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        "quote": "\n\n> [!theorem] 定理（Weyl 特征标公式）\n> 对于$\\mathrm{SU}(2)$的不可约表示$(\\pi_n,V^n)$以及下面的对角元素，\n>\n> $$\n> \\chi_{V^n}\\!\\left(\\begin{pmatrix}e^{i\\theta}&0\\\\0&e^{-i\\theta}\\end{pmatrix}\\right)=\\frac{e^{i(n+1)\\theta}-e^{-i(n+1)\\theta}}{e^{i\\theta}-e^{-i\\theta}}=\\frac{\\sin((n+1)\\theta)}{\\sin\\theta}\n> $$\n\n> [!proof]- 证明\n> 由下列恒等式和方程（9.2）可得此公式：\n>\n> $$\n> \\bigl(e^{in\\theta}+e^{i(n-2)\\theta}+\\cdots+e^{-i(n-2)\\theta}+e^{-in\\theta}\\bigr)(e^{i\\theta}-e^{-i\\theta})=e^{i(n+1)\\theta}-e^{-i(n+1)\\theta}\n> $$\n",
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        "quote": "\n\n> [!theorem] Theorem (Weyl character formula)\n> For the irreducible representation $(\\pi_n,V^n)$ of $\\mathrm{SU}(2)$ and the diagonal element below,\n>\n> $$\n> \\chi_{V^n}\\!\\left(\\begin{pmatrix}e^{i\\theta}&0\\\\0&e^{-i\\theta}\\end{pmatrix}\\right)=\\frac{e^{i(n+1)\\theta}-e^{-i(n+1)\\theta}}{e^{i\\theta}-e^{-i\\theta}}=\\frac{\\sin((n+1)\\theta)}{\\sin\\theta}.\n> $$\n>\n\n> [!proof]- Proof\n> The formula follows from the identity\n>\n> $$\n> \\bigl(e^{in\\theta}+e^{i(n-2)\\theta}+\\cdots+e^{-i(n-2)\\theta}+e^{-in\\theta}\\bigr)(e^{i\\theta}-e^{-i\\theta})=e^{i(n+1)\\theta}-e^{-i(n+1)\\theta},\n> $$\n>\n> together with <a href=\"#equation-9-2\" class=\"cross-reference\" id=\"xref-04-tensor-products-of-representations-1\" data-reference=\"equation-9-2\">equation (9.2)</a>.\n",
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