I've read at different, places (for instance, this answer) that you can choose the vev of an adjoint Higgs to correspond to a generator of the Cartan subalgebra of the symmetry you are trying to break. In the case of the Higgs being an SU(2) triplet, the vev has to be the only traceless diagonal 2-D matrix which happens to be the third pauli matrix:$\begin{pmatrix} \nu &0\\0&-\nu \end{pmatrix}$

It also seems like a generally common thing to have a diagonal vev.

However, in some papers (for example: https://www.researchgate.net/publication/311969493_Higgs_Sector_of_the_Left-Right_Symmetric_Theory), the vev is not diagonal at all. In that particular paper, it is only the bottom left component of the higgs SU(2) triplet that is non-zero: $\begin{pmatrix} 0 &0\\\nu&0 \end{pmatrix}$

I realize that you can choose the Higgs in the adjoint representation to be any linear combination of the Pauli matrices, and this is the case for the two cases above (the first has a component in the third pauli matrix, the second has components in the first and second pauli matrices).

However, if you were to take, for example, the breaking of SU(2), both vevs would break it but in the first case, the generator of the third component of isospin would still be a symmetry (since the vev commutes with the third pauli matrix) which is not the case for the second vev.

How is this possible? Shouldn't vevs in a particular representation be equivalent?

  • $\begingroup$ What are you asking? the vev of Φ is your first, and the vev of Δ is your second. They have different quantum numbers. There is no equivalence transformation between them, if that's what you are after, as their determinants are different. $\endgroup$ – Cosmas Zachos Mar 9 at 19:44
  • $\begingroup$ I don't understand what you mean by the vev of $\Phi$ is your first and the vev of $\Delta$ is your second, they are both vevs of a triplet aren't they? (In the paper, $\Phi$ is a bidoublet of the left/right sector, not a triplet of SU(2)) $\endgroup$ – madcat Mar 10 at 7:13
  • $\begingroup$ You should include the explicit eqn (2) of that Maiezza et al 2017 paper in your question to dramatize what might puzzle you. Φ qua triplet dots a Pauli vector but the spherical basis quasi-triplet of Δ dotting the Pauli vector is $((\delta^{++}+\delta^0)/2, i(\delta^{++}-\delta^0)/2, \delta^+/\sqrt{2})^T$. Apples to apples. V.e.v.s attach to neutral fields. $\endgroup$ – Cosmas Zachos Mar 10 at 10:27

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