Karozo
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At the end of spontaneous symmetry breaking I get these mass terms:
W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )
\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}
So I have M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2}
Is it right? Or there are too many terms and it is enough:
\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}
W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )
\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}
So I have M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2}
Is it right? Or there are too many terms and it is enough:
\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}