{"id":28514,"date":"2024-06-04T04:14:36","date_gmt":"2024-06-04T04:14:36","guid":{"rendered":"https:\/\/science-hub.click\/%E3%82%B7%E3%83%A5%E3%83%AC%E3%83%BC%E3%83%87%E3%82%A3%E3%83%B3%E3%82%AC%E3%83%BC%E6%96%B9%E7%A8%8B%E5%BC%8F%E3%83%97%E3%83%AD%E3%83%91%E3%82%B2%E3%83%BC%E3%82%BF%E3%81%AB%E3%81%A4%E3%81%84%E3%81%A6\/"},"modified":"2024-06-04T04:14:36","modified_gmt":"2024-06-04T04:14:36","slug":"%E3%82%B7%E3%83%A5%E3%83%AC%E3%83%BC%E3%83%87%E3%82%A3%E3%83%B3%E3%82%AC%E3%83%BC%E6%96%B9%E7%A8%8B%E5%BC%8F%E3%83%97%E3%83%AD%E3%83%91%E3%82%B2%E3%83%BC%E3%82%BF%E3%81%AB%E3%81%A4%E3%81%84%E3%81%A6","status":"publish","type":"post","link":"https:\/\/science-hub.click\/?p=28514","title":{"rendered":"\u30b7\u30e5\u30ec\u30fc\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306b\u3064\u3044\u3066\u8a73\u3057\u304f\u89e3\u8aac"},"content":{"rendered":"

\u5c0e\u5165<\/h2>

\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u30fc<\/b>\u3068\u3044\u3046\u7528\u8a9e\u306f\u3001\u30cf\u30df\u30eb\u30c8\u30cb\u30a2\u30f3\u306b\u57fa\u3065\u304f\u6a19\u6e96\u91cf\u5b50\u5316\u306e\u901a\u5e38\u306e\u624b\u9806\u3068\u306f\u5bfe\u7167\u7684\u306b\u3001\u30e9\u30b0\u30e9\u30f3\u30b8\u30a2\u30f3\u3092\u4e2d\u5fc3\u3068\u3057\u305f\u91cf\u5b50\u5316\u3078\u306e\u65b0\u3057\u3044\u30a2\u30d7\u30ed\u30fc\u30c1\u3067\u3042\u308b\u7d4c\u8def\u7a4d\u5206\u306b\u304a\u3051\u308b\u91cf\u5b50\u529b\u5b66\u306e\u5b9a\u5f0f\u5316\u306e\u305f\u3081\u306b\u30011948 \u5e74\u306b\u30d5\u30a1\u30a4\u30f3\u30de\u30f3\u306b\u3088\u3063\u3066\u7269\u7406\u5b66\u306b\u5c0e\u5165\u3055\u308c\u307e\u3057\u305f\u3002<\/p>

\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306f\u975e\u5e38\u306b\u4fbf\u5229\u306a\u6570\u5b66\u7684\u30c4\u30fc\u30eb<\/a><\/span>\u3067\u3042\u308a\u3001Dyson \u306b\u3088\u3063\u3066\u3059\u3050\u306bGreen \u95a2\u6570<\/a><\/span>\u4ee5\u5916\u306e\u4f55\u3082\u306e\u3067\u3082\u306a\u3044\u3068\u8b58\u5225\u3055\u308c\u307e\u3059\u3002\u3053\u306e\u767a\u8a00\u306b\u3088\u308a\u3001\u30c0\u30a4\u30bd\u30f3\u306f 1948 \u5e74\u306b\u3001\u30b7\u30e5\u30a6\u30a3\u30f3\u30ac\u30fc\u304c\u958b\u767a\u3057\u305f<\/a><\/span>\u91cf\u5b50\u96fb\u6c17\u529b\u5b66<\/a><\/span>\u306e\u62bd\u8c61\u7684\u306a\u5b9a\u5f0f\u5316\u3068\u3001\u30d5\u30a1\u30a4\u30f3\u30de\u30f3\u304c\u72ec\u81ea\u306b\u767a\u660e\u3057\u305f\u3001\u56f3\u306b\u57fa\u3065\u3044\u305f\u91cf\u5b50\u96fb\u6c17\u529b\u5b66\u306e\u62bd\u8c61\u7684\u306a\u5b9a\u5f0f\u5316<\/a><\/span>\u3068\u306e\u9593\u306e\u30df\u30c3\u30b7\u30f3\u30b0 \u30ea\u30f3\u30af\u3092\u4f5c\u6210\u3059\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3057\u305f\u3002<\/p>

