Component of a irreducible tensor product Given the operators $\boldsymbol{\alpha}$ and $\boldsymbol{C^{(L)}}$ such that
$$
\boldsymbol{\alpha}=\left(\begin{array}{cc}
0 & \boldsymbol{\sigma}_{p} \\
\boldsymbol{\sigma}_{p} & 0
\end{array}\right) \quad 
$$
where $\sigma^{1}$ are the Pauli matrices
$$
{\sigma}_{x}=\left(\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}\right), \quad \sigma_{y}=\left(\begin{array}{rr}
0 & i \\
-i & 0
\end{array}\right), \quad \sigma_{z}=\left(\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right)
$$
and
$$
\boldsymbol{C^{(L)}}=C_{M}^{(L)}(\theta, \phi)=\left(\frac{4 \pi}{2 L+1}\right)^{1 / 2} Y_{L M}(\theta, \phi)
$$
where $ Y_{L M}(\theta, \phi)$ are the spherical harmonics. We  can construct the irreducible tensor product
$$
\mathbf{X}_{p}^{((1l) K)}=\left[\boldsymbol{\alpha} \mathbf{C}^{(l)}\right]_{Q}^{(K)}=\sum_{p m} C\left(l, 1, m, p ; K, Q\right) \alpha_{p} C_{m}^{\left(l\right)}
$$
where $C\left(l, 1, m, p ; K, Q\right)$ are the Clebsch-Gordan coefficients.
Now in this article Relativistic calculation of atomic structures (eq.6.24) they claim that by orthogonality of the $3 j$ -symbols, we have
$$
\alpha_{Q} C_{0}^{(l)}=\sum_{K=l-1}^{l+1}(-1)^{K+Q}[K]^{1 / 2}\left(\begin{array}{ccc}
l & 1 & K \\
0 &-Q & Q
\end{array}\right) X_{Q}^{((1, l) K)}
$$
But I am not seeing how. Can anyone help me please?
 A: We have
$C(l,1,m,p;K,Q)=(-1)^{l-1+Q}[j_3]^{1/2}\left(\begin{array}{ccc}
l & 1 & K \\
m & p & -Q
\end{array}\right)$
so we have
$\mathbf{X}_{Q}^{((1l) K)}=\underset{m}{\sum} \underset{t}{\sum} (-1)^{l-1+Q}[K_3]^{1/2}\left(\begin{array}{ccc}
l & 1 & K \\
m & t & -Q
\end{array}\right) \alpha_{t} C_{m}^{\left(l\right)}$
The  orthogonality of the 3j -symbols gives
$\underset{j_{3}}{\sum} \underset{m_{3}}{\sum}\left[j_{3}\right]\left(\begin{array}{lll}
j_{1} & j_{2} & j_{3} \\
m_{1} & m_{2} & m_{3}
\end{array}\right)\left(\begin{array}{lll}
j_{1} & j_{2} & j_{3} \\
m_{1}^{\prime} & m_{2}^{\prime} & m_{3}
\end{array}\right)=\delta_{m_{1} m_{1}{\prime}} \delta_{m_{2} m_{2}^{\prime}}\tag 1$
So using $(1)$ we have that
$\underset{K}{\sum} \underset{Q}{\sum}(-1)^{l-1+Q}[K]^{1/2}\left(\begin{array}{ccc}
l & 1 & K \\
0 & p & -Q
\end{array}\right)\mathbf{X}_{Q}^{((1l) K)}=\underset{m}{\sum} \underset{t}{\sum}\underset{K}{\sum} \underset{Q}{\sum} [K]\left(\begin{array}{ccc}
l & 1 & K \\
0 & p & -Q
\end{array}\right)\left(\begin{array}{ccc}
l & 1 & K \\
m & t & -Q
\end{array}\right) \alpha_{t} C_{m}^{\left(l\right)}$
which give us
$\underset{K}{\sum} \underset{Q}{\sum}(-1)^{l-1+Q}[K]^{1/2}\left(\begin{array}{ccc}
l & 1 & K \\
0 & p & -Q
\end{array}\right)\mathbf{X}_{Q}^{((1l) K)}=\alpha_{p} C_{0}^{\left(l\right)}$
Now the $3j$ symbol is not zero only when we have $p=Q$ and so we have
$\underset{K}{\sum} \underset{Q}{\sum}(-1)^{l-1+Q}[K]^{1/2}\left(\begin{array}{ccc}
l & 1 & K \\
0 & p & -Q
\end{array}\right)\mathbf{X}_{Q}^{((1l) K)}=\underset{K}{\sum}(-1)^{l-1+Q}[K]^{1/2}\left(\begin{array}{ccc}
l & 1 & K \\
0 & p & -p
\end{array}\right)\mathbf{X}_{p}^{((1l) K)}$
Finally we use the property of the $3j$ symbol
$\left(\begin{array}{lll}
\mathrm{j}_{2} & \mathrm{j}_{1} & \mathrm{j}_{3} \\
\mathrm{m}_{2} & \mathrm{m}_{1} & \mathrm{m}_{3}
\end{array}\right)=\left(\begin{array}{lll}
\mathrm{j}_{1} & \mathrm{j}_{3} & \mathrm{j}_{2} \\
\mathrm{m}_{1} & \mathrm{m}_{3} & \mathrm{m}_{2}
\end{array}\right)=(-1)^{\mathrm{j}_{1}+\mathrm{J}_{2}+\mathrm{J}_{3}}\left(\begin{array}{lll}
\mathrm{j}_{1} & \mathrm{j}_{2} & \mathrm{j}_{3} \\
\mathrm{m}_{1} & \mathrm{m}_{2} & \mathrm{m}_{3}
\end{array}\right)$
that give us
$\alpha_{Q} C_{0}^{(l)}=\sum_{K=l-1}^{l+1}(-1)^{K+Q}[K]^{1 / 2}\left(\begin{array}{ccc}
l & 1 & K \\
0 &-Q & Q
\end{array}\right) X_{Q}^{((1, l) K)}$
