Electron-Electron Scattering

The electron-electron systems are specified by their initial states, in which there are either two electrons, two positrons, or one electron and one positron. The last case is of special interest, since it permits bound states; these states, however, are not stable. The bound structure which consists of one electron and one positiron is usually call positronium.

As in the photon-electron system, the scatering of two electrons into a final state of exactly two electrons and no photon is only an approximate description of this process. This is so because the scattering of a charge particle through a finite angle will always be accompanied by an undertermined number of very low energy photons. We obtain correct results only in the approximation in which the possible emission of photons in the final state can be consistently neglected. This occures only in lowest order and when the scattering angle is not too small. The radiative corrections to this approximation requires the knowledge of the energy resolution of the experiment which determines the maximum energy of possible photons in the final state. The effect of bremsstrahlung in electron-electron collisions is therefore strictly not separable from that of scattering.

The two graphs for electron-electron scattering are shown in figure 7.11. The additional graph arises because of the identity of the electrons.

Figure 7.11: Feynman diagrams for electron-electron scattering: a) direct and b) exchange.

The $S$-matrix element (with spin labels suppressed) is

$$S_{fi} = \frac{-e^2m^2}{V^2\sqrt{E_1E_2E_1^\prime E_2^\prime}} \left[ i\fr... ...1) \overline{u}(p_2^\prime) (-i\gamma^\mu) u(p_2)} {(p_1-p_1^\prime)^2} \right.$$

$$\left. -i \frac{i\overline{u}(p_1^\prime) (-i\gamma_\mu) u(p_2) \... ...p_2^\prime)^2} \right] (2\pi)^4 \delta^4(p_1^\prime + p_2^\prime - p_1 - p_2) .$$ (7.297)

The relative minus sign between the direct and exchange terms is due to the Fermi statistics, which requires the amplitude to be antisymmetric under interchange of the two identical electrons. By a similar argument the scattering amplitude to or from a state containing two identical Bose particles must be symmetric under their interchange.

No additional normalization factors, such as $1/\sqrt{2}$ or 2, were introduced when the exchange term was added. The rules for constructing differential cross-sections from $S_{fi}$, are not altered by the presence of identical particles in the initial or final states. We must only take care that the factor $1/2$ is included in integrating for a total cross-section when two identical particles appear in the final state. No special factors appear for identical particles in the initial state, since the incident flux is unchanged.

The differential cross-section for scattering of unpolarized electrons can be constructed:

$$d\sigma = \int \frac{\vert S_{fi}\vert^2}{VT} \frac{V}{J_{inc}} \frac{Vd^3p_1^\prime}{(2\pi)^3} \frac{Vd^3p_2^\prime}{(2\pi)^3}$$

$$= \int \frac{e^4}{(2\pi)^2} \frac{m^4}{\vert\vec{v}_1-\vec{v}_2\ver... ...^\prime} \vert\mathcal{M}_{fi}\vert^2 \delta^4(p_1^\prime+p_2^\prime-p_1-p_2) .$$ (7.298)

We now evaluate the invariant matrix element

$$\vert\mathcal{M}_{fi}\vert^2 = \frac{\overline{u}(p_1^\prime)\gamma_\mu u(p_1)\overline{u}(p_2^\... ...\nu u(p_2^\prime)\overline{u}(p_1)\gamma_\nu u(p_1^\prime)}{(p_1-p_1^\prime)^4}$$

$$- \frac{\overline{u}(p_1^\prime)\gamma_\mu u(p_1)\overline{u}(p_2^\... ...overline{u}(p_2)\gamma_\nu u(p_1^\prime)}{(p_1-p_1^\prime)^2(p_1-p_2^\prime)^2}$$

$$- \frac{\overline{u}(p_1^\prime)\gamma_\mu u(p_2)\overline{u}(p_2^\... ...overline{u}(p_1)\gamma_\nu u(p_1^\prime)}{(p_1-p_2^\prime)^2(p_1-p_1^\prime)^2}$$

$$+ \frac{\overline{u}(p_1^\prime)\gamma_\mu u(p_2)\overline{u}(p_2^\... ...u u(p_2^\prime)\overline{u}(p_2)\gamma_\nu u(p_1^\prime)}{(p_1-p_2^\prime)^4} .$$ (7.299)

