It turns out that the lagrangian is fundamentally changed by the addition of the coupling. The field φ_1 becomes "redundant" in the sense that only the value of
B = A - gradient(φ_1)
is relevant in the equation, and we can freely choose φ_1 without affecting the physics of other fields. We say that φ_1 has become redundant.
How is it possible that φ_1 was very relevant before the coupling, but we can discard it after the coupling is done?
/\/\/\/\/\ a
/\/\/\/\ b
A somewhat similar thing happens if we have two springs A and B which exert a constant push force F. Let the length of A be a and the length of B be b. If we squeeze the springs separately, then both a and b are relevant in the lagrangian of the system.
/\/\/\/\/\/\/\/\/\/\/\ a + b
But if we attach the springs together, then only the value of a + b is relevant, not a or b any more. We can freely choose a and b, as long as a + b stays the same.
The Goldstone boson of the uncoupled Higgs model simply does not exist in the coupled model. It would still exist if the coupling were very weak. But the coupling is strong enough, so that the Goldstone field is not relevant at all.
That is what "eating" the Goldstone boson means. The value B ate both the field A and the Goldstone field.
What if the gauge field A is coupled to a fermion?
Let the field A in the Higgs 1964 paper be coupled to a fermion, say, to the electron.
Let us replace A by B in the lagrangian of the fermion, let that be L(B). Let us find a solution ψ for the wave function of the fermion in L(B).
We can make a solution for the old lagrangian with A, by multiplying ψ with a phase factor:
exp(i e φ_1) ψ
for an arbitrary φ_1. What did we show? We showed that we still can choose φ_1 arbitrarily and have a solution, without affecting the physics seen by outside observers. An observer cannot measure the absolute phase factor of ψ. The Goldstone field stays redundant even when A is coupled to a fermion.
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