Why is $langle 0| phi (x) phi (y) |0 rangle = int frac{d^3p}{(2pi)^2} frac{1}{2E_p} e^{-i p (x-y)}$ ? (Peskin and Schroeder equation 2.50)

Question

More specifically, starting from $langle 0| phi (x) phi (y) |0 rangle$  I have arrived at the expression:
$D(x-y)=langle 0| phi (x) phi (y) |0 rangle= {Largeint int} frac{d^3p cdot d^3q}{(2 pi)^3} frac{1}{2E_q} delta^{(3)}( vec p - vec q) e^{i(q cdot y -p cdot x)}={Largeint int} frac{d^3p cdot d^3q}{(2 pi)^3} frac{1}{2E_q} delta^{(3)}( vec p - vec q) e^{i(q_0y^0 - vec q cdot vec y -p_0x^0+ vec p cdot vec x)}=   {Largeint } frac{d^3p}{(2 pi)^3} frac{1}{2E_p} e^{i(q_0y^0  -p_0x^0+ vec p cdot (vec x - vec y))}$
However, I should be getting ${ Largeint} frac{d^3p}{(2pi)^2} frac{1}{2E_p} e^{-i p (x-y)}$
I don't understand why, since the delta function is 3D and thus does not act on $q_0$ to change it into $p_0$ (which would result in Peskin's formula). What am I missing?

Frederic Thomas · Accepted Answer

$p_0$ and $q_0$ are functions of $mathbf{p}$ and $mathbf{q}$. Namely ($m$ is the mass of the particle):
$p_0 = sqrt{mathbf{p}^2 +m^2} quadquad text{and} quad quad q_0 = sqrt{mathbf{q}^2 +m^2}$
(This is not valid for virtual particles however). Therefore when the $delta$-function is applied $q_0rightarrow p_0$, so the problem is solved.
Background : The motion of a free particle is governed by a dispersion relation, for a Schroedinger particle it is $E = frac{(hbar mathbf{k})^2}{2m} = frac{mathbf{p}^2}{2m}$, ($mathbf{k}$ is the wave vector) for a photon it is $E^2=mathbf{p}^2$, and for a relativistic scalar particle (and other types of particles) $E^2=mathbf{p}^2 +m^2$
(speed of light  $c$ set to 1).

Why is $langle 0| phi (x) phi (y) |0 rangle = int frac{d^3p}{(2pi)^2} frac{1}{2E_p} e^{-i p (x-y)}$ ? (Peskin and Schroeder equation 2.50)

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