4 Cross Sections

We must integrate over all possible four-momenta of each particle involved. While doing so we must ensure that all particles are on-shell, i.e. $p_{i}^{2}=m_{i}^{2}$, that their energies are positive $p_{i}^{0}>0$ (this is still Lorentz invariant!), and that momentum is conserved $\sum p_{i}^{\rm in}=\sum p_{i}^{\rm out}$.

Let us begin with the phase space of a single particle. The requirements set out above can be implemented with a $\delta$ function and a $\Theta$ function

The $\delta$ function can be used to solve the $p_{i}^{0}$ integral – almost. We first need to use the following identity (convince yourself that this is true)

Therefore,

where we have used the $\Theta$ function requiring $p_{i}^{0}>0$. The condition that the particle must be on-shell, i.e. $p_{i}^{0}=\sqrt{\vec{p}_{i}^{2}+m_{i}^{2}}$ is now implicit and for the sake of readability keep $p_{i}^{0}$ in the denominator. This object is manifestly Lorentz invariant even if it does not look like it is. Therefore it is called the Lorentz-invariant phase space (LIPS). The total phase space is now

To calculate a cross section, we need to add a flux factor that accounts for the amount of incoming particles. For scattering, this is usually given as (see Appendix B for a short derivation)

Here, $E_{i}$ and $\vec{v}_{i}$ are the energies and velocities of the incoming particles. The second version is equivalent but a bit easier to work with. Note that despite only depending on Lorentz invariant quantities like $p_{1}\cdot p_{2}$ and $m_{i}^{2}$, $\mathcal{F}$ is only invariant for boosts along the beam axis. For boosts along any other axis, it is not invariant which fits well with our intuition of cross-sectional areas.

The cross section is therefore

where the sum averages over initial states and sums over final states. For a decay width, the flux is instead

so that

It is useful to remember the phase space for a $1\to 2$ or $2\to 2$ process. With incoming momentum $P$ and outgoing momenta $p_{1}$ and $p_{2}$,

We can use three of the four $\delta$ functions to perform the integration over $p_{2}$

We require here that $p_{i}^{0}=\sqrt{\vec{p}_{i}^{2}+m_{i}^{2}}$ and that $\vec{p}_{2}=-\vec{p}_{1}$ in the rest frame of $P$ (this is often called the centre-of-mass frame). Next, we use spherical coordinates to write ${\rm d}^{3}p_{1}=|\vec{p}_{1}|^{2}{\rm d}|\vec{p}_{1}|\ {\rm d}\Omega$.

After an annoying but not particularly difficult calculation, we arrive at

The momentum $|\vec{p}_{i}|$ is

We can continue our discussion from Section 3.4 and calculate the cross section. Using the above result for the phase space and $P^{0}=\sqrt{s}$, we have

For simplicity we will set $m=M=0$. Then we can write

And therefore, with $\alpha=e^{2}/4\pi$

The calculation we have just performed is very similar to the one needed for

Although this now involves anti-particles, there is still one single diagram at leading-order and the trace algebra is very similar. Indeed we can re-interpret the incoming $e^{+}$ as an outgoing $e^{-}$ with momentum $-p_{1}$ and the outgoing $\mu^{+}$ as an incoming $\mu^{-}$ with momentum $-p_{3}$. Then we find

This is an example of crossing symmetry. Note in general that there is an additional minus sign for each fermion which swaps from the initial to final state or vice versa. This is because, for example,

In this case there are two minus signs whose combined effect gives just one.

We can therefore recycle our old calculation and write for $m=0$ and $M>0$

where we have rewrite the Mandelstam variable $t$ as

The cross section is

The cross section is

We can now convert the above result to a total cross section by performing the integral over the solid angle. This gives

Now, when an electron and positron annihilate, other fermions may be produced. If these are quarks, they are then observed in the detector as hadrons. The same calculation gives

plus higher-order corrections, where there are $N_{c}$ colours in each of the $n_{f}$ massless flavours of quarks with charge $Q_{i}$. Therefore the ratio

has been used to measure the number of colours to be $N_{c}=3$.

Let us calculate another process in QED, namely Compton scattering

We can follow the same recipe as above and calculate the two diagrams

	$\displaystyle\mathcal{M}$	$\displaystyle=\begin{gathered}\includegraphics{bmlimages/notes-12.svg}\bml@image@depth{12}\bmlDescription{ A diagram with an electron and photon coming in (momenta p1 and p2) and meeting. The electron continues horizontally with momentum p1+p2 and then emits a photon with momentum p4 before leaving with momentum p3. }\end{gathered}\quad+\quad\begin{gathered}\includegraphics{bmlimages/notes-13.svg}\bml@image@depth{13}\bmlDescription{ An electron comes in with momentum p1 and emits a photon with momentum p4 going out. Its line goes straight down with momentum p1-p4 before interacting with an incoming photon of momentum p2. It leaves with momentum p3 }\end{gathered}$		(133)
		$\displaystyle=-ie^{2}\epsilon^{*\mu}(p_{4})\,\epsilon^{\nu}(p_{2})\,\,\bar{u}(p_{3})\Bigg{(}\gamma_{\mu}\frac{\gamma\cdot(p_{1}+p_{2})+m}{(p_{1}+p_{2})^{2}-m^{2}}\gamma_{\nu}+\gamma_{\nu}\frac{\gamma\cdot(p_{1}-p_{4})+m}{(p_{1}-p_{4})^{2}-m^{2}}\gamma_{\mu}\Bigg{)}u(p_{1})\,.$		(134)

