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Consider an $SU(2)\times U(1)_Y\rightarrow U(1)_{EM}$ theory that is broken via a complex triplet with hypercharge 2. The potential is of the form

\begin{align} V(\Phi) = -m^2\Phi^\dagger\Phi + \lambda_1 (\Phi^\dagger\Phi)^2-\lambda_2\sum_a(\Phi^\dagger\mathcal{T}^a\Phi)(\Phi^\dagger\mathcal{T}^a \Phi), \end{align} where the $\mathcal{T}^a $ are the spin 1 matrices from QM & the identity for the hypercharge case. I want the Higgs masses. I received a hint to do the work in the six dimensional real representation, where the real antisymmetric generators take the form \begin{align} iT^a = \begin{bmatrix} -\text{Im } \mathcal{T}^a & -\text{Re} \mathcal{T}^a\\ \text{Re }\mathcal{T}^a& -\text{Im }\mathcal{T}^a\end{bmatrix} \end{align}

The charge matrix $Q = T^3+\frac{Y}{2}$ makes me employ a neutral vacuum (0,0,v,0,0,0). I found the gauge boson mass matrix and did a Weinberg-Salaam-esque diagonalization to find three gauge boson masses & 1 massless photons. I think this means there are three Goldstone bosons (for three broken generators).

If I translate to the real representation \begin{align} \Phi=\begin{bmatrix}\phi_1+i\phi_2\\\phi_3+i\phi_4\\\phi_5+i\phi_6\\\end{bmatrix}\rightarrow\phi=\begin{bmatrix}\phi_1\\\phi_2\\\phi_3\\\phi_4\\\phi_5\\\phi_6\end{bmatrix}. \end{align} and try to use the potential \begin{align} V(\phi) = -m^2\phi^T\phi + \lambda_1 (\phi^T\phi)^2+\lambda_2\sum_a(\phi^TiT^a\phi)(\phi^TT^a\phi), \end{align} the sum does not contribute because the $T^a$ are anti-symmetric. But then this is the linear sigma model. \begin{align} V(\phi) = -m^2\phi^T\phi + \lambda_1 (\phi^T\phi)^2 \end{align} I find the vacuum by finding a stationary point and get $v^2 =\frac{m^2}{4\lambda_1}$ and then expand about the vacuum $(\eta_1,\eta_2,\eta_3+v,\eta_4,\eta_5,\eta_6)$ and hope to get three massive terms. I only get one massive term of course (for $\eta_3$) but that cannot be right.

Does anyone see where I made a mistake?

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