This integration appeared in the reconstruction of cross section

[Plantation]

Homework Statement

Integration appeared in the reconstruction of cross section from invariatn matrix element $\mathcal{M}$.

Relevant Equations

$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) = \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |}$

I am reading the Horatiu Nastase's Introduction to quantum field theory (https://professores.ift.unesp.br/ricardo.matheus/files/courses/2014tqc1/QFT1notes.pdf ) ( Attached file ) or Peskin, Schroeder's quantum field theory book, p.105, (4.77).

Through p.176 ~ p. 177 in the Nastase's Note, he construct the cross section from the invariant matirx element $\mathcal{M}$.
And in p. 176, he wrote :

$$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) \stackrel{?}{=} (k_i^{\perp} = \bar{k}_i^{\perp}) \times \int d \bar{k}^z_A d\bar{k}_B^{z} \delta(\bar{k}_A^{z} + \bar{k}_B^{z} - \Sigma p_f^{z}) \delta ( \bar{E}_A + \bar{E}_B - \Sigma E_f) $$
$$= \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}
\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |} =: \frac{1}{|v_A -v_B|}$$

In the last line we have used that (since $E = \sqrt{k^2 +m^2}$ and there we have $\bar{k}_B^z = \Sigma p_f - \vec{\bar{k}^z}_A$ (?) )

$$ \frac{d\bar{E}_A}{d\bar{k}_A} = \frac{\bar{k}_A}{\bar{E}_A}; \frac{d \bar{E}_B}{d\bar{k}_B} = \frac{\bar{k}_B}{\bar{E}_B} = - \frac{d\bar{E}_B}{d\bar{k}_A}$$

and the fact that $\int dx \delta(f(x) - f(x_0)) = 1/ |f'(x_0)|.$

Why each step of the calculation is true? Can we prove that 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 $\bar{k}_B^z$ , $p_f^z$). And for the finial equality. I found assoicated question : https://www.physicsforums.com/threads/peskin-schroeder-4-77-page-about-cross-sections.913243/ . But I don't understand it more rigorously until now. Perhaps, can any one present concrete (rigoruous) calculation more in detail?

 

[vanhees71]

With which concrete steps do you have problems? Do you understand this strange notation $(k_i^{\perp}=\bar{k}_i^{\perp}) \times$? I don't... I guess, you have to do the phase-space integrals step by step for yourself.

 

[Plantation]

Sorry I am late. Thank you. I mean steps marked by the question symbol!. I'm still trying to integrate. I also don't understand the strange notation $(k_i^{\perp} = \bar{k}_i^{\perp})$.. And what is the definition of $k_i^{\perp}$ ( and $\bar{k}_i^{\perp})$ ? I think that understanding this definiiton is necessary step for the integration. What will be concrete values for each components of $k_i^{\perp}$ ?

 

[Plantation]

Perhaps, 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 this is true, then we can drive the first - marked by question symbol- equality. And can we show that? How?

 

[vanhees71]

Ok, let's see. I think in this online manuscript it's pretty ununderstandable. Peskin-Schroeder is much better. But at the very end he doesn't get the fact right that cross sections are defined as Lorentz-invariant quantities. To that end, and for historical reasons, the cross section is defined in the socalled "lab frame", where one of the initial particles is at rest, $\vec{p}_A$, sitting in the origin of the coordinate system and the other particle is coming in with a rather well-defined momentum, $\vec{p}_B=p_B \vec{e}_3$. The in-states are then defined as wave packets where $\phi_A(\vec{k}_A)$ is a square-integrable normalized function peaking around $\vec{p}_A=0$ and $\phi_B(\vec{k}_B)$ one that is peaking around $\vec{p}_B$ and spatially displaced by $\vec{b}$:
$$|\phi_A \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 E_A(\vec{k}_A)}} \phi_A(\vec{k}_A) \langle \vec{k}_A \rangle$$
and
$$|\phi_B \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 E_B(\vec{k}_B)}} \phi_B(\vec{k}_A) \exp(-\mathrm{i} \vec{k}_B \cdot \vec{b} )\langle \vec{k}_B \rangle.$$
The momentum eigenstates are normalized such that
$$\langle \vec{k}|\vec{k}' \rangle=(2 \pi)^3 2 E(\vec{k}) \delta^{(3)}(\vec{k}-\vec{k}').$$
All four-momenta are onshell, and $E(\vec{k})=\sqrt{m^2 + \vec{k}^2}$ (with $m$ the mass of the corresponding particle).

