{"topic":{"id":6869,"title":"建一个理科主题水楼！","slug":"topic","categoryId":4,"createdAt":"2023-10-29 06:37:10.594279","bumpedAt":"2026-06-05 11:40:40.472164","views":1459,"postsCount":40,"highestPostNumber":43,"likeCount":94,"closed":false,"pinned":false,"pinnedGlobally":false,"pinnedAt":null,"pinnedUntil":null,"isBanner":false,"excerpt":"各种理科主题的科普、讨论、闲聊、灌水、钓鱼、发疯等等都行！","deletedAt":null,"isAnonymous":false,"category":{"id":4,"name":"谈笑风生","slug":"general","color":"25AAE2"},"author":{"id":1707,"username":"wbzd","name":"我不知道","avatarUrl":"/uploads/image/29/1f/291f099d72b6e1ab6e873cd1b0fe4d2f133a9e713c066ea299ae3040b0e2ef27.avif"},"tags":[]},"posts":[{"id":15268,"postNumber":1,"raw":"各种理科主题的科普、讨论、闲聊、灌水、钓鱼、发疯等等都行！","cooked":"

各种理科主题的科普、讨论、闲聊、灌水、钓鱼、发疯等等都行！

\n","createdAt":"2023-10-29 06:37:10.680273","updatedAt":"2023-10-29 06:37:10.680273","replyToPostNumber":null,"replyCount":0,"likeCount":4,"wordCount":29,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":4,"users":[{"id":1798,"username":"sooooog","name":"爱你越久我越背痛"},{"id":1817,"username":"anon1678451","name":null},{"id":1863,"username":"交大椒猫","name":"兴庆宫第一皮革兔"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1707,"username":"wbzd","name":"我不知道","avatarUrl":"/uploads/image/29/1f/291f099d72b6e1ab6e873cd1b0fe4d2f133a9e713c066ea299ae3040b0e2ef27.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15269,"postNumber":2,"raw":"先来个益智小问题\n\n给定一个圆和一条直径，但不给圆心，能否仅用无刻度直尺作图找出其圆心？","cooked":"

先来个益智小问题

给定一个圆和一条直径，但不给圆心，能否仅用无刻度直尺作图找出其圆心？

\n","createdAt":"2023-10-29 06:37:52.646042","updatedAt":"2023-10-29 06:37:52.646042","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":42,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1707,"username":"wbzd","name":"我不知道","avatarUrl":"/uploads/image/29/1f/291f099d72b6e1ab6e873cd1b0fe4d2f133a9e713c066ea299ae3040b0e2ef27.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15270,"postNumber":3,"raw":"东边钓不到，来西边钓是吧","cooked":"

东边钓不到，来西边钓是吧

\n","createdAt":"2023-10-29 06:54:10.230435","updatedAt":"2023-10-29 06:54:10.230435","replyToPostNumber":null,"replyCount":1,"likeCount":8,"wordCount":12,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"🐈️","count":1,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"}],"hasReacted":false},{"reaction":"🤣️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false},{"reaction":"❤️","count":8,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"},{"id":1707,"username":"wbzd","name":"我不知道"},{"id":1817,"username":"anon1678451","name":null},{"id":1835,"username":"Dusk","name":"夕大画家"},{"id":1850,"username":"Reina","name":"星野露比"},{"id":5083,"username":"Puinn","name":null},{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1701,"username":"sincerely","name":"等风也等你","avatarUrl":"/uploads/image/42/f9/42f93baa6a22349c4a71d805c55c5ab69434d8790f4c2a29249c0a1143b952c2.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15271,"postNumber":4,"raw":"你说的对但是教我拓扑","cooked":"

你说的对但是教我拓扑

\n","createdAt":"2023-10-29 06:54:55.716735","updatedAt":"2023-10-29 06:54:55.716735","replyToPostNumber":3,"replyCount":1,"likeCount":1,"wordCount":10,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1707,"username":"wbzd","name":"我不知道","avatarUrl":"/uploads/image/29/1f/291f099d72b6e1ab6e873cd1b0fe4d2f133a9e713c066ea299ae3040b0e2ef27.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15272,"postNumber":5,"raw":"拓扑学不了一点 :pouting_cat:","cooked":"

拓扑学不了一点 😾

\n","createdAt":"2023-10-29 06:55:57.358755","updatedAt":"2023-10-29 06:55:57.358755","replyToPostNumber":4,"replyCount":0,"likeCount":3,"wordCount":8,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1707,"username":"wbzd","name":"我不知道"},{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1701,"username":"sincerely","name":"等风也等你","avatarUrl":"/uploads/image/42/f9/42f93baa6a22349c4a71d805c55c5ab69434d8790f4c2a29249c0a1143b952c2.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15273,"postNumber":6,"raw":"来个有点难度的，证明:\n$$\n\\int_0^{+\\infty}\\frac{\\mathrm dx}{\\sqrt{x^4+25x^2+160}}=\\int_0^{+\\infty}\\frac{\\mathrm dx}{\\sqrt{x^4-95x^2+2560}}\n$$","cooked":"

来个有点难度的，证明:
\n

\n\\int_0^{+\\infty}\\frac{\\mathrm dx}{\\sqrt{x^4+25x^2+160}}=\\int_0^{+\\infty}\\frac{\\mathrm dx}{\\sqrt{x^4-95x^2+2560}}\n

\n","createdAt":"2023-10-29 06:56:39.228862","updatedAt":"2023-10-29 06:56:39.228862","replyToPostNumber":null,"replyCount":2,"likeCount":5,"wordCount":16,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"😅️","count":1,"users":[{"id":1288,"username":"LittleWangInShanghai","name":"小王在上海"}],"hasReacted":false},{"reaction":"❤️","count":5,"users":[{"id":1716,"username":"Mk_XVII_CQC","name":"CQC"},{"id":1746,"username":"Vasavi_Shakti","name":"赞美太阳菜菜酱"},{"id":1817,"username":"anon1678451","name":null},{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1707,"username":"wbzd","name":"我不知道","avatarUrl":"/uploads/image/29/1f/291f099d72b6e1ab6e873cd1b0fe4d2f133a9e713c066ea299ae3040b0e2ef27.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15280,"postNumber":8,"raw":"I is fish","cooked":"

I is fish

\n","createdAt":"2023-10-29 11:25:53.014339","updatedAt":"2023-10-29 11:25:53.014339","replyToPostNumber":6,"replyCount":0,"likeCount":3,"wordCount":3,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1707,"username":"wbzd","name":"我不知道"},{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1724,"username":"Philia","name":"耳玖","avatarUrl":"/uploads/image/c5/ed/c5edfff4ff47377a09cf6d694ecda658436fce7b6d02c364489ecac6551955e1.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15283,"postNumber":9,"raw":":whale:！\n好久不见","cooked":"

🐋！
\n好久不见

\n","createdAt":"2023-10-29 11:29:51.315277","updatedAt":"2023-10-29 11:29:51.315277","replyToPostNumber":7,"replyCount":0,"likeCount":4,"wordCount":5,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":4,"users":[{"id":1707,"username":"wbzd","name":"我不知道"},{"id":1746,"username":"Vasavi_Shakti","name":"赞美太阳菜菜酱"},{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1744,"username":"qianbaidu","name":"千百度","avatarUrl":"/uploads/image/1d/c8/1dc82b9526a1201b0554ef1fc6e81a507b2982d96c8741683f59d1c76d940b57.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15284,"postNumber":10,"raw":"前排顶帖，最优化学不了一点😾","cooked":"

前排顶帖，最优化学不了一点😾

\n","createdAt":"2023-10-29 11:30:44.500571","updatedAt":"2023-10-29 11:30:44.500571","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":14,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"😅️","count":1,"users":[{"id":1288,"username":"LittleWangInShanghai","name":"小王在上海"}],"hasReacted":false},{"reaction":"❤️","count":2,"users":[{"id":1707,"username":"wbzd","name":"我不知道"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1709,"username":"One_Yuan","name":"一块钱","avatarUrl":"/uploads/image/48/38/483867d97065f3344ba0f2600c3c35f7f5933cb85fbabccc13640bb78763f32a.avif","admin":false,"moderator":false,"signatureRaw":"SJTUer","signatureCooked":"

SJTUer

\n"}},{"id":15350,"postNumber":12,"raw":"解答：\n\n令\n$$\nz=\\frac{x^5+40x^3+320x}{x^4+20x^2+80}\n$$\n则有\n$$\n\\frac{\\mathrm dz}{\\sqrt{z^4-95z^2+2560}}=\\frac{\\mathrm dx}{\\sqrt{x^4+25x^2+160}}\n$$\n然后积一下就好了\n\n:nose_steam:","cooked":"

解答：

令
\n

\nz=\\frac{x^5+40x^3+320x}{x^4+20x^2+80}\n

\n则有
\n

\n\\frac{\\mathrm dz}{\\sqrt{z^4-95z^2+2560}}=\\frac{\\mathrm dx}{\\sqrt{x^4+25x^2+160}}\n

\n然后积一下就好了

😤

\n","createdAt":"2023-10-30 02:32:25.314409","updatedAt":"2023-10-30 02:32:25.314409","replyToPostNumber":6,"replyCount":0,"likeCount":3,"wordCount":23,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1047,"username":"anonymous_coward_old","name":"Anonymous Coward Old"},{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1707,"username":"wbzd","name":"我不知道","avatarUrl":"/uploads/image/29/1f/291f099d72b6e1ab6e873cd1b0fe4d2f133a9e713c066ea299ae3040b0e2ef27.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":15373,"postNumber":13,"raw":"这下出口转内销了","cooked":"

这下出口转内销了

\n","createdAt":"2023-10-30 04:07:28.288592","updatedAt":"2023-10-30 04:07:28.288592","replyToPostNumber":null,"replyCount":0,"likeCount":4,"wordCount":8,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":4,"users":[{"id":1411,"username":"lwe","name":null},{"id":1707,"username":"wbzd","name":"我不知道"},{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":20804,"postNumber":14,"raw":"非常好理科水楼","cooked":"

非常好理科水楼

\n","createdAt":"2023-12-30 06:39:35.936793","updatedAt":"2023-12-30 06:39:35.936793","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":7,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1811,"username":"Time_Limit_Exceed","name":null,"avatarUrl":"/uploads/image/d8/9f/d89f1d48f18bbbe6e6743852b213ef5226500b8782ee99f80b02495c9fdd631b.avif","admin":false,"moderator":false,"signatureRaw":"再挑灯火读文章 :bulb: :book: :face_with_monocle:","signatureCooked":"

再挑灯火读文章 💡 📖 🧐

\n"}},{"id":20805,"postNumber":15,"raw":"我不知道！！！","cooked":"

我不知道！！！

\n","createdAt":"2023-12-30 06:44:51.345536","updatedAt":"2023-12-30 06:44:51.345536","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":5,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1811,"username":"Time_Limit_Exceed","name":null,"avatarUrl":"/uploads/image/d8/9f/d89f1d48f18bbbe6e6743852b213ef5226500b8782ee99f80b02495c9fdd631b.avif","admin":false,"moderator":false,"signatureRaw":"再挑灯火读文章 :bulb: :book: :face_with_monocle:","signatureCooked":"

再挑灯火读文章 💡 📖 🧐

\n"}},{"id":22825,"postNumber":16,"raw":"存在不能被任何图灵机 (Turing machine) 计算的函数 $ UC:\\{0,1\\}^*\\to\\{0,1\\} $.\n\n函数 $ UC $ 如下定义：\n $$\n \\forall \\alpha\\in\\{0,1\\}^*, \nUC(\\alpha)=\n\\begin{cases}\n0,M_\\alpha(\\alpha)=1;\\\\\n1,\\text{otherwise}.\n\\end{cases}\n$$\n证明：假设函数 $ UC $ 可被图灵机计算，则存在图灵机 $ M $ 使得 $ \nM(\\alpha)\\equiv UC(\\alpha), \\forall \\alpha\\in\\{0,1\\}^* $\n所以有 $ M(\\lfloor M\\rfloor)=UC(\\lfloor M\\rfloor) $.这与定义矛盾。[]\n注： $ \\{0,1\\}^* $ 为 0 和 1 组成的字符串， $ M_\\alpha $ 为以二进制字符串 $ \\alpha $ 表示的图灵机（数据和指令等价）， $ \\lfloor M\\rfloor $ 为图灵机 $ M $ 的二进制字符串表示\n这个函数的构造方式有点像理发师悖论\n来源：COMPUTATIONAL COMPLEXITY A Modern Approach","cooked":"

存在不能被任何图灵机 (Turing machine) 计算的函数 $ UC:{0,1}^*\\to{0,1} $.

