I've been reading through Matthew Schwartz's book "Quantum Field Theory and the Standard Model" and in chapter 24 there is a section on locality (section 24.4). In it he defines locality in terms of the Lagrangian stating

" We take locality to mean that the Lagrangian is an integral over a Lagrangian density that is a functional of fields and their derivatives evaluated at the same point."

Which is all well and good, but then he further elaborates to say that

"To be clear, this definition is mathematical, not physical: it is a property of our calculational framework, not of observables."

This part confuses me as I thought that the whole motivation was physical, i.e. that objects at different spacetime points should not be able to directly interact with one another? It makes sense to me that, as the Lagrangian density characterises the dynamics of a physical system at a given spacetime point, and locality demands that the dynamics of the system at that given point should depend only on the state of the system at that point (i.e. the field configuration and how it's changing at that point), then clearly the Lagrangian density should depend on no more than the state of the system at that point, i.e. the field configuration and its rate of change (in spacetime)?! Maybe I'm missing something?

  • $\begingroup$ Related: physics.stackexchange.com/q/95018/2451 $\endgroup$
    – Qmechanic
    Commented Sep 3, 2015 at 13:07
  • $\begingroup$ @Qmechanic Thanks for the the link. Unfortunately I've already read that answer and it didn't really help. $\endgroup$
    – Will
    Commented Sep 3, 2015 at 13:17
  • $\begingroup$ Is the author just talking about 24.108 versus 24.109, one is "local" the other is "not local" but as low energy theories they are similar? $\endgroup$
    – Timaeus
    Commented Sep 16, 2015 at 15:51
  • $\begingroup$ @Timaeus Yes, I think so - if one defines a cut-off scale, then we recover a local theory. My issue in particular though, was the statement that the fact that we evaluate the fields and their derivatives at a single spacetime point is a mathematical definition and is not physical. I thought the intuition was physical, i.e. that objects at distinct spacetime points should not be able to interact directly, and that this is implemented mathematically, by requiring that all objects contained in the Lagrangian density are evaluated at the same spacetime point. I am a bit confused! $\endgroup$
    – Will
    Commented Sep 16, 2015 at 16:08
  • $\begingroup$ He's probably just saying that we can perfectly well calculate "nonlocal observables" like $\langle \varphi(x) \varphi(y) \rangle$ even if $x$ and $y$ are spacelike separated, and potentially get nonzero correlation. $\endgroup$
    – tparker
    Commented May 24, 2016 at 20:09

1 Answer 1


For a classical field theory or a classical field and particle theory you want the dynamics of that stationary path. (Or the dynamics of one of the stationary paths.) But you consider all kinds of dynamics, and just reject the ones that don't have a stationary action.

If you know the Euler-Lagrange equations you are aiming for are going to just have/be derivatives at points, it makes sense to start with that, because it makes it easier to get that.

For a quantum field theory, you have to sit back and ask what the observables you are aiming for are. One possibility is that you are computing a scattering matrix. In which case your observables in a sense could be on shell free particle states.

If you want to think about it as a distinction between on shell and off shell interactions that might be what you are aiming for.

Off shell interactions are not what you are looking for as part of your observables. So they are purely about the math you are using to describe relationships between on shell states.

So it might not seem a big deal whether your density is written in terms of derivatives of the field or not. What matters is whether you included all the right on shell states and properly got the relationships between them.

There isn't a location to an on shell state, they are separated more in the sense of being approximately free than in having approximate locations that don't overlap. But the physical distinction and observables is likely the on shell states and their relationships. The mathematics is just how you compute them. So locality could be about how you compute the relationships. And so aiming for local or nonlocal might not be the key compared to having the right on shell states and relationships between them.

I couldn't read page 475 so I could be totally wrong about what the author intended.

  • $\begingroup$ So is it simply that physically, from a quantum perspective, one cannot localise an object to a single point, but in our mathematical description we define locality to mean that the mathematical representations of such objects (e.g. fields) are evaluated at single points and in doing so we ensure that interactions are mediated in a continuous fashion (thus ensuring energy conservation)? Is it correct to say that in general, a local function is one whose value at a single point only depends of the values of its variables at that single point? $\endgroup$
    – Will
    Commented Sep 16, 2015 at 17:50
  • $\begingroup$ @Will No. The physics is the frequencies of the observables you get dynamically. The definition of locality you cite in this context is about how you label the math you use. And energy conservation only happens because of symmetries involved. The symmetries involved have their own (different) kind of locality involved. A physical kind of locality, not this new mathematical one. $\endgroup$
    – Timaeus
    Commented Sep 16, 2015 at 17:57
  • $\begingroup$ In a purely mathematical sense though is it correct to say that, in general, a local function is one whose value at a single point only depends of the values of its variables at that single point? $\endgroup$
    – Will
    Commented Sep 16, 2015 at 18:03

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