So the in-state is
$$|\phi_A, \phi_B \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}_A}{(2 \pi)^3} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}_B}{(2 \pi)^3} \frac{1}{\sqrt{2 E_A(\vec{k}_A) 2 E_B (\vec{k}_B)}} \phi_A(\vec{k}_A) \phi_B(\vec{k}_B) \exp(-\mathrm{i} \vec{b} \cdot \vec{k}_B) |\vec{k}_A,\vec{k}_B \rangle.$$
The S-matrix $\hat{S}=\hat{1} + \mathrm{i} \hat{T}$ has, for a scattering process $A+B \rightarrow \sum_f C_f$, the invariant matrix elements $\mathcal{M}_{fi}$ defined by
$$_\text{out} \langle \vec{p}_1,\ldots,\vec{p}_f |\vec{k}_A,\vec{k}_B \rangle_{\text{in}} = \langle \vec{p}_1 \ldots \vec{p}_f|\hat{T}|\vec{k}_A,\vec{k}_B \rangle=(2 \pi)^4 \delta^{(4)} (k_A+k_B - P),$$
where $P=p_1+p_2 +\ldots p_n$. All four-vectors are meant to be on-shell here, i.e., $k_A^0=\sqrt{m_A^2 + \vec{k}_A^2}$.
The differential cross section is now defined as the transition probability for outgoing particles in the momentum range $\mathrm{d}^3 p_1 \cdots \mathrm{d}^3 p_n$ divided by the flux of the incoming particles. We cannot, of course, simply square $|S_{fi}|^2$ because of the energy-momentum conserving $\delta$ distribution. That's why we have to use the wave packets in the incoming state and integrate over all impact parameters. We have to plug in the incoming states in terms of the integrals over $\vec{k}_A$ and $\vec{k}_B$, and this for both factors $S_{fi}$ and $S_{fi}^*$. For the 2nd factor we need of course other independent integration variables $\vec{k}_A'$ nad $\vec{k}_B'$. This leads to
$$
\begin{split}
\mathrm{d} \sigma = \prod_{f=1}^n \frac{\mathrm{d}^3 p_f}{(2 \pi)^3} &
\frac{1}{2E_f(\vec{p}_f)}
\int_{\mathbb{R}^2}
\mathrm{d}^2
\vec{b}
\int_{\mathbb{R}^3}
\frac{\mathrm{d}^3
k_A}{(2
\pi)^3
\sqrt{2E_A(\vec{k}_A)}}
\int_{\mathbb{R}^3}
\frac{\mathrm{d}^3
k_B}{(2
\pi)^3
\sqrt{2E_B(\vec{k}_B)}}
\\
&\int_{\mathbb{R}^3}
\frac{\mathrm{d}^3
k_A'}{(2
\pi)^3
\sqrt{2E_A(\vec{k}_A')}}
\int_{\mathbb{R}^3}
\frac{\mathrm{d}^3
k_B'}{(2
\pi)^3
\sqrt{2E_B(\vec{k}_A')}}
\\
& \Phi_A(\vec{k}_A) \Phi_B(\vec{k}_B) \Phi_A^*(\vec{k}_A')
\Phi_B^*(\vec{k}_B') \exp[-\mathrm{i} \vec{b} \cdot
(\vec{k}_B-\vec{k}_B')] \\
&(2 \pi)^4 \delta^{(4)}(k_A+k_B-P) (2 \pi)^4 \delta^{(4)}(k_A'-k_B'-P)
\\
&\mathcal{M}_{fi}(k_A,k_B \rightarrow P) \mathcal{M}_{fi}^*(k_A',k_B' \rightarrow P).
\end{split}$$
The integration over $\mathrm{d}^3 \vec{b}$ is trivial,
$$\int_{\mathbb{R}^2} \mathrm{d}^2 \vec{b} \exp[-\mathrm{i} \vec{b} \cdot (\vec{k}_B-\vec{k}_B') = (2 \pi)^2 \delta^{(2)} (\vec{k}_{B \perp}-\vec{k}_{B \perp}').$$
This $\delta$ distribution together with one of the energy-momentum-conserving $\delta$ distributions can now be integrated out by using the integral over $\vec{k}_A'$ and $\vec{k}_B'$:
$$\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k_A'}{(2 \pi)^3} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}_B'}{(2 \pi)^3} (2 \pi)^2 \delta^{(2)} (\vec{k}_{B \perp}-\vec{k}_{B \perp}') (2 \pi)^4 \delta^{(4)} (k_A'+k_B'-P_f)=\int_{\mathbb{R}} \mathrm{d} k_B^{\prime 3} \delta(E_A(\vec{k}_A') + E_B(\vec{k}_B')-P^0)|_{\vec{k}_A'=\vec{P}-\vec{k}_B'}.$$
For the integration over $k_B^{\prime 3}$ we note that
$$\frac{\partial}{\partial k_B^{\prime 3}} (E_A(P-\vec{k}_B')+E_B(\vec{k}_B') = \frac{p_B^{\prime 3}}{E_B(\vec{k}_B')}-\frac{p_A^{\prime 3}}{E_{A}(\vec{k}_A'} :=v_{\text{rel}}(\vec{k}_A',\vec{k}_B').$$
Plugging this in above, we can do also the integal over $p_B^{'3}$:
$$\int_{\mathbb{R}} \mathrm{d} k_B^{\prime 3} \delta(E_A(\vec{k}_A') + E_B(\vec{k}_B')-P^0)|_{\vec{k}_A'=\vec{P}-\vec{k}_B'}=\frac{1}{|v_{\text{rel}}|}.$$
This we can now plug into the expression for $\mathrm{d} \sigma$, where now nicely one of the energy-conserving $\delta$-distributions is gone.