函数 $ UC $ 如下定义：
\n

\n\\forall \\alpha\\in\\{0,1\\}^*, \nUC(\\alpha)=\n\\begin{cases}\n0,M_\\alpha(\\alpha)=1;\\\\\n1,\\text{otherwise}.\n\\end{cases}\n

\n证明：假设函数 $ UC $ 可被图灵机计算，则存在图灵机 $ M $ 使得 $
\nM(\\alpha)\\equiv UC(\\alpha), \\forall \\alpha\\in{0,1}^* $
\n所以有 $ M(\\lfloor M\\rfloor)=UC(\\lfloor M\\rfloor) $.这与定义矛盾。
\n注： $ {0,1}^* $ 为 0 和 1 组成的字符串， $ M_\\alpha $ 为以二进制字符串 $ \\alpha $ 表示的图灵机（数据和指令等价）， $ \\lfloor M\\rfloor $ 为图灵机 $ M $ 的二进制字符串表示
\n这个函数的构造方式有点像理发师悖论
\n来源：COMPUTATIONAL COMPLEXITY A Modern Approach

\n","createdAt":"2024-01-29 11:36:14.539367","updatedAt":"2024-01-29 11:37:51.782405","replyToPostNumber":null,"replyCount":1,"likeCount":5,"wordCount":178,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":5,"users":[{"id":172,"username":"HCCH","name":"HCCH"},{"id":1411,"username":"lwe","name":null},{"id":1811,"username":"Time_Limit_Exceed","name":null},{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":22829,"postNumber":17,"raw":"怎么没人水，那我多水几个","cooked":"

怎么没人水，那我多水几个

\n","createdAt":"2024-01-29 11:42:55.339068","updatedAt":"2024-01-29 11:42:55.339068","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":12,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":22846,"postNumber":18,"raw":"相信大家都在信号与系统中学过 Parseval 方程：\n$$\n(f,f)=\\sum_{i=1}^\\infty{\\left|c_i\\right|^2},\\forall f\\in\\mathscr{H}.\n$$\n其中 $ \\mathscr{H} $ 为全体定义在 $ [a,b] $ 上复值平方可积的函数构成的线性空间。\n课本的证明是最容易想到、最直观的证明：\n$$\n(f,f)=\\Bigl(\\sum_{i=1}^\\infty{c_if_i},\\sum_{i=1}^\\infty{c_if_i}\\Bigr)=\\sum_{i=1}^\\infty{\\left|c_i\\right|^2}.\n$$\n该证明看似简洁，其实并不正确。正交归一函数集 $ \\{f_i\\} $ 的完备性不能保证逐点收敛，所得函数和原函数可能相差一个广义零函数，因此第一个等号不成立。从完备性直接得到的是由积分定义的平均收敛。正确的证明如下：\n$$\n\\begin{align*}\n&\\int_a^b{\\Bigl|f(x)-\\sum_{i=1}^{n}{c_if_i(x)}\\Bigr|^2\\rho(x)\\mathrm{d}x}\\\\\n=&(f,f)-\\sum_{i=1}^{n}{c_i^*(f_i,f)}-\\sum_{i=1}^{n}{c_i(f,f_i)}+\\sum_{i=1}^{n}{|c_i|^2}\\\\\n=&(f,f)-\\sum_{i=1}^{n}{|c_i|^2}.\n\\end{align*}\n$$\n由基函数 $ \\{f_i\\} $ 的完备性得，\n$$\n\\lim_{n\\to\\infty}\\int_a^b{\\Bigl|f(x)-\\sum_{i=1}^{n}{c_if_i(x)}\\Bigr|^2\\rho(x)\\mathrm{d}x}=0,\\\\\ni.e.(f,f)=\\sum_{i=1}^\\infty{\\left|c_i\\right|^2}.\n$$\n\n该方程还有更一般的形式：\n$$\n(f,g)=\\sum_{i=1}^\\infty{(f,f_i)(f_i,g)},\\forall f,g\\in\\mathscr{H}.\n$$\n证明如下：\n$$\n\\begin{align*}\nf(x)&=\\sum_{i=1}^\\infty{c_if_i(x)}=\\sum_{i=1}^\\infty{\\Bigl[\\int_a^b{f_i(x)f_i^*(x')\\rho(x')\\mathrm{d}x'}\\Bigr]f_i(x)}\n\\\\&=\\int_a^b{\\Bigl[\\sum_{i=1}^\\infty{f_i(x)f_i^*(x')\\rho(x')}\\Bigr]f(x')\\mathrm{d}x'}\n\\\\&=\\int_a^b{\\delta(x'-x)\\mathrm{d}x'}\\\\\n&\\Rightarrow\\sum_{i=1}^\\infty{f_i(x)f_i^*(x')\\rho(x')}=\\delta(x-x'),a\\leqslant x,x'\\leqslant b.\\\\\n(f,g)&=\\int_a^b{f^*(x)g(x)\\rho(x)\\mathrm{d}x}\\\\\n&=\\int_a^b{\\Bigl[\\int_a^b{f^*(x)g(x')\\rho(x)\\delta(x-x')\\mathrm{d}x'}\\Bigr]\\mathrm{d}x}\\\\\n&=\\sum_{i=1}^\\infty{\\int_a^b{\\Bigl[\\int_a^b{f^*(x)f_i(x)f_i^*(x')\\rho(x')g(x')\\rho(x)\\mathrm{d}x'}\\Bigr]\\mathrm{d}x}}\\\\\n&=\\sum_{i=1}^\\infty{\\int_a^b{f^*(x)f_i(x)\\rho(x)\\mathrm{d}x}\\int_a^b{f_i^*(x')g(x')\\rho(x')\\mathrm{d}x'}}\\\\\n&=\\sum_{i=1}^\\infty{(f,f_i)(f_i,g)}.\n\\end{align*}\n$$","cooked":"

相信大家都在信号与系统中学过 Parseval 方程：
\n

\n(f,f)=\\sum_{i=1}^\\infty{\\left|c_i\\right|^2},\\forall f\\in\\mathscr{H}.\n

\n其中 $ \\mathscr{H} $ 为全体定义在 $ [a,b] $ 上复值平方可积的函数构成的线性空间。
\n课本的证明是最容易想到、最直观的证明：
\n

\n(f,f)=\\Bigl(\\sum_{i=1}^\\infty{c_if_i},\\sum_{i=1}^\\infty{c_if_i}\\Bigr)=\\sum_{i=1}^\\infty{\\left|c_i\\right|^2}.\n

\n该证明看似简洁，其实并不正确。正交归一函数集 $ {f_i} $ 的完备性不能保证逐点收敛，所得函数和原函数可能相差一个广义零函数，因此第一个等号不成立。从完备性直接得到的是由积分定义的平均收敛。正确的证明如下：
\n

\n\\begin{align*}\n&\\int_a^b{\\Bigl|f(x)-\\sum_{i=1}^{n}{c_if_i(x)}\\Bigr|^2\\rho(x)\\mathrm{d}x}\\\\\n=&(f,f)-\\sum_{i=1}^{n}{c_i^*(f_i,f)}-\\sum_{i=1}^{n}{c_i(f,f_i)}+\\sum_{i=1}^{n}{|c_i|^2}\\\\\n=&(f,f)-\\sum_{i=1}^{n}{|c_i|^2}.\n\\end{align*}\n

\n由基函数 $ {f_i} $ 的完备性得，
\n

\n\\lim_{n\\to\\infty}\\int_a^b{\\Bigl|f(x)-\\sum_{i=1}^{n}{c_if_i(x)}\\Bigr|^2\\rho(x)\\mathrm{d}x}=0,\\\\\ni.e.(f,f)=\\sum_{i=1}^\\infty{\\left|c_i\\right|^2}.\n

该方程还有更一般的形式：
\n

\n(f,g)=\\sum_{i=1}^\\infty{(f,f_i)(f_i,g)},\\forall f,g\\in\\mathscr{H}.\n

\n证明如下：
\n

\n\\begin{align*}\nf(x)&=\\sum_{i=1}^\\infty{c_if_i(x)}=\\sum_{i=1}^\\infty{\\Bigl[\\int_a^b{f_i(x)f_i^*(x')\\rho(x')\\mathrm{d}x'}\\Bigr]f_i(x)}\n\\\\&=\\int_a^b{\\Bigl[\\sum_{i=1}^\\infty{f_i(x)f_i^*(x')\\rho(x')}\\Bigr]f(x')\\mathrm{d}x'}\n\\\\&=\\int_a^b{\\delta(x'-x)\\mathrm{d}x'}\\\\\n&\\Rightarrow\\sum_{i=1}^\\infty{f_i(x)f_i^*(x')\\rho(x')}=\\delta(x-x'),a\\leqslant x,x'\\leqslant b.\\\\\n(f,g)&=\\int_a^b{f^*(x)g(x)\\rho(x)\\mathrm{d}x}\\\\\n&=\\int_a^b{\\Bigl[\\int_a^b{f^*(x)g(x')\\rho(x)\\delta(x-x')\\mathrm{d}x'}\\Bigr]\\mathrm{d}x}\\\\\n&=\\sum_{i=1}^\\infty{\\int_a^b{\\Bigl[\\int_a^b{f^*(x)f_i(x)f_i^*(x')\\rho(x')g(x')\\rho(x)\\mathrm{d}x'}\\Bigr]\\mathrm{d}x}}\\\\\n&=\\sum_{i=1}^\\infty{\\int_a^b{f^*(x)f_i(x)\\rho(x)\\mathrm{d}x}\\int_a^b{f_i^*(x')g(x')\\rho(x')\\mathrm{d}x'}}\\\\\n&=\\sum_{i=1}^\\infty{(f,f_i)(f_i,g)}.\n\\end{align*}\n

