# What is the definition of Locality (in QFT)

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1. Jul 7, 2015

### "Don't panic!"

I am slightly unsure as to whether I have understood the notion of locality correctly. As far as I understand it locality is the statement that if two events occur simultaneously (i.e. at the same time) then no information can be shared between them (they are causally disconnected). Thus a theory is considered local if simultaneous interactions (of the quantities described by the theory) occur at the same point in space.

In QFT I've read that a theory is called local if the Lagrangian density describing it is a function of the fields and their derivatives at a single point in spacetime, i.e. $\mathscr{L}=\mathscr{L}(\phi (x),(\partial_{\mu}\phi) (x))$. Is the reason for this because the action $S$ of the theory is given by $$S=\int d^{4}x\mathscr{L}(\phi (x),(\partial_{\mu}\phi) (x))=\int dt\int d^{3}x\mathscr{L}(\phi (x),(\partial_{\mu}\phi) (x))$$ and so implicitly the fields are evaluated at a single point in time, thus the only way they can obey causality (and therefore be local) is if they are evaluated at a single point in space?

I feel I may be missing something and would appreciate it if someone could give a general definition of locality in physics and then how this notion of locality manifests itself in QFT?

2. Jul 7, 2015

### atyy

There is more than one definition of locality in physics.

If we simply take quantum theory as an operational theory, then the operational notion of locality in relativistic QFT is "no faster than light transmission of classical information". This condition is imposed by the requirement that observables at spacelike separation commute. You will see another definition of locality in relativistic QFT as "cluster decomposition", but under some mild assumptions, the requirement that spacelike operators commute can be derived (I think it also goes the other way, not completely sure).

Quantum theory is a theory about observables, states and Hamiltonians. To use Lagrangians to specify a quantum theory, one can use conditions such as the Osterwalder-Schrader conditions, which tell us which Lagrangians can be translated into relativistic quantum theory, ie. the Osterwalder-Schrader conditions translate the locality conditions that "spacelike operators commute" into locality conditions of the Lagrangian.

Last edited: Jul 7, 2015
3. Jul 7, 2015

### "Don't panic!"

That was my understanding of locality, but I'm unsure as to why the Lagrangian can only be a function of fields (and their derivatives) at a single point in spacetime? Is it just that fields cannot interact with one another directly unless they are at the same point in space (analogous to the classical notion of locality - that objects can only exert forces on one another if they are physically touching, i.e. located at the same point in space, at the same point in time)? Of course, fields can interact indirectly along timelike paths via propagators (2-point correlation functions).

4. Jul 7, 2015

### atyy

I don't have a detailed answer to your question at the moment, but in principle they should come from the Osterwalder-Schrader conditions, which are one set (there are other possibilities, but OS is widely used) of conditions that if satisfied guarantees the Lagrangian corresponds to a relativistic QFT.

5. Jul 7, 2015

6. Jul 7, 2015

### "Don't panic!"

Heuristically, is the point that as the Lagrangian density describes the dynamics of the field each given point in spacetime, its value at a given spacetime point should only depend on the value of the field (and its derivatives) at that point and not the value of field (and its derivatives) at another distinct spacetime point (as otherwise the description of the dynamics of the field at a given spacetime point would require knowledge of behaviour of the field at other points in spacetime which violates locality)?

7. Jul 7, 2015

### atyy

Heuristically, it isn't clear that the manifest locality is needed. However, the theory is easier to handle if it is manifestly local, so heuristically it is a convenient assumption which if it gives a theory that agrees with observation, it is ok. For example, http://isites.harvard.edu/fs/docs/icb.topic473482.files/08-gaugeinvariance.pdf (bottom of p13) comments that maybe it is ok to have nonlocal terms, but we prefer to have a gauge redundancy to have the convenience of a manifestly local theory.

8. Jul 7, 2015

### "Don't panic!"

Why in a local theory, though, does the Lagrangian density have to be function of the fields (and their derivatives) at each point? Why can't it be of the form (for example), $\mathscr{L}=\phi(x)\phi(y)$, where $x^{\mu}$ and $y^{\mu}$ are two distinct spacetime points. What about this violates locality if $x^{\mu}$ and $y^{\mu}$ are separated by a timelike interval?

9. Jul 7, 2015

### WannabeNewton

There is no relationship between the notion of locality you are using in the Lagrangian density sense and the notion of locality tied to causality.

There are field theories that are non-local in the sense that the Lagrangian density contains non-local terms (e.g. 2-point Green's functions) but are still causal. On the other hand you can have local Lagrangian densities that are non-causal; the latter happens a lot in curved space-time with regards to global hyperbolicity.

