feynman rules

Convention:

$f(x) = \frac{1}{2\pi} \int \mathrm{d}q f(q) e^{-iqx}$ and $f(q) = \int \mathrm{d}x f(x) e^{iqx}$ .

Consider

$\frac{i}{2}g \int_{-\infty}^{z^-} \mathrm{d}y^- A^+(y^-) = \frac{i}{2}g \int_{-\infty}^{+\infty} \mathrm{d}y^- A^+(y^-) \theta(z^--y^-)$ .

Using the identity,

$\frac{-1}{2 \pi i} \int \mathrm{d}s \frac{e^{-\frac{i}{2}(z^--y^-)s}}{s+i\epsilon} =\theta(z^--y^-)$ ,

we will have

$\frac{i}{2}g \int_{-\infty}^{z^-} \mathrm{d}y^- A^+(y^-) = -g \int \frac{\mathrm{d}^4q}{(2\pi)^4} \frac{A^+(q)}{q^+ + i\epsilon} e^{-\frac{i}{2}q^+ z^-}$ .

as well as

$\frac{i}{2}g \int^{\infty}_{z^-} \mathrm{d}y^- A^+(y^-) = -g \int \frac{\mathrm{d}^4q}{(2\pi)^4} \frac{A^+(q)}{-q^+ + i\epsilon} e^{-\frac{i}{2}q^+ z^-}$ .

Now, consider

${\bar h}(z^-) \Phi(z^-) \Phi^\dagger (0) h(0)$
$= \int \frac{\mathrm{d}q_1}{(2\pi)^4} \frac{\mathrm{d}q_2}{(2\pi)^4} e^{\frac{-i}{2}(q^+_1+q^+_2)z^-}{\bar h}(q_1) \Phi(q_2) \int \frac{\mathrm{d}q_3}{(2\pi)^4} \frac{\mathrm{d}q_4}{(2\pi)^4} \Phi^\dagger (q_3) h(q_4)$
$= \int \mathrm{d}\omega e^{\frac{-i}{2}\omega z^-} \int \prod_{i=1}^{4} \frac{\mathrm{d}^4q_i}{(2\pi)^4} \delta(\omega-q^+_1-q^+_2){\bar h}(q_1) \Phi(q_2) \Phi^\dagger (q_3) h(q_4)$

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