Lagrangian and Feynman Diagrams

We would like to compute the cross section of fermion-boson scattering process. The Lagrangian is given by \[\begin{eqnarray} \mathcal{L}= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{m^2}{2}\phi^2+i\bar\psi\gamma^\mu\partial_\mu\psi-M\bar\psi\psi+h \phi\bar\psi\psi-\frac{\lambda}{4!}\phi^4 \tag{1}, \end{eqnarray}\] where \(\phi\) represents the neutral scalar particle, and \(\psi_\alpha\) is a four-component spinor field with \(\alpha=1,2,3,4.\) The scattering process we are after is given as \[\begin{eqnarray} \texttip{\phi(k_1)}{Incoming scalar field $\phi$ with momentum $k_1$} \texttip{\psi(p_1)}{Incoming fermion field $\phi$ with momentum $p_1$}\texttip{\longrightarrow}{Scattering with coupling $-ih$} \texttip{\phi(k_2)}{Outgoing scalar field $\phi$ with momentum $k_2$}+ \texttip{\psi(p_2)}{Outgoing fermion field $\psi$ with momentum $p_2$} \tag{2}. \end{eqnarray}\]

The interactive Fenynman diagrams contributing to the process are shown in Fig. 1.

Figure 1: Two Feynman diagrams, with amplitudes \(\mathcal{M}_A\) and \(\mathcal{M}_B\), contributing to the scattering. Hover on the lines and vertices to see more info.

Amplitudes

The amplitudes for the diagrams in Fig. 1 can be written as \[\begin{eqnarray} \mathcal{M}_A&=&-i \overline{U}(p_2)(-ih)\left[i\frac{p\!\!\!/_1+k\!\!\!/_1+M}{(p_1+k_1)^2-M^2}\right](-ih)U(p_1)\nonumber\\ \mathcal{M}_B&=&-i \overline{U}(p_2)(-ih)\left[i\frac{p\!\!\!/_1-k\!\!\!/_2+M}{(p_1-k_2)^2-M^2}\right](-ih)U(p_1)\tag{3}. \end{eqnarray}\] The numerators can be simplified by using the equation of motion for the fermions, namely: \[\begin{eqnarray} (p\!\!\!/_1-M)U(p_1)=0 \tag{4}. \end{eqnarray}\]

Let’s us compute the denominators : \[\begin{eqnarray} (p_1+k_1)^2-M^2&=&p_1^2+k_1^2+2p_1\cdot k_1-M^2=M^2+m^2+2p_1\cdot k_1-M^2\nonumber\\ &=& 2p_1\cdot k_1+m^2 \nonumber\\ (p_1-k_2)^2-M^2&=&p_1^2+k_2^2-2p_1\cdot k_2-M^2=M^2+m^2-2p_1\cdot k_2-M^2\nonumber\\ &=&-2p_1\cdot k_2+m^2\tag{5}. \end{eqnarray}\] Inserting these into Eq. (3), we get \[\begin{eqnarray} \mathcal{M}_A&=&\frac{-h^2}{2p_1\cdot k_1+m^2} \overline{U}(p_2)\left[2 M+k\!\!\!/_1\right]U(p_1)\nonumber\\ \mathcal{M}_B&=&\frac{h^2}{2p_1\cdot k_2+m^2} \overline{U}(p_2)\left[2 M-k\!\!\!/_2\right]U(p_1)\tag{6}. \end{eqnarray}\]

Let’s also consider the process in the high energy limit, i.e., \(E\gg M,m\), that is we will drop the mass terms. In this limit we can simplify the amplitudes:

\[\begin{eqnarray} \mathcal{M}_A&\simeq&\frac{-h^2}{2p_1\cdot k_1} \overline{U}(p_2)k\!\!\!/_1U(p_1)\nonumber\\ \mathcal{M}_B&\simeq&\frac{-h^2}{2p_1\cdot k_2} \overline{U}(p_2)k\!\!\!/_2U(p_1)\tag{7}. \end{eqnarray}\]

