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Elastic scattering of electrons and positrons.

Another class of QED collision processes is the elastic scattering of electrons and positrons. There are three reactions related to each other by crossing symmetry. One of these processes is elastic electron-positron scattering (Bhabha scattering):

$$e^+e^- \to e^+e^-.$$

The corresponding Feynman diagrams are shown in Fig. 7. The other two processes are elastic electron-electron scattering (Møller scattering) and positron-positron scattering,

$$e^-e^- \to e^-e^-, \qquad e^+e^+ \to e^+e^+$$

which we shall discuss presently.

Bhabha scattering is an important process for $e^+e^-$ colliders. On the one hand it is indispensable as a tool to measure the collider luminosity and on the other hand it gives a large background to other reactions which must be known in order to do a sensible data analysis. Another interesting aspect is that in the nonrelativistic limit this reaction coincides with the well known classical scattering of charged particles described by the Rutherford formula.

To calculate the cross section of Bhabha scattering start the CompHEP session and enter this process:

Enter Process:   e1,E1 -> e1,E1

then define the collision energy, for instance $\sqrt {s}=90$ GeV which is a typical operating energy of the colliders LEP and SLC.

Two Feynman diagrams will be generated (see Fig. 7). The first diagram corresponds to the annihilation of the electron-positron pair into a virtual photon which decays into an outgoing electron-positron pair. This represents the direct or $s$-channel amplitude. The second diagram corresponds to the exchange of a virtual photon between the colliding electron and positron. It corresponds to the exchange or $t$-channel amplitude.

To calculate the squared matrix element symbolically use the Squaring and Symbolic calculations facilities. From our previous experience we are used to invoke at this stage the Numerical calculator, expecting to get the total cross section displayed. But in this case we are in for a surprise: CompHEP signals that the total cross section can not been calculated because of a singularity. This is displayed on the screen by the message

Division by zero.

This singularity results from the photon exchange. It is well known from nonrelativistic Rutherford scattering, where one has a factor ${\displaystyle \sin^4(\vartheta/2)}$ in the denominator which vanishes at $\vartheta=0$ and hence gives rise to a divergence in the cross section. Therefore one calculates an integrated cross section for a reduced angular range, cutting out a small region near the pole. Experimentally this procedure is plausible since it is not feasible to detect scattered particles very close to the beam. This consideration applies of course also to electron-positron colliders, where particles in the forward and backward cones of a few degrees opening around the beam are inaccessible.

Let us simulate a collider by applying an angular cut of 5 degrees in the forward and backward directions: $5^\circ <\vartheta <175^\circ$ ($\vartheta$ is the scattering angle of the electron). To do this we choose the option Set angular range and set the limits of $\cos\vartheta$:

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

The package displays the result $\sigma=16.6\times 10^3\,\mbox{pb} = 0.0166\,\mu\mbox{b}$. Applied to the LEP collider operating at a luminosity of $10^{31}\,\mbox{cm}^{-2}\mbox{s}^{-1}$ this rate corresponds to one Bhabha event every 6 seconds. To see the angular distribution we use option Show plot of menu Angular dependence. The corresponding curves at different energies are shown in Fig. 8. The principal feature of the differential cross section is its asymmetry: electrons and positrons are scattered predominantly in the direction of their initial beams. We can also conclude that we have been overcautious when we applied an angular cut to the backward direction since it is evident that there is a pole only at $\vartheta=0$. You can go back to option Set angular range and apply this time a cut only in the forward direction, for example, $5^\circ<\vartheta<180^\circ$ and hence convince yourself that the integrated cross section stays finite.

To get a quantitative picture of the effect of angular cuts on the total cross section of Bhabha scattering let us calculate the energy dependence with the following angular cuts: (i) $5^\circ <\vartheta <175^\circ$ and (ii) $1^\circ <\vartheta <179^\circ$. The resulting curves are shown in Fig. 9.

Our next numerical experiment is designed to demonstrate which one of the Feynman diagrams is responsible for the divergence of the total cross section. To do this return to the step before Squaring and select option View diagrams. In this option CompHEP allows you to delete selected diagrams by pressing the Space bar when the frame around the diagram appears highlighted (the highlighted frame can be moved by using the Arrow keys). Thus we can delete for instance the exchange diagram. Inside the frame the deleted diagram is marked by DEL. Then we return to the options Squaring and Symbolic calculations and start the Numerical calculator. This time the package will integrate over the entire angular range and produce a finite result for the total cross section: $\sigma_{tot}=10.7$ pb. So we conclude that the singularity is produced by the exchange amplitude. We can of course confirm this by repeating the calculation with the direct channel diagram deleted and the exchange diagram restored. We will come back to discuss this point once we have derived the analytical formula for the differential cross section of Bhabha scattering.