Further we use that $\phi_A$ is peaked around $\vec{k}_A=\vec{p}_A=\vec{0}$ and $\phi(B)$ around $\vec{k}_B=\vec{p}_B$. Given that the detector cannot resolve the minute details of the wave packets we can just take $\vec{k}_A=\vec{0}$ and $\vec{k}_B=\vec{p}_A$ in all the kinematical factors except in the integrals over the wave functions, which however both give 1, because $\phi_A$ and $\phi_B$ are normalized to $1$ in the sense of
$$\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k_A}{(2 \pi)^3} |\phi_A(\vec{k}_A)|^2=1,$$
and similarly for $\phi_B$. So finally we get
$$\mathrm{d} \sigma = \prod_{f=1}^n \frac{\mathrm{d}^3 p_f}{(2 \pi)^3} \frac{1}{2 E_f} \frac{1}{2 m_A} \frac{1}{2 E_B(\vec{p}_B)} \frac{1}{|\vec{p}_B|/E_B(\vec{p}_B)} |\mathcal{M_{fi}}|^2 (2 \pi)^4 \delta^{(4)}(p_A + p_B -\sum_f p_f).$$

Now also the ugly-lookin non-covariant "flux factor" can be written in a covariant way. Since we are in the rest frame of particle $A$ we have $u_A=p_A/m_A=(1,0,0,0)$ and thus
$$
|v_{\text{rel}}|=\left |\frac{\vec{p}_B}{E_B(\vec{p}_B)} \right| = \frac{1}{E_B(\vec{p}_B)} \sqrt{E_B^2(\vec{p}_B)-m_B^2} = \frac{1}{E_B(\vec{p}_B) m_A}{\sqrt{(p_A \cdot p_B)^2-m_A^2 m_B^2}} = \frac{I}{E_B(\vec{p}_B) m_A},
$$
and thus finally the differential cross section in manifestly covariant form

$$\mathrm{d} \sigma = \prod_{f=1}^n \frac{\mathrm{d}^3 p_f}{(2 \pi)^3} \frac{1}{2 E_f} \frac{1}{4I} \mathcal{M_{fi}}|^2 (2 \pi)^4 \delta^{(4)}(p_A + p_B -\sum_f p_f).$$

An alternative way to regularize the energy-conserving $\delta$ function is using a finite four-volume and taking the limit after squaring the $S$-matrix. You find the corresponding calculation in Sect. 6.4 of my lecture notes on QFT

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[Plantation]

Thanks for kind explanation, although there seems to be few typos.

And I am somewhat confused since In your derivation of the integral equation ; i.e., Peskin's book (4.77) , it seems that you have used symbols A and B interchangeably.

