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In the Schrodinger equation, we have some potential $V(x)$. But generally, there is some uncertainty in the position with solutions to Schrodinger's equation. Classically, we would say that a particle at position $x$ is associated with the potential at that point -- is there a quantum analog when the position wavefunction isn't localized? What's to say if a particle is in a potential or not if it doesn't have a well-defined position?

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The schrodinger equation tells us that a particle interacts with a potential if its wavefunction has a non-zero amplitude in the region enclosed by the potential. That is to say, the wavefunction's time evolution is independent of the potential energy's value in regions where the wavefunction is zero.

However, the wavefunction may have non-zero amplitude in regions inside and outside the potential simultaneously. Therefore, we can not always say without performing a measurement whether or not the particle is inside or outside the potential.

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  • $\begingroup$ What you've said is correct, but it doesn't really seem to answer my question. I'm wondering how we can say whether or not a particle should interact with a potential if its position isn't well-defined. I have a hunch that the answer lies somewhere in scattering theory, but the text that I've looked at gives a description of the position representation of the wave function scattering, which seems to be even more confusing since I thought that the wave function was just an abstract representation of the state, not representative of the actual position. $\endgroup$ – danielunderwood Mar 9 '18 at 1:05
  • $\begingroup$ I think I understand your question better now. I rewrote my response. $\endgroup$ – creillyucla Mar 12 '18 at 18:09

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