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Let's consider the Feynman diagram of a propagator of a particle. Is it considered as a one leg diagram or a two legs diagram?

Is it considered as an interaction? (I would think that an interaction means that a particle interacts with another particle, so that one needs at least 3 legs)

Do interaction with 2 legs (or one leg) exist in Standard Model?

Qmechanic
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Mathieu Krisztian
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  • A single real particle propagator, or a quantum bubble diagram, a closed one particle disconnected propagator line. So, without interaction, no connections with the sS-matrix. Disconnected diagrams. – MatterGauge Feb 06 '22 at 20:36
  • See here: https://physics.stackexchange.com/q/51993/ – MatterGauge Feb 06 '22 at 20:39
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    @Felicia : sorry, I could not understand because your sentence is not grammatical. In particular, you don't put verb in your sentence so this is impossible to me to understand what is your "point". – Mathieu Krisztian Feb 06 '22 at 21:06
  • Ah! I forgot to write "are examples of one line diagrams, representing free particle propagators without vertices." Sorry... And sS-matrix is the S-matrix. – MatterGauge Feb 06 '22 at 21:17
  • @Felicia : sorry, let's go back to the questions that I raised. Is propagator considered as an interaction : yes or no ? How many legs is it ? – Mathieu Krisztian Feb 06 '22 at 21:30
  • A propagator is a line in Feynman diagrams. Without an interaction, which is the vertex. – MatterGauge Feb 06 '22 at 21:33
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    One and two leg diagrams are not interactions, as Felicia is saying. A one leg diagram corresponds to a "tadpole", which have to vanish for stability reasons (physically they would correspond to processes where particles just pop out of the background; mathematically they vanish when the background fields satisfy the classical equations of motion). A two leg diagram is a propagator, which is not an interaction (one particle goes on, one goes out). Interactions are described by vertices with three or more particles. – Andrew Feb 07 '22 at 03:38
  • @Andrew : thank you. I can understand something with your kind answer. Please put your answer in the "answer", and I will accept it as the official answer. The explanation of Felicia was not clear at all. – Mathieu Krisztian Feb 07 '22 at 12:21
  • @MathieuKrisztian Done – Andrew Feb 07 '22 at 16:49
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    Isn’t it all in the eye of the beholder? And how you decide to split the Lagrangian? One can consider a mass term as a an interaction on a massless theory is one wishes to do that. – Oбжорoв Feb 08 '22 at 15:10

2 Answers2

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One and two leg diagrams are not interactions, as pointed out by Felicia in the comments.

A one leg diagram corresponds to a "tadpole", which have to vanish for stability reasons. Physically, these diagrams would correspond to processes where particles just pop out of the vacuum, and therefore if they didn't vanish they would imply that the vacuum is unstable. Mathematically, these diagrams vanish when the background fields satisfy the classical equations of motion. For instance, in the standard model, tadpole terms involving the Higgs vanish when the Higgs VEV is at a minimum of the potential.

A two leg diagram is a propagator, which is not an interaction -- one particle goes on, one goes out.

Interactions are described by vertices with three or more particles.

Andrew
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To perform perturbation theory, the action $$S~=~S_{\rm free}+S_{\rm int}$$ is split into a free and an interaction part. The free part$^1$ $$S_{\rm free}~=~\frac{1}{2}S^{\rm free}_{k\ell}\phi^k\phi^{\ell}$$ is always quadratic and non-degenerate in the fields $\phi^k$. The inverse matrix $(S_{\rm free}^{-1})^{k\ell}$ is the free propagator, cf. e.g. this Phys.SE post.

  1. An action term linear in the fields produces diagrammatically a 1-vertex (1 leg). It is either a tadpole or a source.

  2. An action term quadratic in the fields may be assigned$^2$ to the free or the interaction part, depending on application. If it belongs to the interaction part it produces diagrammatically a 2-vertex (2 legs). Typical examples are counterterms for wavefunction renormalization or mass-renormalization. This e.g. applies to the standard model.

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$^1$ We use DeWitt condensed notation to not clutter the notation.

$^2$ Note that the free part should be non-degenerate, so one can not assign all quadratic action terms to belong to the interaction part.

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