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Let $\phi$ be a real scalar field and $$\mathcal L = \frac{1}{2}(\partial_\mu \phi)^2 + \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4.$$ Given the standard Feynman diagrams and Feynman rules for $\phi^4$ -theory, let us consider the one particle irreducible (1PI) 4-point correlation function up to $O(\lambda^2)$, that is enter image description here In my lecture notes it is written that the complete correlation function is then given by ($s,t,u$ are Mandelstamm variables) $$i \mathcal{M}_{12 \rightarrow 34}(s, t, u)=-i \lambda+(-i \lambda)^{2}[i V(s)+i V(t)+i V(u)],$$ where $$iV(p^2)= \frac{1}{2}\int \frac{d^{4} k}{(2 \pi)^{4}} \frac{i}{k^{2}-m^{2}+i \epsilon} \frac{i}{(k+p)^{2}-m^{2}+i \epsilon}.$$ I'm confused by this expression because of another set of lecture notes (same uni, the previous year) that state the amplitude of the first diagram in the parenthesis is $$\begin{aligned} F=& (-i \lambda)^{2}(2 \pi)^{4} \delta^{4}\left(p_{1}+p_{2}+p_{3}+p_{4}\right) \prod_{j=1}^{4} \frac{i}{p_{j}+m^{2}-i \epsilon} iV(s), \end{aligned}$$ i.e. they included the momentum conservation and propagators.

I'm struggling to get the terminology right here. $F$ and $\mathcal{M}_{12 \rightarrow 34}$ (I think at least) are essentially the same thing, but what do people usually call "amplitudes"? Do we include momentum conservation and propagators when we write amplitudes down or is it assumed that everybody can fill the gap?

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The first one is for the amputated diagrams, which do not contain any external legs. The second one is the full matrix element plus the delta for the conservation of momentum.

The delta for the conservation of momentum is not fundamental, since it can be added at any moment when evaluating cross sections. From my experience, i'd rather put the conservation of momentum out of the definition of the matrix element for the process.

The important thing is the amputated part. The usage of amputated diagrams is linked to the LSZ Reduction formula which is a non-perturbative formula, and being so it works at any order in perturbation theory. If you check out the formula (for scalar fields but the ideas stays the same for fermionic fields) from Wikipedia

$$ \left\langle p_{1}, \ldots, p_{n} \text { out }\right| q_{1}, \ldots, q_{m} \text { in }\rangle=\prod_{i=1}^{m}\left\{-\frac{i\left(p_{i}^{2}-m^{2}\right)}{(2 \pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}\right\} \prod_{j=1}^{n}\left\{-\frac{i\left(q_{j}^{2}-m^{2}\right)}{(2 \pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}\right\} \int \prod_{i=1}^{n}\left\{\mathrm{~d}^{4} x_{i} e^{i p_{i} \cdot x_{i}}\right\}\left\langle 0\left|\mathrm{~T} \varphi\left(x_{1}\right) \ldots \varphi\left(x_{n}\right)\right| 0\right\rangle $$ You can see that the n-point Green function $$\left\langle 0\left|\mathrm{~T} \varphi\left(x_{1}\right) \ldots \varphi\left(x_{n}\right)\right| 0\right\rangle$$ is multiplied by the factors which are the inverse of the external propagators both for incoming particles $$i\left(p_{i}^{2}-m^{2}\right)$$ and the outgoing ones $$i\left(q_{j}^{2}-m^{2}\right)$$ This cancels the need for the external propagators.

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