Furthermore in the SM we use local Lagrangians but there is nothing fundamentally preventing us from using non-local Lagrangians; in fact they are used in BSM and topological field theory.

10. Jul 7, 2015

### "Don't panic!"

So what is meant by locality in the Lagrangian density sense then?

11. Jul 7, 2015

### WannabeNewton

That the Lagrangian is a functional of the fields and their derivatives evaluated at a single space-time point.

12. Jul 7, 2015

### atyy

There are two notions of locality here

(1) no superluminal communication, ie. spacelike operators commute.

(2) manifest locality in the Lagrangian, which is what you are talking about. Heuristically, it is not clear that (1) requires (2), as the discussion in the link I gave in post #7 says. However, since given (2) we can get (1), we simply use (2) since it is easier. This is why it is said heuristically that the reason we allow gauge redundancy is to maintain manifest locality.

13. Jul 7, 2015

### "Don't panic!"

But what is meant by "local" in this (is it that the fields are functions of spacetime points)? Does this relate to anything physically?

14. Jul 7, 2015

### WannabeNewton

It just means that I don't need to know anything about the fields or their derivatives at other points in order to know the value of the Lagrangian at a given point. If I have something like $\square^{-1}$ in my Lagrangian (which is called 'integrating out the equations of motion') then my Lagrangian is obviously non-local because it contains a 2-point function.

There's nothing inherently physical about it; we can certainly have non-local Lagrangians without breaking any physics-it's just a matter of convenience to have local Lagrangians because of gauge theories. Causality and the cluster decomposition principle are much more physically motivated.

15. Jul 7, 2015

### "Don't panic!"

So locality for a Lagrangian density is the notion that its value at a point is dependent only upon the fields and/or their derivatives at that point (if it is a function of the derivatives of the fields as well then it is only dependent on the behaviour of the fields infinitesimally close to the point at which it is being evaluated)?

(Is the physical intuition behind this that objects can only exert forces on each other if they are in physical contact (at the same point in space) at a given instant in time?)

Is it a general feature mathematically to say that a function is local if it is only dependent on variables at a single point, for example, $f(\mathbf{x})$ is local as it depends on only information at a single point, whereas $f(\mathbf{x},\mathbf{y})$ is non-local as it depends on information at two distinct points?

In the example you gave, i.e. $\Box^{-1}$, is the reason why this is non-local because its description requires an infinite number of derivatives (in the form of a Taylor expansion) and thus requires more and more information about the behaviour of the Lagrangian at points further and further away from the point that you actually want to describe it at? Higher derivatives require that one knows how the fields behave at points further and further away from the point at which you wish to evaluate the Lagrangian. A finite number of derivatives are allowed as this corresponds to knowing how the field behaves at points in the neighbourhood of the point that one is evaluating the Lagrangian at. However, an infinite number of derivatives requires that one know the behaviour of the field on a global scale (which is generally not possible and certainly very inconvenient), thus the Lagrangian would be non-local as it would require information about the fields that it's a function of at arbitrary distances from the point at which one wishes to evaluate it at.

Thus, in general, can we say that to localise a Lagrangian we need to be able to describe it with only the behaviour of the field around that point. If we require information of the fields behaviour at other points to describe the Lagrangian at that point, then the Lagrangian is non-local as it requires one to know the derivatives of the field to arbitrary degree, which implies that one needs to know how the field behaves at an arbitrary distance away from the point that you are evaluating the Lagrangian at?!

Last edited: Jul 7, 2015
16. Jul 7, 2015

### atyy

Don't worry to much about terminology. There are many definitions of locality. For example, in addition to the two discussed above, there is the locality criterion that goes into the statement of the equivalence principle (which involves something like second derivative). So you may also see a term like "ultralocal". Just try to figure out what meaning a person is using in which context.

17. Jul 7, 2015

### "Don't panic!"

But is the general idea of locality, at least as far as QFT is concerned, that given the value of the fields at a given spacetime point along with its values in the neighbourhood of that point (corresponding to the first-, second-,..., finite-order derivatives of the fields at that point, as $\phi (\mathbf{x}+\delta \mathbf{x})=\phi (\mathbf{x})+(\nabla\phi)(\mathbf{x})\cdot\delta\mathbf{x}+\cdots +(\nabla^{n}\phi)(\mathbf{x})\cdot(\delta\mathbf{x})^{n}$, of course in practice we just use first-order derivatives such that the Lagrangian obeys Lorentz invariance), one can describe the dynamics of the physical system at that point. In other words, we don't need to know how the fields behave globally in order to describe the dynamics of the system at each point?!