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?
