# Two-field Symmetry Breaking unitary gauge

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?

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 $$\tfrac{ 1}{2} \left (ev A_\mu- \partial_\mu(\cos\theta ~~\xi_1+ \sin\theta ~~\xi_2 ) \right )^2 +\tfrac{1}{2}\left (\partial_\mu(\sin\theta ~~\xi_1-\cos\theta ~~ \xi_2)\right )^2 .$$