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

Question

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

Answer 1

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

Comments

@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

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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

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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

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@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

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@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

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@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

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...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