Squaring the amplitudes

The total amplitude is given by \[\begin{eqnarray} \mathcal{M}=\mathcal{M}_A+\mathcal{M}_B\tag{8}, \end{eqnarray}\] and we will need to compute its mode-square which will involve mode-squares of the individual amplitudes and the cross terms. We will also average over fermion polarization which will result in trace operations. There are few trace properties of \(\gamma-\)matrices we will make use of: \[\begin{eqnarray} \mathop{\mathrm{Tr}}[I]&=&4\nonumber\\ \mathop{\mathrm{Tr}}[\gamma^\mu\gamma^\nu]&=&4g^{\mu\nu}\\ \mathop{\mathrm{Tr}}[\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma]&=&4[g^{\mu\nu}g^{\rho\sigma}+g^{\mu\sigma}g^{\nu\rho}-g^{\mu\rho}g^{\nu\sigma}]\\ \mathop{\mathrm{Tr}}[\gamma^\mu_1\gamma^\mu_2\cdots\gamma^\mu_{2n+1}]&=&0 \tag{9}, \end{eqnarray}\] The mode-square of the first amplitude becomes \[\begin{eqnarray} \left|\overline{\mathcal{M}_A}\right|^2&=& \frac{h^4}{4(p_1\cdot k_1)^2} \frac{1}{2}\mathop{\mathrm{Tr}}\left[p\!\!\!/_2k\!\!\!/_1p\!\!\!/_1k\!\!\!/_1\right]\nonumber\\ &=& \frac{h^4}{2(p_1\cdot k_1)^2} p_1\cdot k_1 \,p_2\cdot k_2= h^4\frac{p_1\cdot k_2}{p_1\cdot k_1}\tag{10}. \end{eqnarray}\]

Similarly, the mode-square of the second amplitude reads \[\begin{eqnarray} \left|\overline{\mathcal{M}_B}\right|^2&=& \frac{h^4}{4(p_1\cdot k_1)^2} \frac{1}{2}\mathop{\mathrm{Tr}}\left[p\!\!\!/_2k\!\!\!/_2p\!\!\!/_1k\!\!\!/_2\right]\nonumber\\ &=& \frac{ h^4}{(p_1\cdot k_1)^2} p_2\cdot k_2 \,p_1\cdot k_2=h^4\frac{p_1\cdot k_1}{p_1\cdot k_2}\tag{11}, \end{eqnarray}\] where we used conservation of \(4-\)momentum in the last step as follows: \[\begin{eqnarray} p_1+k_1&=&p_2+k_2 \iff p_1-k_2=p_2-k_1\nonumber\\ (p_1+k_1)^2&=&(p_2+k_2)^2 \implies p_1\cdot k_1=p_2\cdot k_2 \tag{12}, \end{eqnarray}\]

Finally one of the cross term can be calculated as \[\begin{eqnarray} \overline{\mathcal{M}_A^*\mathcal{M}_B}&=& \frac{-h^4}{4p_1\cdot k_1\,p_1\cdot k_2} \frac{1}{2}\mathop{\mathrm{Tr}}\left[p\!\!\!/_2k\!\!\!/_1p\!\!\!/_1k\!\!\!/_2\right]\nonumber\\ &=& \frac{h^4}{2p_1\cdot k_1\,p_1\cdot k_2} \left[\,p_2\cdot k_1 \,p_1\cdot k_2+p_2\cdot k_2 \,p_1\cdot k_1-p_2\cdot p_1 \,k_1\cdot k_2 \right]\nonumber\\ &=&\frac{h^4}{2}\left[ \frac{p_1\cdot k_2}{p_1\cdot k_1}+\frac{p_1\cdot k_1}{p_1\cdot k_2} -\frac{p_1\cdot p_2\, k_1\cdot k_2}{p_1\cdot k_1\,p_1\cdot k_2} \right]\tag{13}. \end{eqnarray}\]

Cross-section

Let’s find out which term will have the dominant contribution to the cross-section. To this end, we can treat the problem in the center of mass frame and define:

\[\begin{eqnarray} k_1&=&(\omega,0,0,w)\nonumber\\ p_1&=&(E,0,0,-\omega)\nonumber\\ k_2&=&(\omega,\omega\sin\theta,0,\omega\cos\theta)\nonumber\\ p_1&=&(E,0,0,-\omega)\tag{14}. \end{eqnarray}\] We can observe that the term \(1/p_1\cdot k_2\) will be \(\sim1/ M^2\) at \(\theta=\pm\pi\), and therefore will be the dominating term, since other terms will will behave as \(1/E^2\). So the cross-section will be dominated by the following term