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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?

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

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    $\begingroup$ $i(T_a)_{bc} = i\epsilon_{abc}$ $\endgroup$
    – DanielC
    Commented Dec 12, 2019 at 10:34
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    $\begingroup$ OP's $T$s are anti-Hermitian while the Pauli matrices are Hermitian, so there's for starters an $i$ floating around. $\endgroup$
    – Qmechanic ♦
    Commented Dec 12, 2019 at 11:08

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Okay so basically with the change below everything is correct. An adjoint representation of SU(2) is the following:

$ 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} $

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