We only need to calculate the first two terms since the last two terms can be obtained from the first two terms by the substitution $p_1^\prime \leftrightarrow p_2^\prime$. Averaging over initial state spins and summing over final state spins, we need to evaluate

$$\sum_{s_1,s_2,s_1^\prime,s_2^\prime} \overline{u}(p_1^\prime)_\al... ...ime)_\phi \overline{u}(p_1)_\rho (\gamma_\nu)_{\rho\theta} u(p_1^\prime)_\theta$$

$$= \left(\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\right)_{\theta\alpha}... ...frac{\not{\;\!\!\!p}_2+m}{2m}\right)_{\delta\epsilon} (\gamma_\nu)_{\rho\theta}$$

$$= \textrm{Tr} \left[ \frac{\not{\;\!\!\!p}_1^\prime+m}{2m} \gamma_\... ...p}_2+m}{2m} \gamma^\nu \frac{\not{\;\!\!\!p}_2^\prime+m}{2m} \gamma^\mu \right]$$ (7.300)

and

$$\sum_{s_1,s_2,s_1^\prime,s_2^\prime} \overline{u}(p_1^\prime)_\al... ...ime)_\phi \overline{u}(p_2)_\rho (\gamma_\nu)_{\rho\theta} u(p_1^\prime)_\theta$$

$$= \left(\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\right)_{\theta\alpha}... ...ft(\frac{\not{\;\!\!\!p}_2+m}{2m}\right)_{\delta\rho} (\gamma_\nu)_{\rho\theta}$$

$$= \textrm{Tr} \left[ \frac{\not{\;\!\!\!p}_1^\prime+m}{2m} \gamma_\... ..._2^\prime+m}{2m} \gamma^\mu \frac{\not{\;\!\!\!p}_2+m}{2m} \gamma^\nu \right] .$$ (7.301)

Two of the traces have been evaluated previously and the third trace can be simplified for relativistic energies $E\gg m$ in which we can neglect terms proportional to $m^2$.

$$\textrm{Tr}\left[\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\gamma_\mu \frac{\not{\;\!\!\!p}_1+m}{2m}\gamma_\nu\right] = \frac{1}{m^2} [(p_1^\prime)_\mu (p_1)_\nu + (p_1)_\mu (p_1^\prime)_\nu -g_{\mu\nu}(p_1\cdot p_1^\prime -m^2)] ,$$ (7.302)

$$\textrm{Tr}\left[\frac{\not{\;\!\!\!p}_2^\prime+m}{2m}\gamma^\mu \frac{\not{\;\!\!\!p}_2+m}{2m}\gamma^\nu\right] = \frac{1}{m^2} [(p_2^\prime)^\mu (p_2)^\nu + (p_2)^\mu (p_2^\prime)^\nu -g^{\mu\nu}(p_2\cdot p_2^\prime -m^2)] ,$$ (7.303)

$$\textrm{Tr}\left[\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\gamma_\mu ... ...!\!p}_2^\prime+m}{2m}\gamma^\mu \frac{\not{\;\!\!\!p}_2+m}{2m}\gamma^\nu\right]$$

$$= \frac{2}{m^4} [p_1^\prime\cdot p_2^\prime p_1\cdot p_2 + p_1^\prime\cdot p_2 p_1\cdot p_2^\prime ] ,$$ (7.304)

$$\textrm{Tr}\left[\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\gamma_\mu ... ...!\!p}_2^\prime+m}{2m}\gamma^\mu \frac{\not{\;\!\!\!p}_2+m}{2m}\gamma^\nu\right] = \frac{1}{16m^4} \textrm{Tr}[\not{\;\!\!\!p}_1^\prime \gamma_\mu \... ...\;\!\!\!p}_2^\prime \gamma^\mu \not{\;\!\!\!p}_2 \gamma^\nu] + \mathcal{O}(m^2)$$