Anyway, I still don't understand why such next integral is true :

$$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) $$
$$\stackrel{?}{=} (k_i^{\perp} = \bar{k}_i^{\perp}) \times \int d \bar{k}^z_A d\bar{k}_B^{z} \delta(\bar{k}_A^{z} + \bar{k}_B^{z} - \Sigma p_f^{z}) \delta ( \bar{E}_A + \bar{E}_B - \Sigma E_f) $$
$$ \stackrel{?}{=} \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$

I think that you skipped this step and explained next material :) Can you explain this integration step more in detail?

C.f. 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. And I don't know how to deal with integration whose integrands all consists of delta functions.

And let's return to the final equality in my original question :

$$ \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$
$$\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |} = \frac{1}{|v_A -v_B|} $$

Q. How can we use $\int dx \delta(f(x) - f(x_0)) = 1/ |f'(x_0)|$?

For this final equality, my first attempt is as follows :

From $\frac{d\bar{E}_A}{d \bar{k}_A}= \frac{\bar{k}_A}{\bar{E}_A}$ (19.37), we get
$\frac{\bar{k}_A^{z}}{\bar{E}_A} = \frac{d\bar{E}_A}{d \bar{k}_A^{z}}$. (True?)

And from $\frac{d \bar{E}_B}{d\bar{k}_B} = \frac{\bar{k}_B}{\bar{E}_B} = - \frac{d\bar{E}_B}{d\bar{k}_A}$ (19.37), we get
$\frac{\bar{k}_B^{z}}{\bar{E}_B} = \frac{d \bar{E}_B}{d\bar{k}_B^{z}} = - \frac{d\bar{E}_B}{d\bar{k}_A^{z}}$ (True?)

So formally, $$| \frac{\bar{k}_A^{z}}{\bar{E}_A} - \frac{\bar{k}_B^{z}}{\bar{E}_B}| = |\frac{d}{d \bar{k}_A^{z}}(\bar{E}_A + \bar{E}_B) |$$

Now we try to use (as in our original question ) : $\int dx \delta(f(x) - f(x_0)) = 1/ |f'(x_0)|$.
Let $f(\bar{k}_A^{z}) := \bar{E}_A + \bar{E}_B$ ( $\bar{E}_A \stackrel{?}{=} \sqrt{\bar{k}_A^{2} + m_A^{2}}$ depends on $\bar{k}_A^{z}$ ).
If we can show that there exists a point $x_0$ such that $f(x_0) = \Sigma E_f$, and $f'(x_0) = \frac{\bar{k}_A^{z}}{\bar{E}_A} - \frac{\bar{k}_B^{z}}{\bar{E}_B}$, then we are done.

And is this possible?

Please understand if I don't understand your argument completely.

 

[vanhees71]

Thanks for kind explanation, although there seems to be few typos.

And I am somewhat confused since In your derivation of the integral equation ; i.e., Peskin's book (4.77) , it seems that you have used symbols A and B interchangeably.

Anyway, I still don't understand why such next integral is true :

$$\int d^3\bar{k}_A \int d^3 \bar{k}_B \delta^{(4)}(\Sigma \bar{k}_i - \Sigma p_f) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) $$
$$\stackrel{?}{=} (k_i^{\perp} = \bar{k}_i^{\perp}) \times \int d \bar{k}^z_A d\bar{k}_B^{z} \delta(\bar{k}_A^{z} + \bar{k}_B^{z} - \Sigma p_f^{z}) \delta ( \bar{E}_A + \bar{E}_B - \Sigma E_f) $$
$$ \stackrel{?}{=} \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$

I just did the integral over $\bar{k}_A$ using the $\delta$ distribution. There was a typo the arguments in the $\delta$-distribution over the perpendicular components must be of course $\vec{k}_{B \perp}$, i.e., $\delta^{(2)}(\vec{k}_{B \perp}'-\vec{k}_{B \perp}$. You can also integrate over $\bar{k}_B$ first and then over $\bar{k}_A$. Of course that doesn't change the result.

I think that you skipped this step and explained next material :) Can you explain this integration step more in detail?

I don't know, how to explain this in more detail. You just use the $\delta^{(2)}$ distribution and keep the rest. Then you go on with integration over the $z$-components, which I've also did in detail. For the final integration you need the substitution formula for the energy-conserving $\delta$ distribution. Is taking the derivative the problem?

