I recently learned that a Higgs triplet can be written as a $2 \times 2$ matrix:

\begin{equation} \Delta=\begin{pmatrix} \frac{\Delta^{+}}{\sqrt{2}} & \Delta^{++} \\ \Delta^0 & - \frac{\Delta^{+}}{\sqrt{2}}\end{pmatrix} \end{equation}

Typically, an $SU(2)$ triplet $\phi$ transform like:

\begin{equation} \phi \rightarrow\exp(-i\vec{T}\cdot \vec{\theta})\phi \end{equation}

where $\vec{T}$ are some $3\times3$ matrix representation of the $SU(2)$ generators. How to modify the $\vec{T}$'s such that $\Delta$ transform like a triplet under $SU(2)_L$?


1 Answer 1


Indeed, a triplet $\vec{\phi}$ transforming under the triplet representation matrices T can be doted to a Pauli vector to yield a formal adjoint, so a traceless 2×2 matrix $$\Phi=\tfrac{1}{2}\vec{\phi}\cdot \vec{\tau}~,$$ transforming as the doublet representation, conjugately on both sides. This is dubbed adjoint action, $$ \Phi \mapsto e^{i\vec{\tau}\cdot \frac{\vec{\theta}}{2}}~ \Phi ~e^{-i\vec{\tau}\cdot \frac{\vec{\theta}}{2}}. $$

In your case, \begin{equation} \Delta=\begin{pmatrix} \frac{\Delta^{+}}{\sqrt{2}} & \Delta^{++} \\ \Delta^0 & - \frac{\Delta^{+}}{\sqrt{2}}\end{pmatrix}= \sqrt{2}\Delta^+ \frac{\tau_3}{2} +\Delta^{++} \frac{\tau_1+i\tau_2}{2} + \Delta^0 \frac{\tau_1-i\tau_2}{2} = \sqrt{2}\Delta^+ \frac{\tau_3}{2} +\Delta^{++} \frac{\tau_ +}{2} + \Delta^0 \frac{\tau_-}{2} \\ = \begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+ \end{pmatrix}\cdot \frac{\vec{\tau}}{2}, \end{equation} where you note the normalization $$ \vec{\bar{\phi}}\cdot \vec{ \phi }=2(\Delta^{++~~2}+\Delta^{+~~2}+\Delta^{0~~2}). $$ You can easily see, then, that the Cartesian 3-vector can be unitarily rotated to the more transparent spherical basis vector $\sqrt{2}(\Delta^{++},\Delta^+, \Delta^0)$. The matrix $U^\dagger$ achieving that Cartesian to spherical basis change is given in footnote nb 3 of the WP article or this answer . In an innocuously rephased form, $$ U^\dagger \begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+ \end{pmatrix} = \frac{1}{\sqrt{2}} \left(\begin{matrix} 1 & -i & 0 \\ 0 & 0 & \sqrt{2} \\ 1 & i & 0\end{matrix}\right) ~\begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+ \end{pmatrix}= \sqrt{2}\begin{pmatrix}\Delta^{++}\\\Delta^+ \\ \Delta^0 \end{pmatrix}. $$

  • To confirm the normalization of the half-angle and the signs, take only the 3rd component of θ to be non-vanishing and infinitesimal, so the rotation increment of Δ is a simpler matrix with vanishing diagonals. Check that the increment components of $\vec{\phi}$ are now given by commutators (adjoint) and fully comport with the classical cross-product increment of the triplet representation you specified. It is then evident upon commutation with $\tau_3/2$ that the $T_3$ eigenvalues of the triplet $(\Delta^{++},\Delta^+, \Delta^0)$ are (1,0,-1), so that its $Y=Q-T_3=1$. I normalize the weak hypercharge the modern ("alternative") way, i.e. dropping the superfluous strong denominator of 2.

  • You may find the derivation of T to all orders through this construction in this answer.

  • $\begingroup$ If I understand it correct, in your answer $\Phi$ is given by $\Phi=\begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+\end{pmatrix}$. If that is true, what is the charge of the uppermost component of $\Phi$ i.e. $(\Delta^{++}+\Delta^0)$ and components below it? @CosmasZachos $\endgroup$
    – SRS
    Nov 12, 2018 at 21:57
  • 2
    $\begingroup$ Well, the components of the 3-vector you write are not individually eigenvectors of the charge operator. In this language, obviously, the charge operator is $U\operatorname{diag}(2,1,0) ~U^\dagger$. So, the eigenvector of charge with eigenvalue 2 is $\Phi_{++}= (1,i,0)^T$. It is just a dumb change of basis. Try to see it on the doublet representation. $\endgroup$ Nov 12, 2018 at 22:22

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