EXERCISE 6.18 Show that, individually, the amplitudes 1 and 2 are not gauge invariant but that t

step 1 This question, to start the proof, we have that a hydrogen atom whose electron has initial orbit

EXERCISE 6.18 Show that, individually, the amplitudes 1 and...

EXERCISE 6.18 Show that, individually, the amplitudes $\Re_1$ and $\Re_2$ are not gauge invariant but that their sum indeed satisfies (6.108). It is instructive to work through the calculation of the Compton scattering amplitude in detail. For simplicity, we neglect the mass of the electron, and so the invariant variables for $\gamma(k)+\mathrm{e}(p) \rightarrow \gamma\left(k^{\prime}\right)+\mathrm{e}\left(p^{\prime}\right)$ are $$ \begin{aligned} & s=(k+p)^2=2 k \cdot p=2 k^{\prime} \cdot p^{\prime} \\ & t=\left(k-k^{\prime}\right)^2=-2 k \cdot k^{\prime}=-2 p \cdot p^{\prime} \\ & u=\left(k-p^{\prime}\right)^2=-2 k \cdot p^{\prime}=-2 p \cdot k^{\prime} . \end{aligned} $$ The two invariant amplitudes, (6.103) and (6.104), are $$ \begin{aligned} & \mathscr{R}_1=\varepsilon_v^{\prime *} \varepsilon_\mu e^2 \bar{u}\left(p^{\prime}\right) \gamma^\nu(\not p+k) \gamma^\mu u(p) / s, \\ & \Re_2=\varepsilon_v^{\prime *} \varepsilon_\mu e^2 \bar{u}\left(p^{\prime}\right) \gamma^\mu\left(\not p-k^{\prime}\right) \gamma^\nu u(p) / u . \end{aligned} $$ To obtain the unpolarized cross section, we must average/sum $\left|\Re_1+\mathscr{R}_2\right|^2$ over the initial/final electron and photon spins. Fortunately, this is not as difficult as it first appears. For physical photons, (6.98) applies, and we can make the replacement $$ \sum_T \varepsilon_\mu^{T *} \varepsilon_{\mu^{\prime}}^T \rightarrow-g_{\mu \mathrm{F}^{\prime}} $$ where $T$ denotes transverse. We have a similar completeness relation for the outgoing photon states, $\varepsilon^{\prime}$. Thus, for example, $$ \overline{\left|\mathscr{R}_1\right|^2}=\frac{e^4}{4 s^2} \sum_{s, s^{\prime}}\left(\bar{u}^{\left(s^{\prime}\right)} \gamma^\nu(\not p+k) \gamma^\mu u^{(s)}\right)\left(\bar{u}^{(s)} \gamma_\mu(p+k) \gamma_v u^{\left(s^{\prime}\right)}\right) \text {. } $$ The factor $\frac{1}{4}$ is due to averaging over the initial electron and photon spins. The spinor completeness relation, $(5.47)$, allows the sum over $u \bar{u}$ states to be performed (just as we described for $\mathrm{e}^{-} \mu^{-}$scattering in Section 6.3), and we find $$ \begin{aligned} & \overline{\left|\Re_1\right|^2}=\frac{e^4}{4 s^2} \operatorname{Tr}(\underbrace{\check{p}^{\prime} \gamma^\nu}_{-2 p^{\prime}}(\not p+k) \underbrace{\gamma^\mu p \gamma_\mu}_{-2 \not p}(\not p+k) \gamma_\nu^{\prime}) \\ & =\frac{e^4}{s^2} \operatorname{Tr}\left(p^{\prime} k p k\right) \\ & =\frac{4 e^4}{s^2} 2\left(p^{\prime} \cdot k\right)(p \cdot k) \\ & =2 e^4\left(-\frac{u}{s}\right) \text {. } \\ & \end{aligned} $$ where we have made use of (6.24) and (6.22). Similarly, we obtain $$ \begin{aligned} \overline{\left|\Re_2\right|^2} & =2 e^4\left(-\frac{s}{u}\right), \\ \overline{\pi_1 \Re_2^*} & =0 . \end{aligned} $$ Thus, the spin-averaged Compton amplitude is $$ \overline{\mid \Im U^2}=\overline{\left|9 R_1+9 \pi_2\right|^2}=2 e^4\left(-\frac{u}{s}-\frac{s}{u}\right) . $$

EXERCISE 6.18 Show that, individually, the amplitudes $\Re_1$ and $\Re_2$ are not gauge invariant but that their sum indeed satisfies (6.108).

