Two-field Symmetry Breaking unitary gauge

Question

Let's consider the following theory:

$$L= -\frac{1}{4}F_{\mu \nu}F^{\mu\nu} +{1\over 2} |D_\mu \Phi|^2 +{1\over 2}|D_\mu \chi|^2 + \lambda_1\bigl(|\Phi|^2-\frac{v_1^2}{2}\bigr) +\lambda_2\bigl(|\chi|^2-\frac{v_2^2}{2}\bigr)$$

where $\Phi$ and $\chi$ are complex scalars coupled to a $U(1)$ gauge boson $A_\mu$ through the usual covariant derivative: $$D_\mu \Phi= (\partial_\mu -ieA_\mu)\Phi $$ $$D_\mu \chi= (\partial_\mu -ieA_\mu)\chi $$

Expanding each scalar around its vev, we find $$L= \dots + \frac{1}{2}e^2(v_1+h_1(x))^2(A_\mu-\frac{1}{ev_1}\partial_\mu\xi_1(x))^2 +\frac{1}{2}e^2(v_2+h_2(x))^2(A_\mu-\frac{1}{ev_2}\partial_\mu\xi_2(x))^2$$

Where $h_i(x)$ are the Higgs-like bosons and $\xi_i(x)$ the respective Goldstone bosons. How must gauge invariance be used in this case to reproduce the unitary gauge?

Answer 1

I presume you take $$ \Phi = (v_1+h_1) e^{i\xi_1/v_1},\\ \chi = (v_2+h_2) e^{i\xi_2/v_2}, $$ for dimensional consistency, with $\langle h_i\rangle=0=\langle \xi_i\rangle$. The potential is flat w.r.t. the $\xi_i$s. Ignore the Higgses $h_i$ at first.

The remaining piece of the Lagrangian is $$ \frac{1}{2}e^2[ v_1^2(A_\mu-\frac{1}{ev_1}\partial_\mu\xi_1 )^2 + v_2 ^2(A_\mu-\frac{1}{ev_2}\partial_\mu\xi_2 )^2] . $$ Now define $$ v\equiv \sqrt{v_1^2+v_2^2}, \qquad v_1\equiv v\cos \theta , ~~~~v_2\equiv v\sin \theta , $$ to get $$ \tfrac{ 1}{2} \left (ev A_\mu- \partial_\mu(\cos\theta ~~\xi_1+ \sin\theta ~~\xi_2 ) \right )^2 +\tfrac{1}{2}\left (\partial_\mu(\sin\theta ~~\xi_1-\cos\theta ~~ \xi_2)\right )^2 . $$

So, one combination of the goldstons is absorbable in the unitary gauge, and vanishes from the picture, having given the photon a mass; while the orthogonal one not: it is a surviving massless scalar. Proceed to rename them and reinsert the Higgses...