# Matrix representation of a triplet state

The $SU(2)$ triplet state is typically given in the fundamental representation as a column vector, e.g. $$\vec{\Delta} = \left( \begin{array}{c} \delta^{++} \\ \delta^+ \\ \delta^0 \end{array} \right)$$ where I use this notation to be consistent with the references below.

However, I'm reading through papers (in the context of the type II see-saw mechanism, but I don't think it has much relevance) and most references are using what they call the matrix representation of the $SU(2)$ triplet. My instinct was to assume that they were projecting the fundamental representation onto the Pauli matrices:

$$\Delta \rightarrow \vec{\Delta } \cdot \vec{ \sigma } = \delta ^{++} \sigma _1 + \delta ^+ \sigma _2 + \delta ^0 \sigma _3$$

However, this doesn't quite give me the correct result, which is given by, $$\Delta = \left( \begin{array}{cc} \delta ^+ / \sqrt{2} & \delta ^{ + + } \\ \delta ^0 & - \delta ^+ / \sqrt{2} \end{array} \right)$$

Is there some other way to get this "matrix representation"?

Two sample references where they say they are using the matrix representation can be found here (pg. 124, equation 68) and here (pg. 4 equation 1).

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Typically, the triplet refers to the spin one representation of $SU(2)_L$ of the standard model. Let $I_a$ refer to the generators of $SU(2)_L$. Then the triplet can be organized into a column vector with $I_3$ being diagonal and equal to Diag$(1,0,-1)$. Suppose the triplet has hypercharge $Y$, i.e., the charge associated with $U(1)_Y$. Electroweak symmetry breaking has $SU(2)_L\times U(1)_Y$ broken to $U(1)_{EM}$. The electric charge is given by the formula $Q=I_3+\tfrac12 Y$. If $Q=2$ for the triplet, then we see that $Q=(2,1,0)$ which is written as $(++,+,0)$ and added as a superscript to indicate the charge of a particle in such a triplet. This should explain the notation for $\vec{\Delta}$ that you have given.
Now to go to the matrix representation, one uses the fact that the spin-one representation of $SU(2)$ is also its adjoint representation. Since the three entries in $\vec{\Delta}$ are all $I_3$ eigenstates, in order to write the matrix representation using Pauli sigma matrices, you need to write the Pauli matrices in a basis where they are the analog of eigenstates of $I_3$. Such a basis is given by $\tfrac12\sigma_\pm =\tfrac12( \sigma_1\pm i\sigma_2)$ and $\tfrac12\sigma_3$. So the natural guess would be something like $$\Delta \rightarrow \frac12 (\delta^{++}\ \sigma_+ +\delta^+\ \sigma_3 + \delta^0\ \sigma_-)\ .$$ There will be variations in convention in that $\delta^{++}$ will be paired up with $\sigma_-$ instead of $\sigma_+$ as I have done. I might also be missing some factors of $\sqrt{2}$ but I am sure you get the general idea.
Thanks for your answer! Can you explain what you mean by " you need to write the Pauli matrices in a basis where they are the analogue of eigenstates of $I_3$"? How come $\frac{1}{2}\sigma_{\pm}$ is an analogue of eigenstates of $I_3$? –  JeffDror Feb 7 '14 at 1:34
It is longish story but I will try to be crisp. While acting on states, we look for eigenvectors of $I_3$: $I_3 |m\rangle = m\ | m\rangle$. The analog in the matrix (adjoint) representation is the commutator $[I_3,M]=m\ M$ as we think of $M$ as operators acting on a vector space (two-dimensional here). Of course, standard normalization implies $I_a \leftrightarrow \tfrac12 \sigma_a$ which is the half I put in. –  suresh Feb 7 '14 at 2:23