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  How do the massive gauge bosons eat up the would-be Goldstone bosons?

+ 1 like - 0 dislike
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In the literature one often reads that the would-be Goldstone bosons of the spontaneously broken $SU(2)\times U(1)$ symmetry are *eaten up* by the longitudinal polarisation vectors of the now massive gauge bosons. While I find this explanation pictorially nice, I want to see the maths behind it.

My idea is that the other DOFs of the Higgs $SU(2)$ doublet, let us call them $w,z,\zeta$, will become part of the $W^\pm$ amd $Z$ fields after reparametrisation. Is that true?

The literature usually writes out $(D_\mu\langle\Phi\rangle_0)^\dagger D^\mu\langle\Phi\rangle_0$ first and then reparametrises the fields as
$$W^\pm_\mu=\frac{1}{\sqrt{2}}\left(W^1_\mu\pm iW^2_\mu\right), Z_\mu=\cos\theta_WW^3_\mu-\sin\theta_W B_\mu.$$

Is it that one has to write out the full covariant derivate square of the full Higgs doublet? I mean this:
$$D^\dagger_\mu\begin{pmatrix}w-iz\\ v+H-i\zeta\end{pmatrix}^T D^\mu\begin{pmatrix}w+iz\\ v+H+i\zeta\end{pmatrix}.$$

asked Mar 5 in Theoretical Physics by twening (65 points) [ no revision ]

Isn't it just a variable change in the "now" theory?

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