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I know, the from the Higgs Mechanism, or Spontaneous symmetry breaking, the massless Goldstone boson becomes massive. So in some sense Goldstone bosons are eaten by gauge "boson".

Here I got confused the terminology about Goldstone bosons and Higgs boson. Can i say, that the in the Higgs field, the Goldstone bosons are eaten by higgs boson?

I found some statement about "Higgs bosons"

The Higgs boson is the quantum particle associated with the Higgs field just as the photon is the quantum particle associated with electromagnetic field

I know, the Higgs Mechanism explain the massive gauge boson, in Standard Model, thus the corresponding of above "boson" by higgs boson is plausialbe if the theory we try to explain lies in scalar field is Higgs field.

Is this right?

From @ACuriousMind, I summarized what i learned.

The terminology boson comes from Higgs field. Since Higgs field is scalar field the name boson comes from scalar (spin-0 : boson).

The massive procedure of Higgs boson is related with Higgs potential (In general, we choose Mexican hat-shaped potential, which is related with self-interaction term). And this is no related with gauge theory, (Higgs is not a gauge theory) but the related with the shape of potential. From breaking of potential's symmetry by properly adjusting higgs field, it became massive and this is how higgs boson get mass.

On the other hand, in the standard model, broken symmetry of gauge theory, reduce the massless Goldstone boson to be massive.

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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.

The Higgs mass stems from the self-interaction term $\propto (\phi^\dagger \phi)^2$ in the quartic potential of the Higgs, which produces, among other things, a mass term for the Higgs field $h$ after breaking as $\phi = v + h$ (and some gauge-fixing).

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  • $\begingroup$ why do you say "we are breaking an $SU(2)\in{}SU(2)_L\times{}U(1)_Y$ completely". Isn't it broken all of $SU(2)_L\times{}U(1)_Y$ except a $U(1)_{em}$ which is a combination of generators of $SU(2)_L$ and $U(1)_Y$? Do the broken generators also form a $SU(2)$? $\endgroup$
    – Yossarian
    Commented Jan 25, 2015 at 7:53
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    $\begingroup$ @silrfück: Yes. The $W^\pm$ and the $Z$ still act as if they were $\mathrm{SU}(2)$ bosons, though they are exactly the combinations of which you speak. I am quite sure they form a $\mathrm{SU}(2)$ subgroup of the electroweak group. $\endgroup$
    – ACuriousMind ♦
    Commented Jan 25, 2015 at 14:03

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