I'm working through $SU(2)$ angular momentum coupling and trying to compute the $F$-matrix (recoupling isomorphism) associated with three spin-$\tfrac{1}{2}$ particles coupled to a total spin of $\tfrac{1}{2}$. This corresponds to the standard case:
$$ F^{\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2}}_{\tfrac{1}{2}} $$
which gives the change of basis between two different fusion orderings of the triple tensor product:
$$ \left( \left( \tfrac{1}{2} \otimes \tfrac{1}{2} \right)_{j_{12}} \otimes \tfrac{1}{2} \right)_{\tfrac{1}{2}} \quad \longleftrightarrow \quad \left( \tfrac{1}{2} \otimes \left( \tfrac{1}{2} \otimes \tfrac{1}{2} \right)_{j_{23}} \right)_{\tfrac{1}{2}}. $$
Since $\tfrac{1}{2} \otimes \tfrac{1}{2} = 0 \oplus 1$, the intermediate spins $j_{12}, j_{23} \in \{0, 1\}$, and the space of total spin-$\tfrac{1}{2}$ states is two-dimensional. Hence, the $F$-matrix is $2 \times 2$.
Using the standard formula involving Wigner $6j$ symbols:
$$ [F^{j_1 j_2 j_3}_{j}]_{j_{12}, j_{23}} = (-1)^{j_1 + j_2 + j_3 + j} \sqrt{(2j_{12} + 1)(2j_{23} + 1)} \cdot \begin{Bmatrix} j_1 & j_2 & j_{12} \\ j_3 & j & j_{23} \end{Bmatrix}, $$
and plugging in $j_1 = j_2 = j_3 = j = \tfrac{1}{2}$ and standard values of $6j$ symbols, I obtain:
$$ F = \begin{bmatrix} -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\ \tfrac{\sqrt{3}}{2} & \tfrac{1}{2} \end{bmatrix}. $$
However I have heard that the $ 0=j_{12}=j_{23} $ entry of the $ F $-matrix is usually $ 1 $ over the quantum dimension (the quantum dimension is $ 2 $ in this case), whereas here I have $ -1 $ over the quantum dimension. Did I compute the $ F $-matrix incorrectly? Or is this claim about $ 1 $ over the quantum dimension not quite right?