\n","createdAt":"2024-01-29 14:54:32.024756","updatedAt":"2024-01-29 14:56:31.644771","replyToPostNumber":null,"replyCount":1,"likeCount":3,"wordCount":234,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"😍️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false},{"reaction":"❤️","count":3,"users":[{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":22852,"postNumber":19,"raw":"超，你什么时候改行搞数学了\n这里的 $\\rho$ 是什么","cooked":"

超，你什么时候改行搞数学了
\n这里的 $\\rho$ 是什么

\n","createdAt":"2024-01-29 15:40:02.829573","updatedAt":"2024-01-29 15:45:07.462282","replyToPostNumber":18,"replyCount":1,"likeCount":1,"wordCount":20,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":2,"publicVersion":2,"deletedAt":null,"isAnonymous":false,"user":{"id":1411,"username":"lwe","name":null,"avatarUrl":null,"admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":22853,"postNumber":20,"raw":"我也来写点","cooked":"

我也来写点

\n","createdAt":"2024-01-29 15:46:44.651935","updatedAt":"2024-01-29 15:46:44.651935","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":5,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1411,"username":"lwe","name":null,"avatarUrl":null,"admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":22854,"postNumber":21,"raw":" $ \\rho(x) $ 是定义在 $ [a,b] $ 上不恒为零的非负实函数，在线性空间 $ \\mathscr{H} $ 的内积定义中用作权函数。 $ \\rho(x)\\equiv 1 $ 是一种常见的特殊情况。","cooked":"

$ \\rho(x) $ 是定义在 $ [a,b] $ 上不恒为零的非负实函数，在线性空间 $ \\mathscr{H} $ 的内积定义中用作权函数。 $ \\rho(x)\\equiv 1 $ 是一种常见的特殊情况。

\n","createdAt":"2024-01-29 15:49:19.769417","updatedAt":"2024-01-29 15:50:26.597946","replyToPostNumber":19,"replyCount":0,"likeCount":2,"wordCount":57,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":23054,"postNumber":22,"raw":"Fourier 变换前后形式不变的信号——Gauss 型函数\n$$\ng_{\\alpha}\\left( t \\right) =\\frac{1}{2\\sqrt{\\pi \\alpha}}\\exp \\left\\{ -\\frac{t^2}{4\\alpha} \\right\\} \n$$\n不难发现\n$$\n\\begin{align}\n\\mathscr{F} \\left\\{ g_{\\alpha}\\left( t \\right) \\right\\} &=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{+\\infty}{\\frac{1}{2\\sqrt{\\pi \\alpha}}\\exp \\left\\{ -\\frac{t^2}{4\\alpha} \\right\\} \\exp \\left\\{ -\\mathrm{i}\\omega t \\right\\} \\mathrm{d}t}\n\\\\\n&=\\frac{1}{\\sqrt{2}\\pi}\\int_{-\\infty}^{+\\infty}{\\exp \\left\\{ -\\left( \\frac{t}{2\\sqrt{\\alpha}}+\\sqrt{\\alpha}\\mathrm{i}\\omega \\right) ^2-\\alpha \\omega ^2 \\right\\} \\mathrm{d}\\frac{t}{2\\sqrt{\\alpha}}}\n\\\\\n&=\\frac{1}{\\sqrt{2\\pi}}\\exp \\left\\{ -\\alpha \\omega ^2 \\right\\} \n\\end{align}\n$$\n以此为窗进行的短时 Fourier 变换称为 Gabor 变换：\n$$\n\\begin{align}\n\\mathscr{F} \\left\\{ f\\left( t \\right) g_{\\alpha}\\left( t-b \\right) \\right\\} &=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{+\\infty}{f\\left( t \\right) g_{\\alpha}\\left( t-b \\right) \\mathrm{e}^{-\\mathrm{i}\\omega t}\\mathrm{d}t}\n\\\\\n&=\\frac{1}{2\\pi}\\int_{-\\infty}^{+\\infty}{F\\left( \\xi \\right) \\mathrm{e}^{-\\mathrm{i}\\left( \\omega -\\xi \\right) b}\\mathrm{e}^{-\\alpha \\left( \\omega -\\xi \\right) ^2}\\mathrm{d}\\xi}\n\\\\\n&=\\frac{1}{2\\sqrt{\\pi \\alpha}}\\mathrm{e}^{-\\mathrm{i}\\omega b}\\int_{-\\infty}^{+\\infty}{F\\left( \\xi \\right) g_{\\frac{1}{4\\alpha}}\\left( \\xi -\\omega \\right) \\mathrm{e}^{\\mathrm{i}\\xi b}\\mathrm{d}\\xi}\n\\end{align}\n$$\n时域乘积在频域表现为卷积。 $ g_{\\alpha}\\left( t \\right) $ 是时间窗函数， $ g_{\\frac{1}{4\\alpha}}(\\xi) $ 则是相应的频率窗函数。时间窗宽度和频率窗宽度成反比。","cooked":"

Fourier 变换前后形式不变的信号——Gauss 型函数
\n

\ng_{\\alpha}\\left( t \\right) =\\frac{1}{2\\sqrt{\\pi \\alpha}}\\exp \\left\\{ -\\frac{t^2}{4\\alpha} \\right\\}\n

\n不难发现
\n

\n\\begin{align}\n\\mathscr{F} \\left\\{ g_{\\alpha}\\left( t \\right) \\right\\} &=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{+\\infty}{\\frac{1}{2\\sqrt{\\pi \\alpha}}\\exp \\left\\{ -\\frac{t^2}{4\\alpha} \\right\\} \\exp \\left\\{ -\\mathrm{i}\\omega t \\right\\} \\mathrm{d}t}\n\\\\\n&=\\frac{1}{\\sqrt{2}\\pi}\\int_{-\\infty}^{+\\infty}{\\exp \\left\\{ -\\left( \\frac{t}{2\\sqrt{\\alpha}}+\\sqrt{\\alpha}\\mathrm{i}\\omega \\right) ^2-\\alpha \\omega ^2 \\right\\} \\mathrm{d}\\frac{t}{2\\sqrt{\\alpha}}}\n\\\\\n&=\\frac{1}{\\sqrt{2\\pi}}\\exp \\left\\{ -\\alpha \\omega ^2 \\right\\} \n\\end{align}\n

\n以此为窗进行的短时 Fourier 变换称为 Gabor 变换：
\n

\n\\begin{align}\n\\mathscr{F} \\left\\{ f\\left( t \\right) g_{\\alpha}\\left( t-b \\right) \\right\\} &=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{+\\infty}{f\\left( t \\right) g_{\\alpha}\\left( t-b \\right) \\mathrm{e}^{-\\mathrm{i}\\omega t}\\mathrm{d}t}\n\\\\\n&=\\frac{1}{2\\pi}\\int_{-\\infty}^{+\\infty}{F\\left( \\xi \\right) \\mathrm{e}^{-\\mathrm{i}\\left( \\omega -\\xi \\right) b}\\mathrm{e}^{-\\alpha \\left( \\omega -\\xi \\right) ^2}\\mathrm{d}\\xi}\n\\\\\n&=\\frac{1}{2\\sqrt{\\pi \\alpha}}\\mathrm{e}^{-\\mathrm{i}\\omega b}\\int_{-\\infty}^{+\\infty}{F\\left( \\xi \\right) g_{\\frac{1}{4\\alpha}}\\left( \\xi -\\omega \\right) \\mathrm{e}^{\\mathrm{i}\\xi b}\\mathrm{d}\\xi}\n\\end{align}\n

\n时域乘积在频域表现为卷积。 $ g_{\\alpha}\\left( t \\right) $ 是时间窗函数， $ g_{\\frac{1}{4\\alpha}}(\\xi) $ 则是相应的频率窗函数。时间窗宽度和频率窗宽度成反比。

\n","createdAt":"2024-01-31 14:34:38.987217","updatedAt":"2024-01-31 15:18:29.959781","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":188,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":2,"publicVersion":2,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":23079,"postNumber":23,"raw":"这学期本来打算修一门计算复杂性的课，结果第一节课没选上没听导致第二节课想试听但是完全听不懂遂退了🤡","cooked":"

这学期本来打算修一门计算复杂性的课，结果第一节课没选上没听导致第二节课想试听但是完全听不懂遂退了🤡

\n","createdAt":"2024-02-01 00:17:11.39121","updatedAt":"2024-02-01 00:17:11.39121","replyToPostNumber":16,"replyCount":0,"likeCount":2,"wordCount":49,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1817,"username":"anon1678451","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1819,"username":"Thevenin","name":"交大金猫","avatarUrl":"/uploads/image/ef/7f/ef7f3ea801ae3eebf5777db2453b74070a6ec7cc62905061d848c547190221ff.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":25860,"postNumber":24,"raw":"这里和水源同步更新吧\n如何构造 3 个同分布且总和为恒定常数的随机变量？\n虽然有 3 个变量，但确定其二后另一个唯一确定，所以只有两个自由度。所以问题的关键在于如何让第三个变量也和前两个服从相同分布。\n这让我想到高中地理常用的三角坐标系：二维平面内任意一点具有两个自由度，同时对应等边三角形三条边上的三个坐标。这三个坐标地位均等，所以如果分布对三角形均匀，例如三角形内的均匀分布或者平面内的正态分布，就一定能做到三个坐标同分布。\n![image|328x264](/uploads/image/82/06/8206beb5d6af79e908387b4b43d85a4cdb7a82ea0c873a6a3a064c24334fe601.avif)\n这个问题还可以推广到更高维的情况，关键在于把 $(n-1)$ 维映射到 $n$ 维。\n这个问题的背景源自我的头像，我想让头像每一条线的亮度（RGB 三个值的总和）保持不变，但颜色随机变化，即 RGB 的每个值都在 0~255 之间随机变化。我原打算先给 R 生成 `(0,255)`的随机值，再给 G 生成`(max(0,127-r),min(381-r,255))` 的随机值，最后得到 B，后来意识到这种方法生成的三个变量不满足同分布。如果按照三角坐标系的思路，为了让每个值的范围是 0~255，且平均值处于 127，其中一种方法是把上图正三角形挖去三个角，形成一个正六边形，然后在正六边形中均匀取点，得到三个对应的坐标，如下图。\n![image|609x499](/uploads/image/e5/6f/e56f8ab415a6c14534f2226f4bad377fd659b2e31050ab806454ce06b9792710.avif)\n当然在工程上还有更简单的处理方法，比如按最初方法生成完三个不同分布的随机数后随机交换位置，使之与 RGB 的对应关系也随机。如此同样可以实现最终的同分布。\n原代码来自知乎文章 [Python 画各种有趣的图及源码](https://zhuanlan.zhihu.com/p/431234959)。我改变了线条的长度、粗细和颜色。\n[details=\"代码\"]\n```\n# -*- coding: utf-8 -*-\n\nfrom turtle import *\nimport random\nlength = 10\nangle = 90\nsetup(1280,720)\nup()\n\ngoto(-640,-360)\ndown()\ndef draw_path(path):\n for symbol in path:\n if symbol == 'f':\n colormode(255)\n color(random.randint(0,255),random.randint(0,255),random.randint(0,255))\n fd(length)\n elif symbol == '+':\n lt(angle)\n elif symbol == '-':\n rt(angle)\n\ndef apply_path(rules,path):\n lit = [x for x in path]\n for i in range(len(lit)):\n symbol = lit[i]\n if symbol == 'x':\n lit[i] = rules[symbol]\n elif symbol == 'y':\n lit[i] = rules[symbol]\n path = ''.join(lit)\n return path\n\nrules = {\n 'x':'+yf-xfx-fy+',\n 'y':'-xf+yfy+fx-'\n}\npath = 'x'\nspeed(0)\nfor i in range(7):\n path = apply_path(rules,path)\ndraw_path(path)\ndone()\n```\n[/details]","cooked":"

