And in what sense are they 'non-local'?

  • 3
    $\begingroup$ Higher order in what? Both higher powers of fields and higher powers of derivatives are not non-local. Non-locality always means interaction between fields at points separated by some distance (instead of at a single point). This can be introduced e.g. by non-polynomial functions in the Lagrangian but I am not sure what your higher-order means. In any case, you should spend some effort on making the question precise. $\endgroup$
    – Marek
    Commented Aug 16, 2011 at 8:30
  • $\begingroup$ Marek: But that's exactly my question. I've repeatedly encountered referral to theories with higher order Lagrangians (I mean higher order in derivatives) as 'non-local' despite these not actually having a 'non-local' interaction of the type you are referring to. So what do they mean? $\endgroup$
    – WIMP
    Commented Aug 21, 2011 at 5:42
  • $\begingroup$ @WIMP: Please provide reference to support the claim in the title (v3). $\endgroup$
    – Qmechanic
    Commented Jun 25, 2014 at 20:16

5 Answers 5


The higher the number of derivatives the more initial data you have to provide. If you have some Lagrangian that contains an infinite number of derivatives (or derivatives appearing non-polynomially, such as one over derivative) then you have to provide an infinite amount of initial data which amounts to non-local info, in the sense explained below.

If you think in terms of Taylor expansions around your initial value, then you have to provide the full function (and thus non-local information) if you have an infinite number of derivatives. This is to be contrasted with cases where you provide only the field and its first derivative as initial values (and thus rather local information).

Personally, I would not call any higher-derivative Lagrangian "non-local", but only those theories where the number of derivatives is formally infinite in the Lagrangian.

In any discretization scheme you literally see the non-locality induced by higher derivatives: to define the first derivative you need to know the function on two adjacent lattice points, to define the second derivative on three and to define the n-th derivative on n+1 lattice points. Thus, the more derivatives the more non-locality. If you have an infinite number of derivatives you need to know the function on an infinite set of lattice points.

  • 2
    $\begingroup$ I do not like a discretization ("lattice") explanation. According to it, the first derivative is non local. $\endgroup$ Commented Aug 16, 2011 at 16:01
  • 7
    $\begingroup$ No, the first derivative is not non-local, it is just "less local" than the zeroth derivative. As I said, I would no call any higher derivative theory "non-local", only those where you need infinitely many derivatives - and thus infinitely many lattice points. $\endgroup$ Commented Aug 16, 2011 at 17:25
  • $\begingroup$ Yes, okay, thanks. I still find it confusing to call that 'non-local' but if that's what they mean, then, well, that's what they mean. (I believe what you say is true only for a finite number of derivatives, but then that wasn't my question.) $\endgroup$
    – WIMP
    Commented Aug 21, 2011 at 5:47
  • 4
    $\begingroup$ I also didn't get the reason why or where "non-local" terminology came from. Regarding the number of points needed to define derivatives, if the points are closer and closer to each other as we go to higher derivatives, they will be "local", don't they? $\endgroup$
    – Revo
    Commented Aug 23, 2011 at 2:12
  • $\begingroup$ @VladimirKalitvianski In the lattice we consider local all operators that contain spins at finite distance from each other. That's because in the continuum limit (in the RG sense), they all come to a point. Non-local operators on the lattice are those that extend over an infinite region, roughly speaking. $\endgroup$
    – MannyC
    Commented Apr 25, 2019 at 23:10

Clearly, an interaction involving $\phi(x+h)$ deserved to be called nonlocal. But since $\phi(x+h)=\sum_{k=0}^\infty \phi^{(k)}(x) h^k/k!$, any nonlocal interaction can be expressed as a power series involving arbitrarily many derivatives. Therefore an action (or Lagrangian) is called nonlocal if it involves infinitely many derivatives.

If there are only finitely many derivatives, they are not nonlocal. Indeed, by introcing additional fields for the derivatives togetyher with squares of the differences between the new fields and their definition, one can rewrite these in terms of a new action/Lagrangian which leads to identical equations of motions. (Unfortunately, this doesn't help with quantization, as the corresponding Legendre transformations to the Hamiltonian form is singular.)

