# How to prove the orthogonality of spherical tensor operators?

$$\def\mybra#1{\langle{#1}|}$$ $$\def\myket#1{|{#1}\rangle}$$ $$\def\Tr{\mbox{Tr}}$$ $$\def\mybraket#1{\langle{#1}\rangle}$$ $$\def\B#1#2{B^{#1}_{#2}}$$

$$\def\C#1#2{C^{#1}_{#2}}$$

$$\def\r#1#2{\hat\rho^{#1}_{#2}}$$

I want to expand my density operator $$\hat \rho = \sum_{jm j'm'} \myket{jm}\mybraket{jm|\hat \rho|j'm'} \mybra{j'm'}\tag1$$ with spherical tensor operators, $$\hat\rho = \sum_{k}\sum_{q=-k}^k c^k_q\r kq,\tag2$$ where $$\myket{jm}$$ is eigen vector of angular momentum $$\hat J_z$$, and $$\hat\rho^k_q = \sum_{jmj'm'} \B{jm}{kq j'm'} \myket{jm}\mybra{j'm'}\tag3$$ is spherical operator that satisfies $$[\hat J_z,\r kq]=q\hbar \hat\rho^kq,\quad [\hat J_\pm ,\hat\rho^kq] = \hbar\sqrt{k(k+1)-q(q\pm 1)}\r k{q\pm1}. \tag4$$ Using recursion relations of C-G coefficients, $$\B{jm}{kqj'm'}$$ can be determined to be C-G coefficient except a scale factor, $$\B{jm}{kqj'm'}=f_{{kjj'}} \C{jm}{kqj'm'},\quad \C{jm}{kqj'm'} = \mybraket{kqj'm'|(kj')jm}.\tag5$$ The factor should be independent of $$m,m'$$ because the recursion relation changes $$m,m',q$$.

To determine $$f_{kjj'}$$ and $$c^k_q$$, I should make use of the orthogonality and completeness. For orthogonality, $$\Tr((\r {k_1}{q_1})^\dagger \r {k_2}{q_2}) = \sum f_{k_1j_1j_1'}^* f_{k_2j_2j_2'} \C{j_1m_1}{k_1q_1 j_1'm_1'}\C{j_2m_2}{k_2q_2 j_2'm_2'} \delta_{jj_1'}\delta_{mm_1'}\delta_{jj_2'}\delta_{j_1j_2}\delta_{m_1m_2} \delta_{mm_2'}.\tag6$$ With the restriction $$m_1=m_1'+q_1, m_2=m_2'+q_2$$, the delta's in the sum indicates the L.H.S equals 0 unless $$q_1=q_2$$. And only three summation variables $$j_1,j,m$$ are left, $$\Tr((\r {k_1}{q_1})^\dagger \r {k_2}{q_2}) =\sum_{j_1,j,m} f_{k_1j_1j}^* f_{k_2j_1j} \C{j_1(m+q_1)}{k_1q_1 jm}\C{j_1(m+q_1)}{k_2q_1 jm}.\tag7$$ Using symmetry property of C-G coefficients $$\C{jm}{j_1m_1 j_2m_2}=\sqrt{\frac{2j+1}{2j_2+1}} (-)^{j_1-m_1} \C{j_2(-m_2)}{j_1m_1j(-m)},\tag8$$ I get \begin{aligned} \Tr((\r {k_1}{q_1})^\dagger \r {k_2}{q_2}) &= \sum_{j_1,j,m} f_{k_1j_1j}^* f_{k_2j_1j} \C{j_1(m+q_1)}{k_1q_1 jm}\C{j_1(m+q_1)}{k_2q_1 jm} \\ &= \sum_{j_1,j,m} f_{k_1j_1j}^* f_{k_2j_1j}{\frac{2j_1+1 }{2j+1}} \C{j(-m)}{k_1q_1 j_1(-m-q_1)}\C{j(-m)}{k_2q_1 j_1(-m-q_1)}\\ &= \sum_{j_1,j,m} f_{k_1j_1j}^* f_{k_2j_1j}{\frac{2j_1+1 }{2j+1}} \C{jm}{k_1q_1 j_1(m+q_1)}\C{jm}{k_2q_1 j_1(m+q_1)}. \end{aligned}\tag9 The third equaliy holds because summation over $$m$$ is symmetric about 0. Let $$f_{kj'j}=f_k\sqrt{(2j+1)/(2j'+1)}$$ where $$f_k$$ is another factor to be determined, so $$\Tr((\r {k_1}{q_1})^\dagger \r {k_2}{q_2}) = f_{k_1}^* f_{k_2} \sum_{j_1jm} \C{jm}{k_1q_1 j_1(m+q_1)}\C{jm}{k_2q_1 j_1(m+q_1)}. \tag{10}$$

Now, if $$k_1=k_2$$ I can tell the summation is $$1$$ using the orthonality of C-G coefficients. But I can't tell if it's 0 when $$k_1\neq k_2$$. What's the next step?

Actually this is baked into your sum over $$j$$. Write $$m_1=m+q_1$$ and use $$\sum_{jm} C^{jm}_{k_1q_1,j_1m_1}C^{jm}_{k_1q_1,j_1m_1}=\delta_{k_1k_2}$$ To see this, write your CG as an overlap: $$C^{jm}_{k_1q_1,j_1m_1}=\langle jm \vert k_1q_1,j_1m_1\rangle=\langle k_1q_1,j_1m_1\vert jm\rangle$$ since CG's are real. Then $$\sum_{jm} C^{jm}_{k_1q_1,j_1m_1}C^{jm}_{k_2q_1,j_1m_1}= \sum _{jm}\langle k_1q_1,j_1m_1\vert jm\rangle\langle jm \vert k_2q_1,j_1m_1\rangle\\ =\langle k_1q_1,j_1m_1\vert\left[\sum_{jm}\vert jm\rangle\langle jm \vert \right]\vert k_2q_1,j_1m_1\rangle$$ and the term in bracket is just the expansion of unity so you're left with $$\sum_{jm} C^{jm}_{k_1q_1,j_1m_1}C^{jm}_{k_1q_1,j_1m_1}= \langle k_1q_1,j_1m_1\vert k_2q_1,j_1m_1\rangle=\delta_{k_1k_2}\, .$$
• Thank you, but it confuses me whether I can exchange the sum and dirac noations when $m_1=m+q$, which is dependent of $m$. Commented Oct 19, 2023 at 23:00
• @Luessiaw the CGs are 0 unless $m_1=m+q$ so you can always insert $m_1$ as an independent index: when $m_1=m+q$ the CG is the one you want and when $m_1\ne m+q$ you are adding $0$ to the sum. Commented Oct 19, 2023 at 23:53