I'm taking QFT course in this term. I'm quite curious that in QFT by which part of the mathematical expression can we tell a quantity or a theory is local or non-local?
[Physics] How to tell local and non-local in QFT
field-theorylagrangian-formalismlocalitynon-localityquantum-field-theory
Related Solutions
Lagrangian Formulation. The Lagrangian density for a massive free scalar in the $(+,-,-,-)$ convention is $${\cal L}~=~\frac{1}{2}d_{\mu}\phi ~d^{\mu}\phi-\frac{1}{2}m^2\phi^2.\tag{1} $$ The corresponding Euler-Lagrange equation is the massive Klein-Gordon equation $$ (d_{\mu}d^{\mu}+m^2)\phi~=~0. \tag{2}$$ The momentum is $$\pi ~:=~ \frac{\partial \cal L}{\partial\dot{\phi}}~=~\dot{\phi}.\tag{3}$$
Hamiltonian Formulation. The Hamiltonian density is $$ {\cal H}~=~\frac{1}{2}\pi^2 +\frac{1}{2}({\bf \nabla}\phi)^2+\frac{1}{2}m^2\phi^2. \tag{4}$$ The Hamiltonian equations of motion are $$ \begin{align} \dot{\pi}~=~&({\bf \nabla}^2-m^2)\phi, \cr \dot{\phi}~=~&\pi.\end{align}\tag{5} $$ Note that both $\phi$ and $\pi$ satisfy the massive Klein-Gordon equation (2).
Canonical transformation. $$ \widetilde{\phi}~:=~ \pi, \qquad \widetilde{\pi}~:=~ -\phi. \tag{6}$$ The new Hamiltonian density is precisely OP's Hamiltonian density $$ {\cal H}~=~\frac{1}{2}\widetilde{\phi}^2 +\frac{1}{2}({\bf \nabla}\widetilde{\pi})^2+\frac{1}{2}m^2\widetilde{\pi}^2.\tag{7} $$ Note that both $\widetilde{\phi}$ and $\widetilde{\pi}$ satisfy the massive Klein-Gordon equation (2). So the new Hamiltonian density (7) can be reproduced without any non-local field redefinition.
Now let us return to OP's questions (v3): Yes, $$\begin{align} \widetilde{\phi}~:=~&\sqrt{m^2-{\bf \nabla}^2}\phi, \cr \sqrt{m^2-{\bf \nabla}^2}~:=~&m\sum_{n=0}^{\infty} \begin{pmatrix}1/2 \\ n \end{pmatrix} \left(-\frac{{\bf \nabla}^2}{m^2} \right)^n,\end{align} \tag{8}$$ is a spatially non-local (but temporally local) field redefinition.
Since the field redefinition (8) contains no time-derivatives, Cauchy data still amounts to specify $\widetilde{\phi}$ and its time-derivative on a Cauchy surface. No need for higher time-derivatives.
And Yes, the field redefinition (8) formally leads to the Hamiltonian density (7). However, as mention in section 3, already a local canonical transformation (6) could achieve the same formulation. OP's non-local field redefinition (8) corresponds to a non-local canonical transformation $$\begin{align} \widetilde{\phi}~:=~&(m^2-{\bf \nabla}^2)^{\frac{1}{2}}\phi,\cr \widetilde{\pi}~:=~&(m^2-{\bf \nabla}^2)^{-\frac{1}{2}}\pi.\end{align}\tag{9}$$
OP asks in a comment if such kind of non-local field redefinition (8) is allowed in the quantum field theory in general? It depend on who's the arbitrator. A mathematician would probably focus on whether the detailed derivation/proof makes sense, while a physicist would likely be content if the final result/goal makes sense.
OP considers in an update (v5) the manifestly positive (but non-renormalizable) Hamiltonian density $$\begin{align} {\cal H}_{\beta} ~=~&\frac{1}{2}\left((1+2\beta{\bf \nabla}^2)^{-\frac{1}{2}}\pi\right)^2\cr &+\frac{1}{2}\left((1+\beta{\bf \nabla}^2)^{\frac{1}{2}}{\bf \nabla}\phi\right)^2\cr ~=~&\frac{1}{2}\widetilde{\pi}^2 +\frac{1}{2}\left(\sqrt{\frac{1+\beta{\bf \nabla}^2}{1+2\beta{\bf \nabla}^2}}{\bf \nabla}\widetilde{\phi}\right)^2\cr~\geq~&0, \end{align}\tag{10} $$ with non-local canonical transformation $$ \begin{align} \widetilde{\phi}~:=~&(1+2\beta{\bf \nabla}^2)^{\frac{1}{2}}\phi,\cr \widetilde{\pi}~:=~&(1+2\beta{\bf \nabla}^2)^{-\frac{1}{2}}\pi.\end{align}\tag{11}$$
Yes, there are rigorous ways of defining locality in such contexts, but the precise terminology used unfortunately depends on both the context, and who is making the definition.
