I am reading the Peskin & Schroeder's An introduction to Quantum field theory, p.105~p.106 (Construction of a cross-section from the invariant matrix element (p.104) ) and stuck at understanding some integration. I think that I never seen a pattern like this.
EDIT : I edited this post to try to answer this question on my own. I wish this edited post will be helpful to readers. (And if there is anything that needs to be corrected, then it will be appreciate to inform me.)
First, we are given (their book, p.105, (4.76):)
$$ d\sigma =(\prod_f\frac{d^3p_f}{(2\pi)^3}\frac{1}{2E_f}) \int d^2b(\prod_{i=A,B}\int \frac{d^3k_i}{(2\pi)^3}\frac{\phi_i(\vec{k}_i)}{\sqrt{2E_i}}\int \frac{d^3 \bar{k}_i}{(2\pi)^3}\frac{\phi_i^{*}(\bar{\vec{k}}_i)}{\sqrt{2\bar{E}_i}})$$ $$\times e^{\vec{b} \cdot (\bar{\vec{k}_B} - \vec{k}_B)} ( _{out}\langle \{\vec{p}_f\}|\{\vec{k}_i|\}\rangle_{in})( _{out}\langle \{\vec{p}_f\}|\{\bar{\vec{k}}_i|\rangle_{in})^* \tag{4.76}. $$
And the goal is an expression ( p. 106, (4.78) ) :
$$ d\sigma = (\prod_f\frac{d^3p_f}{(2\pi)^3}\frac{1}{2E_f})\frac{|\mathcal{M}(p_A, p_B \to \{p_f\})|^2}{2E_A 2E_B|v_A-v_B|} \tag{4.78} \int \frac{d^3k_A}{(2 \pi)^3}\int\frac{d^3k_B}{(2\pi)^3} $$ $$\times |\phi_A(\vec{k}_A)|^2|\phi_B(\vec{k}_B)|^2 (2\pi)^4 \delta^{(4)}(k_A + k_B - \Sigma p_f).$$
Through p.105~p.106 he derives this formula but I don't quite understand his argument in part. For example,
"..We can use the scond of these delta functions, together with the $\delta^{(2)}(k_B^{\perp}-\bar{k}_B^{\perp})$, to perform all six of the $\bar{k}$ integrals in $(4.76)$. Of the six (?) integrals, only those over $\bar{k}_A^z$ and $\bar{k}_B^z$ require some work. ( his book p.105 ).."
and
"Now recall that the initial wavepackets are localized in momentum space, centered on $\vec{p}_A$ and $\vec{p}_B$. This means that we can evaluate all factors that are smooth functions of $\vec{k}_A$ and $\vec{k}_B$ at $\vec{p}_A$ and $\vec{p}_B$, pulling them outside except the remaining delta function ( his book p.106, first paragraph ).. "
Second, meanwhile, I found next note (Introduction to quantum field theory by Nastase): https://professores.ift.unesp.br/ricardo.matheus/files/courses/2014tqc1/QFT1notes.pdf
In p.176 ~ p.177 of his note, he notes that ( 19.35 , 19.36 )
$$ \int d^2\vec{b} e^{i\vec{b} \cdot (\vec{\bar{k}}_B-\vec{k}_B)} = (2\pi)^2 \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) $$ $$ ( _{out}\langle \{\vec{p}_f\}|\{\vec{k}_i|\}\rangle_{in}) = i \mathcal{M} (2 \pi)^4 \delta^{(4)}(\Sigma k_i - \Sigma p_f)$$ $$( _{out}\langle \{\vec{p}_f\}|\{\bar{\vec{k}}_i|\rangle_{in})^* = -i \mathcal{M}^* (2 \pi)^4 \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \tag{19.35} $$
Q. What's the definition of $k_B^{\perp}$ ( $\bar{k}_B^{\perp}$)?
EDIT : I think that $k_B^{\perp} := (k_B^{x},k_B^{y})$ , $\bar{k}_B^{\perp}:= (\bar{k}_B^{x}, \bar{k}_B^{y}))$. Accepting this, we can show the first equality in (19.35) ( ; $d^2 b = d b^x d b^y$ ? ).
And ( using $\bar{k}_A = ( \bar{E}_A , \bar{k}_A^{x} , \bar{k}_A^{y} , \bar{k}_A^{z} )$, $\bar{k}_B = ( \bar{E}_B , \bar{k}_B^{x} , \bar{k}_B^{y} , \bar{k}_A^{z} )$ and $p_f = (E_f, p_f^{x}, p_f^{y}, p_f^{z})$ )
$$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) $$ $$\stackrel{?}{=} (k_i^{\perp} = \bar{k}_i^{\perp}) \times \int d \bar{k}^z_A d\bar{k}_B^{z} \delta(\bar{k}_A^{z} + \bar{k}_B^{z} - \Sigma p_f^{z}) \delta ( \bar{E}_A + \bar{E}_B - \Sigma E_f) $$ $$ = \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$ $$\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |} = \frac{1}{|v_A -v_B|} \tag{19.36}$$
Putting everything together, we find the above (4.78).
Q. My question is, why the equalities in (19.36) marked by question symbol are true? Can we prove these more concretely? I think that this integral is one of the hardest one I've ever seen, because partially I don't know what is exact definition for $k_B^{\perp}$ ( or $\bar{k}_B^{\perp}$ ), $\bar{k}_A^z$ ( or $\bar{k}_B^z$ , $p_f^z$).
