Custodial symmetry and Higgs-Kibble

Question

In the context of Higgs mechanism only on $SU(2)_L$ model without the hypercharge, one writes the lagrangian with traces also for the Higgs, i.e. $$ \cdots+\text{Tr}[(D_\mu H)^\dagger D^\mu H)]-\frac{\lambda}{4}\Big(\text{Tr}[H^\dagger H]+\cdots $$ and not only for the $W_\mu$ $$ -\frac{1}{2}\text{Tr}[W_{\mu\nu}W^{\mu\nu}]. $$ I think this happen because of the custodial symmetry ($SU(3)_{\text{custodial}}$) that acts on $H$ as $$ H\rightarrow\gamma H\gamma^\dagger $$ and on $W_\mu$ as $$ W_\mu\rightarrow \gamma W_\mu\gamma^\dagger. $$ In fact it is easy to check that the vector bosons part is invariant with respect to the symmetry, but now my question is: How the same symmetry act on $H^\dagger$?

Answer 1

I am basically repeating my "geeky footnote" answer reviewing the Longhitano magic-hat trick at the heart of your question, namely the recasting of a complex Higgs doublet $\Phi$ into a complex 2×2 matrix $H$, cf. his thesis paper of 1981, into the exponential Gürsey realization for a vector symmetry.

Longhitano starts from the standard Higgs weak left-isodoublet
$$ \Phi = \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix}\equiv \frac{1}{\sqrt 2} \begin{pmatrix} \varphi_1-i\varphi_2 \\ \sigma +i\chi \end{pmatrix}, \\ \Phi \mapsto e^{i\vec{\alpha}\cdot \vec{\tau}/2} \Phi ~.$$ The remnant physical Higgs is $\sigma$, picking up the v.e.v. and splitting off the custodial vector SU(2) triplet of goldstons.

The conjugate doublet is also a left isodoublet,
$$ \tilde \Phi =i\tau_2 \Phi^*= \begin{pmatrix} \phi^{0~~*} \\ -\phi^- \end{pmatrix} ,\\ \tilde \Phi \mapsto e^{i \vec{\alpha}\cdot \vec{\tau}/2}\tilde \Phi ~.$$

Now, your Higgs matrix is defined as a side-by-side juxtaposition of these two left-doublets serving as columns, $$ H\equiv \sqrt{2}(\tilde\Phi, \Phi)= \sqrt {2} \begin{pmatrix} \phi^{0~~*} &\phi^+ \\ -\phi^- & \phi^0 \end{pmatrix}. $$

It is then evident that its transform by left α and right β isorotations is
$$ \bbox[yellow]{ e^{i\vec{\alpha}\cdot \vec{\tau}/2} \sqrt{2}(\tilde\Phi , \Phi )e^{i\vec{\beta}\cdot \vec{\tau}/2} = e^{i\vec{\alpha} \cdot \vec{\tau}/2}\sqrt {2} \begin{pmatrix} \phi^{0~~*} &\phi^+ \\ -\phi^- & \phi^0 \end{pmatrix}e^{i\vec{\beta}\cdot \vec{\tau}/2} = e^{i\vec{\alpha}\cdot \vec{\tau}/2} H e^{i\vec{\beta}\cdot \vec{\tau}/2} }. $$ Visibly, the left and right isorotations are oblivious to each other: a scrambling of the untilded and tilded $\Phi$s effected by the right isorotation does not affect their left-rotation properties.

As is standard in $SU(2)_L\times SU(2)_R$ chiral dynamics, the choice $\vec{\alpha}= -\vec{\beta}$ specifies the vector isospin custodial subgroup, which you parameterize as $\gamma \equiv e^{i\vec{\alpha}\cdot \vec{\tau}/2}$. (Mercifully, you've chosen to ignore the hypercharge, which amounts to a left singlet, but presents as one of the right generators in this language.)

Now, $D_\mu H= \partial_\mu H + ig \frac{\vec{\tau}}{2}\cdot \vec{W}_\mu ~ H$ also transforms like $H$ and the $W$s under the custodial symmetry, despite the left-SU(2) action of the $W$s.

The crucial observation you may well be seeking is that $H^\dagger \mapsto \gamma H^\dagger \gamma^\dagger$ as well, so the bilinear is a custodial invariant (before tracing!), $$ H^\dagger H = 2(\sigma^2+\chi^2+\varphi_1^2+\varphi_2^2) 1\!\! 1 ~, $$ with an obvious SO(4) structure in the coefficient.

As a result, giving σ a v.e.v. preserves the vector SU(2), the SO(3) (not SU(3)!) which scrambles the three goldstons $\chi,\varphi_1,\varphi_2$ among themselves, exactly as it scrambles the three components $W_i$ that eat them. The three broken symmetries are the three axial transformations connecting the scalar σ to these three pseudoscalar goldstons: the combination of the broken pieces of both the left and the right SU(2)s, whereas their unbroken pieces combined into the surviving custodial vector isospin.

It really, really, is the SO(4) σ-model with the left-chiral SU(2) gauged, and fully broken (unlike the Georgi-Glashow model!).

Everything in its action is custodial-SU(2) invariant, so the Ws will stay degenerate forever, in the absence of hypercharge interactions that would mar that degeneracy.