这里和水源同步更新吧
\n如何构造 3 个同分布且总和为恒定常数的随机变量？
\n虽然有 3 个变量，但确定其二后另一个唯一确定，所以只有两个自由度。所以问题的关键在于如何让第三个变量也和前两个服从相同分布。
\n这让我想到高中地理常用的三角坐标系：二维平面内任意一点具有两个自由度，同时对应等边三角形三条边上的三个坐标。这三个坐标地位均等，所以如果分布对三角形均匀，例如三角形内的均匀分布或者平面内的正态分布，就一定能做到三个坐标同分布。
\n $\"image\"$
\n这个问题还可以推广到更高维的情况，关键在于把 $(n-1)$ 维映射到 $n$ 维。
\n这个问题的背景源自我的头像，我想让头像每一条线的亮度（RGB 三个值的总和）保持不变，但颜色随机变化，即 RGB 的每个值都在 0~255 之间随机变化。我原打算先给 R 生成 (0,255)的随机值，再给 G 生成(max(0,127-r),min(381-r,255)) 的随机值，最后得到 B，后来意识到这种方法生成的三个变量不满足同分布。如果按照三角坐标系的思路，为了让每个值的范围是 0~255，且平均值处于 127，其中一种方法是把上图正三角形挖去三个角，形成一个正六边形，然后在正六边形中均匀取点，得到三个对应的坐标，如下图。
\n $\"image\"$
\n当然在工程上还有更简单的处理方法，比如按最初方法生成完三个不同分布的随机数后随机交换位置，使之与 RGB 的对应关系也随机。如此同样可以实现最终的同分布。
\n原代码来自知乎文章 Python 画各种有趣的图及源码。我改变了线条的长度、粗细和颜色。

代码

# -*- coding: utf-8 -*-\n\nfrom turtle import *\nimport random\nlength = 10\nangle  = 90\nsetup(1280,720)\nup()\n\ngoto(-640,-360)\ndown()\ndef draw_path(path):\n    for symbol in path:\n        if symbol == 'f':\n            colormode(255)\n            color(random.randint(0,255),random.randint(0,255),random.randint(0,255))\n            fd(length)\n        elif symbol == '+':\n            lt(angle)\n        elif symbol == '-':\n            rt(angle)\n\ndef apply_path(rules,path):\n    lit = [x for x in path]\n    for i in range(len(lit)):\n        symbol = lit[i]\n        if symbol == 'x':\n            lit[i] = rules[symbol]\n        elif symbol == 'y':\n            lit[i] = rules[symbol]\n    path = ''.join(lit)\n    return path\n\nrules = {\n    'x':'+yf-xfx-fy+',\n    'y':'-xf+yfy+fx-'\n}\npath = 'x'\nspeed(0)\nfor i in range(7):\n    path = apply_path(rules,path)\ndraw_path(path)\ndone()\n

","createdAt":"2024-03-05 06:27:06.353599","updatedAt":"2024-03-05 06:27:06.353599","replyToPostNumber":null,"replyCount":1,"likeCount":3,"wordCount":658,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1381,"username":"greyishsong","name":"greyishsong"},{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":25882,"postNumber":25,"raw":"如果函数 $ L:\\mathbb{R}^n\\to\\mathbb{R} $ 满足如下条件：\n\n1. $ \\boldsymbol{v}\\mapsto L(\\boldsymbol{v}) $ 是凸函数\n2. $ \\displaystyle\\lim_{|\\boldsymbol{v}|\\to \\infty}\\frac{L(\\boldsymbol{v})}{|\\boldsymbol{{v}}|}=\\infty $ （超线性）\n\n则 $ L $ 的 Legendre 变换为\n$$\nL^*(\\boldsymbol{p})=\\sup_{\\boldsymbol{v}\\in\\mathbb{R}^n}\\{\\boldsymbol{p}\\cdot \\boldsymbol{v}-L(\\boldsymbol{v})\\},\\boldsymbol{p}\\in\\mathbb{R}^n.\n$$\nLegendre 变换具有如下性质：\n\n1. $ \\boldsymbol{p}\\mapsto L^*(\\boldsymbol{p}) $ 是凸函数\n2. $ \\displaystyle\\lim_{|\\boldsymbol{p}|\\to \\infty}\\frac{L^*(\\boldsymbol{p})}{|\\boldsymbol{p}|}=\\infty $ （超线性）\n3. $ L^{**}=L $\n\n在理论力学中常见的 Legendre 变换是\n$$\nL^*(\\boldsymbol{p})=\\boldsymbol{p}\\cdot \\boldsymbol{v}-L(\\boldsymbol{v}),\n$$\n其中\n$$\n\\boldsymbol{p}=\\nabla L.\n$$\n如果函数 $ L $ 可导，则两种定义是等价的。","cooked":"

如果函数 $ L:\\mathbb{R}^n\\to\\mathbb{R} $ 满足如下条件：

$ \\boldsymbol{v}\\mapsto L(\\boldsymbol{v}) $ 是凸函数
$ \\displaystyle\\lim_{|\\boldsymbol{v}|\\to \\infty}\\frac{L(\\boldsymbol{v})}{|\\boldsymbol{{v}}|}=\\infty $ （超线性）

则 $ L $ 的 Legendre 变换为
\n

\nL^*(\\boldsymbol{p})=\\sup_{\\boldsymbol{v}\\in\\mathbb{R}^n}\\{\\boldsymbol{p}\\cdot \\boldsymbol{v}-L(\\boldsymbol{v})\\},\\boldsymbol{p}\\in\\mathbb{R}^n.\n

\nLegendre 变换具有如下性质：

$ \\boldsymbol{p}\\mapsto L^*(\\boldsymbol{p}) $ 是凸函数
$ \\displaystyle\\lim_{|\\boldsymbol{p}|\\to \\infty}\\frac{L^*(\\boldsymbol{p})}{|\\boldsymbol{p}|}=\\infty $ （超线性）
$ L^{**}=L $

在理论力学中常见的 Legendre 变换是
\n

\nL^*(\\boldsymbol{p})=\\boldsymbol{p}\\cdot \\boldsymbol{v}-L(\\boldsymbol{v}),\n

\n其中
\n

\n\\boldsymbol{p}=\\nabla L.\n

\n如果函数 $ L $ 可导，则两种定义是等价的。

\n","createdAt":"2024-03-05 09:29:53.059047","updatedAt":"2024-03-06 04:22:59.602015","replyToPostNumber":null,"replyCount":1,"likeCount":3,"wordCount":122,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1819,"username":"Thevenin","name":"交大金猫"},{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":2,"publicVersion":2,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":26029,"postNumber":26,"raw":"好诶！","cooked":"

好诶！

\n","createdAt":"2024-03-06 09:51:08.6769","updatedAt":"2024-03-06 09:51:08.6769","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":3,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1819,"username":"Thevenin","name":"交大金猫","avatarUrl":"/uploads/image/ef/7f/ef7f3ea801ae3eebf5777db2453b74070a6ec7cc62905061d848c547190221ff.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":36584,"postNumber":28,"raw":"基于 Markov 链的解决方法：\n若要生成 n 个均匀分布的随机非负整数，且和为定值，则可以先任取一组满足和条件的数，然后每次随机选择两个数，对其中一个数 +1，另一个数 -1，除非另一个数已经达到 0。重复足够多次后，每个数可以近似为均匀分布，且和始终不变。\n证明：每个数的转移矩阵为\n$$\n\\left( \\begin{matrix}\n\t\\frac{n-1}{n}&\t\t\\frac{1}{n}&\t\t0&\t\t\\cdots\\\\\n\t\\frac{1}{n}&\t\t\\frac{n-2}{n}&\t\t\\frac{1}{n}&\t\t\\cdots\\\\\n\t0&\t\t\\frac{1}{n}&\t\t\\frac{n-2}{n}&\t\t\\cdots\\\\\n\t\\cdots&\t\t\\cdots&\t\t\\cdots&\t\t\\cdots\\\\\n\\end{matrix} \\right) \n$$\n根据 Basic Limit Theorem，显然存在均匀的稳态分布，所以分布会趋于均匀。因为加减同时进行，所以所有数的和不变。\n该方法还可以扩展到矩阵各行列之和固定的情形，只需每次既选择行，又选择列进行加减。\n\n使用 Minecraft 进行蒙特卡洛模拟\nhttps://www.bilibili.com/video/BV19w4m1e7js","cooked":"

基于 Markov 链的解决方法：
\n若要生成 n 个均匀分布的随机非负整数，且和为定值，则可以先任取一组满足和条件的数，然后每次随机选择两个数，对其中一个数 +1，另一个数 -1，除非另一个数已经达到 0。重复足够多次后，每个数可以近似为均匀分布，且和始终不变。
\n证明：每个数的转移矩阵为
\n

\n\\left( \\begin{matrix}\n\t\\frac{n-1}{n}&\t\t\\frac{1}{n}&\t\t0&\t\t\\cdots\\\\\n\t\\frac{1}{n}&\t\t\\frac{n-2}{n}&\t\t\\frac{1}{n}&\t\t\\cdots\\\\\n\t0&\t\t\\frac{1}{n}&\t\t\\frac{n-2}{n}&\t\t\\cdots\\\\\n\t\\cdots&\t\t\\cdots&\t\t\\cdots&\t\t\\cdots\\\\\n\\end{matrix} \\right)\n

\n根据 Basic Limit Theorem，显然存在均匀的稳态分布，所以分布会趋于均匀。因为加减同时进行，所以所有数的和不变。
\n该方法还可以扩展到矩阵各行列之和固定的情形，只需每次既选择行，又选择列进行加减。

