3

I am reading this paper (or this link), and I'm troubled by the custodial transformation. So, I will use the notations and equation labels appearing in this paper.

If we write the Higgs field into the four real components $$ \Phi=\left(\begin{array}{c} -\operatorname{Re} \phi_1 \\ \operatorname{Im} \phi_1 \\ \operatorname{Re} \phi_2 \\ \operatorname{Im} \phi_2 \end{array}\right) \tag{13} $$ where $\phi_1$ and $\phi_2$ are the complex components of $\Phi$ in the ordinary isospinor representation of the field, $$ \left(\begin{array}{l} \phi_1 \\ \phi_2 \end{array}\right) . $$ The Higgs field potential under (13) has the $SO(4)$ symmetry, since $$ \mathrm{SO}(4) \sim \mathrm{SU}(2)_{\text {isospin }} \times \mathrm{SU}(2)_{\text {custodial }}. $$

We denote the isospin generator with $\mathbf{I}$, the custodial generator with $\mathbf{K}$. Then the isospin and custodial transformation on the four vector $\Phi$ are $$ \Phi_i \rightarrow e^{i \epsilon \cdot \mathbf{I}} \Phi_i=\left(e^{i \epsilon \cdot \mathbf{T}}\right)_{i j} \Phi_j \\ \Phi_i \rightarrow e^{i \epsilon' \cdot \mathbf{K}} \Phi_i=\left(e^{i \epsilon' \cdot \mathbf{T}^{\prime}}\right)_{i j} \Phi_j. $$

So, now our goal is to calculate the 4 by 4 matrices $\mathbf{T}$ and $\mathbf{T^\prime}$. The calculation can be done by recasting the transformation law of $\Phi$ in isospinor form. Where, for the isospin transformation, $$ \left(\begin{array}{l} \delta \phi_1 \\ \delta \phi_2 \end{array}\right)=i \epsilon \cdot \tau\left(\begin{array}{l} \phi_1 \\ \phi_2 \end{array}\right) $$ while, for the custodial transformation, $$ \left(\begin{array}{l} \delta \phi_1 \\ \delta \phi_2^* \end{array}\right)=i \epsilon' \cdot \tau\left(\begin{array}{l} \phi_1 \\ \phi_2^* \end{array}\right) \tag{*} . $$

My question is for the $(*)$ equation. Why does the custodial transformation in isospinor format take this form?

Qmechanic
  • 201,751
Daren
  • 1,401

1 Answers1

2

It should be a straight translation from this link answer, but I strongly suspect Carson et al., 1990 flipped a sign, at a deep level in their transition from a complex doublet to the real quartet representation they actually use. (They don't actually use this doublet/matrix representation in practice, and are merely meaning to connect to the readers' shared educational background, adjusting for their errant sign.)

I can explain your (*) equation, provided I insert a minus sign in front of $_1$ in it. You first observe the action of a weak isospin transformation with parameter $\vec \alpha$ on the complex doublet and its conjugate rep, $$ \phi = \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \qquad \phi \mapsto e^{i\vec{\alpha}\cdot \vec{\tau}/2} \phi \\ \tilde \phi =i\tau_2 \phi^*= \begin{pmatrix} \phi^{*}_2 \\ -\phi_1^* \end{pmatrix} \qquad \tilde \phi \mapsto e^{i \vec{\alpha}\cdot \vec{\tau}/2}\tilde \phi ~.$$

Thus, the 2×2 Higgs matrix is defined as a side-by-side juxtaposition of these two left-doublets serving as columns, $$ H\equiv \sqrt{2}(\tilde\phi, \phi)= \sqrt {2} \begin{pmatrix} \phi_2^{*} &\phi_1 \\ -\phi_1^* & \phi_2 \end{pmatrix}. $$

It is then evident that it transforms by left α isospin and also by manifestly commuting right β custodial isorotations K as $$ \bbox[yellow]{ e^{i\vec{\alpha}\cdot \vec{\tau}/2} \sqrt{2}(\tilde\phi , \phi )e^{i\vec{\beta}\cdot \vec{\tau}/2} = e^{i\vec{\alpha} \cdot \vec{\tau}/2}\sqrt {2} \begin{pmatrix} \phi_2^* &\phi_1 \\ -\phi_1^* & \phi_2 \end{pmatrix}e^{i\vec{\beta}\cdot \vec{\tau}/2} = e^{i\vec{\alpha}\cdot \vec{\tau}/2} H e^{i\vec{\beta}\cdot \vec{\tau}/2} ~ . } $$

Taking the adjoint of this equation and looking at the custodial action on the second column of $H^\dagger$, we see that $$ \begin{pmatrix} -\phi_1 \\ \phi_2^* \end{pmatrix} \qquad \mapsto \qquad e^{-i\vec{\beta}\cdot \vec{\tau}/2} \begin{pmatrix} -\phi_1 \\ \phi_2^* \end{pmatrix} \tag{*} $$ with the sign mismatch mentioned.

As indicated, it is trivial to demonstrate the action of I commutes with the action of K, so these two groups do not "know" about each other, i.e. they are in direct product, hence SO(4).

You may also trivially confirm that $$ \operatorname{Tr} H^\dagger H = 4 \phi^\dagger \phi = 4 \Phi^T \Phi, $$ invariant under $\mathrm{SU}(2)_{\text {isospin }} \times \mathrm{SU}(2)_{\text {custodial }}$.

NB Note added

I mapped the custodial $\tau_2 \to T'_2$ on p 2132, and matched correctly my (*), not theirs, on the same page, to their quartet. (Perhaps they meant to have $\small \begin{pmatrix} \phi_2 \\ \phi_1^* \end{pmatrix} \mapsto e^{-i\vec{\beta}\cdot \vec{\tau}/2} \begin{pmatrix} \phi_2 \\ \phi_1^* \end{pmatrix} $, instead?)

Cosmas Zachos
  • 62,595
  • Hi professor, thank you very much! May I ask a related question in Carson et al.'s 1990 paper? In the third equation below equation (13), they say that the isospin transformation for the gauge fields is $\mathbf{A}^a \rightarrow e^{i \epsilon \cdot \mathbf{I}} \mathbf{A}^a=\left(e^{i \epsilon \cdot \Delta}\right)^{a b} \mathbf{A}^b$, where $(\Delta^a)^{bc}=-i\epsilon^{abc}$. I am troubled for this form. In gauge theory, we know the gauge field transforms like $A_\mu^a T^a\rightarrow U (A_\mu^a T^a)U^\dagger+ \frac{i}{g}U(\partial_\mu U^\dagger)$. And it seems that these two things don't match. – Daren Apr 13 '23 at 02:14
  • No hurry, thanks! – Daren Apr 13 '23 at 02:24
  • 1
    As stated at the beginning of sec II, they have fixed the gauge, and monitor the residual global symmetry. – Cosmas Zachos Apr 13 '23 at 11:34
  • Ah, thanks! Clear! – Daren Apr 15 '23 at 09:09