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Consider Schroedinger equation, which I write in the form $$ (\mathscr{L}+V)\psi=0$$where $\mathscr{L}$ is the kinetic and time-derivative operator. Now, imagine I have two point sources 1,2 with potential $V_i(r)=f_i(r) \delta (r-r_i)$. Let $\psi_1$ (resp $\psi_2$) be the solution when I switch on only source 1 (resp 2). It is clear that when both sources are switched on then potential $V=V_1+V_2$. However, unlike the sum of amplitudes, $\psi\neq \psi_1+\psi_2$ , instead it is $$\psi=\psi_1+\psi_2-\frac{1}{\mathscr{L}+V}(V_1\psi_2+V_2\psi_1)$$ Consequently, every other quantities that can be derived from $\psi$ won't respect the usual linearity. I could use similar setup in double-slit experiment , where $V_i$ is infinite except at the location of slit $i$ (The total potential $V$ is infinite except at location of slits 1 and 2). The argument above equally applies for classical theory like Electro-magnetism (as shown here).

It seems from above argument that if EM field (or wavefunction) from multiple sources are given, I cannot always vectorially add them to get the overall field strength at a point. So when should the vector sum of field strength/ sum of amplitudes be valid?

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  • $\begingroup$ Your formalism is unclear, but it looks like $V_i$ is defined to project out any and all solutions except $\psi_i$, i.e. it includes, unstated, some type of a projector to just $\psi_i$, in which case the last term of your 2nd displayed equation vanishes by definition. Can you be more precise in your question? $\endgroup$
    – Cosmas Zachos
    Commented Nov 22, 2023 at 14:19
  • $\begingroup$ @CosmasZachos By projector, do you mean writing $V_1\to V_1\otimes \mathbb{1}$ and $V_2\to \mathbb{1}\otimes V_2$ so that the cross-terms cancel out? $\endgroup$
    – paul230_x
    Commented Nov 22, 2023 at 15:47
  • $\begingroup$ I mean they ensure, e.g., $V_1 \psi_2$ is zero, by construction. $\endgroup$
    – Cosmas Zachos
    Commented Nov 22, 2023 at 15:51
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    $\begingroup$ This new ref, Fig 2 settles the issue. The loop involved cannot be considered as part of $\psi_1$, as originally defined, with a plugged second slit: that plug is the projection. The authors of your first reference use “general” for “generic”. Not an implicit failure of a general proof! There is no paradox here. Once you try to appreciate the definition of your sum potential, the “paradox” collapses…. $\endgroup$
    – Cosmas Zachos
    Commented Nov 24, 2023 at 12:01
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    $\begingroup$ Oh I see it now! Thank you so much for your time! $\endgroup$
    – paul230_x
    Commented Nov 24, 2023 at 12:20

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