Understanding how to derive the differential cross-section formula

[JD_PM]

TL;DR Summary

I am studying how to derive the cross section out of S-matrix elements from the beautiful book Quantum Field Theory by Mandl and Shaw (second edition; chapter 8, section 8.1) and there are several steps I do not get.

Please let me make questions after showing what I am studying.

We first consider two particles (they may be either leptons or photons) with initial (i.e. before collision) four momentum $p_i = (E_i, \mathbf p_i)$, $i=1,2$. These two collide and produce $N$ final particles with momentum $p'_f = (E'_f, \mathbf p'_f)$, $f=1,...,N$. In this post we assume that initial and final particles are in definite polarization states.

I learned in the previous chapter of the book that the S-matrix elements can be written as

$$\langle f| S |i \rangle = \delta_{fi} + \Big[ (2\pi)^4 \delta^{(4)} (P_f-P_i) \Pi_e \Big(\frac{m}{VE}\Big)^{1/2}\Pi_e \Big(\frac{1}{2V \omega}\Big)^{1/2} \Big] \mathscr{M} \ \ \ \ (1)$$

Where the e attached to the products stands for external fermions and photons.

In our case $(1)$ becomes

$$S_{fi} = \delta_{fi} + (2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \Pi_i \Big(\frac{1}{2V E_i}\Big)^{1/2} \Pi_f \Big(\frac{1}{2V E'_f}\Big)^{1/2}\Pi_l (2m_l)^{1/2}\mathscr{M} \ \ \ \ (2)$$

Where $l$ runs over all external leptons involved in the collision.

Eq. $(2)$ applies to an infinite time interval (i.e $T \rightarrow \infty$) and an infinite volume (i.e $V \rightarrow \infty$).

For finite intervals we'd get the exact same as Eq. $(2)$ but $(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i)$ being replaced by $\delta_{TV} (\sum p'_f - \sum p_i)$, as

$$(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) = lim_{T \rightarrow \infty, \ V \rightarrow \infty} \delta_{TV} (\sum p'_f - \sum p_i) \ \ \ \ (3)$$

In deriving the cross section we better take $V$ and $T$ to be finite. In such a case the transition probability per unit time becomes

$$w=|S_{fi}|^2/T \ \ \ \ (4)$$

We notice that $|S_{fi}|^2$ means we have to deal with $\Big( \delta_{TV} (\sum p'_f - \sum p_i) \Big)^2$, which equals to the following

$$\Big( \delta_{TV} (\sum p'_f - \sum p_i) \Big)^2 = TV (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \ \ \ \ (5)$$

With errors which tend to zero as $T \rightarrow \infty$ and $V \rightarrow \infty$. Once we know this $(4)$ becomes

$$w=V(2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \Pi_i \Big(\frac{1}{2V E_i}\Big)^{1/2} \Pi_f \Big(\frac{1}{2V E'_f}\Big)^{1/2} \Pi_l (2m_l)^{1/2} |\mathscr{M}|^2 \ \ \ \ (6)$$

Eq. $(6)$ is the transition rate to one definite final state.

If we wish to obtain the transition rate to a group of final states with momentum in the interval $(\mathbf p'_f, \mathbf p'_f +d\mathbf p'_f)$, $f=1,...,N$, we have to multiply $w$ by the number of these states, which is

$$\Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big)$$

The differential cross section is the transition rate into this group of final states for one scattering center and unit incident flux (which is also known as luminosity; I've read that particle physicists use more the latter term). With our choice of normalization for the states, the volume V contains one scattering center and the incident flux is $\frac{v_{rel}}{V}$, where $v_{rel}$ is the relative velocity of the colliding particles.

At this point we have everything to set up the equation for the differential cross section

$$d \sigma = w \frac{V}{v_{rel}} \Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big) \ \ \ \ (7)$$

Plugging $(6)$ into $(7)$ we get

$$d \sigma = (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \frac{1}{4 E_1 E_2 v_{rel}} \Pi_l (2m_l) \Pi_f \Big(\frac{d^3 \mathbf p'_f}{(2\pi)^3 2 E'_f}\Big) |\mathscr{M}|^2$$

My questions are:

1) I do not see why $(2)$ corresponds to the limit of an infinite time interval and an infinite volume.

2) I do not understand why we have to replace $(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i)$ by $\delta_{TV} (\sum p'_f - \sum p_i)$. Besides, I do not understand equation $(3)$.

3) I do not see how to get equation $(5)$.

4) I first studied differential cross-section from the nice book by Griffiths: Introduction to Quantum Mechanics; (second edition; chapter 11). Then I learned that the larger the infinitesimal patch of cross-sectional area $d\sigma$ is the bigger $d \Omega$ will be, which means we can establish the following proportionality factor $D(\theta)$

$$d \sigma=D(\theta)d \Omega \ \ \ \ (8)$$

Thus (8) and (7) must be equal to each other but I do not see it.Any help is appreciated.

