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GlashowWeinbergSalam problem with mass terms 
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#1
Aug1314, 06:11 PM

P: 4

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] 


#2
Aug1514, 01:37 PM

P: 1,041

The mass terms seem right.
You can always write it as the last + h.c. which is your 2nd term 


#3
Aug1514, 04:16 PM

P: 680

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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