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I have asked a similar unanswered question, focused on different aspects, here.

My intuition regarding statements frequently made about this issue is that, if a theory contains a breaking pattern $G \to H$ and a breaking scale $v$, then at energies $E\gg v$, the model should behave phenomenologically as if the symmetry was intact.

This should imply that the propagating eigenstates are massless, for instance, the exact symmetry eigenstates, the gauge vectors.

Is this true? How does this happen in practice?

I could carry the following simple line of thought. Write the 2-Green Function of the theory, the position-space propagator:

$$ G \sim \left[-g^{\mu \nu}(\Box-m^2)+\left(1-\frac{1}{\xi}\right)\partial^\mu \partial^\nu \right]^{-1}, $$

and expand for

$$ \frac{v}{E} \sim \frac{m}{E} \ll 1, $$

to get

$$ G \sim - \frac{g^{\mu \nu}}{\Box} \left[ 1 + \left(1-\frac{1}{\xi} \right) \frac{\partial^\mu \partial^\nu}{\Box}\right]+\mathcal{O}\left(\frac{m^2}{\Box}\right), $$

which, to lowest order, is the propagator for a massless vector.

Does this have anything to do with anything? If yes, what happens to the longitudinal polarization?

Qmechanic
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GaloisFan
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  • relevant https://physics.stackexchange.com/questions/126497/does-the-higgs-boson-give-mass-to-all-other-particles and duplicate link – anna v Apr 30 '22 at 03:55
  • This review https://mdpi-res.com/d_attachment/symmetry/symmetry-12-01316/article_deploy/symmetry-12-01316.pdf?version=1596710310 The gauge symmetry is exact only when the particles are massless. One needs a subtle breaking of the symmetry, providing the origin of mass without affecting the excellent description of the interactions. – anna v Apr 30 '22 at 03:59
  • @annav thanks for the comment! I understand that the bosons are only massless if the symmetry is exact. My question is a technical one: effectively speaking, does the symmetry looks unbroken at energies way higher than the breaking scale? I'd like to understand this mathematically. – GaloisFan May 02 '22 at 22:47
  • Would not that be true mathematically by construction when the energy is higher than the symmetry breaking? – anna v May 03 '22 at 03:59
  • @annav but that is semantics. I would like to see where it is hidden in the math, even if it works by construction -- how can we see that, for any observable at a high energy scale, we may use the unbroken lagrangian? How does one prove that? How does the transition happen? How does this transition depend on the characteristic scales ratio? – GaloisFan May 05 '22 at 17:16

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