Thanks

 

[vanhees71]

The problem is of course, what to make of $|S_{fi}|^2$, when you have the energy-momentum conserving $\delta^{(4)}(P_f-P_i)$ factor in there. There's no way to simply square the amplitude.

The reason why this $\delta$ distribution is in there is simple from a physical point of view: It occurs, because you put in plane waves for the initial and finial asymptotic free particle states. These are, however, no true states, because they are not square integrable. Thus the physical solution for the problem is to use wave packets (narrow in momentum space), take the square and only at the end take the limit to plane-wave states. This approach you find in Peskin&Schroeder (though I warn you to check every single formula in this book carefully for yourself, because unfortunately there are many typos in it, which is a pity, because it's otherwise a quite got introductory book on relativistic QFT).

The more convenient way to "regularize" the $\delta$ distribution is to impose periodic boundary conditions on a finite volume, e.g., a cube with finite length $L$. Then you still have well-defined momentum operators and the momentum eigen states (plane waves) are square integrable and thus proper states. Then also consider a finite time interval $(-T,T)$ with $T$ large. This tames the $\delta$ distributions for the momenta to actually Kronecker symbols and the energy $\delta$ distribution to a "sinc function". Then you consider not transition probabilities but transition probability density rates and then take the infinite-volume limit and $T \rightarrow \infty$ limit for the squared amplitude/cross sections.

For this approach, see my QFT script:

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

The result of both approaches is the formula you quote next after Eq. (7).

 

[JD_PM]

Thank you for the really helpful insight vanhees71!

I was studying your section on invariant cross sections and found out that you present, as Mandl and Shaw do in section 8.1 of their book, that for finite $T$ and $V$ the $\delta^{(4)}$ distribution is:

$$(2 \pi)^4 \delta_{reg}^{(4)} (P_f - P_i) = \int_{-T/2}^{T/2} dt \int_{V} d^3 \mathbf x \exp[i(P_f - P_i)x] \ \ \ \ (*)$$Mmm this question may sound trivial to you but why does $(*)$ regularalize the $\delta^{(4)}$ distribution? $(*)$ comes a bit out of the blue to me.Besides, the integral at $(*)$ is presented to have the following solution

$$\int_{-T/2}^{T/2} dt \int_{V} d^3 \mathbf x \exp[i(P_f - P_i)x]=2^4 \Big(\frac{2}{T}\Big) \Big(\frac{2}{L}\Big)^3 \Big( \frac{\sin(\Delta P_0 T/2)}{\Delta P_0} \Big)^2 \Pi_{k=1}^3\Big( \frac{\sin(\Delta P_k L/2)}{\Delta P_k} \Big)^2$$

But I neither see why the $\sin$ function shows up nor why they are squared.

Based on the integral of the exponential function $\int dx e^{ax}= \frac {1}{a} e^{ax}$ I would naively (and mistakenly) get

$$\int_{-T/2}^{T/2} dt \int_{V} d^3 \mathbf x \exp[i(P_f - P_i)x]=2^4 \Big(\frac{1}{T}\Big) \Big(\frac{1}{L}\Big)^3 \Big( \frac{1}{\Delta P_0} \Big) \exp[i \Delta P_0 T] \Pi_{k=1}^3\Big( \frac{1}{\Delta P_k} \Big)\exp[i \Delta P_k L]$$
From here on I have more questions about the Mathematics. I'd perfectly understand if you were to see these kind of questions too trivial. If that is the case I'll deal with them and come back to discuss the Physics of the cross-section

 

[vanhees71]

The integrals are of the form
$$\int_{-T/2}^{T/2} \mathrm{d} t \exp(-\mathrm{i} t \Delta P^0)=\left . \frac{\mathrm{i}}{\Delta P^0} \exp(-\mathrm{i} t \Delta P_0)\right|_{t=-T/2}^{t=T/2}=\frac{2}{\Delta P^0} \sin \left (\frac{T \Delta P^0}{2} \right)$$
and the analogous integrals for the three momentum components.

The next formula is taking the matrix element with the regularized $\delta$ distribution plugged in squared and divided by $TL^3$, to get from transition probabilities to (average) transition-rate densities. Then taking the limit $T \rightarrow \infty$ and $L \rightarrow \infty$ leaves you with one energy-momentum conserving $\delta$ distribution.

 

[JD_PM]

The integrals are of the form
$$\int_{-T/2}^{T/2} \mathrm{d} t \exp(-\mathrm{i} t \Delta P^0)=\left . \frac{\mathrm{i}}{\Delta P^0} \exp(-\mathrm{i} t \Delta P_0)\right|_{t=-T/2}^{t=T/2}=\frac{2}{\Delta P^0} \sin \left (\frac{T \Delta P^0}{2} \right)$$
and the analogous integrals for the three momentum components.

