This is a pretty basic QFT question but I haven't found a discussion of it in depth enough for my taste in any textbook.
We want a relativistic theory to be causal, hence we want the commutator of field operators at different points of spacetime to vanish. This is the case for the free (real) scalar field, i.e. $$ [\phi(x), \phi(y)] = 0 .$$ The commutator is often contrasted with the free scalar propagator $$ D(x - y) = \langle 0 \mid \phi(x) \phi(y) \mid 0 \rangle $$ which is not zero for space-like intervals -- it appears as the amplitude to 'anihilate at $y$ a particle created at $x$' is non-zero -- particles can travels faster than time. Usually, this 'paradox' is resolved via the antiparticle interpretation -- in anything physically meaningfull (measurement) it is not possible to single out the case of particle going from $x$ to $y$ from (anti-)particle going from $y$ to $x$ (we can Lorentz transform to a frame exchanging the events in time for space-like intervals). Therefore, it is usually argued, in a measurement the amplitude of one of these cancels againts the amplitude of the other one (we can think of this as any possibly faster-than-light particle is anihilated by equally possible (anti-)particle going back in time).
So far, so good. But for a real scalar field, the field operators are themselves hermitian, hence measurable. This suggests that we can also think about the propagator $$ D(x - y) = \langle 0 \mid \phi(x) \phi(y) \mid 0 \rangle $$ as measuring the vacuum by the operator $\phi(x) \phi(y)$. I see no reason for this to be non-zero in such an interpretation. Doesn't this operator correspond to just measuring the field (in the vacuum state) at the point $x$ and $y$ (at the same time if we are in a frame in which these are in the same time slice)?
(may be the operator of such interpretation should be $\phi(x) + \phi(y)$ but what is then the meaning of the operator $\phi(x) \phi(y)$?)