$$= -\frac{1}{8m^4} \textrm{Tr}[\not{\;\!\!\!p}_1^\prime \gamma_\mu \not{\;\!\!\!p}_1 \not{\;\!\!\!p}_2 \gamma^\mu \not{\;\!\!\!p}_2^\prime]$$

$$= -\frac{1}{2m^4} p_1\cdot p_2 \textrm{Tr}[\not{\;\!\!\!p}_1^\prime \not{\;\!\!\!p}_2^\prime]$$

$$= -\frac{2}{m^4} p_1\cdot p_2 p_1^\prime\cdot p_2^\prime .$$ (7.305)

In the centre-of-mass frame (neglecting terms in $m^2$)

$$E_1 = E_2 = E_1^\prime = E_2^\prime \equiv E ,$$ (7.306)

$$\vert\vec{v}_1\vert = \vert\vec{v}_2\vert \equiv \beta ,$$ (7.307)

$$\vert\vec{v}_1-\vec{v}_2\vert = 2\beta ,$$ (7.308)

where $E$ is the centre-of-mass energy of each electron and $\beta$ its velocity. The dot products become

$$p_1\cdot p_2 = p_1^\prime\cdot p_2^\prime \approx 2E^2 ,$$ (7.309)

$$p_1\cdot p_2^\prime = p_1^\prime\cdot p_2 \approx 2E^2\cos^2(\theta/2) ,$$ (7.310)

$$p_1\cdot p_1^\prime = p_2^\prime\cdot p_2 \approx 2E^2\sin^2(\theta/2) ,$$ (7.311)

$$(p_1^\prime-p_1)^2 \approx -2p_1\cdot p_1^\prime = -4E^2\sin^2(\theta/2) ,$$ (7.312)

$$(p_2^\prime-p_1)^2 \approx -2p_1\cdot p_2^\prime = -4E^2\cos^2(\theta/2).$$ (7.313)

The traces are

$$\textrm{Tr}\left[\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\gamma... ...onumber \\ = 8\left(\frac{E}{m}\right)^4 [1+\cos^4(\theta/2)]$$

and

$$\textrm{Tr}\left[\frac{\not{\;\!\!\!p}_1^\prime+m}{2m}\gamma_\mu ... ...!\!p}_2^\prime+m}{2m}\gamma^\mu \frac{\not{\;\!\!\!p}_2+m}{2m}\gamma^\nu\right] = -8 \left(\frac{E}{m}\right)^4 .$$ (7.314)

The differential cross-section and invariant matrix element thus become

$$d\overline{\sigma} = \frac{e^4m^4}{8(2\pi)^2} \int \frac{d^3... ...thcal{M}}_{fi}\vert^2 \delta^4(p_1^\prime+p_2^\prime-p_1-p_2).$$ (7.315)

and

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 = \frac{1}{2m^4} \le... .../2)} + \frac{1 + \sin^4(\theta/2)}{\cos^4(\theta/2)} \right] ,$$ (7.316)

where the first and third terms are the square of the matrix elements for the two graphs, and the second term is the interference contribution.

Integrating over $d^3p_2^\prime$ using the $\delta$-function and letting $d^3p_1^\prime = E^2dEd\Omega$, we have

$$\frac{d\overline{\sigma}}{d\Omega} = \frac{e^4m^4}{8(2\pi)^2} \int \frac{E^2dE}{E^4\beta} \vert\mathcal{M}_{fi}\vert^2 \delta(E_1^\prime+E_2^\prime-E_1-E_2)$$ (7.317)

$$= \frac{\alpha^2}{8E^2} \left[ \frac{1 + \cos^4(\theta/2)}{\sin^4(\... ...a/2)\cos^2(\theta/2)} + \frac{1 + \sin^4(\theta/2)}{\cos^4(\theta/2)} \right] .$$ (7.318)

We have obtained the high-energy limit of the Møller formula in the centre-or-mass.