使用 Minecraft 进行蒙特卡洛模拟

\n $\"\"$ \n Bilibili\n

\n $\"\"$ \n

一万个末影人搭建一年会发生什么？

https://www.youtube.com/watch?v=LLss-7SPQ-o\nIf you are the original author or copyright holder of this video and want to remove this video for infringement or other reasons, please contact me via private message or send an email to: oh7unz@gmail.com\n\n原視頻連結：https://www.youtube.com/watch?v=LLss-7SPQ-o\n\n原視頻簡介：\nCheck out Starforge Systems and unlock a new gaming experience with their prebuilt PCs: https://starforgepc.com/ElementX\n\n▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬\n\nDid you know: incorrect grammar used in videos increases engagement rates, which in turn increases view count? \n\n\nI make clickbait videos.. that aren't clickbaits, because I code the most unthinkable ideas in Minecraft that looks like clickbaits.\n\nMy contents are for all ages, but most of the people who watch me are teenagers to young adults. I also turn my viewers' ideas into reality.\nI use all sorts of technology! I use AI, such as Chat GPT or Claude.AI, the Java programming language and MCP reborn to modify the video game's code. Artificial Intelligence definitely helps a lot, but only used for the most basic tasks.\n\n▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ \nMy PC Computer specs\n\nRTX 3070 graphics card with Ryzen 5800X CPU.\n\n▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬\n♫Music used:♫\n🎸https://open.spotify.com/artist/6bnZW05ijGE4iuyfiVd3km\n(These are my own music, and you may use them for free if you're a content creator. We just ask to credit the music in your video with a link.)\n\n▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬\n👥Social Media👥\n\nRead before emailing to save your time\n\n● DO NOT reach out if you're offering video editing or thumbnail design services for my channel.\n● If you're looking to download my minecraft mods.... I'm sorry but I won't be sharing them.\n\n✅ EMAIL for business, brand partnerships and for personal: officialelementx47@gmail.com\n✅ Everything else: http://linktr.ee/elementxyt

","createdAt":"2024-07-03 03:27:56.712235","updatedAt":"2024-07-03 03:29:47.970155","replyToPostNumber":24,"replyCount":0,"likeCount":3,"wordCount":245,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1411,"username":"lwe","name":null},{"id":1731,"username":"sJtupid","name":"JacKeiner"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":36585,"postNumber":29,"raw":"看其它内容时经常用到张量，这几篇文章可以补一下基础\n\nhttps://zhuanlan.zhihu.com/p/508715535\n\nhttps://zhuanlan.zhihu.com/p/508715717","cooked":"

看其它内容时经常用到张量，这几篇文章可以补一下基础

https://zhuanlan.zhihu.com/p/508715535

https://zhuanlan.zhihu.com/p/508715717

","createdAt":"2024-07-03 03:28:39.015941","updatedAt":"2024-07-03 03:28:39.015941","replyToPostNumber":null,"replyCount":0,"likeCount":4,"wordCount":27,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"🥳️","count":1,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"}],"hasReacted":false},{"reaction":"❤️","count":4,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"},{"id":1731,"username":"sJtupid","name":"JacKeiner"},{"id":1811,"username":"Time_Limit_Exceed","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":47481,"postNumber":30,"raw":"# 随机测度\n设 $ (\\Omega, \\mathcal{F}, P) $ 是一个概率空间，$ (Z, \\mathcal{B}_Z) $ 是一个可测空间（其中 $ \\mathcal{B}_Z $ 是 $ Z $ 上的 σ-代数）. 一个随机测度 $ \\mu $ 是一个映射：\n\n$$\n\\mu: \\Omega \\times \\mathcal{B}_Z \\to [0, \\infty]\n$$\n\n需要满足以下条件：\n\n1. 对于每个 $ B \\in \\mathcal{B} $，$ \\mu(\\cdot, B) $ 是一个随机变量，但可以取到 $ [0, \\infty] $;\n\n2. 对于每个确定的 $ \\omega \\in \\Omega $，$ \\mu(\\omega, \\cdot) $ 是 $ (Z, \\mathcal{B}_Z) $ 上一个σ-有限的测度（这里的σ-有限指存在 $ \\{U_n\\}_{n=1}^\\infty\\subset\\mathcal{B}_Z $ 使得 $ Z=\\cup_{n=0}^\\infty U_n $ 并且 $ \\mu (U_n,\\omega) < \\infty, \\forall n$ ） .\n\n# Poisson 随机测度\n一个取值为非负整数（包括无穷大）的随机测度 $ \\mu(\\omega,B) $ 如果满足以下条件，则为 Poisson 随机测度：\n1. 对每一个 $ B \\in \\mathcal{B} $，$ \\mu(\\cdot, B) $ 满足参数为 $ \\lambda(B)=\\mathbb{E}[ \\mu(\\omega,B)] $ 的 Poisson 分布，即 $ \\mathbb{P}(\\{\\omega:\\mu(\\omega,B)=n\\})=e^{-\\lambda(B)}\\frac{\\lambda(B)^n}{n!},n=0,1,\\cdots $;\n2. 如果 $ \\{B_j\\}_{j=1}^m\\subset \\mathcal{B}_Z $ 彼此不相交，那么 $ \\{\\mu(\\cdot, B_j)\\}_{j=1}^m $ 彼此独立.\n\n如果在空间 $ Z $ 中随机撒点，那么对空间中一片确定的区域而言，这片区域的点数服从 Poisson 分布；对一次确定的撒点结果而言，空间中不同区域的点数是一个取值为非负整数的测度。所以点数就是一个 Poisson 随机测度。这里的空间中还可以包含时间，例如对 $ Z=\\mathbb{R}_+\\times \\mathbb{R}^d $， $ \\mu: \\Omega \\times \\mathcal{B}_Z \\to\\mathbb{N}\\cup\\{\\infty\\} $ 就可以理解为每时每刻都有一定概率在 $ d $ 维欧氏空间中随机撒点，点数是一个 Poisson 随机测度。","cooked":"

随机测度

设 $ (\\Omega, \\mathcal{F}, P) $ 是一个概率空间，$ (Z, \\mathcal{B}_Z) $ 是一个可测空间（其中 $ \\mathcal{B}_Z $ 是 $ Z $ 上的 σ-代数）. 一个随机测度 $ \\mu $ 是一个映射：

\n\\mu: \\Omega \\times \\mathcal{B}_Z \\to [0, \\infty]\n

需要满足以下条件：

\n
对于每个 $ B \\in \\mathcal{B} $，$ \\mu(\\cdot, B) $ 是一个随机变量，但可以取到 $ [0, \\infty] $;
\n
\n
对于每个确定的 $ \\omega \\in \\Omega $，$ \\mu(\\omega, \\cdot) $ 是 $ (Z, \\mathcal{B}Z) $ 上一个σ-有限的测度（这里的σ-有限指存在 $ {U_n}{n=1}^\\infty\\subset\\mathcal{B}Z $ 使得 $ Z=\\cup{n=0}^\\infty U_n $ 并且 $ \\mu (U_n,\\omega) \\lt \\infty, \\forall n$ ） .
\n

Poisson 随机测度

一个取值为非负整数（包括无穷大）的随机测度 $ \\mu(\\omega,B) $ 如果满足以下条件，则为 Poisson 随机测度：

对每一个 $ B \\in \\mathcal{B} $，$ \\mu(\\cdot, B) $ 满足参数为 $ \\lambda(B)=\\mathbb{E}[ \\mu(\\omega,B)] $ 的 Poisson 分布，即 $ \\mathbb{P}({\\omega:\\mu(\\omega,B)=n})=e^{-\\lambda(B)}\\frac{\\lambda(B)^n}{n!},n=0,1,\\cdots $;
如果 $ {B_j}_{j=1}^m\\subset \\mathcal{B}Z $ 彼此不相交，那么 $ {\\mu(\\cdot, B_j)}{j=1}^m $ 彼此独立.

如果在空间 $ Z $ 中随机撒点，那么对空间中一片确定的区域而言，这片区域的点数服从 Poisson 分布；对一次确定的撒点结果而言，空间中不同区域的点数是一个取值为非负整数的测度。所以点数就是一个 Poisson 随机测度。这里的空间中还可以包含时间，例如对 $ Z=\\mathbb{R}_+\\times \\mathbb{R}^d $， $ \\mu: \\Omega \\times \\mathcal{B}_Z \\to\\mathbb{N}\\cup{\\infty} $ 就可以理解为每时每刻都有一定概率在 $ d $ 维欧氏空间中随机撒点，点数是一个 Poisson 随机测度。

\n","createdAt":"2025-03-26 00:46:43.150862","updatedAt":"2025-03-26 00:46:43.150862","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":434,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":1411,"username":"lwe","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":47482,"postNumber":31,"raw":"在上次约饭时，@anon60541984 提出了头发生长与掉落速度恒定的假设，这里计算头发平均长度从 0 开始逐渐生长，并达到动态平衡的过程。\n设头发的总数为 $ N $，每根头发在单位时间内增长 $ \\iota $，且单位时间内有 $ n $ 根头发脱落和生长。\n若 $ t $ 时刻的头发平均长度为 $ l(t) $ ，则在 $ \\left( t+\\Delta t \\right) $ 时刻有\n$ l\\left( t+\\Delta t \\right) =\\left( l\\left( t \\right) +\\iota \\Delta t \\right) \\left( 1-\\frac{n\\Delta t}{N} \\right) $\n令 $ \\Delta t\\to 0 $，得到一阶常系数微分方程\n$ \\frac{dl}{dt}+\\frac{n}{N}l-\\iota =0 $\n设 $ l(0)=0 $ ，则\n$ l=\\frac{\\iota N}{n}\\left( 1-\\mathrm{e}^{-\\frac{n}{N}t} \\right) $\n头发平均长度最初的增加速度近似等于每根头发生长的速度，在时间足够长后达到稳定值 $ \\frac{\\iota N}{n} $ ，这也与物理直觉相符。","cooked":"

在上次约饭时，@anon60541984 提出了头发生长与掉落速度恒定的假设，这里计算头发平均长度从 0 开始逐渐生长，并达到动态平衡的过程。
\n设头发的总数为 $ N $，每根头发在单位时间内增长 $ \\iota $，且单位时间内有 $ n $ 根头发脱落和生长。
\n若 $ t $ 时刻的头发平均长度为 $ l(t) $ ，则在 $ \\left( t+\\Delta t \\right) $ 时刻有
\n$ l\\left( t+\\Delta t \\right) =\\left( l\\left( t \\right) +\\iota \\Delta t \\right) \\left( 1-\\frac{n\\Delta t}{N} \\right) $
\n令 $ \\Delta t\\to 0 $，得到一阶常系数微分方程
\n$ \\frac{dl}{dt}+\\frac{n}{N}l-\\iota =0 $
\n设 $ l(0)=0 $ ，则
\n$ l=\\frac{\\iota N}{n}\\left( 1-\\mathrm{e}^{-\\frac{n}{N}t} \\right) $
\n头发平均长度最初的增加速度近似等于每根头发生长的速度，在时间足够长后达到稳定值 $ \\frac{\\iota N}{n} $ ，这也与物理直觉相符。

