How can an SU(2) triplet be represented as a 2x2 matrix in the Lagrangian?

[ChrisVer]

Sorry for this "stupid" question... but I am having some problem in understanding how can someone start from let's say an SU(2) triplet and arrive in a 2x2 matrix representation of it in the Lagrangian...
An example is the Higgs-triplet models...I think this happens with the W-gauge bosons too in the EW theory, but I'm currently losing where in the derivation/how this happens. I think for the W's this happens with the complexification?

 

[ChrisVer]

Does it happen by a map $\phi_i \rightarrow \phi_i \sigma^i$?

 

[Orodruin]

The tensor product of two fundamental representations is a singlet representation (the trace of the tensor) and a triplet which is the rank two symmetric tensor. You should therefore not be surprised that you can realize the triplet representation using traceless matrices.

Alternatively, just see it as transforming under the adjoint representation, which is the triplet (the Lie algebra is three dimensional).

 

[vanhees71]

If you have a doublet $\phi$, you can make a triplet out of it, $\phi^{\dagger} \vec{\sigma} \phi$. I'm not sure, whether this is what you mean or need. This combination transforms under the fundamental SO(3).

 

[Orodruin]

If you have a doublet $\phi$, you can make a triplet out of it, $\phi^{\dagger} \vec{\sigma} \phi$. I'm not sure, whether this is what you mean or need. This combination transforms under the fundamental SO(3).

I do not think he has a doublet that he is forming a triplet from. I think he has a fundamental scalar SU(2) triplet such as that which appears in the type-II seesaw.

 

[vanhees71]

Ok, then you can make a doublet via $\tilde{\phi}=\vec{\sigma} \cdot \vec{\phi}$, where $\vec{\phi} \in \mathbb{R}^3$ is the triplet field.

 

[Orodruin]

Ok, then you can make a doublet via $\tilde{\phi}=\vec{\sigma} \cdot \vec{\phi}$, where $\vec{\phi} \in \mathbb{R}^3$ is the triplet field.

This is not a doublet, it is an element of the Lie algebra and transforms under the adjoint representation, which is the triplet representation. The triplet is irreducible.

 

[ChrisVer]

Yup it's like my Post #2...
It's confusing me ...
the $\phi = \begin{pmatrix} \phi_+ \\ \phi_0 \\ \phi_- \end{pmatrix}$ is a triplet of SU(2). That means that it transforms as triplet under SU(2) transformations and so in the adjoint representation (I think with the $\epsilon_{ab}$'s the antisymmetric metric)...
Then also the $\phi \cdot \sigma$ (which is now a 2x2 matrix) transforms in the same rep with epsilons?

 

[Orodruin]

The vector you have written down is just vector containing the components of a vector in the representation. The basis vectors of the adjoint representation are the basis vectors of the Lie algebra, i.e., the Pauli matrices in the case of SU(2). So all $\phi\cdot\sigma$ tells you is the full expression for the triplet, i.e., the full element of the Lie algebra.

 

[samalkhaiat]

Yup it's like my Post #2...
It's confusing me ...
the $\phi = \begin{pmatrix} \phi_+ \\ \phi_0 \\ \phi_- \end{pmatrix}$ is a triplet of SU(2). That means that it transforms as triplet under SU(2) transformations and so in the adjoint representation (I think with the $\epsilon_{ab}$'s the antisymmetric metric)...
Then also the $\phi \cdot \sigma$ (which is now a 2x2 matrix) transforms in the same rep with epsilons?

The triplet transforms by the adjoint map of the Lie algebra $su(2)$ $$\mbox{ad}: \ \ \phi_{i} \to \phi_{i} + \epsilon_{i j k} \phi_{j} \alpha_{k} ,$$ while the Hermitian $2 \times 2$ matrix $\Phi = \tau^{i}\phi_{i}$ transforms in the adjoint map of the group $SU(2)$: $$\mbox{Ad}: \ \ \Phi \to U \Phi U^{\dagger} , \ \ U \in SU(2) .$$ As usual the two maps are related by $$\mbox{Ad} ( e^{\alpha^{i} X^{i}} ) = e^{ \alpha^{i} \mbox{ad}(X^{i}) } .$$

 

[vanhees71]

This is not a doublet, it is an element of the Lie algebra and transforms under the adjoint representation, which is the triplet representation. The triplet is irreducible.

True. Sorry for the confusion.