It is instructive to work through the calculation of the Compton scattering amplitude in detail. For simplicity, we neglect the mass of the electron, and so the invariant variables for $\gamma(k)+\mathrm{e}(p) \rightarrow \gamma\left(k^{\prime}\right)+\mathrm{e}\left(p^{\prime}\right)$ are
$$
\begin{aligned}
& s=(k+p)^2=2 k \cdot p=2 k^{\prime} \cdot p^{\prime} \\
& t=\left(k-k^{\prime}\right)^2=-2 k \cdot k^{\prime}=-2 p \cdot p^{\prime} \\
& u=\left(k-p^{\prime}\right)^2=-2 k \cdot p^{\prime}=-2 p \cdot k^{\prime} .
\end{aligned}
$$

The two invariant amplitudes, (6.103) and (6.104), are
$$
\begin{aligned}
& \mathscr{R}_1=\varepsilon_v^{\prime *} \varepsilon_\mu e^2 \bar{u}\left(p^{\prime}\right) \gamma^\nu(\not p+k) \gamma^\mu u(p) / s, \\
& \Re_2=\varepsilon_v^{\prime *} \varepsilon_\mu e^2 \bar{u}\left(p^{\prime}\right) \gamma^\mu\left(\not p-k^{\prime}\right) \gamma^\nu u(p) / u .
\end{aligned}
$$

To obtain the unpolarized cross section, we must average/sum $\left|\Re_1+\mathscr{R}_2\right|^2$ over the initial/final electron and photon spins. Fortunately, this is not as difficult as it first appears. For physical photons, (6.98) applies, and we can make the replacement
$$
\sum_T \varepsilon_\mu^{T *} \varepsilon_{\mu^{\prime}}^T \rightarrow-g_{\mu \mathrm{F}^{\prime}}
$$
where $T$ denotes transverse. We have a similar completeness relation for the outgoing photon states, $\varepsilon^{\prime}$. Thus, for example,
$$
\overline{\left|\mathscr{R}_1\right|^2}=\frac{e^4}{4 s^2} \sum_{s, s^{\prime}}\left(\bar{u}^{\left(s^{\prime}\right)} \gamma^\nu(\not p+k) \gamma^\mu u^{(s)}\right)\left(\bar{u}^{(s)} \gamma_\mu(p+k) \gamma_v u^{\left(s^{\prime}\right)}\right) \text {. }
$$

The factor $\frac{1}{4}$ is due to averaging over the initial electron and photon spins. The spinor completeness relation, $(5.47)$, allows the sum over $u \bar{u}$ states to be performed (just as we described for $\mathrm{e}^{-} \mu^{-}$scattering in Section 6.3), and we find
$$
\begin{aligned}
& \overline{\left|\Re_1\right|^2}=\frac{e^4}{4 s^2} \operatorname{Tr}(\underbrace{\check{p}^{\prime} \gamma^\nu}_{-2 p^{\prime}}(\not p+k) \underbrace{\gamma^\mu p \gamma_\mu}_{-2 \not p}(\not p+k) \gamma_\nu^{\prime}) \\
& =\frac{e^4}{s^2} \operatorname{Tr}\left(p^{\prime} k p k\right) \\
& =\frac{4 e^4}{s^2} 2\left(p^{\prime} \cdot k\right)(p \cdot k) \\
& =2 e^4\left(-\frac{u}{s}\right) \text {. } \\
&
\end{aligned}
$$
where we have made use of (6.24) and (6.22). Similarly, we obtain
$$
\begin{aligned}
\overline{\left|\Re_2\right|^2} & =2 e^4\left(-\frac{s}{u}\right), \\
\overline{\pi_1 \Re_2^*} & =0 .
\end{aligned}
$$
Thus, the spin-averaged Compton amplitude is
$$
\overline{\mid \Im U^2}=\overline{\left|9 R_1+9 \pi_2\right|^2}=2 e^4\left(-\frac{u}{s}-\frac{s}{u}\right) .
$$

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