I read that the Schrodinger equation for a non-local potential is given by $$-\frac{\hbar^2}{2m}\nabla^2\psi(x)+\int V(x,x')\psi(x')dx'=E\psi(x).$$ In case of a local potential, $$ V(x,x')=V(x)\delta(x-x').$$ In what sense, one is 'local' and the other is 'non-local'? What does a non-local potential mean physically? When does such a potential arise?
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2 Answers
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In your nonlocal potential, the potential that your partical feels at each point in space depends not on the value of some single function, but the sum (Integral) of all the values of a function defined over all space. This comes up a lot in multi-particle QM. Consider trying to find the electron repulsion in the helium atom $1/{r_{12}}$. Both electrons are delocalized over all space. So if I wanted to know the average value of the electron repulsion on electron 1 at point r', I would need to calculate the sum of all the electron repulsion of electron 2 at each point in space, so it becomes a non-local interaction and depends on the shape of the wavefunction for the other electron.
Contrast that with the harmonic oscillator V=$1/2kx^2$. If I want to know the potential energy interaction of the partical at point x', I just need to evaluate V(x).
Does that help?
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$\begingroup$ I recognise this answer is quite old, but I was wondering if you might be able to provide some references for linear Schrodinger equations with nonlocal potentials? $\endgroup$ Commented Jul 12, 2021 at 13:12
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$\begingroup$ In the helium case, the potential is the Coulomb interaction V(r12) = 1/r12, which is local, so this example is not correct. $\endgroup$– ragnarCommented Sep 2, 2022 at 2:13
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$\begingroup$ @Bcpicao. In the 1d harmonic oscillator case, x is not a fixed quantity either. The meaning of “local potential” is a convention, and my understanding of the standard sense of the term is that a local potential is nonzero only if the vector r12 is the same in the bra and ket states. $\endgroup$– ragnarCommented Mar 18, 2023 at 0:19
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1$\begingroup$ @Bcpicao In this example, if you do the integral over r2, you'll get an effective potential $V(r_1)$, which is local as long as you distinguish particles 1 and 2. Specifically, $V(r_1)=\int d^3r_2 \psi^*_2(r_2) \frac{q^2}{|r_2-r_1|} \psi_2(r_2)$. As the OP states, a non-local potential would have the form $V(r_1',r_1)$. Here the prime indicates that the potential felt by particle 1 at position $r_1$ depends on the wave function of particle 1 at other locations $r_1'$. $\endgroup$– ragnarCommented Mar 19, 2023 at 2:04
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$\begingroup$ I have realized, pondering your observations, that I have not understood this concept as well as I thought I had, I have thus retracted by previous comments, thank you. $\endgroup$– BcpicaoCommented Mar 31, 2023 at 18:59
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Let me make some complements. What James Johns said about Helium atom is a quite a good example.
Basically, if we are capable of solving the many body Schrodinger equation directly, then no non-local potential appears. We have no non-local interaction in the nature. However, if we want some effective models or make some approximations, sometimes we have non-local potential.
The key point is that, for usual local potential, no matter how the particle behaves, the interaction never changes. But in the case of non-local potential, the behavior of the particle itself have back effect on the interaction.
For instance, still the Helium atom, if we solve the three body alpha+2electrons directly, then no way to confront a non-local potential. But if we construct an equivalent two body model about one single electron, then we will face a non-local potential.
