# 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

$\exp(a\partial)~f(x)=f(x+a)$ 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.