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Infinite sums of increasingly higher-order derivatives, when present in Lagrangians, are typically taken as a sign of nonlocality. This is supposed to rule out fractional, negative and exotic (for example, $\exp(a\partial)$) derivatives from use in local field theories.

As far as I can tell, the intuition behind this is that for an analytic function, having all of the derivatives at a point means having the whole function. In the answer to this question, for example, Matteo Beccaria points out that, (implicitly for an everywhere analytic $f$), $$ \exp(a \partial)f(x) = f(x+a) $$ Because, $$ \exp(a \partial)f(x) = \sum_{n=0}^{\infty} \frac{a^n \partial^n}{n!}f(x) \\ f(x+a) \approx \sum_{n=0}^{\infty} \frac{\partial^nf(x)}{n!}a^n\\ $$ Now, what bothers me about this is the assumption that $f$ is analytic. In addition to there being functions which are infinitely differentiable but not analytic anywhere, there are even important wavefunctions in physics that are not even infinitely differentiable. (For example, states in the infinite square well.)

A perhaps greater concern of mine with the idea of banning non-analytic functions is that if $f$ is everywhere analytic, the meaningfullness of locality goes out the window, so to speak. Given any finite "snippet" of an analytic function, I can tell you the rest of the function - which means that if $f$ has to be analytic, then the "information in one place is unrelated to information in another place" idea of locality stops working entirely. Another branch of this conceptual weed is that the assertion that $f$ is everywhere analytic is a nonlocal claim! You cannot check whether or not $f$ is analytic, or even analytic anywhere, without scanning over at least a finite patch to check if it agrees with the Taylor expansion.

I can see two possible resolutions to this issue. Maybe there is a way to prove that infinite terms in lagrangians lead to nonlocality, without assuming that $f$ is analytic everywhere. Another possibility is that there is a way to think of locality that doesn't break down in the face of requiring that $f$ be analytic.

What is the resolution to this apparent contradiction?

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You are correct that the usual quantum textbook derivation of the translation action $$ e^{a\partial_x}f(x)=f(x+a) $$ is totally bogus since we cannot assume analyticity. A route that applies to much larger class of functions is via Fourier transforms. Assume that we can write $$ f(x)= \int_{-\infty}^{\infty} \tilde f(k) e^{ikx} dk $$ then we can proceed as follows $$ e^{a\partial_x}f(x) {=} \int_{-\infty}^{\infty} \tilde f(k) \left(\sum_{n=0}^\infty \frac{a^n \partial_x^n}{n!}\right) e^{ikx} dk\\ =\int_{-\infty}^{\infty} \tilde f(k) \left(\sum_{n=0}^\infty \frac{(iak)^n}{n!}\right)e^{ikx} dk\\ = \int_{-\infty}^{\infty} \tilde f(k) e^{iak } e^{ikx} dk\\ = \int_{-\infty}^{\infty} \tilde f(k) e^{ik(x+a)} dk\\ =f(x+a). $$ There are interchanges of sums and integrals here, so there are still conditions on $f$, but this is OK in most QM applications.

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  • $\begingroup$ What about $f(x) = \exp(-x^2-x^{-2})$ at $x=0$? Would that not be a counterexample? (Assume that $f(0) = 0$ is patched in as a case to make $f$ defined everywhere.) $\endgroup$
    – Retracted
    Jul 13, 2020 at 16:22
  • $\begingroup$ I think that $\exp(-x^2-x^{-2})$ has a smooth Fourier transform, so I am not sure what your point is... $\endgroup$
    – mike stone
    Jul 13, 2020 at 16:48
  • $\begingroup$ Yes, I chose a function that the Fourier transform proof would (should?) apply to. However notice that every derivative of my function is zero when evaluated at $x=0$. It is based on a well-known function that is infinitely differentiable but not analytic. $\endgroup$
    – Retracted
    Jul 13, 2020 at 17:14

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