Prof. Dr. Stefan Floerchinger

Researcher and professor for quantum field theory.

Elementary scattering processes

We are now ready to use the formalism of quantum field theory, specifically quantum electrodynamics, to determine actually scattering amplitudes and cross section. The incoming and outgoing states can consist of photons, electrons and positrons but also muons or anti-muons and more generally any charged particles. When the charged particles are scalar bosons, one would use a variant of the theory called scalar electrodynamics, but we are here concerned with charged spin-\(1/2\) particles which are described by standard spinor electrodynamics.

In the following we will bring together several of the elements we have discussed before, such as

  • the Lagrangian of spinor quantum electrodynamics,

  • the idea of perturbation theory as an expansion in the coupling constant \(e\),

  • the graphical representation in terms of Feynman diagrams,

  • solutions to the free Dirac equation for incoming or outgoing electrons and positrons (or muons and anti-muons),

  • the propagators for Dirac femions and for photons,

It might be a good idea to go back and revise these topics if you feel uncertain about them. We will see on the way that we need some additional technical knowledge, specifically

  • how to do spin sums,

  • how to calculate traces of gamma matrices

  • how Mandelstam variables are defined and how one can work with them.

These points will also be discussed in the exercises.

We will then start to look at the elastic scattering of a photon and an electron, a process known as Compton scattering. We will write down the Feynman diagrams and the corresponding algebraic expressions. For another process, namely the scattering of an electron-positron pair to a muon-anti-muon pair we will do this, as well, but then also go on and evaluate the expressions further until we arrive at a nice and compact result for the scattering cross-section.

Compton Scattering

As a first example let us consider Compton scattering \(e^{-} \gamma \to e^{-} \gamma\)

These are two diagrams at order \(e^2\), as shown above. The first diagram corresponds to the expression \[\bar{u}_{s_2}(p_2) (-e\gamma^\nu)\left(-i\frac{-i(\slashed{p}_1+\slashed{q}_1)+m}{(p_1 + q_1)^2+m^2}\right)(-e\gamma^\mu) \; u_{s_1}(p_1) \; \epsilon_{(\lambda_1)\mu} (q_1)\;\epsilon^{*}_{(\lambda_2)\nu}(q_2) .\] Similarly, the second diagram gives \[\bar{u}_{s_2}(p_2) (-e\gamma^\mu)\left(-i\frac{-i(\slashed{p}_1-\slashed{q}_2)+m}{(p_1 - q_2)^2+m^2}\right)(-e\gamma^\nu) \; u_{s_1}(p_1) \; \epsilon_{(\lambda_1)_\mu} (q_1)\;\epsilon^{*}_{(\lambda_2)\nu}(q_2) .\] Combining terms and simplifying a bit leads to \[i\mathcal{T} = -i e^2 \epsilon_{(\lambda_1)\mu}(q_1)\;\epsilon^{*}_{(\lambda_2)\nu} (q_2) \; \bar{u}_{s_2}(p_2) \left[\gamma^\nu \; \frac{-i(\slashed{p}_1+ \slashed{q}_1)+m}{(p_1+q_1)^2+m^2}\gamma^\mu+ \gamma^\mu \frac{-i(\slashed{p}_1-\slashed{q}_2)+m}{(p_1 -q_2)^2+m^2}\gamma^\nu\right]u_{s_1}(p_1).\]

Electron-positron to muon-anti-muon scattering

As another example for an interesting process in QED we consider \(e^{-} e^{+} \to \mu^{-} \mu^{+}.\) From the point of view of QED, the muon behaves like the electron but has a somewhat larger mass. Diagrams contributing to this process are (we keep the polarizations implicit)

The corresponding expression is \[i\mathcal{T} = \bar{v}(p_2)(-e\gamma^\mu) \; u(p_1) \; \left(-i \frac{\eta_{\mu\nu} -\frac{k_\mu k_\nu}{k^2}}{(k^2)}\right) \bar{u}(p_3) \; (-e\gamma^\nu) \; v(p_4),\] with \(k= p_1+p_2 = p_3+p_4.\)