Let me give an example context and definition.
Example context/definition.
For conceptual simplicity, let $\mathcal F$ denote a set of smooth, rapidly decaying functions $f:\mathbb R\to \mathbb R$. A functional $\Phi$ on $\mathcal F$ is a function $\Phi:\mathcal F\to \mathbb R$.
A function (not yet a functional on $\mathcal F$) $\phi:\mathcal F\to\mathcal F$ is called local provided there exists a positive integer $n$, and a function $\bar\phi:\mathbb R^{n+1}\to\mathbb R$ for which \begin{align} \phi[f](x) = \bar\phi\big(x, f(x), f'(x), f''(x), \dots, f^{(n)}(x)\big) \tag{1} \end{align} for all $f\in \mathcal F$ and for all $x\in\mathbb R$. In other words, such a function is local provided it depends only on $x$, the value of the function $f$ at $x$, and the value of any finite number of derivatives of $f$ at $x$.
A functional $\Phi$ is called an integral functional provided there exists a function $\phi:\mathcal F\to\mathcal F$ such that \begin{align} \Phi[f] = \int_{\mathbb R} dx \, \phi[f](x). \tag{2} \end{align} An integral functional $\Phi$ is called local provided there exists some local function $\phi:\mathcal F\to\mathcal F$ for which $(2)$ holds.
What could we have defined differently?
Some authors might not allow for derivatives in the definition $(1)$, or might call something with derivatives semi-local. This makes intuitive sense because if you think of Taylor expanding a function, say, in single-variable calculus, you get \begin{align} f(x+a) = f(x) + f'(x)a + f''(x)\frac{a^2}{2} + \cdots, \end{align} and if you want $a$ to be large, namely if you want information about what the function is doing far from $x$ (non-local behavior), then you need more an more derivative terms to sense that. The more derivatives you consider, the more you sense the "non-local" behavior of the function.
One can also generalize to situations in which the functions involved are on manifolds, or are not smooth but perhaps only differentiable a finite number of times etc., but these are just details and I don't think illuminate the concept.
Example 1 - a local functional.
Suppose that we define a function $\phi_0:\mathcal F\to \mathcal F$ as follows: \begin{align} \phi_0[f](x) = f(x), \end{align} then $\phi_0$ is a local function $\mathcal F\to\mathcal F$, and it yields a local integral functional $\Phi_0$ given by \begin{align} \Phi_0[f] = \int_{\mathbb R} dx\, \phi_0[f](x) = \int_{\mathbb R} dx\, f(x), \end{align} which simply integrates the function over the real line.
Example 2 - another local functional.
Consider the function $\phi_a:\mathcal F\to\mathcal F$ defined as follows: \begin{align} \phi_a[f](x) = f(x+a). \end{align} Is this $\phi_a$ local? Well, for $a=0$ it certainly is since it agrees with $\phi_0$. What about for $a\neq 0$? Well for such a case $\phi_a$ certainly is not because $f(x+a)$ depends both on $f(x)$ and on an infinite number of derivatives of $f$ at $x$. What about the functional $\Phi_a$ obtained by integrating $\phi_a$? Notice that \begin{align} \Phi_a[f] &= \int_{\mathbb R} dx\,\phi_a[f](x) \\ &= \int_{\mathbb R} dx\, f(x+a) \\ &= \int_{\mathbb R} dx\, f(x) \\ &= \int_{\mathbb R} dx\, \phi_0[f](x)\\ &= \Phi_0[f]. \end{align} So $\Phi_a[f]$ is local even though $\phi_a$ is not for $a\neq 0$.
The lesson of this example is this: you may encounter an integral functional $\Phi:\mathcal F\to\mathbb R$ that is defined by integrating over a non-local function $\phi:\mathcal F\to\mathcal F$. However, there might still be a way of writing the functional $\Phi$ as the integral over a different function, say $\phi'$, that is local, in which case we can assert that $\Phi$ is local as well because to verify that a functional is local, you just need to find one way of writing it as the integral of a local function.
Best Answer
A quantity is local if it is a finite linear combination $\sum_k g_k P_k(x)~~$ of products $P_k(x)$ (or other pointwise functions, such as $\sin \Phi(x)~$ for sine-Gordon theory) of field operators or their derivatives at the same point $x$.
A quantum field theory is local if its classical Lagrangian density is local. (By abuse of terminology, an action or a Lagrangian may also be called local if the corresponding Lagrangian density is local.)
Since in QFT fields are only operator-valued distributions, a local quantum field product is not well-defined without a renormalization prescription, which involves an appropriate limit of nonlocal approximations. In 1+1D, normal ordering is sufficient to renormalize the field products, while in 3D and 4D more complicated (mass and wave function) renormalizations are needed to make sense of these products.