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EDIT : I am trying to figure out the effect of symmetry breaking in a $U(1)_Y\times U(1)_Z$ invariant lagrangian where $U(1)_Y$ is local symmetry of the Lagrangian and $U(1)_Z$ is a global symmetry of it. For that I first take a gauge theory with local $U(1)_Y$ invariant Lagrangian: $$(D_\mu\chi)^*(D^\mu\chi)-\frac{\mu^2}{2}(\chi^*\chi)-\frac{\lambda}{4}(\chi^*\chi)^2$$ where $\chi=\chi_1+i\chi_2$ is complex scalar field. After spontaneous symmetry breaking this theory will have no Goldstone bososn in the unitary gauge. If I introduce a fermion $f$ and its conjugate $f^c$ in the theory and add to the Lagrangian a term $h\overline{f^c}f\chi$ with the assignment $Y=-2$ and $Y=1$ for $f$ and $\chi$ respectively then this Lagrangian continues to remain local $U(1)_Y$ invariant.

Next I impose an additional $U(1)_Z$ global symmetry by assigning $Z=-2$ and $Z=1$ for $f$ and $\chi$ respectively. We know when $\chi$ acquires a VEV the symmetry $U(1)_Y\times U(1)_Z$ is broken. The goldstone boson associated with breaking of $U(1)_Y$ disappears in the unitary gauge. How to find out what happens to the goldstone boson related to $U(1)_Z$ global symmetry breaking in this case?

How to start the mathematical analysis? I tried to write $\chi=v+\eta+i\xi$. Then I found that $\xi$ is absorbed in the unitary gauge. But shouldn't there be another real or physical goldstone bososn surviving due to $U(1)_Z$ global symmetry breaking? This question has stuck me while reading this paper THIS PAPER and in trying to figure out how could Majorons be massless. I am stuck to to prove that there will be unabsorbed Goldstone bosons.

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    $\begingroup$ Well, why don't you try working it out yourself first? Please see our homework(-like) questions policy. $\endgroup$
    – JamalS
    Commented Feb 13, 2015 at 7:29
  • $\begingroup$ @ JamalS- I have edited and reframed the question. I believe I have managed to explain where exactly I am stuck. $\endgroup$
    – SRS
    Commented Feb 14, 2015 at 18:42
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    $\begingroup$ This is a rare example of a question being revised in a thorough way to make it much better suited for this site. Good job, @SRS! $\endgroup$
    – Danu
    Commented Feb 15, 2015 at 8:23
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    $\begingroup$ Maybe I'm being silly but I don't quite see what differentiates your two U(1) groups from one another since all particles in your model have the same charges with respect to each. Goldstone's theorem tells you there is a massless boson for each broken generator. In this case you seem to have exactly one broken generator despite there being two groups. You can even envision formulating this on a lattice and since U(1) is abelian the plaquettes factorize and the theory becomes identical to one with a single U(1). Again, perhaps I'm just being stupid, but I can't see what I'd be missing. $\endgroup$
    – Leandro M.
    Commented Feb 20, 2015 at 5:43

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I know the thread is a little bit old, but I'm dealing right now with the same kind of things.

I think the answer is simpler than all this. You have a U(1) global symmetry, not local, so there is no gauge invariance. You can't take the unitary gauge because it's not just a phase, it's a real field and depends on the point you consider.

I hope it helps, and if you already solved it, maybe you can correct me if I'm wrong.

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