In "Cosmic strings and other topological defects" by A. Vilenkin and E. P. S. Shellard, section 2.1.3, they detail some examples of spontaneous symmetry breaking. In the first example, they discuss the Higgs field field transforming as a two-component spinor in the fundamental 2 representation of $SU(2)$:

$\sigma = \begin{pmatrix} \sigma_1\\ \sigma_2\\ \end{pmatrix}$

We then break this symmetry with a quartic potential. The little group $H$ is trivial, as there are no generators that can annihilate the vacuum expectation value. They go on to argue that $\sigma$ may couple to three gauge fields, all of which become massive after the symmetry breaking $SU(2) \to I$. Furthermore, they say it is clear that one massive Higgs boson will remain.

It is not clear to me why in this model there should be a massive Higgs, as earlier they detail the number of massive Higgs fields present will be $n - K$, where $n$ is the dimension of the real representation of G and $K$ is the number of massive gauge fields. In this case, I believe $n$ should be ($N^2 - 1 = 3$), hence no massive Higgs should emerge. Could someone explain please why in $SU(2) \to I$ symmetry breaking there will be a massive Higgs?

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    $\begingroup$ why dont you write the lagrangian of the this model and derive the mass spectrum? $\endgroup$
    – xi45
    Commented May 6, 2020 at 12:17

1 Answer 1


Your complex doublet field σ has 4 degrees of freedom (independent components). The number of (Goldstone bosons of SSB and hence) massive gauge fields is 3. The number of "uneaten" σ components is therefore 4-3=1. So one massive residual Higgs left.

(By contrast, the real 3-dimensional Higgs representation you seem to have erroneously considered in your second paragraph counting is the adjoint one. There, SU(2) → U(1) so K=2, and 3-2=1 massive Higgs again, as you have only two massive gauge bosons. It is a very different model.)

This should, of course, be evident in the standard treatment in most QFT textbooks detailing SSB.


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