Split $X^\mu=x_0^\mu+X'^\mu$. Integrating the constant mode gives <momentum conservation>,
$$
\int d^{26}x_0\,e^{ix_0\cdot\sum_i p_i}
\propto\delta^{26}\!\left(\sum_i p_i\right).
$$
For the nonzero modes, <Wick theorem> and the propagator give
$$
\left\langle\prod_{i=1}^4{}:e^{ip_i\cdot X(z_i)}:\right\rangle
=\exp\!\left[-\sum_{j<k}p_j\cdot p_k
\left\langle X(z_j)X(z_k)\right\rangle\right]
=\prod_{j<k}|z_j-z_k|^{\alpha'p_j\cdot p_k}.
$$
Thus, up to source-independent numerical normalization, the four-point amplitude is
$$
\boxed{
\mathcal A^{(4)}\sim
\frac{g_s^2\delta^{26}(\sum_i p_i)}{\operatorname{Vol}(SL(2,\mathbb C))}
\int\prod_{i=1}^4d^2z_i
\prod_{j<k}|z_j-z_k|^{\alpha'p_j\cdot p_k}}.
$$
The product is the <Koba-Nielsen factor>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 306 / 1 / iii / Solution

For the <Möbius transformation> $z'=(az+b)/(cz+d)$ with $ad-bc=1$,
$$
z_j'-z_k'=\frac{z_j-z_k}{(cz_j+d)(cz_k+d)},
\qquad
d^2z_i'=\frac{d^2z_i}{|cz_i+d|^4}.
$$
The power of $|cz_i+d|$ contributed by every pair containing $i$ is
$$
-\alpha'\sum_{j\ne i}p_i\cdot p_j
=\alpha'p_i^2=4,
$$
where <momentum conservation> and the mass-shell condition $p_i^2=4/\alpha'$ were used. The Koba-Nielsen factor therefore contributes $|cz_i+d|^4$ at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is $SL(2,\mathbb C)$ invariant, and division by its volume removes the residual conformal-gauge redundancy.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 306 / 1 / iv / Solution

{wiki=Momentum#Conservation}

Momentum conservation states that the total momentum of an isolated system is constant. In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.


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 Incoming links: Momentum conservation