C.f. 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. And I don't know how to deal with integration whose integrands all consists of delta functions.

And let's return to the final equality in my original question :

$$ \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$
$$\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |} = \frac{1}{|v_A -v_B|} $$

Q. How can we use $\int dx \delta(f(x) - f(x_0)) = 1/ |f'(x_0)|$?

For this final equality, my first attempt is as follows :

From $\frac{d\bar{E}_A}{d \bar{k}_A}= \frac{\bar{k}_A}{\bar{E}_A}$ (19.37), we get
$\frac{\bar{k}_A^{z}}{\bar{E}_A} = \frac{d\bar{E}_A}{d \bar{k}_A^{z}}$. (True?)

And from $\frac{d \bar{E}_B}{d\bar{k}_B} = \frac{\bar{k}_B}{\bar{E}_B} = - \frac{d\bar{E}_B}{d\bar{k}_A}$ (19.37), we get
$\frac{\bar{k}_B^{z}}{\bar{E}_B} = \frac{d \bar{E}_B}{d\bar{k}_B^{z}} = - \frac{d\bar{E}_B}{d\bar{k}_A^{z}}$ (True?)

So formally, $$| \frac{\bar{k}_A^{z}}{\bar{E}_A} - \frac{\bar{k}_B^{z}}{\bar{E}_B}| = |\frac{d}{d \bar{k}_A^{z}}(\bar{E}_A + \bar{E}_B) |$$

Now we try to use (as in our original question ) : $\int dx \delta(f(x) - f(x_0)) = 1/ |f'(x_0)|$.
Let $f(\bar{k}_A^{z}) := \bar{E}_A + \bar{E}_B$ ( $\bar{E}_A \stackrel{?}{=} \sqrt{\bar{k}_A^{2} + m_A^{2}}$ depends on $\bar{k}_A^{z}$ ).
If we can show that there exists a point $x_0$ such that $f(x_0) = \Sigma E_f$, and $f'(x_0) = \frac{\bar{k}_A^{z}}{\bar{E}_A} - \frac{\bar{k}_B^{z}}{\bar{E}_B}$, then we are done.

And is this possible?

Please understand if I don't understand your argument completely.

 

[Plantation]

I just did the integral over $\bar{k}_A$ using the $\delta$ distribution. There was a typo the arguments in the $\delta$-distribution over the perpendicular components must be of course $\vec{k}_{B \perp}$, i.e., $\delta^{(2)}(\vec{k}_{B \perp}'-\vec{k}_{B \perp}$. You can also integrate over $\bar{k}_B$ first and then over $\bar{k}_A$. Of course that doesn't change the result.

O.K. 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 connected to (4.77) in Peskin's book.

But I think that is true since, for example, $\delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) = \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta(k_B^{x} - \bar{k}_B^{x}) \delta(k_B^{y} - \bar{k}_B^{y})$ is independent from variable $\bar{k}_A^{x}$ ( C.f. $k_B^{\perp} := (k_B^{x},k_B^{y})$ , $\bar{k}_B^{\perp}:= (\bar{k}_B^{x}, \bar{k}_B^{y})$ (True?) ) so that 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?
I don't know, how to explain this in more detail. You just use the $\delta^{(2)}$ distribution and keep the rest. Then you go on with integration over the $z$-components, which I've also did in detail. For the final integration you need the substitution formula for the energy-conserving $\delta$ distribution. Is taking the derivative the problem?

For the final integration ; i.e.,

$$ \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$
$$\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |}$$

, Yes. I don't know how to use the substitution formula for the energy-conserving $\delta$-distribution. Can you explain procedure more in detail? Can you help me?

C.f. My first attempt to derive the final integration is as follows :

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? ; c.f. https://math.stackexchange.com/questions/276583/dirac-delta-function-of-a-function )

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 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 $f'(x_i) \neq 0$?

