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I've been reading chapter 20 in Peskin and Schroeder about the Glashow-Weinberg-Salam Theory for a $U(1)$ gauge symmetry with gauge coupling strengths $g$, and a complex scalar $\phi$, specifically section 2, which Higgs boson and its couplings to the massive gauge boson and fermions. The scalar potential of course is the usual $$V(\phi) = -\mu^2\phi^\dagger \phi + \lambda (\phi^\dagger \phi)^2 \text{ for } \mu^2>0$$

Now suppose we have one of the quiver gauge theories, say a gauge symmetry $U(1)_1 \times U(1)_2$, with gauge coupling strengths $g_1$ and $g_2$, and let's say we know that the complex scalar has the same charge +1 under $U(1)_1$ as well as under $U(1)_2$. Firstly, how would I write out the covariant derivative for this? What happens to the masses of the two gauge bosons/ the Higgs boson? How would I go about writing the couplings between the Higgs and the gauge bosons? Can somebody explain or point me to a textbook which looks at this scenario?

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If I understand your problem, there should be an easy solution. The potential part of the Higgs field stays the same since it only depends on the Higgs. What changes is the kinetic term which would be of the form

$$|D_\mu \phi|^2\longleftarrow D_\mu = \partial_\mu-ig_1A_\mu^1-ig_2A_\mu^2$$

where $A_\mu^1, A_\mu^2$ are the gauge fields of the $U(1)_1\times U(1)_2$ gauge group. Given this, the usual symmetry breaking pattern can be taken where

$$\langle\phi\rangle = \left(\begin{array} &0 \\ v \end{array}\right)$$

From the kinetic term you'll get the masses of the gauge fields

$$|D_\mu\phi|^2 \rightarrow (\partial_\mu-ig_1A_\mu^1-ig_2A_\mu^2)(0\quad v) (\partial^\mu+ig_1A^{1\mu}+ig_2A^{2\mu})\left(\begin{array} &0 \\ v \end{array}\right)=v^2(ig_1A^{1}_\mu+ig_2A^{2}_\mu)^2 = -v^2g_1^2(A_\mu^1)^2-v^2g_2^2(A_\mu^2)^2-2v^2g_1g_2A_\mu^1A^{2\mu}$$

This can be diagonalized

$$v^2(A^{1\mu}\; A^{2\mu})\left(\begin{array}. g_1^2 & g_1g_2 \\ g_1g_2 & g_2^2 \end{array}\right)\left(\begin{array}. A_\mu^1 \\ A_\mu^2\end{array}\right)$$

taking the non-diagonal matrix

$$\det(M-\lambda I) = (g_1^2-\lambda)(g_2^2-\lambda)-g_1^2g_2^2 = \lambda^2-\lambda(g_1^2+g_2^2)=\lambda(\lambda-g_1^2-g_2^2)=0 $$

Therefore

$$\lambda_1\equiv m_1=0\qquad \lambda_2\equiv m_2=g_1^2+g_2^2$$

So you will have one massless gauge boson, linear combination of the two initial ones, and one massive gauge boson, given by another linear combination of the initial ones. The symmetry breaking pattern is $U(1)^2\rightarrow U(1)$

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