# Simultaneous events, space-like separation & QFT

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1. Jul 6, 2015

### "Don't panic!"

First of all, sorry if this is in the wrong forum, wasn't quite sure which one to post it in given the question.

My question is, given two space-time points $x^{\mu}$ and $y^{\mu}$, if the events that occur at these points are simultaneous, i.e. $x^{0}=y^{0}$, are the two events necessarily space-like separated? The reason I ask is that I'm trying to understand the notion of equal-time commutation relations in QFT (in which the commutator is non-zero in the case where $\mathbf{x}=\mathbf{y}$).

If one has two fields $\phi$ and $\phi^{\dagger}$, say, then the commutation relation between them is given by $$[\phi (t,\mathbf{x}),\phi^{\dagger} (t,\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y})$$
Now is the reason for this being equal to a $\delta$-function because of locality? i.e. given that the two fields are evaluated at the same time, then as locality demands that they can only "communicate" if they are separated by a time-like separation, they must necessarily be evaluated at the same spatial point, as if $\mathbf{x}\neq\mathbf{y}$ then there would be a space-like separation between the two fields (as $\Delta s^{2}=(x^{0}-y^{0})^{2}-(\mathbf{x}-\mathbf{y})^{2}=-(\mathbf{x}-\mathbf{y})^{2}<0$), and they would therefore commute (in order to obey locality)?

Last edited: Jul 6, 2015
2. Jul 6, 2015

### Staff: Mentor

Yes. Any pair of events that are simultaneous in any frame must be spacelike separated.

3. Jul 6, 2015

### "Don't panic!"

Is this why the commutator is non-zero only in the case where the fields are evaluated at the same spacetime point (as if they are evaluated at equal times this is the only case in which they can be causally connected?)
Also, is what I put about locality correct?

4. Jul 6, 2015

### Staff: Mentor

"At the same spacetime point" is too strong a condition; fields evaluated at different spacetime points that are timelike or null separated can also have non-zero commutators. The commutator is only required to be zero for fields evaluated at dfferent spacelike separated points.

5. Jul 6, 2015

### "Don't panic!"

Sorry, I meant this only in the case where the two events are simultaneous.

Also, is it correct to say that locality is the condition that two spacetime events, fields etc. can only interact if they are connected by a time-like path?

6. Jul 6, 2015

### Staff: Mentor

First, you need to replace "timelike" with "timelike or null". Events connected by light rays count as causally connected.

Second, "interact" is vague. There is actually not a single precise formulation of locality, because making the vague term "interact" into something precise and mathematical is not a trivial task. From the standpoint of QFT, the best formulation we have is in fact the condition on commutators of field operators: field operators must commute for spacelike separated events; commutators can only be nonzero for timelike and null separated events.

7. Jul 6, 2015

### "Don't panic!"

So is the QFT definition of locality then. In general, is locality the statement that only causally connected events can "influence" ("communicate" with) one another?

8. Jul 6, 2015

### Staff: Mentor

Again, "influence" and "communicate with" are vague. Trying to make them precise is not trivial.

9. Jul 6, 2015

### "Don't panic!"

So was does it mean then to say that physical theories should be local?

10. Jul 6, 2015

### Staff: Mentor

It depends on the theory. In QFT, it means field operators must commute when evaluated at spacelike separated events.

In relativity, one way to look at locality is that all the fundamental equations are local differential equations, i.e., they relate quantities and their derivatives at a single event. For example, the Einstein Field Equation, Maxwell's Equations, etc. all work this way.

11. Jul 6, 2015

### "Don't panic!"

Is this what is meant by local, because differential equations relate quantities that are infinitesimally close to one another?

On a related note, is the reason (apart from the requirement from special relativity that temporal and spatial components are treated on the same level) why the Lagrangian density $\mathscr{L}$ is a function of the appropriate fields, their temporal derivatives and spatial derivatives , i.e. $\mathscr{L}=\mathscr{L}(\phi ,\partial_{t},\nabla\phi)$, because not only do we need to know how the dynamics of the theory change with respect to time, but also how they change in space?

12. Jul 6, 2015

### atyy

There is more than one definition, and Peter Donis is right that making the terms precise is not trivial. However, if one defines "locality" by the requirement that spacelike separated observables commute, then one is defining "locality" to mean that faster than light communication of classical information is not possible. I refer to "classical information", since quantum observables have to do with measurements which are defined to produce classical outcomes. If observables do not commute, then the order in which they are measured does matter for the probabilities of the outcomes because measurement collapses the wave function. Assuming that each observer can freely choose which measurement to make, if spacelike separated observables do not commute, then an observer can freely change the probabilities of outcomes of a spacelike separated observer, and thereby communicate classical information.

http://cds.cern.ch/record/980036/files/197508125.pdf (Section 7 "Messages")

Last edited: Jul 6, 2015
13. Jul 6, 2015

### Staff: Mentor

It's one meaning of "local", yes. There are multiple possible meanings, and which one is most relevant depends on the context. In the context of GR, I would say this meaning of "local" is probably the most relevant.

14. Jul 6, 2015

### Staff: Mentor

Yes.

This is an interpretation (the Copenhagen interpretation), and for QM it's very important to keep interpretation separate from the basics of what the theory's math says. The statement I quoted just above is part of the basics of what the math of QM says, so it's true regardless of which interpretation you adopt. (For example, it's true even if you use the MWI, in which there is no wave function collapse.) Different interpretations will have different arguments for why it's true.

15. Jul 6, 2015

### "Don't panic!"

What about the Lagrangian density of the theory, why is it also dependent on spatial derivatives? (Is what I put correct? As far as I understand it the Lagrangian has to become a density in order for the action to be Lorentz invariant).

16. Jul 6, 2015

### Staff: Mentor

Because in relativity, you can't separate space and time; if you have time derivatives, you have to have space derivatives too, because time derivatives in one frame become a combination of time and space derivatives in another frame, and the Lagrangian density has to be frame-independent.

17. Jul 6, 2015

### atyy

But is there a working version of MWI?

Also, if MWI works, then shouldn't Copenhagen be derivable from it?

18. Jul 6, 2015

### Staff: Mentor

Lots of physicists seem to think so. Anyway, that question (along with your next one) really belongs in the Quantum Physics forum, and it wouldn't be the first thread on that topic there, or even the thousandth.

19. Jul 6, 2015

### atyy

But anyway, the idea is that Copenhagen is really the orthodox interpretation, so it shouldn't be a problem to use wave function collapse, since it is either postulated or derived. Within Copenhagen, the collapse is just a mathematical step, and only the final conclusion is real: ie. there is no faster than light signalling of classical information.

Or do you really think the interpretation of the requirement for commutation at spacelike separation as no faster than light signalling is objectionable in some interpretations, because wave function collapse is used as an intermediate step?

I don't believe that the requirement "spacelike-separated operators must commute" is interpretation independent, if one objects to wave function collapse as being interpretation dependent. In Bohmian mechanics, the operators are not part of the fundamental formalism, and in fact neither are they in MWI (there is only unitary evolution of the wave function).

20. Jul 6, 2015

### Staff: Mentor

atyy, please start a new thread in the Quantum Physics forum if you want to pursue the subthread on QM interpretations, it is off topic here, with the exception of these items:

You're right, it isn't.

In ordinary QM, this is true. But in QFT (which is what the OP of this thread was asking about), we no longer have wave functions; we only have field operators. (How this affects the MWI and other interpretations should be taken to a new thread in the Quantum Physics forum.)