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For instance, if we choose the following scalar field Lagrangian, which is (I hope) Lorentz-invariant, where $l$ is a a length scale, and with a $(-1,1,1,1)$ metric:

$$ \mathfrak{L}(x) \sim l^{-6} \int d^4y \space e^{\frac{- y^2}{ l^2}} (\Phi(x + y)\Phi(x - y) - \Phi(x) \Phi(x)) $$

There are no explicit derivatives of fields in this expression.

Of course, if we develop this expression in orders of $l$, we will find, for instance, at the first order (in $l^0$), a term proportional to:

$$\partial_i \Phi(x) \partial^i \Phi(x).$$

And, of course, we will find higher order derivatives of fields with higher order development in $l$.

But the initial expression does not involve explicit derivative of fields.

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This thing is fine, and it is done explicitly in Wilson's 1974 Rev. Mod. Phys. paper on Renormalization Group (in the section on exact renormalization group). This is usually done by mollifying in k-space, not in x-space, but what you are doing is equivalent. This is a form of regularization, and it is generally the $l\rightarrow\infty$ limit you are interested in. – Ron Maimon Nov 5 '12 at 17:35
@RonMaimon: Thanks for the reference, yes, it is thought like a kind of regularization, so, in my mind, $l$ is a very little length, like Planck length or something like that, so it cannot be $l\rightarrow\infty$. The other problem is that this kind of theory seems to be not causal (see Arnold Neumaier remark) – Trimok Nov 5 '12 at 17:54
"Not causal" is not particularly annoying, he means that there are nonlocal interactions in time, but causality is a macroscopic concept, and failure of microscopic time-ordering is not something you should worry about, since your theory isn't physical with a finite l. You definitely should take the limit $l\rightarrow\infty$. If you want to make space-time discrete in some way, the proper regularization is string theory. – Ron Maimon Nov 5 '12 at 18:20
up vote 3 down vote accepted

Of course one can formulate derivative-free Lorentz invariant actions. But these will typically not be causal. This means that fields at spacelike distance will typically not commute, as reqired for a good interpretation of the fields at a fixed time. Hence the fields lack one of the most important requirements of a relativistic QFT, needed e.g., for good cluster separation properties.

Moreover, your integral is not even classically well-defined. Indeed, $y^2$ is unbounded below in Minkowski space, so that the exponential blows up. You'd need to choose a better behaved function of $y^2$ as density.

For the Euclidean case (definite metric, O(4) invariance rather than Lorentz invariance), your action is sensible for $l>0$. However it is questionable whether for $l>0$ there is an analytic continuation to Minkowski space, as this already fails on the classical level.

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Thanks : For the unbounded problem, I was thinking that a Wick rotation solves it, but I suppose I am wrong... Could you give more details about the non-causal characteristic of these theories (I mean: why are they not causal ?) – Trimok Nov 5 '12 at 15:09
This is discussed at length in the chapter on the cluster decomposition in Weinberg's vol.1 of QFT. – Arnold Neumaier Nov 6 '12 at 7:41
Wick rotation - maybe, but your claim was about Minkowski space, not about Euclidean space. You should perhaps be more precise about what you write. Wick rotating back is the real problem; here even the classical action becomes ill-defined. – Arnold Neumaier Nov 6 '12 at 7:44
OK. Thanks for the reference, and your last remark – Trimok Nov 6 '12 at 11:11

The energy $E$ of a field quantum contains a momentum $\vec{p}$ for all known quanta $$E^2=\vec{p}^2+m^2$$ and the momentum operator is a space derivative $-i\frac{\partial}{\partial\vec{x}}$ (I use units where $c=\hbar=1)$. It is important and unavoidable thing, so the answer is "no", any realistic QFT has to contain the field derivatives.

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Well, there is a subtelty, and the subtelty is in the word "explicit". Of course, there are field derivatives, and in fact, in this baby model, higher order derivatives, but in some sense, there is a kind or resummation, and the field derivatives do not appear explicitely in the formula, but of corse they are here implicitely. – Trimok Nov 5 '12 at 17:59
@Trimok: there is no derivative in your model, strictly speaking. For a regular function $f(x)$ its argument shift can be symbolically represented as $f(x+y) = exp(y\nabla)f(x)$ indeed, but a finite difference is a finite difference. – Vladimir Kalitvianski Nov 5 '12 at 18:17

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