On-shell conditions

The external momenta are on-shell and the spinors \(u(p_1)\) etc. satisfy the Dirac equation, \[(i\slashed{p}_1 + m_e) u(p_1) = 0, \quad\quad\quad (-i\slashed{p}_4 + m_\mu) v(p_4) = 0,\] \[\bar{u}(p_3)(i\slashed{p}_3 + m_\mu)= 0,\quad\quad\quad \bar{v}(p_2)(-i \slashed{p}_2 + m_e) = 0.\] This allows to write \[\begin{split} & \bar v(p_2) \, i \gamma^\mu k_\mu \, u(p_1) = \bar{v}(p_2) \,i (\slashed{p}_1 + \slashed{p}_2 ) \, u(p_1) = \bar{v}(p_2) \, (-m_e + m_e) \, u(p_1) = 0, \\ & \bar{u}(p_3) \, i \gamma^\nu k_\nu \, v(p_4) = \bar{u}(p_3) \, i (\slashed{p}_3 + \slashed{p}_4) \, v(p_4) = \bar{u}(p_3) \, (-m_\mu + m_\mu) \, v(p_4) = 0. \end{split}\] These arguments show that the term \(\sim\) \(k_\mu k_\nu\) can be dropped. This is essentially a result of gauge invariance.

Complex conjugate and squared amplitudes

We are left with \[\mathcal{T} = -\frac{e^2}{k^2} \bar{v}(p_2) \gamma^\mu u(p_1) \, \bar{u}(p_3) \gamma_\mu v(p_4).\] To calculate \(|\mathcal{T}|^2\) we also need \(\mathcal{T}^{*}\). For the fermion chains of spinors and gamma matrices it is easiest to use a hermitian conjugation, \[\mathcal{T}^{*} = -\frac{e^2}{k^2} u^\dagger(p_1) \gamma^{\mu\dagger} \bar{v}^\dagger(p_2) \; v^\dagger(p_4) \gamma^\dagger_\mu \bar{u}^\dagger(p_3).\] Recall that \(\bar{u}(p) = u(p)^\dagger \beta\) with \(\beta % = \begin{pmatrix} % & \mathbb{1} \\ %\mathbb{1} & %\end{pmatrix} = i\gamma^0\). With the explicit representation \[\gamma^\mu = \begin{pmatrix} & -i\sigma^\mu \\ -i\bar{\sigma}^\mu & \end{pmatrix},\] it is also easy to prove \(\beta\gamma^{\mu\dagger}\beta= -\gamma^\mu\). By inserting \(\beta^2 =\mathbb{1}\) at various places we find thus \[\mathcal{T}^{*} = -\frac{e^2}{k^2} \bar{u}(p_1)\gamma^\mu v(p_2) \; \bar{v}(p_4) \gamma_\mu u(p_3)\] Putting together and using \(s=-k^2 = -(p_1+ p_2)^2\) we obtain \[|\mathcal{T}|^2 = \frac{e^4}{s^2} \bar{u}(p_1) \gamma^\mu v(p_2) \; \bar{v}(p_2) \gamma^\nu u(p_1) \; \bar{u}(p_3) \gamma_\nu v(p_4) \; \bar{v}(p_4) \gamma_\mu u(p_3).\]

