- #1

- 63

- 1

## Main Question or Discussion Point

On page 105 of Peskin and Schroeder's book it says that the integral over ##d^2b## in the expression:

$$d\sigma = \left(\Pi_f \frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f}\right) \int d^2b\left(\Pi_{i=A,B} \int \frac{d^3 k_i}{(2\pi)^3}\frac{\phi_i(k_i)}{\sqrt{2E_i}} \int \frac{d^3 \bar{k_i}}{(2\pi)^3}\frac{\phi_i(\bar{k_i})}{\sqrt{2\bar{E_i}}}\right) \times e^{ib(\bar{k_B}-k_B}(<p_f|k_i>)(<p_f|k_i>)^{*}$$

equals ##(2\pi)^2\delta^{(2)}(k_{B}^{\perp}-k_{B}^{\perp})##. I didn't understand why that is so. Can someone explain it to me?

$$d\sigma = \left(\Pi_f \frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f}\right) \int d^2b\left(\Pi_{i=A,B} \int \frac{d^3 k_i}{(2\pi)^3}\frac{\phi_i(k_i)}{\sqrt{2E_i}} \int \frac{d^3 \bar{k_i}}{(2\pi)^3}\frac{\phi_i(\bar{k_i})}{\sqrt{2\bar{E_i}}}\right) \times e^{ib(\bar{k_B}-k_B}(<p_f|k_i>)(<p_f|k_i>)^{*}$$

equals ##(2\pi)^2\delta^{(2)}(k_{B}^{\perp}-k_{B}^{\perp})##. I didn't understand why that is so. Can someone explain it to me?