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unrenormalized vertex diagram and unrenormalized self-energy diagram are as below: Now, let me summarize what Peskin and Schroeder try to express. Feynman propagators for electron $$ S_F=\frac{i}{\require{cancel}\cancel p-m} $$ end photon $$ D_F=\frac{-ig_{\mu\nu}}{q^2} $$ and propagator must hold $$ S_F(p)S_F^{-1}=\mathbb{1} $$ simply differentiate ...


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Your statement that the integral "is actually" $\propto g^{\mu\nu}\Pi$ is incorrect because you can clearly see the $q^\mu q^\nu$ in the numerator of the integrand. The correct expression that you have in your final equation is not a choice, it is the result of calculating the loop using dimensional regularization. Finally, indeed $q_\mu \Pi_2^{\mu\nu}$ is ...


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The CP phase in the strong sector comes from a topological term (a total derivative) $$\require{cancel} \mathcal L_{\cancel{CP}} \propto \theta_\mathrm{QCD} \epsilon^{\mu \nu \rho \sigma} G_{\mu \nu} G_{\rho \sigma}, $$ where $G_{\mu \nu}$ is the gluon field-strength. Such an operator can never be produced as an effective operator from weak interactions, ...


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First I'll address the last question you mentioned regarding commutativity in the path integral. One benefit of the path integral formulation is that the integrand behaves like an ordinary complex or Grassman number---if you're computing the trace of a product of matrices in Einstein notation, you can rearrange terms in the product as long as you keep track ...


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Do particles with exactly zero energy exist? No. if something has no mass it cannot be said to "exist" since it cannot possibly have energy or momentum and thus cannot participate in interactions or be detected. A photon has no mass, but it does have energy-momentum, it does participate in interactions, and it can be detected. It exists. ...


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The emission of massless particles (e.g. photons) with zero momentum (or momentum tending towards zero) in the rest frame of a charged particle is called collinear emission. Collinear emission is somewhat problematic for massless particles, because it results in a so-called IR divergence that cannot be removed by renormalization (cf. UV divergences). The ...


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The concept of a particle with exactly zero energy is rigorously meaningless. The issue is that the quantum field is not an operator, it is an operator-valued distribution. Therefore, strictly speaking, you can't apply $\phi(x)$, $a(p)$ or $a^\dagger(p)$ to anything, but you have to smear these things out. Strictly speaking, $\phi(x)$ doesn't even mean ...


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Contrary to what the language implies, spontaneous symmetry breaking in usual vacuum QFT is not something that "occurs at a scale". Rather, the symmetry breaking scale is the scale below one may not pretend that the symmetry is whole (and e.g. the gauge bosons massless), see this answer. Speaking intuitively, the symmetry breaking scale is the scale below ...


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I don't think it is exactly the same regulator: In the first method, you integrate $\int_{-\infty}^\infty dk^0 \int^\Lambda d^3k$, but in the second calculation you integrate $\int^\Lambda d^4 k_E$.


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Yes, $\mu$ can be anything. Usually in renormalization, we measure (or define) the coupling constant $g$ at scale $\mu$, and then use this information to predict the coupling constant $g'$ at another scale $\mu'$. We require that $g'$ at $\mu'$ is independent of $\mu$. What I mean is that you should get the same $g'$ at $\mu'$ even if you use another $g$ ...


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If physics isn't an issue, you can add arbitrarily many terms. Once the physics comes in though, you will encounter a few restrictions : As said by Gennaro, it is assumed that the Poincaré symmetry applies. Higher derivative terms (second derivatives and above) are generally bad news. They can cause vacuum instability (energies can be arbitrarily ...


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1.why is renormalization even necessary? It is because, experiments force us to do that. You suppose to get finite values i.e, for charge or for mass of a particle. 2.if we always get a single finite result, why do we care that there are infinities in the equations at the level of amplitudes? Actually we don't. Your integrals in the loop ...


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Actually there are no infinities in physics. These "infinities" come from integrating over loop momenta when the momentum goes to infinity. But we don't know what actually happens to particles when they have infinite momentum. Maybe they turn into strings, maybe loop quantum gravity and black holes become relevant, maybe there's infinitely small unicorns ...



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