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In Physics From Symmetry Section 4.2 Restriction, the Author said that

It is sometimes claimed that the constraint to first-order derivatives is a consequence of our demand to get a local theory, but this only rules out an infinite number of derivatives.

How one can reason that locality implies the restriction to first-order derivatives?

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As the quote said, you cannot conclude that the Lagrangian must involve only first order derivatives based on locality. What you cannot have is an infinite number of derivatives.

To give an explicit example:

Start from the local theory of 2 scalar fields

\begin{equation} \mathcal{L} = -\frac{1}{2} (\partial \psi)^2 - \frac{1}{2} (\partial \phi)^2 - \frac{1}{2} M^2 \psi^2 - \frac{1}{2} m^2 \phi^2 + g \psi \phi^2 \end{equation}

Now we integrate out $\psi$ (in practice, we replace it by the formal solution of its equation of motion). The equation of motion for $\psi$ is \begin{equation} \left(\square + M^2\right) \psi = g \phi^2 \end{equation} which has the formal solution \begin{equation} \psi = \frac{g\phi^2}{\square + M^2} \end{equation} Note that the operator $(\square + M^2)^{-1}$ actually involves an infinite number of derivatives: \begin{equation} \frac{1}{\square + M^2} = \frac{1}{M^2} \left(1 - \frac{\square}{M^2} + \frac{\square^2}{M^4} + \cdots \right) \end{equation} In fact this operator is non-local. The notation $(\square + M^2)^{-1} \psi$ is really just shorthand for a convolution with the Green's function \begin{equation} \frac{g\phi^2}{\square + M^2} = g \int {\rm d^4} y G(x,y) \phi^2(y) \end{equation} where $G(x,y)$ is the propagator associated with the Klein-Gordan operator for $\psi$. The Green's function is not a delta function, and so the integration over all of space is a non-local function of $\phi^2$.

If we plug this expression back into the original Lagrangian, we get \begin{equation} \mathcal{L} = -\frac{1}{2} (\partial \phi)^2 - \frac{1}{2} m^2 \phi ^2 + \frac{g \phi^4}{\square + M^2} + ... \end{equation} This Lagrangian is a non-local function of $\phi$, because of the operator $(\square + M^2)^{-1}$.

Now in this case, the non-locality is "mild" in the sense that we know it actually descends from a local theory, and the non-locality is only an artifact of not working with all of the relevant degrees of freedom. However in general, if you write down a theory with an infinite number of derivatives, you aren't guaranteed to be able to introduce a new field so that the theory is local, so normally we require when writing Lagrangians for a local field theory that the Lagrangian is expressed in terms of a terms with a finite number of derivatives. (But this finite number can be larger than one).

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