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?
[Physics] Non-local potential
non-localitypotentialquantum mechanicsschroedinger equation
Related Solutions
The answer in short is that when you are dealing with a complex potential, the direct consequence is that you are dealing with complex energy eigenvalues & hence a non-hermitian Hamiltonian.Therefore, for a complex potential, the eigenstates of H are not stationary anymore and that itself negates the initial assumption in the question.
To expand on this, let me reproduce the calculations from "Complex potential well" -- Max Lewandowski, Universität Postdam, 2011 [ http://users.physik.fu-berlin.de/~pelster/Projects/lewandowski.pdf ]
Best Answer
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?