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At the end of spontaneous symmetry breaking I get these mass terms:

$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp i W_{\mu}^{2} \bigr )$$

$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

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:

$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$

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  • $\begingroup$ "Or there are too many terms and it is enough:" What did you mean? $\endgroup$
    – Andrew McAddams
    Commented Aug 14, 2014 at 9:20
  • $\begingroup$ Does $\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$ give mass only to $M_{W^-}$ o $M_{W^+}$ or both? $\endgroup$
    – Karozo
    Commented Aug 14, 2014 at 10:08

1 Answer 1

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Notation $W^{-}, W^{+}$ may confuse in a sense that it may seem that here are two different particles which aren't connected by charge conjugation. But of course, $W^{+}$ is only $(W^{-})^{\dagger}$, so it is an antiparticle to $W^{-}$. So term $( W^{-} \cdot W^{+} )$ is simple $|W|^{2}$ (which is standard for the mass-term), and, of course, both of particle and antiparticle have the equal masses.

Also before making substitution $$ \tag 1 W^{\pm}_{\mu} = \frac{1}{\sqrt{2}}(W_{\mu}^{1} \mp iW_{\mu}^{2}) $$ you can see that both of fields $W^{1}, W^{2}$ have equal masses. So of course that their linear combinations $(1)$ also have equal masses.

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  • $\begingroup$ You've copied the error of missing $i$ in front of $W^2_\mu$, didn't you? $\endgroup$
    – Luboš Motl
    Commented Aug 14, 2014 at 11:14
  • $\begingroup$ @LubošMotl : yes, you're right. $\endgroup$
    – Andrew McAddams
    Commented Aug 14, 2014 at 11:15
  • $\begingroup$ Thank you a lot, I have a last question. A term like $\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $ gives a mass of $ M_{W^+}=g \frac{v}{2}$ or a mass of $ M_{W^+}=g \frac{v}{2\sqrt{2}}$. $\endgroup$
    – Karozo
    Commented Aug 14, 2014 at 14:13
  • $\begingroup$ @Karozo : The mass traditionally fixes by the equation of free motion (which coincides with Klein-Gordon equation) $(\partial^{2} + m^{2})W = 0$. If you "neglect" the Higgs-interaction term you will give, by using Euler-Lagrange equation for $W$ or for $W^{\dagger}$. So for your notation the mass will be equal to $\frac{g v}{2\sqrt{2}}$ (note that I don't know the multiplying factor of the kinetic term in your notation). $\endgroup$
    – Andrew McAddams
    Commented Aug 14, 2014 at 14:24
  • $\begingroup$ Here I use the convention $$ L = -\frac{1}{4}G_{\mu \nu}^{a}G^{\mu \nu}_{a} - \frac{1}{4}F^{\mu \nu}F_{\mu \nu} - |D_{\mu} \varphi |^{2}, $$ $$ G_{\mu \nu}^{a} = \partial_{\mu}W_{\nu}^{a} - \partial_{\nu}W_{\mu}^{a} + \varepsilon^{abc}W_{\mu}^{b}W_{\nu}^{c}, \quad F_{\mu \nu} = \partial_{\mu}B_{\nu} - \partial_{\nu}B_{\mu} , $$ $$ D_{\mu} = \partial_{\mu} - \frac{ig}{2}(\tau \cdot A_{\mu}) - \frac{ig_{1}}{2}B_{\mu}. $$ $\endgroup$
    – Andrew McAddams
    Commented Aug 14, 2014 at 14:34

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