 

[vanhees71]

The argument of the $\delta$ distribution is (writing a prime instead of a bar for typographical reasons
$$f(k_A^{\prime z})=\sqrt{\vec{k}_A^{\prime 2}+m_A^2} + \sqrt{\vec{k}_B^{\prime 2}+m_B^2},$$
where $\vec{k}_{B \perp}^{\prime}=\vec{k}_{B \perp}$ and $k_B^{\prime z}=\sum_f p_f^z - k_A^{\prime z}$. Thus you get
$$\frac{\mathrm{d}f(k_A^{\prime z})}{\mathrm{d} k_A^{\prime z}} = \frac{k_A^{\prime z}}{\sqrt{\vec{k}_A^{\prime 2}+m_A^2}} - \frac{k_B^{\prime z}}{\sqrt{\vec{k}_B^{\prime 2}+m_B^2}}.$$

 

[Plantation]

The argument of the $\delta$ distribution is (writing a prime instead of a bar for typographical reasons
$$f(k_A^{\prime z})=\sqrt{\vec{k}_A^{\prime 2}+m_A^2} + \sqrt{\vec{k}_B^{\prime 2}+m_B^2},$$
where $\vec{k}_{B \perp}^{\prime}=\vec{k}_{B \perp}$ and $k_B^{\prime z}=\sum_f p_f^z - k_A^{\prime z}$. Thus you get
$$\frac{\mathrm{d}}{\mathrm{d} k_A^{\prime z}} = \frac{k_A^{\prime z}}{\sqrt{\vec{k}_A^{\prime 2}+m_A^2}} - \frac{k_B^{\prime z}}{\sqrt{\vec{k}_B^{\prime 2}+m_B^2}}.$$

O.K. Thank you. And.. I think that I also reached to such a step, as I wrote. I still don't know why

$$ \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$
$$\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A} -\frac{\bar{k}_B^z}{\bar{E}_B} |}$$

To verify this, I am trying 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 physics, p456)

Can we use this theorem? How?

 

[vanhees71]

But you just take the modulus and divide by the result of the derivative, note that $E_A(\vec{k}_A)=\sqrt{m_A^2+\vec{k}_A^2}$.

 

[Plantation]

But you just take the modulus and divide by the result of the derivative, note that $E_A(\vec{k}_A)=\sqrt{m_A^2+\vec{k}_A^2}$.

By taking modulus, what does you exactly means? Shall we find the roots of $$f(\bar{k}_A^{z}) := (\sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A} $$ directly, and then brutally force them into the formula $$ \int \delta[f(x)] dx = \Sigma_{i=1} \frac{1}{|f'(x_i)|}$$ ( $f(x_i)=0, f'(x_i) \neq 0$) ? Is there a way to deduce our desired result faster by noting something?

 

[vanhees71]

You need
$$\frac{\mathrm{d} f(k_A^{\prime z})}{\mathrm{d} k_A^{\prime z}} = \frac{k_A^{\prime z}}{E_A'}-\frac{k_B^{\prime z}}{E_B'}=v_{\text{rel}},$$
and then the final integral simply is $1/|v_{\text{rel}}|$.

 

[Plantation]

You need
$$\frac{\mathrm{d} f(k_A^{\prime z})}{\mathrm{d} k_A^{\prime z}} = \frac{k_A^{\prime z}}{E_A'}-\frac{k_B^{\prime z}}{E_B'}=v_{\text{rel}},$$
and then the final integral simply is $1/|v_{\text{rel}}|$.

O.K. Looking closerly, in (4.77) (Peskin's book), $\frac{1}{|\frac{\bar{k}_A^{z}}{\bar{E}_A}- \frac{\bar{k}_B^{z}}{\bar{E}_B}|} = \frac{1}{|\frac{d}{d \bar{k}_A^{z}} f(\bar{k}_A^z)|}$ seems a 'function' which depends on variable $\bar{k}_A^{z}$.

On the other hand, if we may use $\int \delta[f(x)] dx = \Sigma_{i=1} \frac{1}{|f'(x_i)|}$, then $\Sigma_{i=1} \frac{1}{|f'(x_i)|}$ is a 'number'.

Where does this discrepancy occurs? Perhaps, the notation $\frac{1}{|\frac{\bar{k}_A^{z}}{\bar{E}_A}- \frac{\bar{k}_B^{z}}{\bar{E}_B}|}$ is a 'implicit' notation
indicating for $\Sigma_{i=1} \frac{1}{|f'(x_i)|}$ ?

Am I pointing out well?

 

[vanhees71]

It's precisely this formula, which is used. I don't know, how else I can explain it.

 

[Plantation]

It's precisely this formula, which is used. I don't know, how else I can explain it.

O.K. Anyway, thank you :)