# How does the linearity of the Schroedinger equation reflect the interactions?

There is a common lore that linear equations describe non-interacting systems, why non-linearities correspond to non-trivial interactions. My (loose) question is how is that compatible with the Schroedinger equation, which is always linear.

Take a classical particle described by the EOM $$m\ddot{x}(t)+V'\big(x(t)\big)=0$$ If $$V(x)$$ is quadratic the EOM is linear which physically corresponds to the absence of interactions in the system. Interactions are described by potentials that are not quadratic. This generalizes to many degrees of freedom.

Quantum-mechanically the particle is described by the wavefunction $$\psi(x,t)$$ which obeys the linear Schroedinger equation $$\left[i\hbar \partial_t -\frac{\hbar^2 m^2}{2}\partial_x^2+V(x)\right]\psi(x,t)=0$$

Again, this generalizes to many degrees of freedom and the equation always stays linear. Of course there is a correspondence between the quantum problem and the classical problem so that in the $$\hbar\to0$$ limit they must ultimately describe the same physics. The correspondence is somewhat obscured by the fact that one typically asks different questions in the classical and quantum cases -- classically one looks for trajectories while in QM one usually looks for eigenstates of the Hamiltonian.

My question is surely not well-defined, but I would like to get the most straightforward and intuitive explanation for how the Schroedinger equation, which is always linear, manages to exactly describe interacting systems.

Of course if I were to formulate everything precisely, including the definition of interaction, there would be no contradictions or tensions. However I do not see what is a good way to do this. I hope that my confusion will resonate with some people and they could resolve it. For example quadratic Largangian == free theory is clear in the context of a field theory, but perhaps I use it outside the intended scope.

• Schrödinger equation is not always linear. Check Kerr effect. You can always choose an appreciate potential. There is even a Wikipedia page for it en.wikipedia.org/wiki/Nonlinear_Schr%C3%B6dinger_equation – Paradoxy Feb 23 at 11:03
• The Schrodinger equation is always linear in quantum mechanics. There are proposals to study a non-linear Schrodinger equation for various reasons, but these proposals intentionally modify quantum mechanics. – Andrew Feb 23 at 12:26
• @Paradoxy The interaction may not be linear in the fields but the Schrodinger equation is still linear: a potential in $1/r$ isn’t linear but the TISE remains linear as a differential equation. – ZeroTheHero Feb 23 at 13:01
• – Chiral Anomaly Feb 23 at 14:25
• @Paradoxy In quantum mechanics, the Schrodinger equation is always linear in $\psi$ for fundamental reasons. But you are right that in optics, when the Schrodinger equation appears as an approximation to Maxwell's equations, additional corrections can appear which are not linear in $\psi$. – Andrew Feb 24 at 15:46

There are different equivalent formulations of quantum mechanics, and how an interaction is represented in each formulation is different.

In the Schrodinger picture, operators are time independent and the state (or wavefunction) $$\Psi$$ depends on time. The Schrodinger equation governs the dynamics of the wavefunction:

$$\begin{equation} i \hbar \frac{\partial \Psi}{\partial t} = H \Psi \end{equation}$$ Note that the Schrodinger equation is always linear in the wave function $$\Psi$$, even in more complicated situations such as when you generalize to multi-particle states, or field theory. Interactions appear via terms in the Hamiltonian operator $$H$$ which are higher-than-quadratic-order in the position operator $$x$$ and momentum operator $$p$$.

For example, consider the following Hamiltonian:

$$\begin{equation} H = \frac{p^2}{2 m} + \frac{m\omega^2}{2}x^2 + \lambda x^3 \end{equation}$$

• If $$\lambda = 0$$, then the Hamiltonian is the same as a harmonic oscillator. When we generalize this case to field theory, we would say there are no interactions. Note that the classical equations of motion following from this Hamiltonian (the equations of motion for a harmonic oscillator) are linear in $$x$$. We can solve the Schrodinger equation exactly in this case.
• If $$\lambda \neq 0$$, then when generalizing to field theory we would say there is a cubic interaction. Note that the classical equations of motion are not linear. We cannot usually solve the Schrodinger equation in cases like this. However note that the Schrodinger equation is still linear in the state $$\Psi$$.

