The Goldstone boson equivalence theorem tells us that the amplitude for emission/absorption of a longitudinally polarized gauge boson is equal to the amplitude for emission/absorption of the corresponding Goldstone boson at high energy. I'm wondering what's the physical meaning of this theorem. Is there any relation between equivalence theorem and Higgs mechanism ?


Probably after 3 years you've answered this question, but it might be useful for others. I think you got it. At high energies, you can choose to view the spectrum as a heavy vector gauge boson with 3 degrees of freedom (3 independent polarizations) or view the spectrum as a massless vector gauge boson (2 degrees of freedom) with an extra massless Goldstone boson (1 degree of freedom) for a total of 3 degrees of freedom. The Higgs mechanism is exactly this process. You have a global $SU(2)_L \times U(1)_Y$ symmetry that in the standard model which gets broken to $U(1)_{EM}$. $SU(2)$ has 3 generators, and $U(1)$ has 1 generator. So at the end of the day, it would seem as if you lose 3 degrees of freedom, but not really, as you also produce 3 massless Goldstone bosons. These in turn give mass terms to W+ W- and Z bosons. At high energies, you can think of W+ W- and Z as massive, or as massless with these 3 goldstone bosons.

Practically, the advantage of all this is that although in a cross section calculation, thinking of massless Goldstone bosons adds more diagrams to a given process, they are much easier to calculate.

  • $\begingroup$ Any comment on the last part of the question (regarding the connection between equivalence principle (not GBET) and the higgs' mechanism) ?? $\endgroup$ – negligible_singularity Nov 17 '16 at 10:21
  • $\begingroup$ @Gabriel Good answer, but you got the masses wrong. In GBET both gauge and Goldstone bosons are massive. A more precise statement of Goldstone Boson Equivalence theorem reads: at high energies, the amplitude for a longitudally polarised massive gauge boson equals the amplitude of the eaten Goldstone boson with the same mass. $\endgroup$ – Ramtin Jan 11 at 9:27

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