# QFT- Computing matrix elements when momentum is NOT conserved

In QFT, when computing cross sections one calculates the $$T$$ matrix elements (probability amplitudes that the interaction Hamiltonian takes us from the initial state $$|\mathbf{k}_A,\mathbf{k}_B\big>$$ to the final state $$|\mathbf{p}_1,\mathbf{p}_2...\big>$$):

$$\big<\mathbf{p}_1\mathbf{p}_2...|iT|\mathbf{k}_A\mathbf{k}_B\big>=(2\pi)^4\delta^4(k_A+k_B-\sum p_f)i\mathcal{M}(k_A,k_B\rightarrow p_f)$$

(This is discused in Peskin Eq. 4.73.) In free space 4-momentum is conserved, and the quantity $$\mathcal{M}$$ appears in the expression of the differential cross section as: $$d\sigma=\frac{1}{2E_AE_B|v_A-v_B|}|\mathcal{M}(p_A,p_B\rightarrow \{p_f\})|^2(2\pi)^4\delta^4(k_A+k_B-\sum p_f),$$ where $$p_A$$ and $$p_B$$ are the values where the initial momenta distributions are concentrated. $$E_A$$ and $$E_B$$ ($$v_A$$ and $$v_B$$) are the energies (velocities) of the intial particles.

What happens when momentum in a given direction (say, the $$z$$ axis) is not conserved? I imagine we would get something like $$\big<\mathbf{p}_1\mathbf{p}_2...|iT|\mathbf{k}_A\mathbf{k}_B\big>=(2\pi)^3\delta(E_A+E_B-\sum E_f)\delta^2(\mathbf{p}_{\perp,A}+\mathbf{p}_{\perp,B}-\sum \mathbf{p}_{\perp,f})i\mathcal{M},$$ where the subscript $$\perp$$ indicates the components of momenta perpendicular to the $$z$$-axis. How can we define cross sections in that case?

Momentum is not conserved when the lagrangian density depends explicitly on $$x$$, that is when the interaction vertices are no longer constant but functions of spacetime points. This typically happen when coupling to an external source. When integrating any vertex in the $$x$$-space Feynman's rule, one finds not only the $$e^{\pm ik x}$$ factors from the fourier transform of the propagators or from the external legs, but also the explicit $$x$$-dependence from the vertex. Fourier transforming the explicit $$x$$-dependence at the vertex we can turn its dependence in an extra $$e^{i Q x}$$ factor. Therefore, the momentum conservation at each vertex gets an extra entry $$(2\pi)^4\delta^4(\sum_i k_i + Q)$$ at each non-constant vertex, exactly like adding an external source.
It's convenient to visualize this by adding an external leg as for an external current to any such a vertex, so that the leg is keeping track of the momenta $$Q_i$$ at each vertex where it enters. In this way you can use the usual rules for Feynman's diagrams to calculate the amplitude. In particular, since they are external legs, in any connected Feynman diagrams they will definitely show up in the overall conserving $$4$$-momentum $$(2\pi)^4\delta^4(\sum_{i}Q_i+\sum_i k_j)$$. That's not the only place: the momentum $$Q_i$$ appear as well in the internal propagator once the momentum conservation at each vertex has been resolved. Finally, to get the final result one has to integrate over the $$Q_i$$, which removes (only) one delta function (for each connected term of the amplitude where they enter), as the OP suggested. If momentum in the z-direction is not conserved then the last equation of the OP is correct.
To get the cross-section formula one should repeat the usual story with one less delta function in principle. But I think it's simpler to just recycle the usual formula by working with the 4 delta functions by keeping the external sources until the end avoiding integrating to early in the $$Q_i$$, and then integrate the usual cross-section formula over the $$Q_i$$ (more than once because of squaring the amplitude etc...) only as last step.