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Recently I have done some simulations with a linear system that behaves similarly to the wave equation. Now I have a bit of trouble understanding when interaction is possible in linear theories, especially in the context of quantum mechanics. If we, for instance, consider a linear theory like the maxwell equations we have no interaction between two light waves (as far as I recount).

I have read the general answer here, however I still do not get how one would formulate the interaction between e.g. two electrons described by a wave function. If we take the time dependent Schrödinger Equation:

$$i \hbar \frac{\partial}{\partial t} | \psi \rangle = \hat{H} | \psi \rangle $$

How would the Hamiltonian be formulated to incorporate interaction and still be linear? The often seen expression

$$\hat{H} = \hat{T} + \hat{V}$$ where $$\hat{V} = \frac{e^2}{4 \pi \epsilon_{0}} \frac{1}{\left| \vec{r}_{1} - \vec{r}_{2} \right|} $$

does not make so much sense to me here, because I do not understand how $\vec{r}_1 , \vec{r}_2$ would be defined at different times in the time dependent Schrödinger equation without involving the wave function itself in a non linear fashion.

Qmechanic
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amh23
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1 Answers1

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Potential energy operator is a linear operator. In position representation it amounts to multiplication of the wavefunction by the potential energy:

$$\langle \vec r_1,\vec r_2|\hat V|\psi\rangle=V(\vec r_1,\vec r_2)\psi(\vec r_1,\vec r_2).\tag1$$

When we talk about the complete Schrödinger's equation in position representation, inclusion of this term still retains the linearity of the equation:

$$-\frac1{2m}(\Delta_1+\Delta_2) \psi+V\psi=i\hbar\frac{\partial\psi}{\partial t}.\tag2$$

Although the expression for $V(\vec r_1,\vec r_2)$ is non-linear in $\vec r_1,\vec r_2$, this fact is irrelevant: the equation is linear in $\psi$. Note how the only way this function enters $(2)$ is as the first power of the function, acted on by linear operators. This in particular results in the fact that any linear combination of any solutions of $(2)$ is also a solution.

Ruslan
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