The irreducible representation of rank $n$ spinor in 3D I was reading Ref. 1, where it is asserted that the irreducible resolution of a rank $n$ spinor can be written as totally symmetric spinors.
$$\Psi^{n}=\Psi^{\{n\}}+\zeta \Psi^{\{n-2\}}+\zeta \zeta \Psi^{\{n-4\}}+\cdots+\zeta \zeta \zeta \Psi^{\{n-6\}}+\cdots \tag{2.39}$$
where $$\zeta_{\alpha \beta}=\left(\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right)\tag{2.28}$$
is the spinor metric tensor.
A example given is a rank $2$ tensor，where
$$\Psi^{\{2\}}=\frac{1}{2}\left(\Psi^{\alpha \beta}+\Psi^{\beta \alpha}\right)$$
and
$$\Psi^{\{0\}}=\frac{1}{4}\left(\Psi^{\alpha \beta}-\Psi^{\beta \alpha}\right) \zeta_{\alpha \beta}.$$
Further more, the book says this spinor corresponds to a vector $\boldsymbol a$, whose components are
$$a_{1}=\frac{\mathrm{i}}{\sqrt{2}}\left(\Psi_{11}-\Psi_{22}\right), \quad a_{2}=-\frac{1}{\sqrt{2}}\left(\Psi_{11}+\Psi_{22}\right), \quad a_{3}=-\mathrm{i} \sqrt{2} \Psi_{12}\tag{2.40}$$
The symmetrised tensor $\Psi^{\{2\}}$ is easy to interpret, but I wonder where does the expression for $\Psi^{\{1\}}$, as well as the component for the vector $\boldsymbol a$ come from?
References:

*

*Laurent Baulieu et al, "From Classical to Quantum Fields", page 20.

 A: (2.40) might be more transparent in the spherical basis of the rotation group when you raise the spinor indices.
You (and the rest of us M Jourdains speaking prose all our lives) are looking at $1/2\otimes 1/2=0\oplus 1$ spin compositions, where spin 0 is the rotation singlet, without indices (the rotation matrix in this trivial irrep is the identity!!) the antisymmetrizing symplectic metric ζ having eaten them all up.
The rest (rotation triplet), spin 1, the symmetric piece, is a traceless 2x2 matrix state vector often depicted as
$$
\uparrow\uparrow; \qquad (\uparrow\downarrow+\downarrow \uparrow)/\sqrt{2}; \qquad ; \downarrow\downarrow
$$
in elementary QM, spherical basis; or else
$$
|1,1\rangle = a_1^{\dagger~2}|0\rangle/ \sqrt{2}; \qquad 
|1,0\rangle = a_1^{\dagger}a_2^{\dagger}|0\rangle  \qquad 
|1,-1\rangle = a_2^{\dagger~2}|0\rangle/ \sqrt{2}
$$ in the Jordan-Schwinger symmetric (boson) oscillator realization.
You may apply the above recursively, to get the n-fold tensor product of spin 1/2s. For perspicacity and inclusion of the above case, take even n=2m,
$$
(1/2)^{\otimes n}= [1](n/2)\oplus [n-1](n/2-1)\oplus \\ ...\oplus  [3n!/(m-1)!(m+2)!](1)\oplus [n!/m!(m+1)!](0),
$$
where the parentheses denote the spin s of the direct summand in the Clebsch decomposition of this n-fold product, while the square brackets the multiplicity of the summand multiplet,
$$
M(s,2m,1/2)=\frac{(1+2s)n!}{(m-s)!(m+s+1)!} ~,
$$
given by   Catalan's Triangle, cf WP.
This is more complete information than your schematic formula (2.39).