Spin sums and averages

To proceed further, we need to specify also the spins of the incoming and outgoing particles. The simplest case is the one of unpolarized particles so that we need to average the spins of the incoming electrons, and to sum over possible spins in the final state. Summing over the spins of the \(\mu^{+}\) can be done as follows (exercise) \[\sum_{s=1}^2 v_s(p_4) \bar{v}_s(p_4) = -i\slashed{p}_4 -m_\mu,\] and similarly for \(\mu^{-}\) \[\sum_{s=1}^2 u_s(p_3) \bar{u}_s(p_3) = -i\slashed{p}_3 +m_\mu.\] We can therefore write \[\sum_\text{spins} \bar{u}(p_3) \gamma_\nu v(p_4) \, \bar{v}(p_4) \gamma_\mu u(p_3) = \text{tr}\left\{(-i\slashed{p}_3+m_\mu)\gamma_\nu (-i\slashed{p}_4-m_\mu)\gamma_\mu\right\}.\] Spins of the electron and positron must be averaged instead, \[\begin{split} \frac{1}{2}\sum_{s=1}^2 u(p_1) \bar{u}(p_1) & = \frac{1}{2}(-i\slashed{p}_1+m_e),\\ \frac{1}{2}\sum_{s=1}^2 v(p_2) \bar{v}(p_2) & = \frac{1}{2}(-i\slashed{p}_2-m_e). \end{split}\] This leads to \[\frac{1}{4}\sum_{\text{spins}}|\mathcal{T}|^2 = \frac{e^4}{4s^2} \text{tr}\left\{(-i\slashed{p}_1+m_e)\gamma^\mu(-i\slashed{p}_2-m_e) \gamma^\nu\right\} \times \text{tr}\left\{(-i\slashed{p}_3+m_\mu)\gamma_\nu(-i \slashed{p}_4 -m_\mu)\gamma_\mu\right\}.\] In order to proceed further, we need to know how to evaluate traces of up to four gamma matrices.

Traces of gamma matrices

We need to understand how to evaluate traces of the form \[\text{tr}\{\gamma^{\mu_1}\cdots \gamma^{\mu_n}\}.\] To work them out we can use \(\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}\), \(\gamma_5^2 =\mathbb{1}\) and \(\{\gamma^\mu,\gamma_5\} = 0\). Also, \(\text{tr}\{\mathbb{1}\}=4\). First we prove that traces of an odd number of gamma matrices must vanish, \[\begin{split} \text{tr}\{\gamma^{\mu_1}\cdots \gamma^{\mu_n}\}&= \text{tr}\{\gamma_5^2\;\gamma^{\mu_1} \gamma_5^2 \cdots \gamma_5^2 \gamma^{\mu_n}\}\\ &=\text{tr}\{(\gamma_5 \gamma^{\mu_1}\gamma_5)\cdots(\gamma_5 \gamma^{\mu_1} \gamma_5)\}\\ &=\text{tr}\{(-\gamma_5^2\gamma^\mu_1) \cdots (-\gamma_5^2\gamma^{\mu_n})\}\\ &=(-1)^n \text{tr}\{\gamma^{\mu_1}\cdots\gamma^{\mu_n}\}. \end{split}\] This implies what we claimed.

