# 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?

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

$$\alpha_{Q} C_{0}^{(l)}=\sum_{K=l-1}^{l+1}(-1)^{K+Q}[K]^{1 / 2}\left(\begin{array}{ccc} l & 1 & K \\ 0 &-Q & Q \end{array}\right) X_{Q}^{((1, l) K)}$$