Let us therefore now obtain the analytical formula of the squared matrix element. Repeating all steps described in Section 0.1.1 you can get the squared matrix element in the following form:

$$\vert M\vert^2 \;=\; 64\pi^2\alpha^2\left[ \frac{m_e^4}{t^2} \right. \left(\frac{s^2+u^2}{2m_e^4}+4\frac{t}{m_e^2}-4\right) + \frac{m_e^4}{s^2} \left(\frac{t^2+u^2}{2m_e^4}+4\frac{s}{m_e^2}-4\right)$$

$$+ \left.\frac{m_e^4}{st} \left(\frac{u}{m_e^2}-2\right)\left(\frac{u}{m_e^2}-6\right)\right]$$ (7)

where $s$ is the square of the CMS energy and $t$ is the squared 4-momentum transfer which in the $t$-channel amplitude is carried by the exchanged photon. The range of variation of $t$ is given by the inequality

$$-(s-4m_e^2) \le t \le 0$$

(see Section ) and therefore it is evident that there is a $1/t^2$ singularity at $t=0$ (forward scattering) from the photon propagator of the exchange amplitude. Note also that another way of seeing the physical meaning of $t$ is that it is the invariant mass-squared of the virtual photon which is zero for a real photon, i.e. for a photon exchanged at long range.

Let us rewrite now the cross section in terms of the CMS scattering angle $\vartheta$. For the Mandelstam variables we have the following relations:

$$t\;=\; -(s-4m_e^2) \sin^2\frac{\vartheta}{2}, \qquad\mbox{and}\qquad u\;=\; -(s-4m_e^2) \cos^2\frac{\vartheta}{2}.$$ (8)

In the nonrelativistic limit $s\to 4m_e^2$, $\vert t\vert\ll m_e^2$ and $\vert u\vert\ll m_e^2$. Therefore we can neglect the second and the third terms in the squared matrix element (7) and get the differential cross section in the well known form of the Rutherford formula:

$$\frac{d\sigma}{d\cos\vartheta} \;\approx\; \frac{\pi \alpha^2 m_e^2}{8 p^{*4}} \mbox{cosec}^4\frac{\vartheta}{2},$$

where $p^*=\sqrt{s/4 -m_e^2}$ is the CMS momentum.

It is interesting to note that only the exchange amplitude contributes to this formula: you can check that the first term in Eq. (7) results from this squared diagram. To do this delete the direct channel diagram before the Squaring step and then perform the Symbolic calculations. You conclude that in the nonrelativistic limit the direct channel amplitude contributes to the cross section neither as a squared amplitude itself nor through the interference with the exchange amplitude. Remarkable is also that we have obtained in the nonrelativistic limit the formula, which is derived in nonrelativistic quantum mechanics assuming spinless particles. The reason for this is that the spin makes a significant contribution only in the relativistic domain.

The exchange amplitude remains the dominant part of the interaction at small angles at all energies. The contribution of the direct channel amplitude becomes significant at relativistic energies only at large scattering angles. Let us calculate the angular dependence for each squared amplitude separately. To do this you have to return to Squaring, delete all squared diagrams by pressing the Del key and then restore the diagram that you want for further calculation by pressing the Space bar. In Fig. 10 we have put together the curves representing the angular dependances at $\sqrt {s}=90$ GeV for each of the three squared amplitudes separately and for the sum of their contributions. You can see that for large scattering angles the contributions of the individual squared amplitudes, including the interference term, are of the same order. For backward scattering the contributions of the squared direct channel and exchange amplitudes are equal. This follows directly from Eq. (8) if we put $\vartheta = \pi$ and consider the ultrarelativistic case, $m_e=0$. The interference term is $u^2/st$ in the ultrarelativistic approximation, and hence it vanishes in the backward direction.

It remains to calculate the total cross section which is defined by the integral

$$\sigma \;=\; \frac{1}{16\pi s (s-4m_e^2)} \int dt \; \vert M\vert^2 \Theta_{cut},$$

where the function $\Theta_{cut}$ imposes the necessary cuts to keep the cross section finite. In the ultrarelativistic approximation, integrating over an angular range of $\vartheta_0\le\vartheta\le\pi$, we get

$$\sigma\;=\;\frac{4\pi\alpha^2}{s}\left(\frac{1}{\sin^2(\vartheta_0/2)} +\log\sin^2\frac{\vartheta_0}{2}+\frac{4}{3}\right)$$

which at $\sqrt {s}=90$ GeV is good to within 1%.

Let us now consider the elastic scattering of electrons, $e^-e^-\to e^-e^-$ (Møller scattering). As we have already mentioned this reaction is related to Bhabha scattering by crossing symmetry. To get its squared matrix element you can either retrace the procedures described above to get the analytical expression of $\vert M\vert^2$ or, simpler, you apply the crossing principle and make the substitution $s\Longleftrightarrow u$ in Eq. (7). Again, as for the two-photon annihilation process, the cross section of Møller scattering includes the statistical factor of 1/2 to account for the two identical particles in the final state.

$\star\star$

Problem. Using the formulæ for the squared matrix elements, show that at high energies ($s\gg m_e^2$) the cross sections of Bhabha scattering and of elastic electron scattering are related as

$$d\sigma_{e^+e^-}\;\approx\;\cos^4\frac{\vartheta}{2} \; d\sigma_{e^-e^-}.$$