スレーター積分
{\displaystyle {\begin{aligned}c^{k}(\ell ,m,\ell ',m')&=c^{k}(\ell ,-m,\ell ',-m')\\&=(-1)^{m-m'}c^{k}(\ell ',m',\ell ,m)\\&=(-1)^{m-m'}{\sqrt {\frac {2\ell +1}{2k+1}}}c^{\ell }(\ell ',m',k,m'-m)\\&=(-1)^{m'}{\sqrt {\frac {2\ell '+1}{2k+1}}}c^{\ell '}(k,m-m',\ell ,m).\\\sum _{m=-\ell }^{\ell }c^{k}(\ell ,m,\ell ,m)&=(2\ell +1)\delta _{k,0}.\\\sum _{m=-\ell }^{\ell }\sum _{m'=-\ell '}^{\ell '}c^{k}(\ell ,m,\ell ',m')^{2}&={\sqrt {(2\ell +1)(2\ell '+1)}}\cdot c^{k}(\ell ,0,\ell ',0).\\\sum _{m=-\ell }^{\ell }c^{k}(\ell ,m,\ell ',m')^{2}&={\sqrt {\frac {2\ell +1}{2\ell '+1}}}\cdot c^{k}(\ell ,0,\ell ',0).\\\sum _{m=-\ell }^{\ell }c^{k}(\ell ,m,\ell ',m')c^{k}(\ell ,m,{\tilde {\ell }},m')&=\delta _{\ell ',{\tilde {\ell }}}\cdot {\sqrt {\frac {2\ell +1}{2\ell '+1}}}\cdot c^{k}(\ell ,0,\ell ',0).\\\sum _{m}c^{k}(\ell ,m+r,\ell ',m)c^{k}(\ell ,m+r,{\tilde {\ell }},m)&=\delta _{\ell ,{\tilde {\ell }}}\cdot {\frac {\sqrt {(2\ell +1)(2\ell '+1)}}{2k+1}}\cdot c^{k}(\ell ,0,\ell ',0).\\\sum _{m}c^{k}(\ell ,m+r,\ell ',m)c^{q}(\ell ,m+r,\ell ',m)&=\delta _{k,q}\cdot {\frac {\sqrt {(2\ell +1)(2\ell '+1)}}{2k+1}}\cdot c^{k}(\ell ,0,\ell ',0).\end{aligned}}}