\n\"\u30b7\u30e5\u30ec\u30fc\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306b\u3064\u3044\u3066\u8a73\u3057\u304f\u89e3\u8aac\"<\/figure>

\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf<\/h2>

\u5c0e\u5165<\/span><\/h3>

1\u6b21\u5143<\/a><\/span>\u3067\u8cea\u91cf<\/a><\/span>m<\/i><\/span>\u306e\u975e\u76f8\u5bfe\u8ad6\u7684\u7c92\u5b50\u3092\u8003\u3048\u307e\u3059\u3002\u305d\u306e\u30cf\u30df\u30eb\u30c8\u30cb\u30a2\u30f3\u6f14\u7b97\u5b50\u306f<\/a><\/span>\u6b21\u306e\u3088\u3046\u306b\u8a18\u8ff0\u3055\u308c\u307e\u3059\u3002 <\/p>
$$ {\\hat{H} \\ = \\ \\frac{\\hat{p}^2}{2m} \\ + \\ V(\\hat{q}) } $$<\/div><\/td><\/tr><\/table>

\u30b7\u30e5\u30ec\u30c7\u30a3\u30f3\u30ac\u30fc\u8868\u73fe\u3067\u306f\u3001\u3053\u306e\u7c92\u5b50\u306f\u30b1\u30c3\u30c8\u306b\u3088\u3063\u3066\u8a18\u8ff0\u3055\u308c\u307e\u3059\u3002

$$ {| \\psi(t) \\rangle} $$<\/div>\u3053\u308c\u306f\u30b7\u30e5\u30ec\u30fc\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u306b\u5f93\u3044\u307e\u3059\u3002 <\/p>
$$ {i \\hbar \\ \\frac{d | \\psi(t) \\rangle}{dt} \\ = \\ \\hat{H} \\ | \\psi(t) \\rangle} $$<\/div><\/td><\/tr><\/table>

\u6700\u521d\u306e\u77ac\u9593<\/a><\/span>t<\/i> 0<\/sub><\/span>\u3067\u521d\u671f\u6761\u4ef6\u3092\u56fa\u5b9a\u3059\u308b\u3068\u3057\u307e\u3059\u3002

$$ {| \\psi(t_0) \\rangle } $$<\/div> \u3001\u305d\u3057\u3066\u6f14\u7b97\u5b50\u304c
$$ {\\ \\hat{H}} $$<\/div>\u306f\u6642\u9593\u306b\u4f9d\u5b58\u3057\u306a\u3044\u305f\u3081\u3001\u5f8c\u7d9a\u306e\u6642\u9593t<\/i> > t<\/i> 0<\/sub><\/span>\u306b\u304a\u3051\u308b\u30b7\u30e5\u30ec\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u306e\u89e3\u306f\u6b21\u306e\u3088\u3046\u306b\u66f8\u304f\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002 <\/p>
$$ {| \\psi(t) \\rangle \\ = \\ e^{-i\\hat{H} (t-t_0) \/\\hbar} \\ | \\psi(t_0) \\rangle} $$<\/div><\/td><\/tr><\/table>

\u3053\u306e\u65b9\u7a0b\u5f0f\u3092\u4f4d\u7f6e\u8868\u73fe\u306b\u6295\u5f71\u3057\u3066\u307f\u307e\u3057\u3087\u3046\u3002 <\/p>
$$ { \\langle q | \\psi(t) \\rangle \\ = \\ \\langle q |e^{-i\\hat{H} (t-t_0) \/\\hbar} \\ | \\psi(t_0) \\rangle} $$<\/div><\/td><\/tr><\/table>

\u305d\u3057\u3066\u3001\u53f3\u5074\u306e\u9805\u306b\u9589\u5305<\/a><\/span>\u95a2\u4fc2\u3092\u633f\u5165\u3057\u307e\u3059\u3002 <\/p>
$$ {1 \\ = \\ \\int dq_0 \\ | q_0 \\rangle \\ \\langle q_0 |} $$<\/div><\/td><\/tr><\/table>

\u5f7c\u306f\u6765\u307e\u3059\uff1a <\/p>
$$ { \\langle q | \\psi(t) \\rangle \\ = \\ \\int dq_0 \\ \\langle q |e^{-i\\hat{H} (t-t_0) \/\\hbar} \\ | q_0 \\rangle \\ \\langle q_0 | \\psi(t_0) \\rangle} $$<\/div><\/td><\/tr><\/table>

