Component of a irreducible tensor product

Question

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?

Answer 1

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

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