conventions - Is the adjoint representation of $SU(2)$ the same as the triplet representation?

Is the triplet representation of $SU(2)$ the same as its adjoint representation? Where the convention for the adjoint representation used is the one used in particle physics, where the structure constants are real and antisymmetric:

$$ \mathrm{ad}(t^b_G)_{ac} = i f^{abc} $$

I was under the impression that is was, but I see two different forms of the generators in the triplet representations used, one being just the real skew symmetric generators of the $SO(3)$ rotation group, which agrees with the adjoint representation, and the other being:

$$ T^1 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \quad T^2 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{matrix}\right)$$

These two representations do not agree, I assume that my idea about the adjoint reperesentation of $SU(2)$ being its triplet representation is wrong, but why?

Answer

It is just matter of a missed factor $i\sqrt{2} $ due to different conventions. The antiHermitian matrices $i\sqrt{2} T^k $ can be transformed into the real antisymmetric matrices $L_k $ (which therefore are also complex antiHermitian) by means of a suitable unitary matrix $U$, $$L_k = U i\sqrt{2}\:T^kU^\dagger\quad k=1,2,3\:.$$ This is because both triples of matrices are irreducible representations of the Lie algebra of $SU (2) $ with the same value of the Casimir operator $\sum_k (\sqrt{2} T^k)^2 = -\sum_k (L_k)^2 =2I$ (so that $2= j(j+1)$ with $j=1$ which is the spin of the representation). As is known, up to unitary equivalences there is only one unitary irreducible representation of $SU (2) $ for every value of the spin, essentially due to Peter-Weyl theorem.