\u3068\u3044\u3046\u4e8b\u5b9f\u3092\u8003\u616e\u3059\u308b\u3068\u3001

$$ {\\langle q | \\psi(t) \\rangle = \\psi(q,t)} $$<\/div> \u3001\u524d\u306e\u5f0f\u306f\u6b21\u306e\u5f62\u5f0f\u3067\u8a18\u8ff0\u3055\u308c\u307e\u3059\u3002 <\/p>
$$ { \\psi(q,t) \\ = \\ \\int dq_0 \\ \\langle q |e^{-i\\hat{H} (t-t_0) \/\\hbar} |q_0 \\rangle \\ \\psi(q_0,t_0) } $$<\/div><\/td><\/tr><\/table>
\n\"\u30b7\u30e5\u30ec\u30fc\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306b\u3064\u3044\u3066\u8a73\u3057\u304f\u89e3\u8aac\"<\/figure>

\u610f\u5473<\/a><\/span><\/span><\/h3>

\u30b7\u30e5\u30ec\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u306e\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u3092<\/strong>\u6b21\u306e\u3088\u3046\u306b\u5b9a\u7fa9\u3057\u307e\u3059\u3002 <\/p>

$$ {{K(q,t|q_0,t_0) \\ = \\ }} $$<\/q><\/div><\/tr><\/table><\/center>

\u6ce2\u52d5\u95a2\u6570\u306f\u7a4d\u5206\u65b9\u7a0b\u5f0f\u306b\u5f93\u3063\u3066\u767a\u5c55\u3057\u307e\u3059\u3002 <\/p>

$$ { \\psi(q,t) \\ = \\ \\int dq_0 \\ K(q,t|q_0,t_0) \\ \\psi(q_0,t_0) } $$<\/div><\/td><\/tr><\/table><\/center>

\u6c17\u3065\u3044\u305f<\/span><\/h3>

\u03c8( q<\/i> , t<\/i> ) \u306f<\/span>\u30b7\u30e5\u30ec\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u306e\u89e3\u3067\u3042\u308b\u305f\u3081\u3001\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u3082\u3053\u306e\u65b9\u7a0b\u5f0f\u306e\u89e3\u306b\u306a\u308a\u307e\u3059\u3002 <\/p>
$$ {i \\hbar \\ \\frac{\\partial K(q,t|q_0,t_0) }{\\partial t} \\ = \\ – \\ \\frac{\\hbar^2}{2m} \\ \\Delta_q \\ K(q,t|q_0,t_0) \\ + \\ V(q) \\ K(q,t|q_0,t_0)} $$<\/div><\/td><\/tr><\/table>

\u521d\u671f\u6761\u4ef6<\/i>\u3082\u691c\u8a3c\u3059\u308b\u5fc5\u8981\u304c\u3042\u308a\u307e\u3059\u3002 <\/p>
$$ {\\lim_{t \\to t_0} K(q,t|q_0,t_0) \\ = \\ \\delta(q-q_0)} $$<\/div><\/td><\/tr><\/table>

\u3053\u306e\u5834\u5408\u3001\u6570\u5b66\u8005\u306f\u30b7\u30e5\u30ec\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u306e\u521d\u7b49\u89e3<\/i>\u306b\u3064\u3044\u3066\u8a9e\u308a\u3001\u7269\u7406\u5b66\u8005\u306f\u4ee3\u308f\u308a\u306b\u30b0\u30ea\u30fc\u30f3\u95a2\u6570<\/i>\u3068\u3044\u3046\u540d\u524d\u3092\u4f7f\u7528\u3057\u307e\u3059\u3002<\/p>

\u81ea\u7531\u7c92\u5b50\u306e\u4f1d\u64ad\u4f53\u306e\u8868\u73fe<\/h2>

\u30d5\u30fc\u30ea\u30a8\u5909\u63db\u306b\u3064\u3044\u3066\u306e\u6ce8\u610f\u4e8b\u9805<\/span><\/h3>

\u4ee5\u4e0b\u306e\u95a2\u4fc2\u3092\u601d\u3044\u51fa\u3057\u3066\u304f\u3060\u3055\u3044\u3002 <\/p>

$$ { \\hat{\\psi}(p) \\ = \\ \\int \\frac{dq}{\\sqrt{2 \\pi \\hbar}} \\ e^{\\, – \\, i p q\/\\hbar} \\ \\psi(q) } $$<\/div><\/dd><\/dl>
$$ { \\psi(q) \\ = \\ \\int \\frac{dp}{\\sqrt{2 \\pi \\hbar}} \\ e^{\\, + \\, i p q\/\\hbar} \\ \\hat{\\psi}(p) } $$<\/div><\/dd><\/dl>

