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The Higgs mass does not stem from eating Goldstone bosons, since the Higgs is not a gauge field. Since we are breaking an $\mathrm{SU}(2) \subset \mathrm{SU}(2)_L \times \mathrm{U}(1)_Y$ completely, we have three Goldstone bosons, which are eaten by three of the four electroweak gauge bosons to form the massive $W^\pm,Z$ with the photon remaining massless. ...

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I think that you have to neglect the s-quark mass not its momenta in your expression. Anyway, if you perform the standard procedure for calculating this integral you will find an expression which is a function of all the possible external momenta and the metric tensor as well. Then you can carry the limit $p_s \to 0$ you will find the formula which are you ...

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It is mainly measurement and detector errors that make up the width in the plots you show. The Monte Carlo simulates the detector resolution and folds in the theoretical values when it says that the width agrees. The real width is expected to be much smaller. In this we see that the real width is only given as a bound by the experiments the CMS ...

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The spatial wavefunction is $Y_L^m(\theta,\phi)$. When exchanging the two particles, the spatial wavefunction becomes to $Y_L^m(\pi-\theta,\pi+\phi)$. Mathematically, we have $Y_L^m(\pi-\theta,\pi+\phi)=(-1)^L Y_L^m(\theta,\phi)$. If $L$ is odd, the spatial part is antisymmetric, otherwise symmetric. BTW, you may post this question as a comment of the ...

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