How come Goldstone boson, PQWQ axion, be able to have mass at all? Quote:

Goldstone's theorem: For every spontaneously broken continuous symmetry, there is a massless particle created by the symmetry current. 

However, under $U(1)_{PQ}$ symmetry, I read that PQWW axion can obtain mass from $G\tilde{G}$. 
Both these sentence make sense, but, being a Goldstone boson, there seemed to be a contraction around PQWW axion.
I read that this might be connect with the approximation from t' Hooft's determinental interaction. Could you explain to me what the determinental  interaction was?
Does that mean Goldstone's theorem only works for the first order of t' Hooft determinental interaction?
Does PQWW boson has mass or not?
Could you explain to me why a Goldstone could every obtain a mass?
 A: Like the pions, the axion is a pseudo-Golstone boson, so it has a mass. The $U(1)_{PQ}$, as a chiral symmetry, is broken by quantum corrections by the axion's coupling to gluons, and so is no longer an exact symmetry. Goldstone's theorem them does not apply. This is completely analagous to the so-called '$U(1)$ problem', where the $\eta '$ was seen to be too heavy to be a Goldstone boson. The resolution of this puzzle by 't Hooft made physical the CP-violating $G\tilde{G}$ term in the Lagrangian, which is what suggests a QCD axions in the first place.
A: Two conditions for pseudo goldstone boson: 


*

*The broken symmetry is not gauged, otherwise the (pseudo)
goldstone boson will be eaten by the gauge field.  

*Something else (chiral anomaly in this case) breaks the symmetry
even before the spontaneous symmetry breaking is taking place.


To visualize, think about a tilted Mexican hat potential. The lateral move in the tilted Mexican hat groove would be uphills (non-zero mass), as opposed to being flat (massless) for the usual Mexican hat potential. The non-zero mass of the pseudo goldstone boson would be very small if there is only a tiny bit tilting.