Now for even numbers \[\text{tr}\{\gamma^\mu\gamma^\nu\} = \text{tr}\{\gamma^\nu\gamma^\mu\} = \tfrac{1}{2} \text{tr}\{\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu\} = \eta^{\mu\nu} \text{tr}\{\mathbb{1}\} = 4\eta^{\mu\nu}.\] From this it also follows that \[\text{tr}\{\slashed{p}\slashed{q}\} = 4p \cdot q.\] Now consider \(\text{tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\}.\) This idea is to commute \(\gamma^\mu\) to the right using \(\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}\). Thus \[\begin{split} \text{tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\} &= -\text{tr}\{\gamma^\nu\gamma^\mu\gamma^\rho\gamma^\sigma\} + 2\eta^{\mu\nu}\;\text{tr}\{\gamma^\rho \gamma^\sigma\}\\ &= \text{tr}\{\gamma^\nu\gamma^\rho\gamma^\mu\gamma^\sigma\}-2\eta^{\rho\mu} \text{tr}\{\gamma^\nu\gamma^\sigma\}+ 2\eta^{\mu\nu}\;\text{tr}\{\gamma^\rho\gamma^\sigma\}\\ &= -\text{tr}\{\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^\mu\} + 2\eta^{\sigma\mu}\;\text{tr}\{\gamma^\nu\gamma^\rho\}- 2\eta^{\rho\mu}\;\text{tr}\{\gamma^\nu \gamma^\sigma\}+ 2\eta^{\mu\nu}\; \text{tr}\{\gamma^\rho\gamma^\sigma\}. \end{split}\] But by the cyclic property of the trace \[\text{tr}\{\gamma^\nu \gamma^\rho \gamma^\sigma \gamma^\mu\} = \text{tr}\{\gamma^\mu \gamma^\nu\gamma^\rho \gamma^\sigma\}\] which is also on the left hand side. Bringing it to the left and dividing by 2 gives \[\begin{split} \text{tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\} &= \eta^{\sigma\mu}\;\text{tr}\{\gamma^\nu \gamma^\rho\}- \eta^{\rho\mu}\;\text{tr}\{\gamma^\nu\gamma^\sigma\}+ \eta^{\mu\nu}\text{tr}\{\gamma^\rho \gamma^\sigma\}\\ &= 4(\eta^{\sigma\mu}\eta^{\nu\rho}-\eta^{\rho\mu}\eta^{\nu\sigma}+ \eta^{\mu\nu}\eta^{\rho\sigma}). \end{split}\] This is the result we were looking for. Clearly by using this trick we can in principle evaluate traces of an arbitrary number of gamma matrices.

Result so far

Coming back to \(e^{-}e^{+} \to \mu^{-}\mu^{+}\) we find \[\begin{split} \frac{1}{4} \sum_\text{spins} |\mathcal{T}|^2 = & \frac{4 e^4}{s^2}\left[-p_1^\mu p_2^\nu -p_1^\nu p_2^\mu + (p_1 \cdot p_2 - m_e^2) \eta^{\mu\nu}\right]\\ & \times \left[-(p_3)_\nu(p_4)_\mu -(p_3)_\mu(p_4)_\nu + (p_3 \cdot p_4 - m_\mu^2) \eta^{\mu\nu}\right]\\ = & \frac{8 e^4}{s^2} \left[(p_1 \cdot p_4)(p_2 \cdot p_3)+(p_1 \cdot p_3)(p_2 \cdot p_4) -m_\mu^2(p_1 \cdot p_2) -m_e^2(p_3 \cdot p_4) +2m_e^2 m_\mu^2\right]. \end{split}\] This looks already quite decent but it can be simplified even further in terms of Mandelstam variables.

Mandelstam Variables

The Mandelstam variables for a \(2\to 2\) process

are given by \[\begin{split} s &=-(p_1+p_2)^2 = -(p_3+p_4)^2, \\ t & = -(p_1-p_3)^2 = -(p_2-p_4)^2, \\ u &= -(p_1-p_4)^2 = -(p_2-p_3)^2. \end{split}\] Together with the squares \(p_1^2\), \(p_2^2\), \(p_3^2\), \(p_4^2\), the Mandelstam variables can be used to express all Lorentz invariant bilinears in the momenta. Incoming and outgoing momenta are on-shell such that \(p_1^2 + m_1^2 =0\) etc. The sum of Mandelstam variables is \[s+t+u = -(p_1^2+ p_2^2+ p_3^2+ p_4^2) = m_1^2+ m_2^2+ m_3^2+m_4^2.\] Using these variables for example through \[p_1 \cdot p_4 = -\frac{1}{2}\left[(p_1-p_4)^2-p_1^2-p_4^2\right] = \frac{1}{2}\left[u-m_e^2+ m_\mu^2\right],\] one finds for \(e^{-}e^{+} \to \mu^{-}\mu^{+}\) \[\frac{1}{4}\sum_\text{spins}|\mathcal{T}|^2 = \frac{2e^4}{s^2}\left[t^2+ u^2+ 4s(m_e^2+ m_\mu^2)-2(m_e^2+ m_\mu^2)^2\right].\]

