# Tag Info

Yes, they are using the substitution of dirac delta $$\delta' f(x)=-\delta(x)f'(x)$$ And the calculation follows as $$(\partial^2+m^2)D_R(x-y)=(\partial^2\theta(x^0-y^0))\langle 0 |[\phi(x),\phi(y)]|0\rangle +\theta(x^0-y^0)(\partial^2+m^2)\langle 0 |[\phi(x),\phi(y)]|0\rangle + ... 3 Peskin & Schroeder, An Intro to QFT, are using that^1$$i\Delta(x-y)~:=~\langle 0 | [\phi(x), \phi(y)] |0\rangle \tag{K} $$vanishes for space-like vectors, see below eq. (2.53) on p. 28. In particular for equal times x^0=y^0, we have$$i\Delta(0,{\bf x}-{\bf y})~=~0.\tag{L} Therefore at the physics level of rigor ...