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      "label": "> [!theorem] <a class=\"object-permalink\" href=\"#theorem-14-1\">定理 14.1</a> · 海森堡不确定性定理 (Heisenberg uncertainty theorem)",
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        "quote": "\n\n> [!theorem] <a class=\"object-permalink\" href=\"#theorem-14-1\">定理 14.1</a> · 海森堡不确定性定理 (Heisenberg uncertainty theorem)\n>\n> （海森堡）设 $A$ 和 $B$ 为两个厄米算符，则对任意态 $| \\psi \\rangle$\n>\n> <span id=\"equation-14-6\" class=\"block-anchor\"></span>\n>\n> $$\n> \\Delta A \\Delta B \\geq \\frac {1}{2} | \\langle [ A, B ] \\rangle |\\tag{14.6}\n> $$\n>\n> 其中 $[ A , B ] = A B - B A$ 是两个算符的对易子\n\n> [!proof]- 证明\n>\n> 设\n>\n> $$\n> | \\psi_ {1} \\rangle = (A - \\langle A \\rangle I) | \\psi \\rangle , \\quad | \\psi_ {2} \\rangle = (B - \\langle B \\rangle I) | \\psi \\rangle ,\n> $$\n>\n> 使得 $\\Delta A = \\| \\psi _ { 1 } \\|$ 和 $\\Delta B = \\| \\psi _ { 2 } \\|$。利用柯西–施瓦茨不等式，\n>\n> $$\n> \\begin{aligned}\n> \\Delta A   \\Delta B &= \\| \\psi_ {1} \\|   \\| \\psi_ {2} \\| \\geq \\big | \\langle \\psi_ {1} | \\psi_ {2} \\rangle \\big | \\\\[0.4em]\n>  &\\geq \\big | \\mathrm{Im}   \\langle \\psi_ {1} | \\psi_ {2} \\rangle \\big | \\\\[0.4em]\n>  &= \\Big | \\frac {1}{2 i} \\big (\\langle \\psi_ {1} | \\psi_ {2} \\rangle - \\langle \\psi_ {2} | \\psi_ {1} \\rangle \\big) \\Big |.\n> \\end{aligned}\n> $$\n>\n> 现在\n>\n> $$\n> \\begin{aligned}\n> \\langle \\psi_ {1} | \\psi_ {2} \\rangle &= \\langle (A - \\langle A \\rangle I) \\psi | (B - \\langle B \\rangle I) \\psi \\rangle \\\\[0.4em]\n>  &= \\langle \\psi | A B | \\psi \\rangle - \\langle A \\rangle \\langle B \\rangle .\n> \\end{aligned}\n> $$\n>\n> 因此\n>\n> $$\n> \\Delta A \\Delta B \\geq \\frac {1}{2} \\big | \\langle \\psi | A B - B A | \\psi \\rangle \\big | = \\frac {1}{2} \\big | \\langle [ A, B ] \\rangle \\big |.\n> $$\n",
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        "quote": "\n\n> [!theorem] <a class=\"object-permalink\" href=\"#theorem-14-1\">Theorem 14.1</a> · Heisenberg uncertainty theorem\n>\n> (Heisenberg) Let $A$ and $B$ be two hermitian operators, thenfor any state $| \\psi \\rangle$\n>\n> <span id=\"equation-14-6\" class=\"block-anchor\"></span>\n>\n> $$\n> \\Delta A \\Delta B \\geq \\frac {1}{2} | \\langle [ A, B ] \\rangle |\\tag{14.6}\n> $$\n>\n> where $[ A , B ] = A B - B A$ is the commutator of the two operators\n\n> [!proof]- Proof\n>\n> Let\n>\n> $$\n> | \\psi_ {1} \\rangle = (A - \\langle A \\rangle I) | \\psi \\rangle , \\quad | \\psi_ {2} \\rangle = (B - \\langle B \\rangle I) | \\psi \\rangle ,\n> $$\n>\n> so that $\\Delta A = \\| \\psi _ { 1 } \\|$ and $\\Delta B = \\| \\psi _ { 2 } \\|$ . Using the Cauchy–Schwarz inequality,\n>\n> $$\n> \\begin{aligned}\n> \\Delta A   \\Delta B &= \\| \\psi_ {1} \\|   \\| \\psi_ {2} \\| \\geq \\big | \\langle \\psi_ {1} | \\psi_ {2} \\rangle \\big | \\\\[0.4em]\n>  &\\geq \\big | \\mathrm{Im}   \\langle \\psi_ {1} | \\psi_ {2} \\rangle \\big | \\\\[0.4em]\n>  &= \\Big | \\frac {1}{2 i} \\big (\\langle \\psi_ {1} | \\psi_ {2} \\rangle - \\langle \\psi_ {2} | \\psi_ {1} \\rangle \\big) \\Big |.\n> \\end{aligned}\n> $$\n>\n> Now\n>\n> $$\n> \\begin{aligned}\n> \\langle \\psi_ {1} | \\psi_ {2} \\rangle &= \\langle (A - \\langle A \\rangle I) \\psi | (B - \\langle B \\rangle I) \\psi \\rangle \\\\[0.4em]\n>  &= \\langle \\psi | A B | \\psi \\rangle - \\langle A \\rangle \\langle B \\rangle .\n> \\end{aligned}\n> $$\n>\n> Hence\n>\n> $$\n> \\Delta A \\Delta B \\geq \\frac {1}{2} \\big | \\langle \\psi | A B - B A | \\psi \\rangle \\big | = \\frac {1}{2} \\big | \\langle [ A, B ] \\rangle \\big |.\n> $$\n",
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