= Solution
{c}
For <Møller scattering>, both the $t$-channel and $u$-channel <photon> exchange <tree-level Feynman diagrams> contribute, as shown in the two remaining lower panels. Exchanging the identical final <electrons> reverses the sign of the <scattering amplitude>, so the <fermionic sign> fixes
$$
\mathcal M=e^2\left(\frac{J_t}{t}-\frac{J_u}{u}\right),\qquad
J_t=[\bar u_3\gamma^\mu u_1][\bar u_4\gamma_\mu u_2],\quad
J_u=[\bar u_4\gamma^\mu u_1][\bar u_3\gamma_\mu u_2].
$$
The two direct squares follow from the preceding distinct-lepton calculation:
$$
\frac14\sum|J_t|^2=2(s^2+u^2),\qquad
\frac14\sum|J_u|^2=2(s^2+t^2).
$$
The interference cannot be discarded because the final <particles> are identical. Summing their <spins> joins all four external <Dirac spinors> into a single <trace>:
$$
\begin{aligned}
\sum J_tJ_u^*&=\operatorname{tr}(\not p_3\gamma^\mu\not p_1\gamma^\nu\not p_4\gamma_\mu\not p_2\gamma_\nu)\\
&=-32(p_3\cdot p_4)(p_1\cdot p_2)=-8s^2.
\end{aligned}
$$
The result from part (i) was used with relabelled contracted indices. Its negative sign combines with the explicit subtraction of the two exchange <scattering amplitudes>, so the interference contribution to the averaged square is positive:
$$
-\frac{2e^4}{4tu}\operatorname{Re}\sum J_tJ_u^*=\frac{4e^4s^2}{tu}.
$$
Thus
$$
\boxed{\langle|\mathcal M|^2\rangle=2e^4\left[\frac{s^2+u^2}{t^2}+\frac{s^2+t^2}{u^2}+\frac{2s^2}{tu}\right].}
$$
This is symmetric under $t\leftrightarrow u$, as required by the identical final <electrons>. There is no additional $1/2!$ in the squared <scattering amplitude>. Such a factor is used when integrating the <cross section> over a final-state region that counts both orderings of the identical pair; alternatively one integrates over one representative half of that region.
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