# Trouble deriving expression for differential scattering cross section from $S$-matrix

I am following the derivation of the scattering cross-section from Peskin and Schroeder textbook. On page 105, we get an expression for the differential cross-section:

$$d\sigma = \left(\prod_f \frac{d^3p}{(2\pi)^3}\frac{1}{2E_f}\right) \int d^2b \left(\prod_{i=A,B} \int\frac{d^3k_i}{(2\pi)^3} \frac{\phi_i(\bf{k_i})}{\sqrt{2E_i}} \int \frac{d^3\bar{k}_i}{(2\pi)^3}\frac{\phi_i^*(\bar{\bf{k}}_i)}{\sqrt{2\bar{E}_i}}\right)\times e^{i\vec{b}(\bar{\textbf{k}}_B-\textbf{K}_B)} \left(_{\text{out}}\langle\{\textbf{p}_f\}|\{\textbf{k}_i\}\rangle_{\text{in}}\right) \left(_{\text{out}}\langle\{\textbf{p}_f\}|\{\bar{\textbf{k}}_i\}\rangle_{\text{in}}\right)^*. \tag{4.76}$$ We also have two $$\delta$$-functions available to use, derived earlier: $$(2\pi)^2\delta ^{(2)}(k_{B\perp} - \bar{k}_{B\perp}).$$ From the $$d^2b$$ integral and $$(2\pi)^4\delta ^{4}(\sum\bar{k}_i - \sum p_f).$$ From the complex conjugate part and 4 momentum conservation.

The text states

we can use these delta functions to perform all 6 of the integrals over $$\bar{k}$$. Of the 6 integrals, only those over $$\bar{k}_A^z$$ and $$\bar{k}_B^z$$ require some work.

The Problem

I wanted to work out the integral over $$\bar{k}^x$$, since it's implied it is easy. However, I'm stumped!

My Attempt

First, I separated the parts out of (4.76) I think are important: $$I = \prod_{i=A,B}\int \frac{d^3\bar{k}_i}{(2\pi)^3} \frac{\phi ^*_i(\bar{\textbf{k}}_i)}{\sqrt{2\bar{E}_i}}$$

Taking the x-components and removing the constants to the front: $$I^x = \frac{1}{\sqrt{4\bar{E}_A\bar{E}_B}} \int \phi_A(\bar{\textbf{k}}_A^{x})^*\phi_B(\textbf{k}_B^{x})^*\frac{d\bar{k}_A^x}{2\pi}\frac{d\bar{k}_B^x}{2\pi}$$

Sticking in the delta functions, cancelling factors of $$\pi$$ and ignoring the constants I pulled out at the front (is it right to remove the $$\phi$$?) leaves me: $$I^x = \int d\bar{k_A^x}d\bar{k_B^x}\left( \phi_A(\bar{\textbf{k}}_A^{x})^*\phi_B(\bar{\textbf{k}}_B^{x})^* \delta^{(1)}(k_{B\perp} - \bar{k}_{B\perp})\delta^{(1)}(\sum\bar{k}_i^x - \sum p_f^x)\right)$$

$$I^x = \int d\bar{k_A^x}d\bar{k}_B^x \left(\phi_A(\bar{\textbf{k}}_A^{x})^*\phi_B(\bar{\textbf{k}}_B^{x})^* \delta^{(1)}(k_{B\perp} - \bar{k}_{B\perp})\delta^{(1)}(\bar{k}_A^x + \bar{k}_B^x- \sum p_f^x)\right)$$

Now, I am stuck. I think the delta functions are meant to yield factors of energy, but I don't know how to use them to do that. I am also unsure what the first delta function means with the perpendicular sign.

• Those $\phi$ depend on the momentum, they cannot be removed from the integrands. Commented Jan 9, 2021 at 21:09
• @Triatticus yes, I suppose you're right. I'll change it to include them inside. Does adding them in help? Commented Jan 9, 2021 at 21:41
• I also follow his text recently, P.S. wrote on (4.77) $$\frac{1}{|\frac{\overline{k}^z_A}{\overline{E}_A}-\frac{\overline{k}^z_B}{\overline{E}_B}|}\equiv \frac{1}{|v_A-v_B|}$$ I think what he meant is really a equal sign and $$v_A, v_B$$ is not something like $$\frac{k_A^x+k_A^y+k_A^z}{E_A}$$ because in his next page, P.S. wrote something like $$E_AE_B|v_A-v_B|=|E_Bp^z_A-E_Ap^z_B|$$ which make sense only when $$v_A$$ is just a short hand of $$v_A=\frac{\overline{k}^z_A}{\overline{E}_A}$$ Commented Oct 19, 2022 at 6:37

• This is really helpful! I am wondering why the $\delta$-function you sub in for the complex conjugated matrix element chunk is $$\delta ^{(4)}(k_A + k_B - k_A^' - k_B^')$$ , instead of $$\delta ^{(4)}(k_A^' +k_B^' - \sum p_f )$$ ? This seems to follow from your expression above with the matrix element. Commented Jan 10, 2021 at 10:52
• This is simply a property of the delta distribution, that is $\delta(x-y)\delta(z-y) = \delta(x-y)\delta(z-x)$ Commented Jan 10, 2021 at 14:53
• I do have one last question: why are you able to say that: $d^3 p_1 = d|\vec{p_1}||\vec{p_1}|^2d\Omega$ ? Commented Jan 11, 2021 at 16:56