  • 5
    $\begingroup$ This is the only correct answer. Higher order theories are not nonlocal, I don't know anyone who calls them nonlocal--- they give rise to local equations of motion (of high order), and local simulation methods. $\endgroup$
    – Ron Maimon
    Commented May 14, 2012 at 1:49
  • 1
    $\begingroup$ @Arnold: I am fine with the fact that any non-local interacn can be expressed as a power series involving arbitrarily many derivatives, but how can you say that any infinite series of higher order derivatives, you give you such an interaction? How can we generally assume, any infinite power series of higher order derivatives would give a non-local interaction? $\endgroup$
    – user7757
    Commented Dec 21, 2012 at 13:33
  • $\begingroup$ @ramanujan_dirac: Non-local is just defined this way. There is no other way to tell local from nonlocal. $\endgroup$ Commented Sep 12, 2013 at 10:18

Disillusionment with systems described by higher order Lagrangians harks back to a 1950 paper by Pais and Uhlenbeck, in which they showed that such systems were prone to pathologies, including states with negative energy and states with negative norm. There's a more recent discussion of this in arXiv:hep-th/0408104.

  • $\begingroup$ I know the Pais paper. But my question wasn't why people don't like these theories. I really just wanted to know why they're called 'non-local' as different authors seem to mean very different things with the word. $\endgroup$
    – WIMP
    Commented Aug 21, 2011 at 5:48

This is the first time I write an answer here. Tell me if it's bad or if I did something wrong.

I think I can contribute something here:

My mental image is the following:

A simple derivative is $$ f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\,, $$ for the second derivative $$ f''(x) = \lim_{h\to 0} \frac{f(x+2h) -f(x+h) + f(x)}{h^2}\,,$$ and so on....

Now if you have a function like $e^{\Theta\partial_x^2}$ acting on f(x) you have an infinite amount of derivatives, this means you will go out to $f(x + \infty h)$ (don't hurt me, this is just for the mental image) and the limit $\lim_{h\to 0} f(x+\infty h)$ gives an idea that the value of $$e^{\Theta \partial_x^2}f(x)$$ does depend on more than the point x (or a neighbourhood of it), but maybe whole $\mathbb{R}$. This could be done more rigorous I think.

Every single term of the Taylor expansion $$ e^{\Theta\partial_x^2} = \sum_{i=0}^{\infty} \frac{(\Theta\partial_x^2)^i}{i!}$$ is local, since it only contains a finite amount of derivatives. The 'last' term with infinitely many derivatives is nonlocal. With this I want to say that truncating the expansion of a nonlocal term can make the nonlocal effects vanish.

The other way of seeing the nonlocality is by Fourier Transforming and inverse Fourier Transforming:

$$ e^{\Theta \partial_x^2} f(x) = e^{\Theta \partial_x^2} \int \frac{\mathrm{d}p}{\sqrt{2\pi}}e^{i x p} \tilde{f}(p) \\ = \int \frac{\mathrm{d}p}{\sqrt{2\pi}}e^{-\Theta p^2} e^{i x p} \tilde{f}(p) \\ \int \frac{\mathrm{d}p}{\sqrt{2\pi}}e^{-\Theta p^2} e^{i x p} \int\frac{\mathrm{d}y}{\sqrt{2\pi}} e^{-i y p} f(y) \\ = \int \frac{\mathrm{d}p\,\mathrm{d}y}{2\pi} e^{-\Theta p^2 + i p (x-y)} f(y) \\ = \frac{1}{2\pi} \sqrt{\frac{\pi}{\Theta}} \int \mathrm{d}y\, e^{-\frac{(x-y)^2}{4 \Theta}} f(y)\,. $$ Thus the nonlocal operator $e^{\Theta \partial_x^2}$ is equivalent to the convolution with a gaussian kernel which is clearly nonlocal since it is an integral over $\mathbb{R}$. If you can solve the $p$ integral this should be possible of every nonlocal operator?


Just for info: there is a so called shift operator $\text{e}^{h\frac{d}{dx}}$ that shifts a function argument to another point, for example: $\phi(x+h)=\text{e}^{h\frac{d}{dx}}\phi(x)$. It is obvious that it contains an infinite number of differential operators for any function $\phi(x)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.