\u30c7\u30a3\u30e9\u30c3\u30af\u8a18\u6cd5\u3092\u4f7f\u7528\u3057\u3001\u30a4\u30f3\u30d1\u30eb\u30b9\u306e\u9589\u5305\u95a2\u4fc2\u3092\u4f7f\u7528\u3057\u307e\u3059\u3002 <\/p>

$$ {1 \\ = \\ \\int dp \\ | p > \\ < p |} $$<\/div><\/dl>

2 \u756a\u76ee\u306e\u95a2\u4fc2\u306f\u6b21\u306e\u3088\u3046\u306b<\/a><\/span>\u8a18\u8ff0\u3055\u308c\u307e\u3059\u3002 <\/p>

$$ { < q | \\psi > \\ = \\ \\int \\frac{dp}{\\sqrt{2 \\pi \\hbar}} \\ e^{\\, + \\, i p q\/\\hbar} \\

\\ = \\ \\int dp \\ < q | p > \\

} $$<\/p><\/p><\/div><\/dl>

\u6b21\u306e\u5f0f\u3092\u5c0e\u304d\u51fa\u3057\u307e\u3059\u3002 <\/p>

$$ { < q | p > \\ = \\ \\frac{e^{\\, + \\, i p q\/\\hbar}}{\\sqrt{2 \\pi \\hbar}} \\ } $$<\/div><\/dl>
\n\"\u30b7\u30e5\u30ec\u30fc\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306b\u3064\u3044\u3066\u8a73\u3057\u304f\u89e3\u8aac\"<\/figure>

\u81ea\u7531\u7c92\u5b50\u306e\u4f1d\u64ad\u4f53\u306e\u8868\u73fe<\/span><\/h3>

\u53f3\u5074\u306e\u81ea\u7531\u7c92\u5b50\u306e\u5834\u5408\u3001\u30cf\u30df\u30eb\u30c8\u30cb\u30a2\u30f3\u6f14\u7b97\u5b50\u306f\u4f4d\u7f6e\u306b\u4f9d\u5b58\u3057\u307e\u305b\u3093\u3002 <\/p>

$$ {\\hat{H} \\ = \\ \\frac{\\hat{p}^2}{2m}} $$<\/div><\/dd><\/dl>

\u6b21\u306b\u3001\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf (\u3053\u306e\u5834\u5408\u306fK<\/i> 0<\/sub><\/span>\u3068\u3057\u307e\u3059) \u306f\u6b21\u306e\u3088\u3046\u306b\u66f8\u304b\u308c\u307e\u3059\u3002 <\/p>

$$ { K_0 (q,t|q_0,t_0) \\ = \\ } $$<\/q><\/div><\/dl>

\u6b21\u306b\u3001\u30a4\u30f3\u30d1\u30eb\u30b9\u306e\u9589\u5305\u95a2\u4fc2\u3092\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306e\u5b9a\u7fa9\u306b 2 \u56de\u633f\u5165\u3057\u307e\u3057\u3087\u3046\u3002 <\/p>

$$ {K_0 (q,t|q_0,t_0) \\ = \\ \\int dp \\int dp_0 \\ \\ <\/q> \\ <\/q>} $$<\/q><\/p_0|q_0><\/q><\/p|e^{-i\\hat{p}^2(t-t_0)><\/q><\/div><\/dl>

\u30b1\u30c3\u30c8| p<\/i> 0<\/sub> ><\/span>\u306f\u5b9a\u7fa9\u306b\u3088\u308a\u30a4\u30f3\u30d1\u30eb\u30b9\u6f14\u7b97\u5b50\u306e\u56fa\u6709\u72b6\u614b\u3067\u3042\u308b

$$ {\\hat{p}} $$<\/div> \u3001 \u6211\u3005\u306f\u6301\u3063\u3066\u3044\u307e\u3059\uff1a <\/p>
$$ {\\hat{p}\\, | p_0 > \\ = \\ p_0 \\, |p_0 >} $$<\/div><\/dl>