EDIT : So far, I think that $k_B^{\perp} := (k_B^{x},k_B^{y})$ , $\bar{k}_B^{\perp}:= (\bar{k}_B^{x}, \bar{k}_B^{y}))$. Correct?
For the first equality in (19.36), perhaps, does next formula holds true?
$$ \int d \bar{k}_A^{x} \int d \bar{k}_A^{y} \int d \bar{k}_B^x \int d \bar{k}_B^{y} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) = 1 \operatorname{or} ( k_i^{\perp}= \bar{k}_i^{\perp})$$ ? If so, how? If this is true, then we maybe drive the first equiality in (19.36) (?) So far, I don't understand what the strange notation $(k_i^{\perp} = \bar{k}_i^{\perp})$ exactly means. Here, it seems that the definition of $k_B^{\perp}$ and $\bar{k}_B^{\perp}$ plays key role.
EDIT : I think that the strange notation $(k_i^{\perp} = \bar{k}_i^{\perp})$ in (19.36) is negligible.
For the first equality in (19.36), I think that we may use $$\int_{-\infty}^{\infty} \delta(x_i-x)dx_i = 1 $$ Note that $$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) =\int d \bar{k}^z_A d\bar{k}_B^{z} \delta(\bar{k}_A^{z} + \bar{k}_B^{z} - \Sigma p_f^{z}) \delta ( \bar{E}_A + \bar{E}_B - \Sigma E_f) $$ $$\times \int d \bar{k}_A^{x} \int d \bar{k}_A^{y} \int d \bar{k}_B^x \int d \bar{k}_B^{y} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) $$ So, if $$ \int d \bar{k}_A^{x} \int d \bar{k}_A^{y} \int d \bar{k}_B^x \int d \bar{k}_B^{y} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) =1 $$, we obtain $$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) =\int d \bar{k}^z_A d\bar{k}_B^{z} \delta(\bar{k}_A^{z} + \bar{k}_B^{z} - \Sigma p_f^{z}) \delta ( \bar{E}_A + \bar{E}_B - \Sigma E_f) $$ , which is the first equality in the (19.36) above. But I think that is true since, for example, $\delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) = \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta(k_B^{x} - \bar{k}_B^{x}) \delta(k_B^{y} - \bar{k}_B^{y})$ ( C.f. Here we use $k_B^{\perp} := (k_B^{x},k_B^{y})$ , $\bar{k}_B^{\perp}:= (\bar{k}_B^{x}, \bar{k}_B^{y})) $ is independent from variable $\bar{k}_A^{x}$, this can be pulled outside the $\bar{k}_A^{x}$-integral ; i.e., $$ \int d \bar{k}_A^{x} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp})$$ $$= \int d \bar{k}_A^{x} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta(k_B^{x} - \bar{k}_B^{x}) \delta(k_B^{y} - \bar{k}_B^{y}) $$ $$ = \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta(k_B^{x} - \bar{k}_B^{x}) \delta(k_B^{y} - \bar{k}_B^{y}) \int d \bar{k}_A^{x} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) $$ $$ = \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta(k_B^{x} - \bar{k}_B^{x}) \delta(k_B^{y} - \bar{k}_B^{y}) \times 1$$ And so on..
This arguemnt really works?
And for the final equality in (19.36), my frist attempt is,
I want to use that $$ \delta[f(x)] = \Sigma_{i} \frac{\delta(x-x_i)}{|f'(x_i)|}$$ if $f(x_i)=0, f'(x_i) \neq 0$ ; (C.f. Boas, mathematical methods in the physical sciences, p456) From this, we can show that $$ \int \delta[f(x)] dx = \Sigma_{i=1} \frac{1}{|f'(x_i)|}$$(True?) And, let $$f(\bar{k}_A^{z}) := (\sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A} $$ Note that (by simply taking derivative) $$ \frac{d}{d \bar{k}_A^{z}} f(\bar{k}_A^z) = \frac{\bar{k}_A^{z}}{\bar{E}_A}- \frac{\bar{k}_B^{z}}{\bar{E}_B} $$ Then from this, how can we deduce the above final equality in (19.36) using $$ \int \delta[f(x)] dx = \Sigma_{i=1} \frac{1}{|f'(x_i)|}$$ ? What will be roots $x_i$ of $f( \bar{k}_A^{z})$ such that $f'(x_i) \neq 0$?
EDIT : For this issue, let's look (19.36) more closely.
First, $\int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$ is a 'number', which can be calculated by $ \int \delta[f(x)] dx = \Sigma_{i=1} \frac{1}{|f'(x_i)|}$.
Second, on the other side, $\frac{1}{|\frac{\bar{k}_A^{z}}{\bar{E}_A}- \frac{\bar{k}_B^{z}}{\bar{E}_B}|} = \frac{1}{|\frac{d}{d \bar{k}_A^{z}} f(\bar{k}_A^z)|}$ is a 'function', which depends on variable $\bar{k}_A^{z}$.
Where does this discrepancy occurs? I guess that, the notation $\frac{1}{|\frac{\bar{k}_A^{z}}{\bar{E}_A}- \frac{\bar{k}_B^{z}}{\bar{E}_B}|} $ means an 'implicit' notation indicating for $\Sigma_{i=1} \frac{1}{|f'(x_i)|}$. True?
