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 scattering of electrons and positrons. QED processes Energy spectrum
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Electron-positron annihilation and pair creation.

The processes of electron-positron annihilation into photon pairs and of pair creation by photons are of interest both theoretically and experimentally. Electron-positron annihilation takes places with high rates at $e^+e^-$ colliders such as LEP, SLC and TRISTAN. It is therefore important to account for this reaction correctly in order to carry out reliable analyses of experimental data. Moreover this is also important for the development of new, higher energy electron-positron colliders and for the planning of new experiments because this process is a large source of background. Theoretically the process is well understood ever since it was first calculated by P.A.M. Dirac in 1930. We shall see later that the study of $e^+e^-$ annihilation into photons is also a first step towards the understanding of a similar process, the annihilation of quark-antiquark pairs into gluons, which has acquired importance in elementary particle physics in recent years.

When you enter the annihilation process in the CompHEP session you will see that only one Feynman diagram is generated (Fig. 4). This should come as some surprise since you probably know that by crossing you should expect each of the two Compton diagrams to give rise to one Feynman diagram for the annihilation process (see, for instance, [2] Section 7.8, [1] Section 6.15). The reason for this is simple: the two diagrams which you expect have only different 4-momenta and polarization vectors attached to the final state photons, topologically however they are identical. The single diagram displayed by CompHEP is therefore equivalent to two diagrams with labelled external lines. When you let CompHEP carry out the squaring of the amplitude and then inspect the squared diagrams, you will see that there are two squared diagrams and not three. One of these represents the two topologically identical diagrams with different photon momenta and polarizations, the second one is the expected interference term.

Now carry out the analytical calculations of the squared matrix element applying the familiar procedure of invoking option Symbolic calculations. We can explore the annihilation process by performing a numerical experiment using the option Numerical calculator.

Let us begin by choosing a CMS energy of 64 GeV, characteristic for the electron-positron collider TRISTAN. The total cross section will be displayed as $715\, \mbox{pb}$. Assuming the nominal luminosity of TRISTAN, $L=3.7\times 10^{31}\mbox{cm}^{-2}\mbox{s}^{-1}=3.7\times
10^{-5}\mbox{pb}^{-1}\mbox{s}^{-1}$, such a cross section means the production of one photon pair every 40 seconds. This is a high rate, more than an order of magnitude greater than for any other annihilation process at this energy. Therefore the two-photon annihilation could turn out to be a very intensive source of background for other physical processes. However let us look at the photon angular distribution using option Angular dependence. You will see the curve reproduced here in Fig. 5. The main feature of this distribution is the forward/backward symmetry which is always observed when there are two identical particles in the final state. Furthermore we note a very strong peaking at $\vartheta_\gamma = 0$ and $\pi$. Experimentally this implies that most of the photons escape in the beam pipe.

To get a more quantitative picture about the narrow peaks at the edges let us store the results in the form of a table using option Save result in a file and then view the table with the help of function key F5. The differential cross section equals $\sim 10^{5}\,\mu\mbox{b}$ at $\cos \vartheta_\gamma = \pm 1$ and drops by nine orders of magnitude between $\cos\vartheta_\gamma =1$ and $\cos \vartheta_\gamma = 0.9$, and stays flat between $-0.9$ and $0.9$. Next let us cut out a typical pair of cones around the beam pipe, where particles remain undetected (see section [*]), using option Set angular range:

f-angle   between   in - e1    and    out - A
cos(f):   min = -0.996200      max = 0.996200

This corresponds to an angular cut of $5^\circ <\vartheta _\gamma <175^\circ $. As a result the total cross section is reduced to $168\,\mbox{pb}$. This means that about 80% of the photons escape undetected in the beam pipe or, conversely, that the number of photons actually seen in the detector is significantly less than the total number of photons produced by annihilation. Nevertheless this rate is still large and such events can be a significant background to other reactions.

Let us calculate now the energy dependence of the total cross section for three cases: for the whole angular range and for the angular intervals $-0.99985 <\cos\vartheta_\gamma < 0.99985$ ( $1^\circ <\vartheta _\gamma <179^\circ $) and $-0.9962< \cos\vartheta_\gamma < 0.9962$ ( $5^\circ <\vartheta _\gamma <175^\circ $). To get the corresponding plots use the menu Parameter dependence and select the options Total Cross Section and Energy. The package will prompt you for the energy range you want to examine, for the scale (logarithmic or linear) and for the number of points (21 by default). Let us choose the following values:

 Parameter    : Energy     min  10                     max  100

                    Scale  norm             Number of points  21

As soon as these parameters are entered, the numerical calculation is done automatically. When the calculation is completed you can inspect the results using options Show plot and Save result in a file. The latter option produces tables in the form of files with the names tab_N.txt, where $\mbox{\sf N} = 1,\, 2,\, 3,\,\ldots$. Here in Fig. 6 we have combined the three curves in one plot. From these curves we infer that the total cross section has the smooth behaviour characteristic of collisions of structureless particles. The drop with energy is independent of the angular range. Presently we shall see that it is close to a $1/s$ law.

