How $W^{\pm}$ bosons get their electric charge $\pm 1$ as opposed to $Z_0$ that have neutral electric charge?

In the 'Standard Model' book by Y. Grossman and Y. Nir, in chapter 7 (the leptonic standard model) on page 93 after defining the charge of the broken symmetry generator, i.e $$Q=T_3+Y$$

they say that the $$W_{\mu}^{\pm}$$ are eigenstates of $$T_{3}$$ with eigenvalues $$\pm1$$, and have Y = 0. Hence, their electromagnetic charges are: $$q(W_{\mu}^+ ) = +1; q(W_{\mu}^{-} ) = -1$$

Both $$W_{3\mu}$$ and $$B_{\mu}$$ are eigenstates of $$T_3$$ with eigenvalue 0, and have Y = 0. Hence, they are neutral under $$U(1)_{EM}$$ and so is any linear combination of them: $$q(Z^0_{\mu}) = 0; q(A^0_{\mu}) = 0$$

My question is how they found those eigenvalues. What are the matrices represent $$W_{\mu}^{\pm}$$,$$W_{3\mu}$$ and $$B_{\mu}$$ I need to diagonalise and find those values?

• Write down the weak isotriplet of the W s, the three T generators of SU(2) in the triplet representation, and see how these three matrices act on your triplet. As an isosinglet, $B_\mu$ is defined to be a number not acted upon by these matrices. You may represent Y as the 3x3 identity matrix. Commented Jun 15, 2019 at 14:49