You can check explicitly that this fulfils the Ward identity by replacing $\epsilon(k)$ with $k$ (see tutorial sheet). We can square the amplitude, summing over fermion spins and photon polarisations and find

Here we have used the identities

from the tutorial to simplify the algebra before completing the trace. We have also used the rules

that follow from momentum conservation. For completeness, we also give the amplitude squared for $e^{-}(p_{1})e^{+}(p_{2})\to\gamma(p_{3})\gamma(p_{4})$ as

	$\displaystyle\mathcal{M}$	$\displaystyle=\begin{gathered}\includegraphics{bmlimages/notes-14.svg}\bml@image@depth{14}\bmlDescription{ An electron comes in with momentum p1 and emits a photon with momentum p3 going out. The electron line continues downwards where it joins with a positron coming in (momentum p2) that has emitted a photon with momentum p4. }\end{gathered}\quad+\quad\begin{gathered}\includegraphics{bmlimages/notes-15.svg}\bml@image@depth{15}\bmlDescription{ An electron comes in with momentum p1 and emits a photon with momentum p4 going out. The electron line continues downwards where it joins with a positron coming in (momentum p2) that has emitted a photon with momentum p3. The outgoing photons are visibly swapped. }\end{gathered}$		(140)
		$\displaystyle=-ie^{2}\epsilon^{*\mu}(p_{3})\,\epsilon^{*\nu}(p_{4})\,\,\bar{v}(p_{2})\Bigg{(}\gamma_{\nu}\frac{\gamma\cdot(p_{1}-p_{3})+m}{(p_{1}-p_{3})^{2}-m^{2}}\gamma_{\mu}\gamma_{\mu}\frac{\gamma\cdot(p_{1}-p_{4})+m}{(p_{1}-p_{4})^{2}-m^{2}}\gamma_{\nu}\Bigg{)}u(p_{1})\,.$		(141)

And for $|\mathcal{M}|^{2}$, we have

This result could also have been obtained by setting $p_{3}\leftrightarrow-p_{2}$ and swapping the sign in (134).

Working in the rest frame of the incoming electron, we have

We can evaluate the two scalar products we need as $p_{1}\cdot p_{2}=m\omega$ and $p_{1}\cdot p_{4}=m\omega^{\prime}$. Using (137b) which arises from $m^{2}=p_{3}^{2}=(p_{1}+p_{2}-p_{4})^{2}$, we can write

And therefore find for $\omega^{\prime}$ as a function of $\theta$

With this, we can write $|\mathcal{M}|^{2}$ very simple as

Note that this still depends on $m$ via the relation between $\omega^{\prime}$ and $\theta$. Adding the flux factor and phase space, we arrive at

Let us think about this result means by considering various limits

non-relativistic

where $\omega\ll m$ where $\omega^{\prime}\approx\omega$ and

$\displaystyle\frac{{\rm d}\sigma}{{\rm d}\Omega}\Bigg{|}_{\omega\ll m}=\frac{\alpha^{2}}{2m^{2}}\big{(}2-\sin^{2}\theta\big{)}=\frac{\alpha^{2}}{2m^{2}}\big{(}1+\cos^{2}\theta\big{)}\,.$

(148)

This is the Thomson cross section for the scattering of classical electromagnetic radiation by a free electron, which is symmetrical in the scattering angle, i.e. the photon is just as likely to scatter backwards as forwards.

high-energy

where $\omega\gg m$, we have $\omega^{\prime}\approx m/(1-\cos\theta)$ and

$\displaystyle\frac{{\rm d}\sigma}{{\rm d}\Omega}\Bigg{|}_{\omega\gg m}=\frac{\alpha^{2}}{2m^{2}\omega}\frac{1}{1-\cos\theta}$

(149)

and the cross section is strongly peaked for small angles. This leads to a logarithmic enhancement when you perform the angular integration. These collinear logarithms arise whenever massless particles are emitted; this will be discussed in more detail in the phenomenology course.

small angle at high energy

Since $\omega>\omega^{\prime}$, we have in the high-energy limit $1-\cos\theta>m/\omega$. Therefore, if $\theta\ll 1$, i.e. small angle scattering,

$\displaystyle\frac{{\rm d}\sigma}{{\rm d}\Omega}\Bigg{|}_{\omega\gg m,\theta\ll 1}=\frac{\alpha^{2}}{m^{2}}\,.$

(150)

The forward (small scattering angle) Compton scattering cross section is then a valuable method to measure the QED coupling $\alpha$.