quantities, Observables, are represented by hermitian linear operators. Equation of movement is then (for a non-relativistic massive particle) :

$$ m \dfrac{d^2\hat X(t)}{dt^2} = - \dfrac{\partial V(\hat X)}{\partial \hat X}(t) \tag{1}$$

with the quantization conditions :

$[\hat X(t),m \dfrac{d\hat X(t')}{dt}]_{|t=t'} =i \hbar$

The equation $(1)$ is an equation between operators, so we have :

$$\forall |\psi\rangle, \quad m \dfrac{d^2\hat X(t)}{dt^2} |\psi\rangle = - \dfrac{\partial V(\hat X)}{\partial \hat X}(t) |\psi\rangle\tag{2}$$

Here $|\psi\rangle$ is a constant state (not depending on time).

Equation $(2)$ clearly arises because, in Quantum mechanics, one is using linear operators.

Now, this does not mean that equation $(1)$ is a linear equation relatively to the position operator $\hat X(t)$. This is generally not the case, except for very particular cases (free particle, harmonic oscillator)

We may also use an energy integral equation which is also an equation between operators :

$$ \frac{m}{2} \dot {\hat X(t)}^2 + V(\hat X)(t) = E \tag{3}$$

where $E$ is a constant matrix (not depending on time). We have then :

$$\forall |\psi\rangle, \quad \frac{m}{2} \dot {\hat X(t)}^2|\psi\rangle + V(\hat X)(t)|\psi\rangle = E |\psi\rangle\tag{4}$$

As before, this equation is "linear" in $\psi$, but does not correspond to a linear equation of movement for $\hat X(t)$, except if the potential is zero or at most quadratic in $\hat X$


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