What's wrong with this QFT thought experiment?

Question

In quantum field theory, the propagator $D(x-y)$ doesn't vanish for space-like separation. In Zee's book, he claims that this means a particle can leak out of the light-cone. Feynman also gives this interpretation.

What is wrong, then, with this thought experiment:

Bob and Alice synchronize their watches and space-like separate themselves by some distance. Bob tells Alice that at exactly 3 o'clock, he intends to test his new particle oven, which claims to make a gazillion particles. Since Bob's oven makes so many particles, there is a pretty good chance that Alice will detect a particle right at 3 o'clock. If she detects such a particle, Bob will have transmitted a piece of information (whether or not the oven works) instantaneously - way faster than the speed of light.

Answer 1

The problem with the thought experiment is, as the comments already stated, that the propagator is not a physical quantity per se.

The (Feynman) propagator is defined as the time-ordered VEV of two fields (from Wikipedia) $$ D_F(x-y) = i \langle 0 | T \, \Phi(x) \Phi(y) | 0 \rangle $$ The fact that this does not vanish over spacelike intervals is one of the puzzling features of QFT. One must not forget, however, that bra and ket in the above equation are the vacuum state. What the propagator describes (in this fashion) is a vacuum fluctuation.

In order to make contact to real particles, one has to invoke the LSZ reduction mashinery. What you actually want is for $\langle x | y \rangle$ to exist over spacelike separations, where $| x \rangle$ is the position eigenstate at Bob's oven and $| y \rangle$ is the position eigenstate at Alice's detector. In order to evalute this bracket, the LSZ formula will inevitably give you equation-of-motion operators (like $\square + m^2$) acting on the field operators that will, in the end, reside between vacuum states, which gives the link to the propagator. $$\langle x | y \rangle = (\square_x + m^2) (\square_y + m^2) \langle 0 | T\, \Phi(x) \Phi(y) | 0 \rangle$$ (Note: The LSZ formula is usually given with momentum eigenstates in mind, I'm not sure whether I adopted it correctly to the position eigenstate case)

These operators, acting on the propagator, which might not vanish for $x-y$ spacelike, ensure that the actual matrix element DOES vanish for all spacelike intervals. And since it is the matrix elements that carry the physical information, QFT does not violate special relativity on which it is based.

Answer 2

Propagators $D_F(x-y)$ are just field correlations between the fields at the points $x$ and $y$.

On the other way, transmitting information means transmitting energy. Choosing for simplicity the example of a scalar field $\Phi$, we could consider the stress-energy operators $T_{\mu\nu}(\Phi)$. These operators are quadratic in the first derivative of fields. To respect causality, one demands that measurement of local physical quantities like stress-energy at some point x has no dependence with other local measurement at a point $y$ not causally connected to x, that is we demand that the stress-energy operators $[T_{\mu\nu}(\Phi(x)), T_{\lambda\nu}(\Phi(y))] = 0$ for a space-like interval $(x-y)^2 <0$.

This is, in fact, the case, with the usual quantization of a scalar bosonic field, where $[\Phi(x), \Phi(y)] = 0$ for a space-like interval $(x-y)^2 <0$

One interesting analogy is considering entangled states. The word "entanglement" means in fact some special and strong correlations between sub-systems of the entangled system. But, if the sub-systems are spatially separated, there is no possibility to send a signal from one sub-system to an other sub-system, there is no "spooky action at a distance". The correlations are just some numbers which correspond to joint probabilities laws of finding $2$ ore more sub-systems in some conjoint state.