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In relativistic quantum field theories (QFT),

$$[\phi(x),\phi^\dagger(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$

On the other hand, even for space-like separation

$$\phi(x)\phi^\dagger(y)\ne0.$$

Many texts (e.g. Peskin and Schroeder) promise that this condition ensures causality. Why isn't the amplitude $\langle\psi| \phi(x)\phi^\dagger(y)|\psi\rangle$ of physical interest?

What is stopping me from cooking up an experiment that can measure $|\langle\psi| \phi(x)\phi^\dagger(y)|\psi\rangle|^2$? What is wrong with interpreting $\langle\psi| \phi(x)\phi^\dagger(y)|\psi\rangle \ne 0$ as the (rather small) amplitude that I can transmit information faster than the speed of light?

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Recall that commuting observables in quantum mechanics are simultaneously observable. If I have observables A and B, and they commute, I can measure A and then B and the results will be the same as if I measured B and then A (if you insist on being precise, then by the same I mean in a statistical sense where I take averages over many identical experiments). If they don't commute, the results will not be the same: measuring A and then B will produce different results than measuring B and then A. So if I only have access to A and my friend only has access to B, by measuring A several times I can determine whether or not my friend has been measuring B or not.

Thus it is crucial that if A and B do not commute, they are not spacelike separated. Or to remove the double negatives, it is crucial that A and B must commute if they are spacelike separated. Otherwise I can tell by doing measurements of A whether or not my friend is measuring B, even though light could not have reached me from B. Then with the magic of a lorentzian spacetime I could end up traveling to my friend and arriving before he observed B and stop him from making the observation.

The correlation function you wrote down, the one without the commutator, is indeed nonzero. This represents the fact that values of the field at different points in space are correlated with one another. This is completely fine, after all there are events that are common to both in their past light cone, if you go back far enough. They have not had completely independent histories. B U T the point is that these correlations did not arise because you made measurements. You cannot access these correlations by doing local experiments at a fixed spacetime point, you can only see these correlations by measuring field values at spatial location x and then comparing notes with your friend who measured field values at spatial location y. You can only compare notes when you have had time to travel to get close to each other. The vanishing commutator guarantees that your measurements at x did not affect her measurements at y.

It is dangerous to think of fields as creating particles at spacetime locations, because you can't localize a relativistic particle in space to a greater precision than its compton wavelength. If you are thinking of fields in position space it is better to think of what you are measuring as a field and not think of particles at all.

(Actually I should say that I don't think you could actually learn that your friend was measuring B at y by only doing measurements at A. But the state of the field would change, and the evolution of the field would be acausal. I think this is a somewhat technical point, the main idea is that you don't want to be able to affect what the field is going OVER THERE outside the light cone by doing measurements RIGHT HERE because you get into trouble with causality)

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    $\begingroup$ This is a fantastic answer. Two points: 1) I don't think you can travel to my friend and stop him from making the observation because he is still space-like separated, right? 2) I think you can learn that your friend was measuring B if [A,B]=/=0, at least in principle. For example, you could imagine setting up some crazy experiment which prepared the quantum field in some eigenstate. Your friend could be doing something that brings the field out of its eigenstate. Then, if you found that your field wasn't in its original eigenstate, you know (superluminally) that your friend did something. $\endgroup$ – hwlin Feb 9 '13 at 0:13
  • $\begingroup$ ...but from what I understand, the fields in QFT are not actually observables. So it seems to me at best a heuristic to say that this is why they commute at spacelike separations...and therefore not an actual explanation. I would be interested to know if there is something I am missing here; or if not, I would be interested in another argument that does explain why locality requires $[\phi(t, \vec{x}), \phi(t, \vec{y})]=0$. $\endgroup$ – doublefelix Dec 15 '17 at 22:16

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