# Adjoint representation of $SU(2)$

I'm trying to understand how the $$SU(2)$$ representations work. We know that the fundamental representation of $$SU(2)$$ is $$\frac{1}{2} \sigma^{\alpha}$$ where $$\sigma^{\alpha}$$ the Pauli matrices. These are 2x2 matrices that follow the Lie algebra. How can we derive a 3x3 representation (and basis) for $$su(2)$$?

When I try the following basis I don't get the same algebra as the Pauli matrices, shouldn't that be the case?

$$T_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}0&i&0\\i&0&i\\0&i&0\end{pmatrix}$$ $$T_2 = \begin{pmatrix} i&0&0\\0&0&0\\0&0&-i\end{pmatrix}$$ $$T_3 = \frac{1}{\sqrt{2}}\begin{pmatrix} 0&1&0\\-1&0&1\\0 &-1&0\end{pmatrix}$$

• $i(T_a)_{bc} = i\epsilon_{abc}$ – DanielC Dec 12 '19 at 10:34
• OP's $T$s are anti-Hermitian while the Pauli matrices are Hermitian, so there's for starters an $i$ floating around. – Qmechanic Dec 12 '19 at 11:08

$$T_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}$$ $$\;\;\;\;\;\;T_2 =\frac{1}{\sqrt{2}} \begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix}$$ $$\;\;\;\;\;\;T_3 =\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$