\u884c\u5217\u8981\u7d20\u306f\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002 <\/p>

$$ { \\ = \\ e^{-ip_0^2(t-t_0)\/ (2m\\hbar)} \\ } $$<\/p|><\/p|e^{-i\\hat{p}^2(t-t_0)><\/div><\/dl>

< p<\/i> | p<\/i> 0<\/sub> > = \u03b4( p<\/i> \u2212 p<\/i> 0<\/sub> )<\/span> \u3001\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306b\u3064\u3044\u3066\u306f\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002 <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\int dp \\ \\ e^{-ip^2(t-t_0)\/ (2m\\hbar)} \\ <\/q>} $$<\/q><\/p|q_0><\/q><\/div><\/dl>

\u30d5\u30fc\u30ea\u30a8\u5909\u63db\u3067\u4ee5\u524d\u306b\u793a\u3057\u305f\u5f0f\u3092\u8003\u616e\u3059\u308b\u3068\u3001\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002 <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\int dp \\ \\frac{e^{\\, + \\, i p q\/\\hbar}}{\\sqrt{2 \\pi \\hbar}} \\ \\times \\ e^{-ip^2(t-t_0)\/ (2m\\hbar)} \\ \\times \\ \\frac{e^{\\, – \\, i p q_0\/\\hbar}}{\\sqrt{2 \\pi \\hbar}}} $$<\/div><\/dd><\/dl>

\u3053\u308c\u306f\u66f8\u304d\u63db\u3048\u3089\u308c\u307e\u3059: <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\int \\frac{dp}{2 \\pi \\hbar} \\ \\exp \\left[ \\, \\frac{i p (q-q_0)}{\\hbar} \\ – \\ \\frac{ip^2(t-t_0)}{2m\\hbar} \\, \\right] } $$<\/div><\/dd><\/dl>

\u6307\u6570<\/a><\/span>\u5f15\u6570\u306f\u6b21\u306e\u3088\u3046\u306b\u66f8\u304d\u63db\u3048\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002 <\/p>

$$ {\\frac{i p (q-q_0)}{\\hbar} \\ – \\ \\frac{ip^2(t-t_0)}{2m\\hbar} \\ = \\ – \\ \\frac{i (t-t_0)}{2m\\hbar} \\ \\times \\ \\left[ \\ p^2 \\ – \\ \\frac{2mp(q-q_0)}{(t-t_0)} \\ \\right]} $$<\/div><\/dd><\/dl>

\u30d5\u30c3\u30af\u306f\u5b8c\u5168\u306a\u6b63\u65b9\u5f62<\/a><\/span>\u306e\u59cb\u307e\u308a\u3067\u3059\u3002 <\/p>

$$ {p^2 \\ – \\ \\frac{2mp(q-q_0)}{(t-t_0)} \\ = \\ \\left[ \\ p \\ – \\ \\frac{m(q-q_0)}{(t-t_0)} \\ \\right]^2 \\ – \\ \\frac{m^2(q-q_0)^2}{(t-t_0)^2}} $$<\/div><\/dd><\/dl>

\u3057\u305f\u304c\u3063\u3066\u3001\u6307\u6570\u95a2\u6570\u306e\u5f15\u6570\u306f\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002 <\/p>

$$ {- \\ \\frac{i (t-t_0)}{2m\\hbar} \\ \\times \\ \\left[ \\ \\left( \\ p \\ – \\ \\frac{m(q-q_0)}{(t-t_0)} \\ \\right)^2 \\ – \\ \\frac{m^2(q-q_0)^2}{(t-t_0)^2} \\right]} $$<\/div><\/dd><\/dl>
$$ {= \\ – \\ \\frac{i (t-t_0)}{2m\\hbar} \\ \\left( \\ p \\ – \\ \\frac{m(q-q_0)}{(t-t_0)} \\ \\right)^2 \\ + \\ \\frac{i m(q-q_0)^2}{2 \\hbar (t-t_0)} } $$<\/div><\/dd><\/dl>

\u6700\u5f8c\u306e\u9805\u306f\u30a4\u30f3\u30d1\u30eb\u30b9\u304b\u3089\u72ec\u7acb\u3057\u3066\u304a\u308a\u3001\u7a4d\u5206<\/a><\/span>\u3092\u96e2\u308c\u3001\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306f\u6b21\u306e\u3088\u3046\u306b\u66f8\u304b\u308c\u307e\u3059\u3002 <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\exp \\left( \\frac{i m(q-q_0)^2}{2 \\hbar (t-t_0)} \\right) \\ \\times \\ \\int \\frac{dp}{2 \\pi \\hbar} \\ \\exp \\left[ \\, – \\ \\frac{i (t-t_0)}{2m\\hbar} \\ \\left( \\ p \\ – \\ \\frac{m(q-q_0)}{(t-t_0)} \\ \\right)^2 \\, \\right] } $$<\/div><\/dd><\/dl>

