Higher-Order Feynman Diagrams for $e^{+}e^{-} \rightarrow \mu^{+}\mu^{-}$

Question

In his book "Modern Particle Physics" Mark Thomson has the following Figures for $e^{+}e^{-}\rightarrow \mu^{+}\mu^{-}$:

(i) Concerning the first Feynman diagram in Figure 6.2: Couldn't the intermediate loop also contain $\mu^{-}\mu^{+}$?

(ii) Concerning the second Feynman diagram in Figure 6.2: This is how I understand what is drawn. Could somebody please tell me whether my understanding is correct?

(iii) Concerning the third Feynman diagram in Figure 6.2: Here I drew my understanding of the diagram:

So basically an $e^{-}$ emits a photon $\gamma$ but is still an $e^{-}$ and this emitted photon is absorbed by a muon $\mu^{-}$ which still stays a muon $\mu^{-}$. Is that diagram correct? :-)

Answer 1

In order to turn Feynman diagrams to calculations, one needs, in addition to having the correct quantum numbers on the virtual particles, the coupling constants at the vertices.

There are three fundamental forces in elementary particle interactions, and the three coupling constants are very different in orders of magnitude. So when one talks of "higher orders" one has to evaluate the couplings at the vertices within the content of "higher" . Weak interaction loops are not forbidden but are much depressed by the couplings that one ignores them when electromagnetic couplings are available, in order counting.

I want to include the comment for the second question ,which is formulated as if the internal lines are real particles, they are virtual particles, just place holders for the transmission of quantum numbers. Their mass is off mass shell, that is why they are defined as virtual. As Nihar Karve says in his comment to the question:

Once you get to higher order diagrams, I'd recommend moving away from the notion that a Feynman diagram represents anything other than a diagrammatic representation of a formal mathematical process - there is no photon that is physically emitted by the electron –

As for the last diagram, there are still four electromagnetic vertices, so it is the same order of magnitude as the others.