# Exact meaning of locality and its implications on the formulation of a QFT

+ 3 like - 0 dislike
1534 views

As far as I understand it, locality in physics is the statement that interactions can only occur between physical objects if the spacetime interval separating them is null or time-like. Thus, if the two events $(t,\mathbf{x})$ and $(t',\mathbf{y})$ occur simultaneously, i.e. $t=t'$, one cannot affect the other (unless $\mathbf{x}=\mathbf{y}$) as they will be separated by a space-like interval.

From reading notes on QFT however I have gathered that a physical theory is local if interactions between the quantum fields (contained in the theory) occur at the same point in spacetime. Why is this so, why can't they be time-like separated?

Is it because theories in QFT are described by Lagrangian densities $\mathscr{L}(\phi (x),(\partial_{\mu}\phi) (x))$ which describe the physics at each spacetime point and as the action $S$ of the theory is the integral of $\mathscr{L}$ over spacetime $$S=\int d^{4}x\mathscr{L}=\int dt\int d^{3}x\mathscr{L}(\phi (t,\mathbf{x}),(\partial_{\mu}\phi) (t,\mathbf{x}))$$ the interactions occur at the same point in time and thus for locality to be obeyed the fields must interact at the same spatial point also (as simultaneous events in which $\mathbf{x}\neq\mathbf{y}$ are always separated by a space-like path)?

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will

+ 2 like - 0 dislike

Locality is a physical requirement we impose (for good reasons). Locality is implemented in the theory by using fields, with local interactions in a Lagrangian density (ie, the Lagrangian only depends on products of fields and derivatives at a single point). I would definitely not say that locality occurs because Lagrangian densities show up in field theory, I would rather say that we use local Lagrangian densities because we want to implement locality.

We typically want locality because, especially when we also demand Lorentz invariance, locality is deeply related to causality. As you say, we typically don't want to allow for spacelike separated interactions because we could send signals back in time.

We typically don't want timelike separated interactions because they are just acausal--that would allow a field value in the future to interact with the field values now. Furthermore, if your theory is Lorentz invariant, then if you allow for timelike separated interactions you also have to allow for spacelike separated interactions.

There is research into non-local theories, and there is some evidence that quantum gravity is non-local. However, standard QFTs, in particular the Standard Model, are local in the above sense.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
answered Jul 7, 2015 by (135 points)
Ah ok, so is the reason why we consider Lagrangians as functions of fields (and their derivatives) at a single point in spacetime because we are considering interactions that are simultaneous (i.e. the interaction between the fields occur at the same point in time and this immediately requires that they occur at the same point in space, as otherwise they will be spacelike separated)? Or is it simple as you said that if we allow timelike interactions then spacelike interactions will necessarily also be allowed, which violates locality?!

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
... I find it slightly confusing because in general aren't timelike separations allowed in the principle of locality? What would you say was the definition of locality? Are there any good notes on the subject that you know of.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
The physics is that we want interactions to occur at the same spacetime point. It's more or less common sense, if we don't demand that then the theory could describe something here instantaneously appearing over there. To me that's enough to focus on local theories. In particular I wouldn't want two points at different times to interact directly, information could flow back in time. When you add in lorentz invariance as well, the case gets even stronger for focusing on local theories, because you'll probably break causality if the interactions aren't local.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
Timelike separated events can causally influence each other, but that doesn't mean fields at those points interact directly. Direct interactions between points of any separation, timelike, space like, or null, is inconsistent with locality. The only source I really know on this is weinberg vol 1, he derives the need for a local lagrangian density from the "cluster decomposition principle" (basically experiments at different places should be uncorrelated).

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
For what it's worth maybe I should mention that fields at timelike and null separations don't commute with each other, and this is consistent with locality. But that's different from having points at timelike or null separations directly interacting in the lagrangian.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
@Andew So is locality the requirement that fields can only interact directly at the same point in spacetime (intuitively, in a classical sense can this be viewed as the fields needing to be in "physical contact" ("touching") in order to exert any influence on one another)? Would it be correct to say that fields separated by timelike paths can indirectly influence one another (through propagators)?

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
Yes that's exactly correct. Incidentally the retarded (not feynman) propagator (that's the one you use in classical physics) is the commutator of the fields. So the fact that the commutator of two timelike or null separated events is nonzero is directly related to the fact that classically information can propagate from one point to the other.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
"For what it's worth maybe I should mention that fields at timelike and null separations don't commute with each other, and this is consistent with locality. But that's different from having points at timelike or null separations directly interacting in the lagrangian" - How do the two concepts differ? Is it that the Lagrangian density describes the state of the physical state of a system at a given point in spacetime and therefore, for the description to be local, the Lagrangian should only be dependent on values at that given spacetime point?

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
In classical physics the analogue statement is that there is a difference between having a non-local equation of motion (/ hamiltonian / lagrangian), and having the propagator / Green's function be non-local. The Lagrangian density describes the dynamics of the field--how should the field evolve. The commutator describes whether the field here was influenced by (or is correlated with) by the field there--that can occur either because there was an instanteous interaction (non-locality), or because information propagated from there to here (locally).

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
what about $[\phi(x), \phi(y)] = 0$ if space-like separated?

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user innisfree
@Andrew Ok, so the Lagrangian density describes the dynamics of the field at each point in spacetime and therefore, to be local, its value at a given spacetime point should only depend on the value of the field (and its derivatives) at that given point (and not how it behaves at any other point). The value of the field at a given spacetime point can be influenced by other fields that are causally connected to it, but the dynamics that it follows at each spacetime point only depend on the value of the field (and its derivatives) at that point. Is this correct?

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
@Andrew ... So if I've understood things correctly, a Lagrangian density of the form $\mathscr{L}(\phi (x),(\partial_{\mu}\phi)(x))$ is local as the value of $\mathscr{L}$ at each spacetime point $x^{\mu}$ is only dependent on the values of the field $\phi$ (and its derivatives at that point)...

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
...If the Lagrangian density was (for example) of the form $\mathscr{L}(\phi (x),(\partial_{\mu}\phi)(x))=\phi (x+y)$, then the theory would be non-local, as the value of the Lagrangian density at each spacetime point $x^{\mu}$ would also depend on its value at another (distinct) spacetime point $y^{\mu}$.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
@Will yes that is an excellent summary!

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
@innisfree Do you mind clarifying? I'm not sure if this is relevant but the retarded propagator vanishes outside the lightcone, which is directly related to the statement that the commutator is zero for space like separations.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
The propagator decays exponentially outside the lightcone (it doesn't vanish). for locality, one requires the above commutator to vanish.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user innisfree
The feynman propagator doesn't vanish outside the lightcone, the retarded propagator does. It's the retarded propagator that is equivalent to the commutator.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Andrew
@Andrew Thanks, and thanks for your help :)

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user Will
@Andrew ah, sorry, yes we are in agreement.

This post imported from StackExchange Physics at 2015-11-01 19:32 (UTC), posted by SE-user innisfree

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.