\n","createdAt":"2025-03-26 00:48:14.608165","updatedAt":"2025-03-26 00:48:14.608165","replyToPostNumber":null,"replyCount":1,"likeCount":2,"wordCount":233,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":2,"users":[{"id":4668,"username":"IO","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":47483,"postNumber":32,"raw":"[quote=\"被遗忘的人，post:5046, topic:324380, full:true, username:KindergartenKiller\"]\n长度为 1 的木棍扫过角度为 \\frac{\\pi}{3}π3\\frac{\\pi}{3} 的墙角，请问扫过的面积？\n[/quote]\n\n对这题，我出于对平面几何的朴素认识决定用微积分算。\n首先，这个运动过程前后对称，所以只需要研究木棍从紧贴上沿运动到与墙角形成等边三角形的过程。\n其次，木棍从紧贴上沿到运动到与下沿垂直的过程中，扫过的图形显然为直角三角形，这部分很容易求解，所以只需要计算从直角三角形到等边三角形过程中扫过的不规则图形面积。\n在这里，我把不规则图形拆成从上到下的一个个三角形（见蓝色部分），彼此不重合。所以只需要计算每个三角形的面积再积分。\n![image|558x500](/uploads/image/95/6f/956f303099d3024b4e67960b774f37a277c89ae11a83fe3f112c69c502c54703.avif)\n如何计算三角形的面积呢？如下图，我们首先规定木棍的长度为 $ l $ ，相对上沿的夹角为 $ \\theta $ ，那么根据 $ \\bigtriangleup DAE $ 的正弦定理，另外两条边的长度满足\n$ \\frac{a}{\\sin \\theta}=\\frac{b}{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right)}=\\frac{l}{\\sin \\frac{\\pi}{3}} $\n![image|582x499](/uploads/image/65/f9/65f95396ba0cc09d18d46cd99479f485442b65e282014b5f10ede32ead896436.avif)\n解得 $ a=\\frac{2}{\\sqrt{3}}l\\sin \\theta ,b=\\frac{2}{\\sqrt{3}}l\\sin \\left( \\theta +\\frac{\\pi}{3} \\right) $\n当夹角 $ \\theta $ 变化一个小量 $ \\Delta \\theta $ 时，两条边长的变化量为\n$ \\Delta a\\approx \\frac{2}{\\sqrt{3}}l\\cos \\theta \\Delta \\theta ,\\Delta b\\approx -\\frac{2}{\\sqrt{3}}l\\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\Delta \\theta $\n根据 $ \\bigtriangleup DAE \\text{∽} \\bigtriangleup IEG$ ，得 $ IE $ 长度\n$ \\Delta a^*\\approx \\Delta a\\frac{b}{a}\\approx \\frac{2}{\\sqrt{3}}l\\cos \\theta \\frac{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right)}{\\sin \\theta}\\Delta \\theta $\n同样根据相似三角形，$ \\bigtriangleup DFH $ 和 $ \\bigtriangleup EIH $ 的高之比为 \n$ \\frac{h_1}{h_2}=\\frac{\\Delta b}{\\Delta a^*}\\approx \\frac{-\\frac{2}{\\sqrt{3}}l\\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\Delta \\theta}{\\frac{2}{\\sqrt{3}}l\\cos \\theta \\frac{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right)}{\\sin \\theta}\\Delta \\theta}=-\\frac{\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right)} $\n所以\n$ h_1=\\frac{\\sqrt{3}}{2}a\\frac{-\\frac{\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right)}}{1-\\frac{\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right)}}=\\frac{\\sqrt{3}}{2}a\\frac{-\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta} $\n$ \\Delta S_1=\\frac{1}{2}\\Delta bh_1\n\\\\=\\frac{1}{2}\\frac{2}{\\sqrt{3}}l\\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\Delta \\theta \\frac{\\sqrt{3}}{2}\\frac{2}{\\sqrt{3}}l\\sin \\theta \\frac{-\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta}\n\\\\\n=\\frac{1}{\\sqrt{3}}l^2\\frac{-\\sin \\theta \\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta}\\Delta \\theta $\n$ S_1=\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{3}}{\\frac{1}{\\sqrt{3}}l^2\\frac{\\sin \\theta \\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta}d\\theta}\n\\\\=\\frac{l^2}{\\sqrt{3}}\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{3}}{\\frac{\\sin ^2\\theta \\cos ^2\\left( \\theta +\\frac{\\pi}{3} \\right)}{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right) \\cos \\theta -\\sin \\theta \\cos \\left( \\theta +\\frac{\\pi}{3} \\right)}d\\theta}\n\\\\\n=\\frac{5l^2}{288}\\left( 2\\pi -3\\sqrt{3} \\right) \n$\n扫过的总面积为 $ \\frac{5l^2}{144}\\left( 2\\pi -3\\sqrt{3} \\right) +\\frac{\\sqrt{3}}{4}l^2=\\left( \\frac{7}{48}\\sqrt{3}+\\frac{5}{72}\\pi \\right) l^2 $","cooked":"

\n\t\t\t\t\t

\n\t\t\t\t\t\t

\n\t\t\t\t\t\t\t

\n\t\t\t\t\t\t\t\t\n\t\t\t\t\t\t\t

\n\t\t\t\t\t\t\t被遗忘的人，post:5046, topic:324380, full:true, username:KindergartenKiller:\n\t\t\t\t\t\t

\n\t\t\t\t\t

\n
长度为 1 的木棍扫过角度为 \\frac{\\pi}{3}π3\\frac{\\pi}{3} 的墙角，请问扫过的面积？
\n

\n\t\t\t\t

对这题，我出于对平面几何的朴素认识决定用微积分算。
\n首先，这个运动过程前后对称，所以只需要研究木棍从紧贴上沿运动到与墙角形成等边三角形的过程。
\n其次，木棍从紧贴上沿到运动到与下沿垂直的过程中，扫过的图形显然为直角三角形，这部分很容易求解，所以只需要计算从直角三角形到等边三角形过程中扫过的不规则图形面积。
\n在这里，我把不规则图形拆成从上到下的一个个三角形（见蓝色部分），彼此不重合。所以只需要计算每个三角形的面积再积分。
\n $\"image\"$
\n如何计算三角形的面积呢？如下图，我们首先规定木棍的长度为 $ l $ ，相对上沿的夹角为 $ \\theta $ ，那么根据 $ \\bigtriangleup DAE $ 的正弦定理，另外两条边的长度满足
\n$ \\frac{a}{\\sin \\theta}=\\frac{b}{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right)}=\\frac{l}{\\sin \\frac{\\pi}{3}} $
\n $\"image\"$
\n解得 $ a=\\frac{2}{\\sqrt{3}}l\\sin \\theta ,b=\\frac{2}{\\sqrt{3}}l\\sin \\left( \\theta +\\frac{\\pi}{3} \\right) $
\n当夹角 $ \\theta $ 变化一个小量 $ \\Delta \\theta $ 时，两条边长的变化量为
\n$ \\Delta a\\approx \\frac{2}{\\sqrt{3}}l\\cos \\theta \\Delta \\theta ,\\Delta b\\approx -\\frac{2}{\\sqrt{3}}l\\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\Delta \\theta $
\n根据 $ \\bigtriangleup DAE \\text{∽} \\bigtriangleup IEG$ ，得 $ IE $ 长度
\n$ \\Delta a^\\approx \\Delta a\\frac{b}{a}\\approx \\frac{2}{\\sqrt{3}}l\\cos \\theta \\frac{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right)}{\\sin \\theta}\\Delta \\theta $
\n同样根据相似三角形，$ \\bigtriangleup DFH $ 和 $ \\bigtriangleup EIH $ 的高之比为
\n$ \\frac{h_1}{h_2}=\\frac{\\Delta b}{\\Delta a^}\\approx \\frac{-\\frac{2}{\\sqrt{3}}l\\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\Delta \\theta}{\\frac{2}{\\sqrt{3}}l\\cos \\theta \\frac{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right)}{\\sin \\theta}\\Delta \\theta}=-\\frac{\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right)} $
\n所以
\n$ h_1=\\frac{\\sqrt{3}}{2}a\\frac{-\\frac{\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right)}}{1-\\frac{\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right)}}=\\frac{\\sqrt{3}}{2}a\\frac{-\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta} $
\n$ \\Delta S_1=\\frac{1}{2}\\Delta bh_1
\n\\=\\frac{1}{2}\\frac{2}{\\sqrt{3}}l\\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\Delta \\theta \\frac{\\sqrt{3}}{2}\\frac{2}{\\sqrt{3}}l\\sin \\theta \\frac{-\\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta}
\n\\
\n=\\frac{1}{\\sqrt{3}}l^2\\frac{-\\sin \\theta \\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta}\\Delta \\theta $
\n$ S_1=\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{3}}{\\frac{1}{\\sqrt{3}}l^2\\frac{\\sin \\theta \\cos \\left( \\theta +\\frac{\\pi}{3} \\right) \\tan \\theta}{\\tan \\left( \\theta +\\frac{\\pi}{3} \\right) -\\tan \\theta}d\\theta}
\n\\=\\frac{l^2}{\\sqrt{3}}\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{3}}{\\frac{\\sin ^2\\theta \\cos ^2\\left( \\theta +\\frac{\\pi}{3} \\right)}{\\sin \\left( \\theta +\\frac{\\pi}{3} \\right) \\cos \\theta -\\sin \\theta \\cos \\left( \\theta +\\frac{\\pi}{3} \\right)}d\\theta}
\n\\
\n=\\frac{5l^2}{288}\\left( 2\\pi -3\\sqrt{3} \\right)
\n$$
\n扫过的总面积为 $ \\frac{5l^2}{144}\\left( 2\\pi -3\\sqrt{3} \\right) +\\frac{\\sqrt{3}}{4}l^2=\\left( \\frac{7}{48}\\sqrt{3}+\\frac{5}{72}\\pi \\right) l^2 $

\n","createdAt":"2025-03-26 00:50:48.015429","updatedAt":"2025-03-26 00:50:48.015429","replyToPostNumber":null,"replyCount":0,"likeCount":3,"wordCount":574,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":3,"users":[{"id":1411,"username":"lwe","name":null},{"id":4668,"username":"IO","name":null},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":47546,"postNumber":33,"raw":"有点像逻辑斯蒂增长模型，生物种群 S 型增长","cooked":"

有点像逻辑斯蒂增长模型，生物种群 S 型增长

\n","createdAt":"2025-03-29 09:38:41.00183","updatedAt":"2025-03-29 09:38:41.00183","replyToPostNumber":31,"replyCount":1,"likeCount":2,"wordCount":20,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"wave_gif","count":1,"users":[{"id":1288,"username":"LittleWangInShanghai","name":"小王在上海"}],"hasReacted":false},{"reaction":"❤️","count":2,"users":[{"id":1288,"username":"LittleWangInShanghai","name":"小王在上海"},{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":424,"username":"ascah","name":null,"avatarUrl":"/uploads/image/5e/97/5e97b59e771d24d1195b20454be3dd5ec6a74b6581b637ec88711aa70f6372bb.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":47547,"postNumber":34,"raw":"这是上帝的旨意。","cooked":"

这是上帝的旨意。

\n","createdAt":"2025-03-29 09:58:26.766694","updatedAt":"2025-03-29 09:58:26.766694","replyToPostNumber":33,"replyCount":0,"likeCount":1,"wordCount":8,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":7352,"username":"Love_FDU","name":"🌲➕🖐️"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1288,"username":"LittleWangInShanghai","name":"小王在上海","avatarUrl":"/uploads/image/63/d2/63d25f4f2f6693f9ab450f32bd8d04e2d29f96a980b51555a585423f113c4a58.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":49629,"postNumber":35,"raw":"似乎 KaTeX 没有 `\\boldsymbol`，只有 `\\vec` 吧？比如 $\\vec{a}$ 等．\n\n~~现在这样有点像黑箱，得猜公式里面有啥~~","cooked":"

