In Wavefunction in quantum mechanics and locality, wavefunction is constrained by $H = \sqrt{m^2 - \hbar^2 \nabla^2} $, and taylor-expanding $H$ results in:
$$ H = \dots = m\sqrt{1 - \hbar^2/m^2 \cdot \nabla^2} = m(1-\dots) $$
While the person who asked this question accepted the answer, I was not able to understand fully.
Why would $\nabla^{200}$ in the taylor-expansion of $H$ be so problematic (200 can be replaced by any arbitrary number) - resulting in non-locality? Isn't this just some exponentiation of dot product of gradient?
Answer
Non-locality comes from presence of infinite many terms in that expansion. To see that, lets assume we are applying the non-polynomial function $f(\vec{z})$ of $i\nabla$ on any function $\psi(\vec{x})$: $f(i\nabla) \psi(\vec{x})$. Assuming that $f(z)$ is "nice enough" to have the Fourier representation $f(\vec{z}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{i \vec{k}\cdot \vec{z}}$ for suitable transform function $F(\vec{k})$. Then:
$f(i \nabla) \psi(x) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{i \vec{k}\cdot (i \nabla)} \psi(\vec{x}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{- k\cdot \nabla} \psi(\vec{x}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) \psi(\vec{x}-\vec{k})$
This is the superposition of values of $g$ calculated at points different than $\vec{x}$. that is the non-locality.
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