I just realised that I don't remember how to conclude theoretically that $W^{\pm}$ bosons have electric charges of $\pm 1$. After some searching I'm quite surprise that I couldn't find a satisfying answer, most literature that I came across simply just stated the result like it is trivially true (which it probably is).

So I came up with my own reasoning which I'm not really confident that it's right. So if someone could check the correctness, spot the error or suggest a more standard way of answering this question that would be great. So electric charge is defined by \begin{equation} Q = T_3 + \frac{Y}{2}, \end{equation} \begin{equation} T_3 = \frac{1}{2}\begin{pmatrix} 1 & 0 \\[0.3em] 0 & -1 \end{pmatrix} \end{equation} and $W$ Bosons live in the adjoint representation of $SU(2)_L$. That is \begin{equation} \textbf{W}_{\mu} = \begin{pmatrix} W^3_{\mu} & \sqrt{2} W^+_{\mu} \\[0.3em] \sqrt{2} W^-_{\mu} & -W^3_{\mu} \end{pmatrix} \end{equation} will transform as $\mathbf{W}_{\mu} \rightarrow \mathbf{W'}_{\mu} = \exp(i\alpha_i T_i)\mathbf{W}_{\mu}\exp(-i\alpha_i T_i)$ under global transformation $\exp(i\gamma_i \lambda_i)\exp(i\alpha_i T_i)\exp(i\beta Y)$ in $SU(3)_C \times SU(2)_L \times U(1)_Y$. So if I consider applying a global transformation $\exp(iQ\theta)$ from $U(1)_{EM}$ subgroup, where $\theta$ is some constant, onto a particle with known electric charge such as an electron, I would get $e^- \rightarrow e^{-i\theta}e^-$. At the same time $\mathbf{W}_{\mu}$ will transform as \begin{eqnarray} \mathbf{W}_{\mu} \rightarrow \mathbf{W'}_{\mu} &=& \exp(i\theta T_3)\mathbf{W}_{\mu}\exp(-i\theta T_3) \\ &=& \begin{pmatrix} e^{i\theta/2} & 0 \\[0.3em] 0 & e^{-i\theta/2} \end{pmatrix}\begin{pmatrix} W^3_{\mu} & \sqrt{2} W^+_{\mu} \\[0.3em] \sqrt{2} W^-_{\mu} & -W^3_{\mu} \end{pmatrix}\begin{pmatrix} e^{-i\theta/2} & 0 \\[0.3em] 0 & e^{i\theta/2} \end{pmatrix} \\ &=& \begin{pmatrix} W^3_{\mu} & \sqrt{2} e^{+i\theta} W^+_{\mu} \\[0.3em] \sqrt{2} e^{-i\theta} W^-_{\mu} & -W^3_{\mu} \end{pmatrix} \end{eqnarray} So $W^{\pm}_{\mu}$ transform according to $W^{\pm}_{\mu} \rightarrow e^{\pm i\theta}W^{\pm}_{\mu}$. Since the coefficient of $\theta$ in the exponential are the same as that of electron transformation I conclude that the relative strength of interaction with photon of $W^{\pm}$ is the same of that of electron and so the magnitude of electric charges must be the same. Therefore $W^{\pm}$ have electric charges of $\pm 1$.

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    $\begingroup$ What you wrote is correct: $W^{\pm}$ acquire their charges through the electroweak mixing. I don't see a question, though :) $\endgroup$ – Prof. Legolasov Jan 13 '15 at 5:48
  • $\begingroup$ Thank you for your answer. The question was just whether the idea is right? (it's at the beginning of second paragraph) Sorry I didn't make it more apparent. $\endgroup$ – user113988 Jan 13 '15 at 5:57
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    $\begingroup$ What would be better is if you edit the question to just ask how to theoretically determine the charges, then put the argument for figuring it out in an answer. (You can answer your own questions.) While you're at it, it wouldn't hurt to expand on where you searched. $\endgroup$ – David Z Jan 13 '15 at 6:09

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