To get the analytical answer for the squared matrix element you must repeat all steps which we have made in the case of Compton scattering for both of the squared amplitudes and then add the results to get the final expression:

\begin{displaymath}
\vert M\vert^2 \;=\; 16 \pi^2 \alpha^2
\left[-4 \left(\frac...
...ht)
+\frac{m_e^2-t}{m_e^2-u}+\frac{m_e^2-u}{m_e^2-t} \right].
\end{displaymath} (6)

You can check that the squared matrix elements have the same functional form in the two cases of Compton scattering and of the annihilation process. To get one formula from the other one has to exchange the two Mandelstam variables, $s$ and $t$ and take account of some additional factors. One of these is connected with the Fermi statistics of electrons and positrons. As there is one fermion in the final state of Compton scattering and there are no fermions in the case of annihilation, those two matrix elements have opposite signs (see section [*]). There is furthermore a factor of $1/2$ which arises in the annihilation channel because of the presence of two identical particles, the two photons, in the final state (see section [*]). Such very strong similarity between the matrix elements of the processes under consideration is based on the fact that the same set of Feynman diagrams contribute. This is a consequence of a fundamental property called crossing symmetry.3In the case of $2\to 2$ processes we can say that crossing symmetry relates the process $A+B\to C+D$ to the process $\bar D +B\to C+\bar
A$, where $\bar A$ and $\bar D$ denote the antiparticles of $A$ and $D$. These processes are represented by the same set of Feynman diagrams. However you have to remember that in CompHEP Feynman diagrams are constructed without permutation over final identical particles (this operation is performed in later steps of the calculation).

The differential cross section of annihilation is given by

\begin{displaymath}
\frac{d\sigma}{d t}\;=\; \frac{1}{16\pi s(s-4m_e^2)} \vert M\vert^2.
\end{displaymath}

(see the normalization in Section [*]).

It remains to integrate the differential cross section at fixed energy over the kinematical range of $t$,

\begin{displaymath}
-\frac{s}{2} - \frac{1}{2}\sqrt{s(s-4m^2_e)} \le t-m_e^2
\le -\frac{s}{2} + \frac{1}{2}\sqrt{s(s-4m^2_e)}
\end{displaymath}

(see section [*]) and hence get the following formula for the total cross section:

\begin{displaymath}\sigma_{tot}\;=\;
\frac{2\pi\alpha^2}{s^2 (s-4m^2_e)}
\left...
...sqrt{s-4m^2_e}}\right)
-(s+4m^2_e)\sqrt{s(s-4m^2_e)} \right]. \end{displaymath}

At high energies, $\sqrt{s}\gg m_e$, this simplifies to the asymptotic formula

\begin{displaymath}
\sigma_{tot}\;\sim\;\frac{2\pi\alpha^2}{s}\left(\log\frac{s}{m^2_e}-1\right)
\end{displaymath}

and you see that the total cross section decreases with increasing energy almost as fast as $1/s$ (cf. Fig. 6).

Finally consider the third reaction which is related by crossing symmetry to Compton scattering and to electron-positron annihilation, i.e the creation of electron-positron pairs in $\gamma\gamma$ collisions, $\gamma\gamma\to e^+e^-$. This is the process reversed in time with respect to the annihilation reaction. There are two Feynman diagrams representing the scattering amplitude. According to crossing symmetry the squared matrix element has the same dependence on the kinematical variables as the squared matrix element of the annihilation process but it is twice as large because of the absence of the combinatorial factor of $1/2$ that was needed in the case of annihilation to account for the two identical particles in the final state. We leave it to the reader to verify this statement by doing the next problem.

$\star\star$
Problem. Using the CompHEP REDUCE output obtain analytical answer for the squared matrix element of $\gamma\gamma\to e^+e^-$ process and compare it with that of the annihilation channel.



Footnotes

... symmetry.3
for a detailed study of crossing symmetry consult, for instance, [2].

Elastic
 scattering of electrons and positrons. QED processes Energy spectrum
 of backward scattered photons in the Compton process. Contents