I am trying to do a Higgs triplet extension of the standard model but I don't know how to find the corresponding vacuum expectation value for the Higgs triplet. I read a paper where they found the expectation value for doublet to be $$<\phi>_0 =\frac{1}{\sqrt{2}}\Big{(}\begin{matrix}0\\v\end{matrix}\Big{)}$$and the triplet expectation value is found to be $$<\Delta>_0\Big{(}\begin{matrix}0&0\\v_T& 0\end{matrix}\Big{)}$$

but I was expecting a 3x1 matrix for the VEV of the Higgs triplet, why is it a 2x2 matrix

  • $\begingroup$ When you write the triplet as a 3-vector, you see that any of the 3 components can pick up the v.e.v. The other two components may then rotate into each other, that is SU(2) ~ SO(3) breaks down to SO(2), not completely, as in the doublet representation. In this model, to the extent they assign a charge to the components of the triplet, they choose the neutral component , so left, lower one, to align with the (neutral) vacuum. $\endgroup$ Apr 24, 2018 at 0:16

1 Answer 1


The "triplet" representation is the adjoint representation of the $\mathfrak{su}(2)$-algebra, i.e. the representation of the algebra upon itself through the commutator. Although the algebra is three-dimensional, it is customary to identify it with the traceless Hermitian algebra of 2x2 matrices spanned by the Pauli matrices. It's still three-dimensional since the conditions of being Hermitian and traceless eliminate the additional degrees of freedom a 2x2 matrix has.

  • $\begingroup$ Even then I don't understand why VEV for triplet can't be something like this :$$<\Delta>_0=\Big{(}\begin{matrix}0&v_T\\0&0\end{matrix}\Big{)}$$ $\endgroup$
    – MSB
    Apr 22, 2017 at 20:50
  • 1
    $\begingroup$ @MSB Who says it can't? The doublet can also have $(v,0)$ instead of $(0,v)$, which form you prefer is essentially a gauge choice. $\endgroup$
    – ACuriousMind
    Apr 22, 2017 at 21:15

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