When the gauge symmetry of our Lagrangian breaks spontaneously through the Higgs mechanism, we usually find that $n$ Higgs degrees of freedom become massless through the vacuum expecation value (vev), where $n$ is the number of broken generators. This means if our original gauge group has $N$ generators, the group that leaves the vev invariant has only $N'=N-n$ generators. The $n$ broken generatore become massive through the Higgs vev and in turn we can see that the mass terms for $n$ Higgs degrees of freedom vanish if we put in the vev. Often one says the Goldstone boson get eaten by gauge bosons, which then become massive.
What exactly happens to these massless degrees of freedom in the Higgs potential after symmetry breaking, i.e. after we expand the Higgs fields about the vev? **The mass-terms= quadratic terms vanish as noted above if we put in the vev, but, in general, there are several quartic terms possible. These would describe interactions of the Goldstone bosons with the Higgs bosons and surely there must be a good way to see that these vanish, too or have no influence for some reason?
Best Answer
First, note that, strictly speaking, there is no such thing as spontaneous symmetry breaking in Higgs mechanism. I mean, that below and under the Higgs scale (i.e., at scale, at which non-zero Higgs VEV appears) the lagrangian can be rewritten in a gauge invariant way. How is it possible? The answer is that there are different physical states (i.e., eigenstates of hamiltonian) below and under the Higgs scale. Under the Higgs scale eigenstates of hamiltonian form also representations of the gauge group. But below it eigenstates of hamiltonian don't form representations of the gauge group. These eigenstates are linear combination of transverse and longitudinal degrees of freedom.
So, in some sence, nothing happens with three scalar fields in Higgs doublet. They just don't appear as physical states below the Higgs scale.