Differential cross section

From the squared matrix element we can calculate the differential cross section in the center of mass frame. For relativistic kinematics of \(2\to2\) scattering and the normalization conventions we employ here one has in the center of mass frame \[\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2 s}\frac{|\mathbf{p}_3|}{|\mathbf{p}_1|} \frac{1}{4} \sum_\text{spins}|\mathcal{T}|^2.\] Let us express everything in terms of the energy \(E\) of the incoming particles and the angle \(\theta\) between the incoming \(e^{-}\) electron momenta and outgoing \(\mu^{-}\) muon. \[\begin{split} |\mathbf{p}_1|=\sqrt{E^2 -m_e^2},\quad \quad\quad & s=4E^2, \\ |\mathbf{p}_3| = \sqrt{E^2 -m_\mu^2},\quad\quad\quad & t=m_e^2+ m_\mu^2-2E^2+2\mathbf{p}_1 \cdot \mathbf{p}_3, \\ \mathbf{p}_1 \cdot \mathbf{p}_3 = |\mathbf{p}_1||\mathbf{p}_3|\cos\theta, \quad\quad\quad& u = m_e^2 + m_\mu^2 - 2E^2 -2\mathbf{p}_1 \cdot \mathbf{p}_3. \end{split}\] With these relations we can express \(d\sigma/d\Omega\) in terms of \(E\) and \(\theta\) only.

Ultrarelativistic limit

Let us concentrate on the ultrarelativistic limit \(E\gg m_e, m_\mu\) so that we can set \(m_e = m_\mu = 0\). One has then \(|\mathbf{p}_1|=|\mathbf{p}_3|\) and \[t^2 + u^2 = 8E^4(1+ \cos^2\theta), \quad\quad\quad \frac{2(t^2+ u^2)}{s^2}= 1+ \cos^2 \theta,\] which leads to \[\frac{d\sigma}{d\Omega} = \frac{e^4}{64\pi^2 s}(1+\cos^2 \theta) = \frac{\alpha^2}{4s}(1+\cos^2 \theta).\] In the last equation we used \(\alpha = e^2/ (4\pi)\).

Electron-Muon Scattering

We can also consider the scattering process \(e^{-} \mu^{-} \to e^{-}\mu^{-}\),

\[i \mathcal{T} = \bar{u}(q_3) (-e\gamma^\mu) u(q_1) \left(-i\frac{\eta_{\mu\nu} -\frac{k_\mu k_\nu}{k^2}}{k^2}\right)\bar{u}(q_4)(-e\gamma^\nu) u(q_2).\] By a similar argument as before the term \(\sim k_\mu k_\nu\) drops out, \[\mathcal{T} = -\frac{e^2}{(q_1-q_3)^2} \bar{u}(q_3) \gamma^\mu u(q_1) \bar{u}(q_4) \gamma_\mu u(q_2)\quad\quad\quad (e^{-}\mu^{-} \to e^{-}\mu^{-}).\]

Comparison to electron to muon scattering

Compare this to what we have found for \(e^{-} e^{+} \to \mu^{-}\mu^{+}\) \[\mathcal{T} = - \frac{e^2}{(p_1+p_2)^2} \bar{v}(p_2) \gamma^\mu u(p_1) \bar{u}(p_3) \gamma_\mu v(p_4),\] where the conventions were according to

There is a close relation and the expressions agree if we put \[\begin{split} q_1 = +p_1, \quad\quad\quad & u(q_1) = u(p_1), \\ q_2 = -p_4, \quad\quad\quad & u(q_2) = u(-p_4) \to v(p_4), \\ q_3 = -p_2, \quad\quad\quad & \bar{u}(q_3) = \bar{u}(-p_2) \to \bar{v}(p_2), \\ q_4 = +p_3, \quad\quad\quad & \bar{u}(q_4) = \bar{u}(p_3). \end{split}\]

