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It seems I have finally decided on the question. When expanding the interaction process in a perturbation theory series, it is also necessary to take into account the contribution of the diagrams in which no interaction occurs? As in the first two diagrams in the figure. enter image description here

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    $\begingroup$ I think it depends on what you're trying to compute. Usually we calculate the S-matrix without the trivial part corresponding to no scattering, but of course for the "full" S-matrix they should be included (this point is sometimes glossed over) $\endgroup$ Jan 21, 2021 at 8:58

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Consider the Dyson series for the S-matrix: $$ S = \lim_{t\to\infty} U(-t, +t) = \mathcal{T}\exp\left(-i\int_{-\infty}^{\infty}\mathrm{d}^4 x \ \mathcal{H}(x)\right) $$ Expanding out the first few terms, we see that this is $$ S = \mathcal{T}\left[\color{red}1-i\int \mathrm{d}^4x \ \mathcal{H}(x) + \frac{-i^2}{2!}\iint \mathrm{d}^4x \ \mathrm{d}^4y \ \mathcal{H}(x)\mathcal{H}(y)+\dots \right] $$

Usually, when people talk about computing the S-matrix, they are really talking about the transfer matrix $T$, in which the trivial $1$ part is discarded, corresponding to ignoring the processes in which there is no interaction (zero vertex diagrams). In fact, most people take it a step further and really talk about evaluating the "amplitude" $\mathcal{M}_{fi}$, which is defined as $$ S_{fi} = \langle f|S|i\rangle = \delta_{fi}-iT_{fi} = \delta_{fi}-i(2\pi)^4\delta^4(p_f-p_i)\mathcal{M}_{fi} $$

For the "full" S-matrix defined by the Dyson formula though, the trivial diagrams must be included.

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The answer is no, there is no meaning to the first two , because in the crossection for the interaction ,the term "interaction" means there is energy-momentum exchange. The Feynman diagrams are a pictorial representation of how that exchange happens, breaking that down to the perturbative terms. This link may help you

Each Feynman diagram represents a term in the perturbation theory expansion of the matrix element for an interaction.Normally, a full matrix element contains an infinite number of Feynman diagrams.

feynmadiag

There is no term with no vertex.

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  • $\begingroup$ You have to start with free field theory. Only then can you turn on the interaction and read the corrections. And, of course, a diagram without interaction has its own amplitude, since ALL stories must be considered. It’s obvious to me that you’re not good at theory. $\endgroup$ Jan 21, 2021 at 7:41
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    $\begingroup$ I gave you a link for my statements, where is yours? Free field theory describes nothing, it is mathematics, if it exists at all. Physics is about measurements and obervations. $\endgroup$
    – anna v
    Jan 21, 2021 at 8:19
  • $\begingroup$ physics.stackexchange.com/questions/606461/… $\endgroup$ Jan 21, 2021 at 9:08
  • $\begingroup$ this answer is about non interacting, if there is no interaction there is no measurement to check , I can find no zero order here web.mit.edu/tabbott/www/papers/feynman-diagrams.pdf feynman diagram,. Here it talks of "unlinked diagrams" eduardo.physics.illinois.edu/phys582/… might enlighten you , if you can wade through it. $\endgroup$
    – anna v
    Jan 21, 2021 at 9:42
  • $\begingroup$ @anna v To calculate the total scattering matrix, it is also necessary to take into account the contribution of the non-interacting component. $\endgroup$ Jan 21, 2021 at 10:12

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