# Why infinite order derivative in Lagrangian density implies non-local?

There is a homework in field theory. It says that negative order of derivative( such as $\frac{1}{\nabla^2}$), fraction order of derivative ( such as $\nabla^{2/3}$ ) and infinite order derivative in general cannot occur in a local field theory.

It's easy to prove : $$\frac{1}{\nabla^2} \phi(x)= -\int d^3k \frac{1}{k^2} \tilde\phi(k) e^{ikx} \propto \int d^3y \frac{1}{|\mathbf{x}-\mathbf{y}|}\phi(y)$$ So it's nonlocal.

In the same way, $$\nabla^{2/3} \phi(x)\propto \int d^3k k^{2/3}\tilde{\phi}(k )\propto \int d^3y \frac{1}{|\mathbf{x}-\mathbf{y}|^{8/3}}\phi(y)$$ Also nonlocal.

But I can't prove why infinite order derivative will imply nonlocal? For example $e^{\nabla^2}\phi(x)$ should depends only on quantities on point $x$. I also try to argue $$\sum_{n=0}^{\infty} (\nabla^2)^{n}=\frac{1}{1-\nabla^2}$$ But I think it's not true,since $$\sum_{n=0}^{\infty} (\nabla^2)^{n} \phi(x)=\int d^3k \sum_{n=0}^\infty k^{2n} \tilde{\phi}(k)$$ only when $k<1$, above quantities can be equal to $\int d^3k \frac{1}{1-k^2} \tilde{\phi}(k)$.

So is all infinite order derivative theory imlpys nonlocal or there exist infinite infinite order derivative theory which is nonlocal?

Give me a concrete example of infinite order derivative theory which is nonlocal.

• – Qmechanic May 10 '17 at 19:02
• If $G(x,t)$ is the propagator, then for any distance $d>0$ there should exist a $T>0$ such that $G(x,t) = 0$ if $|x|>d$ and $t<T$. – Count Iblis May 10 '17 at 20:53

## 1 Answer

$\exp(a\partial)~f(x)=f(x+a)$ gives $f$ translated by a, as it summarizes its Taylor expansion in a around a=0. f then actually depends on its value at a shifted point.

• +1 ! you wrote Lagrange's celebrated shift operator, the very essence of nonlocality. – Cosmas Zachos May 10 '17 at 18:41
• Nice example! But there is still one thing I cannot fully understand. In principle， any order derivative in one point should also a local information in this point. Maybe we assume the function is analytical, such that infinite order derivative has all information of the function in analytical region. – user153663 May 10 '17 at 19:12
• @fff123123 a finite-order derivative contains local information about "infinitesimally close" neighbors, but the shift operator contains the dangerous nonlocal information. You do run into problems with stability with generic Lagrangians involving higher order derivatives, though (see Ostragradski's theorem). – Alex Nelson May 10 '17 at 19:17