One-photon exchange between the two electrons. The virtual photon carries momentum transfer $t = (p_1 - p_1')^2$.
The computational chain
From diagram to number—every step of the amplitude calculation.
The diagram
Two electrons exchange a single virtual photon. The photon carries momentum $k = p_1 - p_1'$ with $k^2 = t$.
Feynman rules expression
$$i\mathcal{M}_t = \frac{ie^2}{t}\left[\bar{u}(p_1')\gamma^\mu u(p_1)\right]\left[\bar{u}(p_2')\gamma_\mu u(p_2)\right]$$
Each electron line contributes a current $\bar{u}\gamma^\mu u$. The photon propagator $-ig_{\mu\nu}/t$ connects them, contracting the Lorentz indices.
Square the amplitude and sum over spins
$$\overline{|\mathcal{M}_t|^2} = \frac{1}{4}\frac{e^4}{t^2}\,\mathrm{Tr}\left[(\not{p}_1'+m)\gamma^\mu(\not{p}_1+m)\gamma^\nu\right]\,\mathrm{Tr}\left[(\not{p}_2'+m)\gamma_\mu(\not{p}_2+m)\gamma_\nu\right]$$
Average over initial spins (factor 1/4 from two spin-1/2 particles), sum over final spins using the completeness relations $\sum_s u^s \bar{u}^s = \not{p}+m$. Each current squared becomes a trace over gamma matrices.
Evaluate the first trace
$$\mathrm{Tr}\left[(\not{p}_1'+m)\gamma^\mu(\not{p}_1+m)\gamma^\nu\right] = 4\left(p_1'^\mu p_1^\nu + p_1^\mu p_1'^\nu - g^{\mu\nu}(p_1 \cdot p_1' - m^2)\right)$$
Expand the product. The cross terms with a single $m$ vanish (trace of an odd number of gamma matrices is zero). Apply the trace identities: $\mathrm{Tr}[\gamma^\mu\gamma^\nu] = 4g^{\mu\nu}$ and $\mathrm{Tr}[\gamma^\alpha\gamma^\mu\gamma^\beta\gamma^\nu] = 4(g^{\alpha\mu}g^{\beta\nu} - g^{\alpha\beta}g^{\mu\nu} + g^{\alpha\nu}g^{\beta\mu})$.
Evaluate the second trace
$$\mathrm{Tr}\left[(\not{p}_2'+m)\gamma_\mu(\not{p}_2+m)\gamma_\nu\right] = 4\left(p_{2\mu}' p_{2\nu} + p_{2\mu} p_{2\nu}' - g_{\mu\nu}(p_2 \cdot p_2' - m^2)\right)$$
Same structure as the first trace with $p_1 \to p_2$, $p_1' \to p_2'$, and indices lowered.
Contract the two traces
$$\overline{|\mathcal{M}_t|^2} = \frac{e^4}{4t^2}\cdot 16\left[2(p_1'\cdot p_2')(p_1\cdot p_2) + 2(p_1'\cdot p_2)(p_1\cdot p_2') - \ldots\right]$$
Multiply the two trace tensors and contract the $\mu,\nu$ indices. Nine terms arise from the $3\times 3$ product. Many simplify using $g^{\mu\nu}g_{\mu\nu} = 4$ and $g^{\mu\nu}p_\mu = p^\nu$.
Substitute Mandelstam variables (massless limit)
$$\overline{|\mathcal{M}_t|^2} = \frac{2e^4(s^2+u^2)}{t^2}$$
In the massless limit ($m \to 0$), express all dot products in terms of $s = 2p_1\cdot p_2$, $t = -2p_1\cdot p_1'$, $u = -2p_1\cdot p_2'$. After collecting terms, the result is compact: the numerator is $s^2 + u^2$, and the $1/t^2$ denominator gives the characteristic forward scattering peak.