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Considering the double slit experiment with a charged particle, after the particle passes through the slits, do the two portions of the wavefunctions feel the electromagnetic attraction of the other portion?

More generally, can the potential term in the Schrödinger equation be dependant on the wavefunction itself?

On the one hand it is easy to imagine an electron wave splitting up, each part carrying a proportion of the charge. On the other hand, the wavefunction is a probability distribution, which implies that there is no interaction.

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Wavefunctions are mathematical constructs that live on pieces of paper that were scribbled on by physicists. We usually do not believe that these symbols interact with themselves. Apologies, I couldn't resist the urge to point out the flaw of mistaking a human description of reality for reality. The correct answer to your question is: no, one can not construct generally useful models of reality outside of limited problem sets in molecular and solid state physics by using semi-classical mean field approximations of a non-relativistic wave function. For certain problems, however, it works. – CuriousOne Aug 12 '14 at 20:19
up vote 11 down vote accepted

"More generally, can the potential term in the Schrödinger equation be dependent on the wavefunction itself?"

The answer is negative: The resulting Schroedinger-like equation would turn out to be non-linear. It would not be associated with a unitary time evolutor (the self-adjoint generator, the Hamiltonian operator, would not be defined) against some basic postulate of quantum mechanics.

However non-linear Schroedinger-like equations where the potential depends on the wavefunction itself have great physical interest, but they describe systems of many particles. I am referring, in particular, to the so-called Gross-Pitaevskii equation.

In the double slit experiment, the "self-interaction" of the wavefunction is a linear phenomenon, the superposition principle applied to the two parts of the same wavefunction emitted by the two sources (the slits). Non-linearity arises as soon as you compute the probability amplitude just in view of the mathematical procedure.

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+1, the answer (at least part of which) i never gave :)) – Nikos M. Aug 12 '14 at 20:39
...well, thanks! – Valter Moretti Aug 12 '14 at 20:41

This answer was supposed to answer sth else the OP was not asking, in any case i will leave it here

Yes it is called self-energy, for example the interaction of an electron (represented as a wavefunction in quantum field theory) with its own electric field, which is calculated using techniques known as renormalization (among other jargon like off-shell and on-shell mass etc.)

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I believe the OP's question is about a possible self-interaction term entering the Schrodinger equation in non-relativistic quantum mechanics. In this context this answer is incorrect, see V. Moretti's answer. Furthermore, it is at best imprecise and at worst totally misleading to claim that "an electron [is] represented as a wavefunction in quantum theory". In QFT the electron may be represented as an excitation of a quantum (Dirac) field, for example. A wavefunction is just a list of probability amplitudes for different states, in both QFT and non-relativistic QM. – Mark Mitchison Aug 12 '14 at 20:04
@MarkMitchison, on the other hand i think this is what the OP's question asks for (the OP himself can select whatener he wants). Now the electron in QFT is a wavefunction (in 2nd quantization), as such it is not misleading much so incorrect, we can discuss naming conventions if needed, but i dont think we will disagree much there. That is what 2nd quantization is all about taking a (non-relativistic) wavefunction and interpret as a particle field (the fact remains of a wavefunction representing a particle or particles), comment accepted – Nikos M. Aug 12 '14 at 20:10
The OP asks: "do the two portions of the wavefunctions feel the electromagnetic attraction of the other portion?". To which you have answered "yes", quite unambiguously in your first sentence. I think this answer is wrong, whether we are talking about QFT or not. About the meaning of the term "wavefunction", of course it's down to definitions. But in my experience thinking of the electron field as a "quantised wavefunction" leads to all kinds of confusion. – Mark Mitchison Aug 12 '14 at 20:18
@MarkMitchison, ah yes, you are right sorry, read it too fast and carried away by the title of the question, in any case i will leave the answer even if it (unfortunately) has negative votes, since by the title one can reach this question while asking for what i tried to answer – Nikos M. Aug 12 '14 at 20:25
...indeed, this is one of the many kinds of confusion I was referring to! :) – Mark Mitchison Aug 12 '14 at 20:32

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