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.

  • $\begingroup$ Those $\phi$ depend on the momentum, they cannot be removed from the integrands. $\endgroup$
    – Triatticus
    Commented Jan 9, 2021 at 21:09
  • $\begingroup$ @Triatticus yes, I suppose you're right. I'll change it to include them inside. Does adding them in help? $\endgroup$ Commented Jan 9, 2021 at 21:41
  • $\begingroup$ 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 $$v_A=\frac{\overline{k}^z_A}{\overline{E}_A}$$ $\endgroup$
    – Li Chiyan
    Commented Oct 19, 2022 at 6:37

1 Answer 1


I did this calculation some time ago myself. Since I am not sure what is your problem precisely I just give you my old notes. As far as I remember they should be fairly detailed and approximately correct. If you have any specific questions you can ask me. Also if someone notices that there is something wrong with my derivation (which there very well may be), please tell me!

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  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jan 10, 2021 at 10:52
  • 1
    $\begingroup$ This is simply a property of the delta distribution, that is $\delta(x-y)\delta(z-y) = \delta(x-y)\delta(z-x)$ $\endgroup$
    – jkb1603
    Commented Jan 10, 2021 at 14:53
  • $\begingroup$ Got it, thanks! It took me a while but I've finished the derivation now. Your notes were incredibly helpful. $\endgroup$ Commented Jan 10, 2021 at 14:54
  • $\begingroup$ Good. Congratulations :) $\endgroup$
    – jkb1603
    Commented Jan 10, 2021 at 14:56
  • $\begingroup$ 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 $ ? $\endgroup$ Commented Jan 11, 2021 at 16:56

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