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Consider the process $Xq\rightarrow Yq$ at tree level via exchange of a photon. It is depicted in the following Feynman diagram. In various literature it is said that for a nearly on-shell photon this amplitude factorises into the product of two amplitudes of the subprocesses $X\gamma\rightarrow Y$, $q\rightarrow q\gamma$ and the denominator squared of the photon propagator. But there is one subtlety that I do not understand about this.

Feynman diagram

The tensorial structure admits to writing the matrix element as the product $M_\mu\Delta^{\mu\nu}N_\nu=M_\mu(-g^{\mu\nu})N_\nu\frac{i}{k^2}$. The polarisation sum of massless vector bosons is $\sum_i\epsilon^\mu_i\epsilon^{\nu\ast}_i=-g^{\mu\nu}+\epsilon^\mu_+\epsilon^{\nu\ast}_-+\epsilon^\mu_-\epsilon^{\nu\ast}_+$, which allows us to substitute the metric for this sum. If the photon is nearly on-shell, the terms with $\epsilon_\pm$ nearly vanish, leading to the expression $\sum_iM_\mu\epsilon^\mu_i\epsilon^{\nu\ast}_iN_\nu\frac{i}{k^2}$. Now this almost looks like the product of two distinct matrix elements, except there is the sum over $i$, which couples both factors. If we calculate the amplitude and define $M_{\mu\nu}:=M^\dagger_\mu M_\nu$, $N_{\alpha\beta}:=N^\dagger_\alpha N_\beta$, we find that $$|M|^2_{Xq\rightarrow Yq}=\frac{1}{k^4}\sum_{i,j}M_{\mu\nu}\epsilon^{\mu\ast}_i\epsilon^\nu_j\times N_{\alpha\beta}\epsilon^\alpha_i\epsilon^{\beta\ast}_j.$$ But this is not necessarily equal to the product $\frac{1}{k^4}|M|^2_{X\gamma\rightarrow Y}\times|M|^2_{q\rightarrow q\gamma}=\frac{1}{k^4}\sum_iM_{\mu\nu}\epsilon^{\mu\ast}_i\epsilon^\nu_i\times\sum_jN_{\alpha\beta}\epsilon^\alpha_j\epsilon^{\beta\ast}_j$ that you find in literature (e.g. Peskin & Schroeder pp.578).

I suppose there must be a profound error in my thinking somewhere but I just cannot see where. I hope somebody can explain.

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What you found is precisely that the amplitude factorizes since the correct statement of factorization is

$$ \lim_{P^2\rightarrow0} M(1,\cdots,n) = \sum_{ i\in \text{helicities}} M(1,\cdots,k,P^i)\frac{i}{P^2} M(P^{-i}, k+1,\cdots,n)\,,$$

which includes a sum over helicities/polarizations (or more generally states). For a more detailed discusion see Sec. 3.4 of the classic TASI Lectures by Lance Dixon.

Also, note that factorization works at the level of the amplitude, so you don't need to take the absolute value squared.

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