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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.

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  • $\begingroup$ Thank you for the link. I guess not really as I have trouble to understand the notation in this answer. How can the variables $\vec{r}_1, \vec{r}_2 $ of $ | \psi \rangle$ be the same as the $\vec{r}_1, \vec{r}_2$ in the interaction potential? My guess would have been that in $\hat{V}$ they represent something like the "position" of the two electrons (however that would be defined) but as a variable of the wave function the assign a complex value to every point in space. $\endgroup$
    – amh23
    Commented Sep 2, 2020 at 19:26
  • $\begingroup$ Ok I guess now I see it. As one applies $\hat{V}$ with $\psi$ this involves the wave function at each point in space... correct? $\endgroup$
    – amh23
    Commented Sep 2, 2020 at 19:37
  • $\begingroup$ Well, that's how the wave function $\psi(\vec{r}_1,\vec{r}_2,t)$ works. There is not "the position" of the two electron, because they can be everywhere (with certain probabilities as given by $\psi$). $\endgroup$
    – Thomas Fritsch
    Commented Sep 2, 2020 at 19:37

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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.

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  • $\begingroup$ Thanks now I get it, I had some trouble to understand that only in $V \psi$ the current state of the electrons comes into play and not in $V$ alone. $\endgroup$
    – amh23
    Commented Sep 2, 2020 at 19:38
  • $\begingroup$ @amh23 in Schrödinger's picture, the time evolution is encoded in the wavefunction. If the potential explicitly changes with time, then the potential operator does too. But in your case the potential doesn't explicitly depend on time: only on particle positions. And the positional state is completely encoded in the wavefunction. $\endgroup$
    – Ruslan
    Commented Sep 2, 2020 at 19:39
  • $\begingroup$ Thank you again this resolves the knot I had in my head :) $\endgroup$
    – amh23
    Commented Sep 2, 2020 at 19:42

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