\u30d1\u30eb\u30b9\u306e\u5909\u6570\u3092<\/a><\/span>\u5909\u66f4\u3057\u3001\u4ed6\u306e\u30d1\u30e9\u30e1\u30fc\u30bf\u306f\u56fa\u5b9a\u3057\u307e\u3059\u3002 <\/p>

$$ {p \\ \\longrightarrow \\ k \\ = \\ p \\ – \\ \\frac{m(q-q_0)}{(t-t_0)} \\ \\Longrightarrow \\ dp \\ \\longrightarrow \\ dk \\ = \\ dp } $$<\/div><\/dd><\/dl>

\u3053\u308c\u306f\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059: <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\frac{1}{2 \\pi \\hbar} \\ \\exp \\left( \\frac{i m(q-q_0)^2}{2 \\hbar (t-t_0)} \\right) \\ \\times \\ \\int dk \\ \\exp \\left[ \\, – \\ \\frac{i (t-t_0) k^2}{2m\\hbar} \\, \\right]} $$<\/div><\/dd><\/dl>

\u6b63\u78ba\u306b\u8a08\u7b97\u3055\u308c\u305f\u30ac\u30a6\u30b9\u7a4d\u5206\u304c\u6b8b\u308a\u307e\u3059\u3002 <\/p>

$$ {\\int dk \\ e^{- \\alpha k^2} \\ = \\ \\sqrt{\\frac{\\pi}{\\alpha}}} $$<\/div><\/dd><\/dl>

\u79c1\u305f\u3061\u306f\u6b21\u306e\u3088\u3046\u306b\u63a8\u6e2c\u3057\u307e\u3059\u3002 <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\frac{1}{2 \\pi \\hbar} \\ \\sqrt{\\frac{2\\pi m \\hbar}{i(t-t_0)}} \\ \\exp \\left( \\frac{ + i m(q-q_0)^2}{2 \\hbar (t-t_0)} \\right)} $$<\/div><\/dd><\/dl>

\u3057\u305f\u304c\u3063\u3066\u3001\u81ea\u7531\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306e\u6700\u7d42\u7684\u306a\u5f0f\u306f\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002 <\/p>

$$ {K_0(q,t|q_0,t_0) \\ = \\ \\sqrt{\\frac{m}{2 \\pi i \\hbar (t-t_0)}} \\ \\exp \\left( \\frac{ + i m(q-q_0)^2}{2 \\hbar (t-t_0)} \\right)} $$<\/div><\/td><\/tr><\/table><\/center>

\u6c17\u3065\u3044\u305f<\/span><\/h3>

d<\/i>\u6b21\u5143\u30e6\u30fc\u30af\u30ea\u30c3\u30c9\u7a7a\u9593<\/a><\/span>\u306e\u81ea\u7531\u7c92\u5b50\u306e\u5834\u5408\u3001\u6b21\u306e\u3053\u3068\u3092\u540c\u69d8\u306b\u8a3c\u660e\u3067\u304d\u307e\u3059\u3002<\/p>

<\/center><\/div>

\u53c2\u8003\u8cc7\u6599<\/h2>
  1. Propagator \u2013 allemand<\/a><\/li>
  2. Propagator \u2013 anglais<\/a><\/li>
  3. Propagilo \u2013 esp\u00e9ranto<\/a><\/li>
  4. Propagador \u2013 espagnol<\/a><\/li>
  5. \u05e4\u05e8\u05d5\u05e4\u05d2\u05d8\u05d5\u05e8 \u2013 h\u00e9breu<\/a><\/li>
  6. Propag\u00e1tor \u2013 hongrois<\/a><\/li><\/ol><\/div>\n
    \n

    \n \u30b7\u30e5\u30ec\u30fc\u30c7\u30a3\u30f3\u30ac\u30fc\u65b9\u7a0b\u5f0f\u30d7\u30ed\u30d1\u30b2\u30fc\u30bf\u306b\u3064\u3044\u3066\u8a73\u3057\u304f\u89e3\u8aac\u30fb\u95a2\u9023\u52d5\u753b\n <\/h2>\n
    \n \n
    \n
    \n