$∑_{|m_1|≤j_1,|m_2|≤j_2,|m_1+m_2|≤j_1+j_2}c_1(\{j_1,m_1\},\{j_2,m_2\})$

hbghlyj

$j_1,j_2\inN,$
设\[c_1(\{j_1,m_1\},\{j_2,m_2\}):=\frac{(2 j_1)! (2 j_2)! (j_1+j_2-m_1-m_2)! (j_1+j_2+m_1+m_2)!}{(2 j_1+2 j_2)! (j_1-m_1)! (j_1+m_1)! (j_2-m_2)! (j_2+m_2)!}\]
证明
\[\tag1\label1\sum_{\begin{subarray}{rl}
|m_1|&\le j_1\\
|m_2|&\le j_2\\
|m_1+m_2|&\le j_1+j_2\end{subarray}} c_1(\{j_1,m_1\},\{j_2,m_2\}) =2 j_1+2 j_2+1\]

1. c1[{j1_,m1_},{j2_,m2_}]:=((2 j1)! (2 j2)! (j1+j2-m1-m2)! (j1+j2+m1+m2)!)/((j1-m1)! (j1+m1)! (j2-m2)! (j2+m2)! (2j1+2j2)!);
2. With[{j1=4,j2=5},Sum[If[Abs[m1+m2]<=j1+j2,c1[{j1,m1},{j2,m2}],0],{m1,-j1,j1},{m2,-j2,j2}]-2j1-2j2-1]

Copy the Code

tommywong

artofproblemsolving.com/community/c728438h176 … binomial_coefficient
$\displaystyle \sum_{n_i\ge m_i\atop n_1+n_2+\dots+n_k =n}
\binom{n_1}{m_1}\binom{n_2}{m_2}\dots\binom{n_k}{m_k}=
\binom{n+k-1}{\sum m_i+k-1}$
$\displaystyle \sum_{m_2=-j_2}^{j_2}
\binom{j_1-m_1+j_2-m_2}{j_1-m_1}\binom{j_1+m_1+j_2+m_2}{j_1+m_1}=\binom{2j_1+2j_2+1}{2j_1+1}$

$\displaystyle \sum_{|m_1|\le j_1\atop {|m_2|\le j_2\atop |m_1+m_2|\le j_1+j_2}}
=\sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2}$

$\displaystyle \sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2}
\frac{(2 j_1)! (2 j_2)! (j_1+j_2-m_1-m_2)! (j_1+j_2+m_1+m_2)!}{(2 j_1+2 j_2)! (j_1-m_1)! (j_1+m_1)! (j_2-m_2)! (j_2+m_2)!}$
$\displaystyle =\frac{(2 j_1)! (2 j_2)!}{(2 j_1+2 j_2)!}\sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2}
\binom{j_1-m_1+j_2-m_2}{j_1-m_1}\binom{j_1+m_1+j_2+m_2}{j_1+m_1}$
$\displaystyle =\frac{(2 j_1)! (2 j_2)!}{(2 j_1+2 j_2)!}\sum_{m_1=-j_1}^{j_1}\binom{2j_1+2j_2+1}{2j_1+1}$
$\displaystyle =\frac{(2j_1+1)(2 j_1)! (2 j_2)!}{(2 j_1+2 j_2)!}\binom{2j_1+2j_2+1}{2j_1+1}$
$=2j_1+2j_2+1$

hbghlyj

类似的：$j_1,j_2\inN,j_1\ge j_2$
设\[c_2(\{j_1,m_1\},\{j_2,m_2\}):=\frac{(2 j_2)! (2 j_1-2 j_2+1)! (j_1-m_1)! (j_1+m_1)!}{(2 j_1+1)! (j_2-m_2)! (j_2+m_2)! (j_1-j_2-m_1-m_2)! (j_1-j_2+m_1+m_2)!}\]
证明
\[\tag2\label2\sum_{\begin{subarray}{rl}
|m_1|&\le j_1\\
|m_2|&\le j_2\\
|m_1+m_2|&\le j_1-j_2\end{subarray}} c_2(\{j_1,m_1\},\{j_2,m_2\}) =2 j_1-2 j_2+1\]
\[\tag3\label3\sum_{\begin{subarray}{rl}
|m_1|&\le j_1\\
|m_2|&\le j_2\\
|m_1+m_2|&\le j_1-j_2\end{subarray}}(-1)^{j_2+m_2}\sqrt{c_1(\{j_1,m_1\},\{j_2,m_2\})c_2(\{j_1,m_1\},\{j_2,m_2\})} =0\]

