Glashow-Weinberg-Salam problem with mass terms

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SUMMARY

The discussion centers on the mass terms derived from spontaneous symmetry breaking in the context of the Glashow-Weinberg-Salam model. The mass terms for the W bosons are expressed as \( M_{W^+} = g \frac{v}{2} \) and \( M_{W^-} = g \frac{v}{2} \), confirming their equivalence. Participants agree that the mass Lagrangian can be simplified to \( \mathcal{L}_{mass} = \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-} W^{\mu}{}^{+} \) without losing essential information. The consensus is that the additional terms are redundant due to the commutation of W fields.

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  • Understanding of spontaneous symmetry breaking in particle physics
  • Familiarity with the Glashow-Weinberg-Salam model
  • Knowledge of gauge bosons and their mass generation
  • Proficiency in Lagrangian mechanics in quantum field theory
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  • Study the implications of spontaneous symmetry breaking in the Standard Model
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The discussion is beneficial for theoretical physicists, students of particle physics, and researchers focusing on the Standard Model and mass generation mechanisms in quantum field theory.

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}}^{+}
 
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The mass terms seem right.
You can always write it as the last + h.c. which is your 2nd term
 
Well, the W fields commute so there really is no point in writing it as two terms or with hc., just add a 2.
 

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