Crossing symmetry

Recall that \[(i\slashed{p}+m)\;u(p) = 0 \quad\quad\quad \text{but} \quad\quad\quad (-i\slashed{p}+ m)\;v(p) = 0.\] However one sign arises from the spin sums \[\begin{split} \sum_{s=1}^2 \; u_s(p)\bar{u}_s(p) & =-i\slashed{p}+m, \\ \sum_{s=1}^2 \; v_s(p)\;\bar{v}_s(p) & = -i\slashed{p}-m = -\sum_s u_s(-p)\;\bar{u}_s(-p). \end{split}\] Because it appears twice, the additional sign cancels for \(|\mathcal{T}|^2\) after spin averaging and one finds indeed the same result as for \(e^{-}e^{+} \to \mu^{-}\mu^{+}\) but with \[\begin{split} s_q = & -(q_1+q_2)^2 = -(p_1-p_4)^2 = u_p, \\ t_q = & -(q_1 -q_3)^2 = -(p_1+p_2)^2 = s_p,\\ u_q = & -(q_1 -q_4)^2 = -(p_1 -p_3)^2 =t_p. \end{split}\] We can take what we had calculated but must change the role of \(s\), \(t\)and \(u\) ! This is an example of crossing symmetries.

Electron-muon scattering in the massless limit

Recall that we found for \(e^{-}e^{+} \to \mu^{-}\mu^{+}\) in the massless limit \(m_e = m_\mu = 0\) simply \[\frac{1}{4}\sum_\text{spins} |\mathcal{T}|^2 = \frac{2e^4}{s^2} \left[t^2+u^2 \right].\] For \(e^{-}\mu^{-} \to e^{-}\mu^{-}\) we find after the replacements \(u \to s\), \(s\to t\), \(t \to u\), \[\frac{1}{4} \sum_\text{spins}|\mathcal{T}|^2=\frac{2e^4}{t^2}\left[u^2+ s^2\right].\]

More on Mandelstam variables

To get a better feeling for \(s\), \(t\) and \(u\), let us evaluate them in the center of mass frame for a situation where all particles have mass \(m\).

\[\begin{split} p_1^\mu & =(E,\mathbf{p}),\quad\quad p_2^\mu =(E,-\mathbf{p}), \\ p_3^\mu & =(E,\mathbf{p}'), \quad\quad p_4^\mu=(E, -\mathbf{p}'). \\ \end{split}\] While \[s = - (p_1+p_2)^2 = (2E)^2\] measures the center of mass energy,  \[t = - (p_1-p_3) = - 2 \mathbf{p}^2 [1-\cos(\theta)]\] is a momentum transfer that vanishes in the soft limit \(\mathbf{p}^2 \to 0\) and in the colinear limit \(\theta \to 0.\) Similarly, \[u= - (p_1 - p_4)^2 = - 2 \mathbf{p}^2 [1+\cos(\theta)]\] vanishes for \(\mathbf{p}^2 \to 0\) and for backward scattering \(\theta \to \pi\).

\(s\)-, \(t\)- and \(u\)-channels

One speaks of interactions in different channels for tree diagrams of the following generic types,

Electron-muon scattering

For the cross section we find for \(e^{-}\mu^{-} \to e^{-}\mu^{-}\) in the massless limit \[\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2 s} \frac{1}{4} \sum_{\rm spins} |\mathcal{T}|^2 = \frac{\alpha^2[4+(1+\cos\theta)^2]}{2s(1-\cos\theta)^2}\] This diverges in the colinear limit \(\theta \to 0\) as we had already seen for Yukawa theory in the limit where the exchange particle becomes massless.

Note that by the definition \(s \geq 0\) while \(u\) and \(t\) can have either sign. Replacements of the type used for crossing symmetry are in this sense always to be understood as analytic continuation.

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