1. c2[{j1_,m1_},{j2_,m2_}]:= (((2 j1-2 j2+1)! (2 j2)! (j1-m1)! (j1+m1)!)/((1+2 j1)! (j2-m2)! (j1-j2-m1-m2)! (j2+m2)! (j1-j2+m1+m2)!));
2. With[{j1=4,j2=3},Sum[If[Abs[m1+m2]<=j1-j2,c2[{j1,m1},{j2,m2}],0],{m1,-j1,j1},{m2,-j2,j2}]-1-2j1+2j2]
3. With[{j1=4,j2=3},Sum[If[Abs[m1+m2]<=j1-j2,(-1)^(j2+m2)Sqrt[c1[{j1,m1},{j2,m2}]c2[{j1,m1},{j2,m2}]],0],{m1,-j1,j1},{m2,-j2,j2}]]

Copy the Code

模仿@tommywong来证明\eqref{2}:
$\displaystyle\sum_{m_2=-j_2}^{j_2}\sum_{m_1=-j_1}^{j_1}
c_2(\{j_1,m_1\},\{j_2,m_2\})$
$\displaystyle =\frac1{\binom{2 j_1+1}{2j_2}}\sum_{m_2=-j_2}^{j_2}\sum_{m_1=-j_1}^{j_1}\binom{j_1-m_1}{j_2+m_2}\binom{j_1+m_1}{j_2-m_2}$
$\displaystyle =\frac1{\binom{2 j_1+1}{2j_2}}\sum_{m_2=-j_2}^{j_2}\binom{2j_1+1}{2j_2+1}$
$\displaystyle =\frac1{\binom{2 j_1+1}{2j_2}}(2j_2+1)\binom{2j_1+1}{2j_2+1}$
$=2j_1-2j_2+1$

那么如何证明\eqref{3}

tommywong

吓？點解喺根號入面嘅？唔識喎😲

hbghlyj

tommywong 发表于 2025-3-8 10:22
吓？點解喺根號入面嘅？唔識喎😲

在平方根中是上面定义的两个函数 $c_1, c_2$ 的乘积
我在 mathematica 中测试过，\eqref{3}是正确的

hbghlyj

等式 \eqref{3} 化简得
\[\sum_{\begin{subarray}{rl}
|m_1|&\le j_1\\
|m_2|&\le j_2\\
|m_1+m_2|&\le j_1-j_2\end{subarray}}(-1)^{j_2+m_2}\frac{(2 j_2)!}{(j_2-m_2)! (j_2+m_2)!} \sqrt{\frac{(2 j_1-2 j_2+1)! (j_1+j_2-m_1-m_2)!(j_1+j_2+m_1+m_2)!}{(j_1-j_2-m_1-m_2)! (j_1-j_2+m_1+m_2)!(2 j_1+2j_2)!(2 j_1+1)}}=0\]

1. With[{j1=4,j2=3},Sum[If[Abs[m1+m2]<=j1-j2,(-1)^(j2+m2) (2 j2)!  /((j2-m2)! (j2+m2)!) \[Sqrt](((2 j1-2 j2+1)! (j1+j2-m1-m2)!(j1+j2+m1+m2)!)/((j1-j2-m1-m2)! (j1-j2+m1+m2)!(2 j1+2j2)!(2 j1+1) )),0],{m1,-j1,j1},{m2,-j2,j2}]]

Copy the Code

还能化简吗