Cross Section Formula in Peskin and Schroeder

  • #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?
 

Answers and Replies

  • #2
170
64
Well, ##b## is the parameter impact, which is perpendicular to the beam direction (let's call it ##z##). Then, ##\vec{b}\cdot\vec{k}=\vec{b}\cdot\vec{k^{\perp}}## and, since the only dependence in ##b## is in the exponential, this is the typical definition of ##\delta(\vec{k})##.
 
  • #3
63
1
Not quite get it. Why doesn't it reduce to merely ##(2\pi)^2\delta^{(2)}(k_B-\bar{k_B})##? Or is that the same thing as ##(2\pi)^2\delta^{(2)}(k_{B}^{\perp}-\bar{k_{B}^{\perp}})##? If so, why?
 
  • #4
170
64
Well, ##\vec{k}## is a 3d vector, does ##\delta^2(\vec{k})## make any sense to you?
Let's simplify things, your problem is with the following integral:
$$\int dx dy e^{i\vec{r}\cdot \vec{k}}, \qquad \text{with }\vec{r}=x\hat{x}+y\hat{y}$$
Try to do it with all the details and you'll see that it gives ##(2\pi)^2\delta^2(\vec{k}^\perp)## with ##\vec{k}^\perp=k_x \hat{x}+k_y\hat{y}##.
 
  • #5
63
1
I understand now, should've kept track of all my vectors. Thanks a lot for helping me out.
 
  • #6
63
1
Can I bother you a little more? I got a bit further on the deduction of the formula, but now I'm stuck at:

$$ \int \bar{dk_{A}^{z}}\delta(\sqrt{ \bar{k_{A}^{2}}+m_{A}^{2}}+\sqrt{ \bar{k_{B}^{2}}+m_{B}^{2}}-\sum E_f)|_{\bar{k_{B}^{z}}=\sum p_{f}^{z}-\bar{k_{A}^{z}}} $$
$$ = \frac{1}{|\frac{\bar{k_{A}^{z}}}{\bar{E_A}}-\frac{\bar{k_{B}^{z}}}{\bar{E_B}}|}$$

Is there any property of the delta function that could lead to this? I really can't see how to get this result.
 
Last edited:
  • #7
170
64
Can I bother you a little more? I got a bit further on the deduction of the formula, but now I'm stuck at:

$$ \int \bar{dk_{A}^{z}}\delta(\sqrt{ \bar{dk_{A}^{2}}+m_{A}^{2}}+\sqrt{ \bar{dk_{B}^{2}}+m_{B}^{2}}-E_f)|_{\bar{k_{B}^{z}}=\sum p_{f}^{z}-\bar{k_{A}^{z}}} $$
$$ = \frac{1}{|\frac{\bar{k_{A}^{z}}}{\bar{E_A}}-\frac{\bar{k_{A}^{z}}}{\bar{E_B}}|}$$

Is there any property of the delta function that could lead to this? I really can't see how to get this result.
Yes, it's one of the most important relations of the ##\delta##, at least when computing cross-sections and decay widths:
$$\int \delta(f(x))dx = \sum_{x_0}\frac{1}{f'(x_0)}\int\delta(x-x_0)dx$$
where ##x_0## are all the solutions of the equation ##f(x)=0##, and ##f(x)## is a continuously diferentiable function that fullfills ##f'(x)\neq 0##.

Let me advise you that ##\delta## functions come all the time in QFT and QM, so if you are not very used to it, maybe would be a nice idea to go first into another book. Anyway, we are here to help.
 
  • #8
63
1
Thanks again!
 
  • #9
170
64
$$\int \delta(f(x))dx = \sum_{x_0}\frac{1}{f'(x_0)}\int\delta(x-x_0)dx$$
Sorry I forget an absolute value in the equation, it should be
$$\int \delta(f(x))dx = \sum_{x_0}\frac{1}{|f'(x_0)|}\int\delta(x-x_0)dx$$
 

Related Threads for: Cross Section Formula in Peskin and Schroeder

Replies
7
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
3
Views
873
  • Last Post
Replies
11
Views
3K
Replies
8
Views
1K
Replies
6
Views
3K
Replies
7
Views
944
  • Last Post
Replies
17
Views
2K
Top