Cross Section Formula in Peskin and Schroeder

[Wledig]

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?

 

[Gaussian97]

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 $\delta(\vec{k})$.

 

[Wledig]

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?

 

[Gaussian97]

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 $\vec{k}^\perp=k_x \hat{x}+k_y\hat{y}$.

 

[Wledig]

I understand now, should've kept track of all my vectors. Thanks a lot for helping me out.

 

[Wledig]

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.

 

[Gaussian97]

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.

 

[Wledig]

Thanks again!

 

[Gaussian97]

$$\int \delta(f(x))dx = \sum_{x_0}\frac{1}{f'(x_0)}\int\delta(x-x_0)dx$$

Sorry I forget an absolute value in the equation, it should be
$$\int \delta(f(x))dx = \sum_{x_0}\frac{1}{|f'(x_0)|}\int\delta(x-x_0)dx$$