# Commutation relations $[\phi(x), \phi^\dagger (y)]=-i \Delta(x,y)$ for spacelike separations $(x-y)^2<0$ [duplicate]

In my QFT course, we studied the Klein Gordon equation, and we found that the field operators verifies : $[\phi(x), \phi^\dagger (y)]=-i \Delta(x,y)$

With $$\Delta(x,y)=i \int d\widetilde{k} (e^{-ik(x-y)}-e^{+ik(x-y)})$$

Where $d\widetilde{k}=\frac{d^3k}{2k^0(2\pi)^3}$.

And we said that : if $(x-y)^2<0$ then $\Delta(x,y)=0$

But I don't know how to prove this ? I tried to manipulate the integral to show this but nothing very useful...

How to prove this ?

And as I understand the property, it means that if we have $x$ and $y$ that have space-like separation, then they commutes.

So in more physical way, it means that if I measure $\phi$ in $x$ and then in $y$ or in $y$ and then in $x$ in a faster way than light could reach the distance between $x$ and $y$, I will have the same result. Which is logic because in thoose configurations, the field "can't be aware" of a previous measurement as $(x-y)$ lives outside of the light cone.

So my two questions :

1. How to prove that $\Delta(x,y)=0$ for $(x-y)^2<0$.
2. Is my physical interpretation of this entirely correct?
• Relevant (or duplicate?): physics.stackexchange.com/q/281414, physics.stackexchange.com/q/53148, physics.stackexchange.com/q/112220 and links therein. Apr 2, 2017 at 14:12
• This question is e.g. explained in Peskin & Schroeder, QFT, p. 28. Apr 2, 2017 at 15:00
• Sorry you can close the topic. I searched before my post but I didn't find thoose topics in link. Apr 2, 2017 at 15:01
• Apr 2, 2017 at 15:09