似乎 KaTeX 没有 \\boldsymbol，只有 \\vec 吧？比如 $\\vec{a}$ 等．

~~现在这样有点像黑箱，得猜公式里面有啥~~

\n","createdAt":"2025-04-21 10:49:34.248176","updatedAt":"2025-04-21 10:50:44.74133","replyToPostNumber":25,"replyCount":1,"likeCount":1,"wordCount":36,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":1817,"username":"anon1678451","name":null}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":5905,"username":"PipaQinse233","name":"PipaQinse233","avatarUrl":"/uploads/image/f7/1b/f71bbda87a32de08df517fe7ee4a14e3f63bfd1b2105d3ddb8ba1f4c2eaa1c29.avif","admin":false,"moderator":false,"signatureRaw":"[right]——發自我的手機[/right]","signatureCooked":"

——發自我的手機

"}},{"id":49630,"postNumber":36,"raw":"刷新一下试试？","cooked":"

刷新一下试试？

\n","createdAt":"2025-04-21 11:07:29.232681","updatedAt":"2025-04-21 11:07:29.232681","replyToPostNumber":35,"replyCount":1,"likeCount":0,"wordCount":7,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":49645,"postNumber":37,"raw":"正常显示啦\n\n---\n\n~~坏了有错别字~~","cooked":"

正常显示啦

~~坏了有错别字~~

\n","createdAt":"2025-04-21 13:34:58.566913","updatedAt":"2025-09-30 09:46:36.68742","replyToPostNumber":36,"replyCount":0,"likeCount":0,"wordCount":14,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":2,"publicVersion":2,"deletedAt":null,"isAnonymous":false,"user":{"id":5905,"username":"PipaQinse233","name":"PipaQinse233","avatarUrl":"/uploads/image/f7/1b/f71bbda87a32de08df517fe7ee4a14e3f63bfd1b2105d3ddb8ba1f4c2eaa1c29.avif","admin":false,"moderator":false,"signatureRaw":"[right]——發自我的手機[/right]","signatureCooked":"

——發自我的手機

"}},{"id":66437,"postNumber":38,"raw":"# 量子 Fourier 变换（1）\n众所周知，经典的离散 Fourier 变换（DFT）为\n\n$ y_j \\equiv \\frac{1}{\\sqrt{N}} \\sum_{k=0}^{N-1} x_j e^{2\\pi i j k / N} $\n\n其中 $\\frac{1}{\\sqrt{N}}$ 的系数是为了使得正变换和逆变换在形式上更加对称。特别地，当 $N=2^n$ 时，可以使用高效的 FFT 算法来减少运算次数（但代价是需要级联，在空间足够的情况下花费时间更长）；当 $N$ 为一般合数时也可以使用 FFT 算法，只是不彻底。\n\n如果把经典 DFT 的输入 $ (x_1,\\ldots, x_N) $ 看作一个 $ N $ 维向量，那么量子 Fourier 变换在形式上就与经典 DFT 非常相似。设量子态 $ \\lvert \\psi \\rangle $ 所在空间的标准正交基为 $ \\lvert 0 \\rangle, \\ldots, \\lvert N - 1 \\rangle $，量子 Fourier 变换算子 $ F_N $ 对基向量 $ \\lvert j \\rangle $ 的作用为\n$$\nF_N\\lvert j \\rangle = \\frac{1}{\\sqrt{N}} \\sum_{k=0}^{N-1} e^{2\\pi i j k / N} \\lvert k \\rangle\n$$\n即 $ F_N $ 的每个矩阵元满足\n$$\n\\big(F_N\\big)_{jk} = \\frac{1}{\\sqrt{N}} \\, e^{\\tfrac{2\\pi i}{N} \\, jk}, \n\\quad j,k = 0,1,\\ldots,N-1.\n$$\n\n这看起来并不复杂。那么如何用量子系统来实现这个变换呢？最简单且最常用的量子系统是量子比特（qbit），只有 $ \\lvert 0\\rangle $ 和 $ \\lvert 1\\rangle $ 两个基矢态，其状态可以在这组基下展开成 $ \\lvert \\psi \\rangle = a\\lvert 0\\rangle+b\\lvert 1\\rangle $ ，其中 $ a,b $ 为复数，且满足 $ |a|^2+|b|^2=1 $. \n\n那么如果把 $ n $ 个量子比特组合起来，其状态就是所有量子比特状态的张量积 $ \\lvert \\psi_1 \\rangle \\otimes \\lvert \\psi_2 \\rangle \\otimes \\cdots \\otimes \\lvert \\psi_n \\rangle $ （以下简写为 $ \\lvert \\psi_1 \\rangle \\lvert \\psi_2 \\rangle \\cdots \\lvert \\psi_n \\rangle $ 或 $ \\lvert \\psi_1 \\psi_2 \\cdots \\psi_n \\rangle $ ，张量的定义可参见 [张量基础知识](https://xjtu.app/t/topic/6869/29)），状态空间相应的正交基为 $ \\lvert 0\\cdots 00 \\rangle ,\\lvert 0\\cdots 01 \\rangle ,\\ldots,\\lvert 1\\cdots 11 \\rangle $ 总共 $2^n $ 个，也就是只需要 $n$ 个量子比特就足以实现对 $N=2^n$ 个输入参数的量子 Fourier 变换的模拟。\n\n不过严格来说，经典 DFT 的每个输入都是任意相位和幅值的复数，自由度为 $ 2\\cdot 2^n $ ，而量子 Fourier 变换由于整体相位不可观测和参数归一化的要求，自由度只有 $ 2\\cdot 2^n-2 $.","cooked":"

量子 Fourier 变换（1）

众所周知，经典的离散 Fourier 变换（DFT）为

$ y_j \\equiv \\frac{1}{\\sqrt{N}} \\sum_{k=0}^{N-1} x_j e^{2\\pi i j k / N} $

其中 $\\frac{1}{\\sqrt{N}}$ 的系数是为了使得正变换和逆变换在形式上更加对称。特别地，当 $N=2^n$ 时，可以使用高效的 FFT 算法来减少运算次数（但代价是需要级联，在空间足够的情况下花费时间更长）；当 $N$ 为一般合数时也可以使用 FFT 算法，只是不彻底。

如果把经典 DFT 的输入 $ (x_1,\\ldots, x_N) $ 看作一个 $ N $ 维向量，那么量子 Fourier 变换在形式上就与经典 DFT 非常相似。设量子态 $ \\lvert \\psi \\rangle $ 所在空间的标准正交基为 $ \\lvert 0 \\rangle, \\ldots, \\lvert N - 1 \\rangle $，量子 Fourier 变换算子 $ F_N $ 对基向量 $ \\lvert j \\rangle $ 的作用为
\n

\nF_N\\lvert j \\rangle = \\frac{1}{\\sqrt{N}} \\sum_{k=0}^{N-1} e^{2\\pi i j k / N} \\lvert k \\rangle\n

\n即 $ F_N $ 的每个矩阵元满足
\n

\n\\big(F_N\\big)_{jk} = \\frac{1}{\\sqrt{N}} \\, e^{\\tfrac{2\\pi i}{N} \\, jk}, \n\\quad j,k = 0,1,\\ldots,N-1.\n

这看起来并不复杂。那么如何用量子系统来实现这个变换呢？最简单且最常用的量子系统是量子比特（qbit），只有 $ \\lvert 0\\rangle $ 和 $ \\lvert 1\\rangle $ 两个基矢态，其状态可以在这组基下展开成 $ \\lvert \\psi \\rangle = a\\lvert 0\\rangle+b\\lvert 1\\rangle $ ，其中 $ a,b $ 为复数，且满足 $ |a|^2+|b|^2=1 $.

那么如果把 $ n $ 个量子比特组合起来，其状态就是所有量子比特状态的张量积 $ \\lvert \\psi_1 \\rangle \\otimes \\lvert \\psi_2 \\rangle \\otimes \\cdots \\otimes \\lvert \\psi_n \\rangle $ （以下简写为 $ \\lvert \\psi_1 \\rangle \\lvert \\psi_2 \\rangle \\cdots \\lvert \\psi_n \\rangle $ 或 $ \\lvert \\psi_1 \\psi_2 \\cdots \\psi_n \\rangle $ ，张量的定义可参见张量基础知识），状态空间相应的正交基为 $ \\lvert 0\\cdots 00 \\rangle ,\\lvert 0\\cdots 01 \\rangle ,\\ldots,\\lvert 1\\cdots 11 \\rangle $ 总共 $2^n $ 个，也就是只需要 $n$ 个量子比特就足以实现对 $N=2^n$ 个输入参数的量子 Fourier 变换的模拟。

不过严格来说，经典 DFT 的每个输入都是任意相位和幅值的复数，自由度为 $ 2\\cdot 2^n $ ，而量子 Fourier 变换由于整体相位不可观测和参数归一化的要求，自由度只有 $ 2\\cdot 2^n-2 $.

