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2

The procedure is quite simple: the symmetry factor counts how many of the same diagrams you would be counting if you didn't consider that the single particles in your diagrams are identical. To do this, ask yourself "how many of this same diagram can I make?" where "same diagram" means "same external particles, same connections, and ...


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In particular, for the first equation we need small $|q|$, $|p_1|≃p_F$, and $|p_2|≃p_F$. Don't we just need for the second equation small $|q_1|$, small $|q_2|$, and $|p|≃p_F$? This is correct. But notice, while small $|q|$ requires fine-tuning in all 3 directions in k-space, the condition $|p_1|\simeq p_f$ requires fine-tuning only in one direction - ...


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You're right. A "Born" and "tree level" are the same thing. It's not very common to say Born anymore, but the reason why the call it like that in the reference is likely due to the more standard quantum mechanical definition. In nonrelativistic quantum mechanics one usually calls the "Born approximation" as the approximation at ...


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If time is left-to-right in your diagram, then momentum conservation implies that the internal fermion has momentum $p+k=p'+k'$. The convention of reversing arrows for antiparticles is just to make it easy to check that the discrete quantum numbers that change sign for antiparticles -- like charge -- are conserved at each node. But conservation of energy and ...


1

You understanding is essentially correct. The whole truth is quite complicated, but essentially well-understood (see for example John Collins' book on renormalization). But let me give you some other rough considerations. Take your formula $$ D = d - E(d-2)/2 - \sum V_n [g_n]. $$ If $[g_n]$ is negative, since for any amplitude the amount of $n$-type vertices ...


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Hints: Let us for simplicity put mass $m=1$ and go to Euclidean signature as e.g. my Phys.SE answer here. The free propagator is $$\Delta(t,t^{\prime})~=~\frac{1}{2\omega_0}e^{-\omega_0 |t-t^{\prime}|},\tag{A} $$ while the Feynman rule for the 2-vertex insertion is $$-\frac{\omega_1^2}{\hbar}\int_{[0,T]}\!dt. \tag{B} $$ The connected Feynman vacuum ...


5

If you actually calculate the one-loop correction, with a neutrino mass in the propagator as a placeholder for the moment. You will end up seeing the finite correction due to the resumed loop effect is only proportional to the mass. Now here comes the problem, the SM doesn't give the mechanism of neutrino masses in the first place. Hence the loop effect, ...


2

I don't know how many details do you want, but a simple argument to understand why neutrinos are massless in the SM up to any order in perturbation theory is the following. From a Lorentz perspective, fermion masses require both left- and right-handed spinor components to form the $m_D\,\overline\Psi\Psi$ mass term. In the case of neutrinos, they are present ...


1

In the post just one certainly important aspect of Quantum field theory is described. One starts up with a linear theory and in order to see real physics coupling between the fields is needed. And as non-linear theories only in the rarest cases can be solved, one applies perturbation theory. However, Quantum field theory is more than that. There are a whole ...


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Perhaps you are confusing field operators in the Heisenberg picture and field operators in the interaction picture? We usually compute Green's functions in perturbative QFT by moving to interaction picture where the field operators obey the free field equation like a free field does in Heisenberg picture, and the interaction term acts on states like in ...


1

This type of radiative mass diagrams can be found in models like the Scotogenic, proposed by E. Ma, linking dark matter and neutrino mass. In simple words, it is just a 1-loop realization of the Weinberg operator: two leptons and two SM-Higgs bosons. You can find the details of the calculation of the same type of diagram for this model in Appendix A of this ...


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I looked at my personal notes and I've found a somewhat related principle, and only for vertices, not propagators (but unfortunately I don't know the source of these notes...) : Let $T^n_{S_I}[\{\phi_i\}_i]$ be a term in the interaction action involving $n$ fields. Express it as an inverse Fourier transform of the fields $\phi_i$. You then have $T^n_{S_I}[\{\...


4

Yes this theory is Borel summable even after taking the box $\Lambda$ to infinity. To learn about these things look at the pedagogical paper "Constructive field theory in zero dimension" by Rivasseau. Note that in principle to show a model is Borel summable, the naive procedure is to: 1) define the Borel transform, i.e., the function with power ...


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One particle is a source of an external (retarded) field for another one, and vice versa. Retarded fields are described with differential equations.


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There's no QFT without Green functions and much more, but something like your question is addressed by Mattuck's "drunken man propagator", followed by "the classical quasi particle propagator", in the introductory chapters of his book "A Guide To Feynman Diagrams In The Many-Body Problem".


1

Are you sure you want to sum the amplitude itself over spins? It is much more standard to perform a sum over spins after squaring the amplitude. And then it is immaterial whether the interaction contains $\gamma_\mu$, $\gamma_5$ or $\gamma^\mu\gamma_5$. The process you mention would lead to \begin{align} \mathcal{A} \propto \frac{(\bar{u}_2 \gamma_5 u_1)(\...


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There are already excellent answers posted here. Let me add two aspects: This was actually measured for the first time a few years ago at LHC https://www.nature.com/articles/nphys4208 (I was peripherally involved in a comparable CMS measurement) It makes sense to talk about attraction/repulsion only if the initial state particle at least survives the ...


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I am fully disclosing off the bat here that I do not know of a plausible way/gimmick to attribute an "attractive-versus-repulsive-force" feature to these amplitudes, a subject discussed amply on this site, as the relevant 2-to-2 scattering cross sections cannot be used for these purposes (C-conjugation). You might have to strain to define this ...


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