# In QFT, why does a vanishing commutator ensure causality?

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?

• ...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 $[\phi(t, \vec{x}), \phi(t, \vec{y})]=0$. – doublefelix Dec 15 '17 at 22:16