In the Heisenberg picture, operators are time dependent and the state $$\Psi$$ is time-independent. For an operator $$\mathcal{O}$$ without explicit time dependence (such as the position or momentum operators), the equations of motion

$$\begin{equation} \frac{{\rm d}\mathcal{O}}{{\rm d}t} = \frac{i}{\hbar}[H,\mathcal{O}] \end{equation}$$

If we take our example above, and use the fact that $$[x,p]=i \hbar$$, then we would find the position and momentum have the Heisenberg equations of motion

$$\begin{eqnarray} \frac{{\rm d} x}{{\rm d}t} &=& \frac{i}{\hbar}[H,x] = \frac{1}{m} p \\ \frac{{\rm d} p}{{\rm d}t} &=& \frac{i}{\hbar}[H,p] = - m \omega^2 x - 3 \lambda x^2 \end{eqnarray}$$

This system of equations probably more closely matches your intuition. When $$\lambda=0$$, the Heisenberg equations of motion are linear (in the position and momentum operator), and non-interacting. When $$\lambda \neq 0$$, the Heisenberg equations of motion are not linear and are interacting.

Finally I'll mention the Feynman path integral approach. There the basic quantity is the path integral

$$\begin{equation} Z = \int \mathcal{D} x e^{\frac{i}{\hbar} S[x]} \end{equation}$$ where $$S$$ is the classical action. The measure $$\mathcal{D} x$$ is constructed via a limiting process and represents an integral over all classical paths.

In our example, the action is $$\begin{equation} S = \int {\rm d} t \left( \frac{m}{2} \dot{x}^2 - \frac{m}{2} \omega^2 x^2 - \lambda x^3 \right) \end{equation}$$

When $$\lambda=0$$ (free theory), the path integral is Gaussian and can be done exactly. When $$\lambda \neq 0$$ (interacting theory), the path integral is not Gaussian and usually can't be done exactly. Usually at this point a physicist would expand the exponential in powers of $$\lambda$$, assuming it is small, and use perturbation theory and Feynman diagrams.

This is not an exhaustive list of all possible formulations of quantum mechanics, but these are the most common, and at least should give you the idea that a physical idea (like whether a particle has interactions) can be mathematically represented in different ways. "A particular equation being linear or not linear" is a statement about the formalism and can depend on how you formulate the problem, whereas "a particle having interactions or not" is a statement about physics and cannot.

• The most illuminating part of you answer for me is the reference to the Heisenberg picture where EOM are clearly non-linear. Interestingly, one can do a similar trick in reverse in the classical mechanics. It is possible to introduce a "wavefunction" that will convert usual dynamical EOM into linear equations on the "wavefunction". This is known as the Koopman - von Neumann formulation of classical mechanics. – Weather Report Feb 24 at 14:45
• @WeatherReport Absolutely. Also relevant is the Hamiltonian-Jacobi equation, which is a kind of wave equation describing classical particles that can be understood as the WKB limit of the Schrodinger equation. Ie, we write the wavefunction as $\psi=A(x)e^{iS/\hbar}$, assume $A$ is slowly varying, and apply the Schrodinger equation. – Andrew Feb 24 at 15:20
• Indeed, but the Hamilton-Jacobi equation is still non-linear so it does not directly resolve my imaginary conundrum. By the way, the Hamilton-Jacobi formulation seems close in spirit with the Koopman-von Neumann formalism and probably there is a simple relation between the "classical wavefunction" and the action $S$, but I have not seen this discussed. – Weather Report Feb 25 at 7:29
• @WeatherReport The Hamilton-Jacobi equation is not linear in the action $S$, but the Schrodinger equation also is not, since the Schrodinger equation is linear in $\psi$ which is non-linearly related to $S$ via $\psi = A e^{i S/\hbar}$. – Andrew Feb 25 at 7:41