- #1
Karozo
- 4
- 0
At the end of spontaneous symmetry breaking I get these mass terms:
[itex]W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )[/itex]
[itex]\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}}^{+} [/itex]
So I have [itex] M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} [/itex]
Is it right? Or there are too many terms and it is enough:
[itex]\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} [/itex]
[itex]W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp W_{\mu}^{2} \bigr )[/itex]
[itex]\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}}^{+} [/itex]
So I have [itex] M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} [/itex]
Is it right? Or there are too many terms and it is enough:
[itex]\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} [/itex]