I had forgot that

$$\sin(\theta) = \frac{ \exp(i\theta) - \exp(-i\theta) }{2i}$$

Now it is clear, Thank you!

 

[JD_PM]

About my section 4).

I was wondering what's the analogy between these two equations for the differential cross section

$$d \sigma = w \frac{V}{v_{rel}} \Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big) \ \ \ \ (7)$$

$$d \sigma=D(\theta)d \Omega \ \ \ \ (8)$$

My point is that if someone were to show me Eqs. (7) and (8) without telling me what they are, I'd only be able to identify (8) as the differential cross section.

 

[vanhees71]

For that you'd need to give a bit more context. What's discussed in (8) is obviously some special case of (7). I guess it's for a 2->2 scattering process? Of course the trick is to integrate out the remaining energy-momentum conserving $\delta$ distribution, i.e., to do the "phase space integrals". A very helpful summary about the most important cross section formulae can be found in the Review of Particle Physics, which is anyway a great source:

http://pdg.lbl.gov/

The summary article on cross-section formulae is at

http://pdg.lbl.gov/2019/reviews/rpp2019-rev-kinematics.pdf

 

[JD_PM]

For that you'd need to give a bit more context. What's discussed in (8) is obviously some special case of (7). I guess it's for a 2->2 scattering process? Of course the trick is to integrate out the remaining energy-momentum conserving $\delta$ distribution, i.e., to do the "phase space integrals".

Oh I've just noticed I was basically comparing quantum scattering to classical!

On the one hand, Eq. (7) is the differential cross section for a scattering process in which two particles (they may be leptons or photons) collide and produce $N$ final particles.

On the other hand, Eq. (8) is the differential cross section for a (classical) scattering process in which a particle (for instance, a proton fired at a heavy nucleus) collides and scatters at an angle $\theta$

So 4) is now clear :)

I had a look at the Cross-section Formulae for Specific Processes; it will be really useful, thanks for sharing.

Footnote: with classical I mean that it is not possible to create/destroy particles out of the collision of several particles.

 

[JD_PM]

I have a question about the Mathematics

I was trying to understand how to prove that

$$\lim_{y \rightarrow \infty} \frac{\sin^2(xy)}{yx^2}=\pi \delta(x)$$

In your script you suggest that we do the following Inverse Fourier Transform:

$$\int dx \frac{\sin^2(xy)}{yx^2} \exp(-ipx) = \frac{\pi}{2} \Theta(2y - |p|) \Big(2-\frac{|p|}{y}\Big)$$

But, how did you get such a result? I could not solve the integral so I used Mathematica and it seems it does not process it (I may have typed something in the wrong way though)

I used another website to compute the integral, Integral Calculator, but did not solve my doubt.

 

[vanhees71]

To prove the formula, just prove the inverse, i.e., for $y>0$ set
$$\tilde{f}(p)=\frac{\pi}{2} \Theta(2y-|p|) \left (2-\frac{|p|}{y} \right)$$
and evaluate
$$f(x)=\int_{\mathbb{R}} \frac{\mathrm{d} p}{2 \pi} \exp(\mathrm{i} p x) \tilde{f}(p).$$

 

[JD_PM]

I think that I got the general idea; Given the Fourier Transform (FT)

$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(k) e^{ikx} dk$$

And the Inverse Fourier Transform (IFT)

$$F(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{ikx} dx$$

If we plug the IFT into the FT we are going to get the Dirac-Delta function times an integral. Actually, we get the sifting property of the Dirac-Delta function. Explicitly:

$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dk e^{ikx} \int_{-\infty}^{\infty} f(x')e^{-ikx'} dx'=\int_{-\infty}^{\infty} dx' f(x') \frac{1}{2\pi}\int_{-\infty}^{\infty} dk e^{ik(x-x')}=\int_{-\infty}^{\infty} dx' f(x') \delta(x-x')$$

Source: MIT OpenCourseWare

 

[JD_PM]

So in our particular problem we actually expect $\int_{-a}^{a} dx' f(x')=\pi$

So we plug the IFT into the FT and get

$$f(x)=\int \frac{dp}{2 \pi} e^{ipx} \tilde f(p)=\frac 1 4 \int dp \Theta(2y - |p|)\Big(2-\frac{|p|}{y}\Big) e^{ipx}=\frac 1 4 \delta(x) \int dp \Theta (2y - |p|)\Big(2-\frac{|p|}{y}\Big)$$

So taking into account that $y \rightarrow \infty$, (and assuming I did not make mistakes at this point) we have to show that

$$2 \int dp \Theta (2y - |p|)=\pi$$

Do you agree at this point? Now I am trying to figure out how to compute integrals of the Jacobi Theta Functions.

PS: the integrals are assumed to run $[-\infty, \infty]$