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14

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 ...

2

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 ...

1

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 ...

1

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 ...

1

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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