\n","createdAt":"2025-09-30 09:34:18.207857","updatedAt":"2025-09-30 11:11:22.325279","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":609,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false}],"wiki":false,"version":3,"publicVersion":3,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":66439,"postNumber":39,"raw":"# 量子 Fourier 变换（2）\n根据量子力学公设，封闭量子系统的演化可用酉变换 $ U $ 来描述。作用于量子比特的酉算符 $ U $ 就是量子门，可以视为经典逻辑门的拓展。\n\n接下来介绍搭建 Fourier 变换需要用到的量子门。\n\n* Hadamard 门：\n$$\nH\\equiv\\frac{1}{\\sqrt{2}}\n\\begin{bmatrix}\n1 & 1 \\\\\n1 & -1\n\\end{bmatrix}\n$$\n\n* Pauli-Z 算子：\n\n$$\nZ \\equiv\n\\begin{bmatrix}\n1 & 0 \\\\\n0 & -1\n\\end{bmatrix}\n$$\n\n* $\\hat{z}$ 旋转算子：\n$$\nR_z(\\theta) \\;\\equiv\\; e^{-i\\theta Z/2} \n= \\cos\\!\\left(\\tfrac{\\theta}{2}\\right) I \\;-\\; i \\sin\\!\\left(\\tfrac{\\theta}{2}\\right) Z \n= \n\\begin{bmatrix}\ne^{-i\\theta/2} & 0 \\\\\n0 & e^{i\\theta/2}\n\\end{bmatrix}\n$$\n\n* 受控 U 门：\n符号如图，但这里只考虑 $U$ 作用于单个量子比特的情形：![image|300x300, 50%](/uploads/image/b6/05/b605d8bdc991d645eace5ed09c5e77ad82edd21de8d60549a07a58ae33e23a29.avif)\n若 $U$ 是一个量子门，则受控 $U$ 门的第一个量子比特是控制比特，第二个量子比特是受控比特。受控 $U$ 门不会改变控制比特，且当控制比特为 $\\lvert 0 \\rangle$ 时，受控比特保持不变；控制比特为 $\\lvert 1 \\rangle$ 时，受控比特受到 $U$ 作用。写成矩阵形式就是\n$$\nC_U \\;=\\;\n\\begin{bmatrix}\nI & 0 \\\\\n0 & U\n\\end{bmatrix}\n$$\n\n为了使用常见的量子门来搭建出 Fourier 变换的实现电路，我们需要把量子 Fourier 变换的公式改写为张量积形式，其中用 $ j_1j_2\\ldots j_n $ 来表示 $j$ 的逐位二进制， $ 0. j_1j_2\\ldots j_n $ 表示逐位二进制小数：\n\n$$\n\\begin{align}\nF_N\\lvert j \\rangle \n&=\\frac{1}{2^{n/2}} \\sum_{k=0}^{2^n - 1} e^{\\tfrac{2\\pi i j k}{2^n}} \\lvert k \\rangle \\\\\n\n&= \\frac{1}{2^{n/2}} \\sum_{k_1=0}^1 \\cdots \\sum_{k_n=0}^1 \ne^{\\tfrac{2\\pi i j}{2^n} \\sum_{l=1}^n k_l 2^{\\,n-l}} \n\\lvert k_1 \\cdots k_n \\rangle \\\\\n\n&= \\frac{1}{2^{n/2}} \\sum_{k_1=0}^1 \\cdots \\sum_{k_n=0}^1 \ne^{2\\pi i j \\sum_{l=1}^n {k_l}/{2^l}} \n\\lvert k_1 \\cdots k_n \\rangle \\\\\n\n&= \\frac{1}{2^{n/2}} \\bigotimes_{l=1}^n \n\\left( \\sum_{k_l=0}^1 e^{2\\pi i j k_l / 2^l} \\lvert k_l \\rangle \\right) \\\\\n\n&= \\frac{1}{2^{n/2}} \\bigotimes_{l=1}^n \n\\left( \\lvert 0 \\rangle + e^{2\\pi i j / 2^l} \\lvert 1 \\rangle \\right) \\\\\n\n&= \\frac{1}{2^{n/2}} \n\\left( \\lvert 0 \\rangle + e^{2\\pi i j_n} \\lvert 1 \\rangle \\right)\n\\left( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_{n-1} j_n} \\lvert 1 \\rangle \\right)\n\\cdots\n\\left( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2 \\cdots j_n} \\lvert 1 \\rangle \\right).\n\\end{align}\n$$\n\n接下来就能画出实现量子 Fourier 变换的电路图了！这里的 $ R_k $ 门为\n$$\nR_k \\;\\equiv\\;\n\\begin{bmatrix}\n1 & 0 \\\\\n0 & e^{2\\pi i / 2^k}\n\\end{bmatrix}\n$$\n其实就是上文 $\\hat{z}$ 旋转算子的特殊情况，并且乘上一个不产生影响的整体相位。\n![tmp_file44877803b040f18df5bb4c987c1c4408aa6836d1|690x170](/uploads/image/db/c9/dbc95026b76a18674f805bf886be1dad4bc23e971dee6e8fbedfd142636e55be.avif)\n考虑输入态为 $ \\lvert j_1 j_2 \\cdots j_n \\rangle $ 时发生的变化。经过第一个 Hadamard 门后，系统状态为\n$$\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1} \\lvert 1 \\rangle \\Big)\n\\otimes\n\\lvert j_2 \\cdots j_n \\rangle\n$$\n再经过受控 $R_2$ 门，得到的状态是：\n$$\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2} \\lvert 1 \\rangle \\Big)\n\\otimes\n\\lvert j_2 \\cdots j_n \\rangle\n$$\n依此类推，系统在第一个量子比特上的操作结束时状态为\n$$\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2 \\cdots j_n} \\lvert 1 \\rangle \\Big)\n\\otimes\n\\lvert j_2 \\cdots j_n \\rangle.\n$$\n同样第 $k$ 个比特操作后的状态为\n$$\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_k j_{k+1} \\cdots j_n} \\lvert 1 \\rangle \\Big).\n$$\n所以所有比特的最终状态为\n$$\n\\frac{1}{2^{n/2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2 \\cdots j_n} \\lvert 1 \\rangle \\Big)\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_2 \\cdots j_n} \\lvert 1 \\rangle \\Big)\n\\cdots\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_n} \\lvert 1 \\rangle \\Big)\n$$\n颠倒输出顺序，就得到上文 Fourier 变换结果的张量积形式\n$$\n\\frac{1}{2^{n/2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i 0.j_1} \\lvert 1 \\rangle \\Big)\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i 0.j_2 j_1} \\lvert 1 \\rangle \\Big)\n\\cdots\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i 0.j_n j_{n-1} \\cdots j_1} \\lvert 1 \\rangle \\Big).\n$$","cooked":"

量子 Fourier 变换（2）

根据量子力学公设，封闭量子系统的演化可用酉变换 $ U $ 来描述。作用于量子比特的酉算符 $ U $ 就是量子门，可以视为经典逻辑门的拓展。

接下来介绍搭建 Fourier 变换需要用到的量子门。

\n
Hadamard 门：
\n
$\nH\\equiv\\frac{1}{\\sqrt{2}}\n\\begin{bmatrix}\n1 & 1 \\\\\n1 & -1\n\\end{bmatrix}\n$
\n
\n
Pauli-Z 算子：
\n

\nZ \\equiv\n\\begin{bmatrix}\n1 & 0 \\\\\n0 & -1\n\\end{bmatrix}\n

\n
$\\hat{z}$ 旋转算子：
\n
$\nR_z(\\theta) \\;\\equiv\\; e^{-i\\theta Z/2} \n= \\cos\\!\\left(\\tfrac{\\theta}{2}\\right) I \\;-\\; i \\sin\\!\\left(\\tfrac{\\theta}{2}\\right) Z \n= \n\\begin{bmatrix}\ne^{-i\\theta/2} & 0 \\\\\n0 & e^{i\\theta/2}\n\\end{bmatrix}\n$
\n
\n
受控 U 门：
\n符号如图，但这里只考虑 $U$ 作用于单个量子比特的情形： $\"image\"$
\n若 $U$ 是一个量子门，则受控 $U$ 门的第一个量子比特是控制比特，第二个量子比特是受控比特。受控 $U$ 门不会改变控制比特，且当控制比特为 $\\lvert 0 \\rangle$ 时，受控比特保持不变；控制比特为 $\\lvert 1 \\rangle$ 时，受控比特受到 $U$ 作用。写成矩阵形式就是
\n
$\nC_U \\;=\\;\n\\begin{bmatrix}\nI & 0 \\\\\n0 & U\n\\end{bmatrix}\n$
\n

为了使用常见的量子门来搭建出 Fourier 变换的实现电路，我们需要把量子 Fourier 变换的公式改写为张量积形式，其中用 $ j_1j_2\\ldots j_n $ 来表示 $j$ 的逐位二进制， $ 0. j_1j_2\\ldots j_n $ 表示逐位二进制小数：

\n\\begin{align}\nF_N\\lvert j \\rangle \n&=\\frac{1}{2^{n/2}} \\sum_{k=0}^{2^n - 1} e^{\\tfrac{2\\pi i j k}{2^n}} \\lvert k \\rangle \\\\\n&= \\frac{1}{2^{n/2}} \\sum_{k_1=0}^1 \\cdots \\sum_{k_n=0}^1 \ne^{\\tfrac{2\\pi i j}{2^n} \\sum_{l=1}^n k_l 2^{\\,n-l}} \n\\lvert k_1 \\cdots k_n \\rangle \\\\\n&= \\frac{1}{2^{n/2}} \\sum_{k_1=0}^1 \\cdots \\sum_{k_n=0}^1 \ne^{2\\pi i j \\sum_{l=1}^n {k_l}/{2^l}} \n\\lvert k_1 \\cdots k_n \\rangle \\\\\n&= \\frac{1}{2^{n/2}} \\bigotimes_{l=1}^n \n\\left( \\sum_{k_l=0}^1 e^{2\\pi i j k_l / 2^l} \\lvert k_l \\rangle \\right) \\\\\n&= \\frac{1}{2^{n/2}} \\bigotimes_{l=1}^n \n\\left( \\lvert 0 \\rangle + e^{2\\pi i j / 2^l} \\lvert 1 \\rangle \\right) \\\\\n&= \\frac{1}{2^{n/2}} \n\\left( \\lvert 0 \\rangle + e^{2\\pi i j_n} \\lvert 1 \\rangle \\right)\n\\left( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_{n-1} j_n} \\lvert 1 \\rangle \\right)\n\\cdots\n\\left( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2 \\cdots j_n} \\lvert 1 \\rangle \\right).\n\\end{align}\n

接下来就能画出实现量子 Fourier 变换的电路图了！这里的 $ R_k $ 门为
\n

\nR_k \\;\\equiv\\;\n\\begin{bmatrix}\n1 & 0 \\\\\n0 & e^{2\\pi i / 2^k}\n\\end{bmatrix}\n

\n其实就是上文

\\hat{z}

旋转算子的特殊情况，并且乘上一个不产生影响的整体相位。
\n $\"tmp_file44877803b040f18df5bb4c987c1c4408aa6836d1\"$
\n考虑输入态为 $ \\lvert j_1 j_2 \\cdots j_n \\rangle $ 时发生的变化。经过第一个 Hadamard 门后，系统状态为
\n

\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1} \\lvert 1 \\rangle \\Big)\n\\otimes\n\\lvert j_2 \\cdots j_n \\rangle\n

\n再经过受控

R_2

门，得到的状态是：
\n

\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2} \\lvert 1 \\rangle \\Big)\n\\otimes\n\\lvert j_2 \\cdots j_n \\rangle\n

\n依此类推，系统在第一个量子比特上的操作结束时状态为
\n

\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2 \\cdots j_n} \\lvert 1 \\rangle \\Big)\n\\otimes\n\\lvert j_2 \\cdots j_n \\rangle.\n

\n同样第

k

个比特操作后的状态为
\n

\n\\frac{1}{\\sqrt{2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_k j_{k+1} \\cdots j_n} \\lvert 1 \\rangle \\Big).\n

\n所以所有比特的最终状态为
\n

\n\\frac{1}{2^{n/2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_1 j_2 \\cdots j_n} \\lvert 1 \\rangle \\Big)\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_2 \\cdots j_n} \\lvert 1 \\rangle \\Big)\n\\cdots\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i \\, 0.j_n} \\lvert 1 \\rangle \\Big)\n

\n颠倒输出顺序，就得到上文 Fourier 变换结果的张量积形式
\n

\n\\frac{1}{2^{n/2}}\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i 0.j_1} \\lvert 1 \\rangle \\Big)\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i 0.j_2 j_1} \\lvert 1 \\rangle \\Big)\n\\cdots\n\\Big( \\lvert 0 \\rangle + e^{2\\pi i 0.j_n j_{n-1} \\cdots j_1} \\lvert 1 \\rangle \\Big).\n

\n","createdAt":"2025-09-30 09:40:39.46829","updatedAt":"2025-09-30 09:40:39.46829","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":894,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":72096,"postNumber":40,"raw":"# 量子光学简介(1)\n众所周知，量子力学中，位置算符是\$\\hat{x}\\psi(x)=x\\psi(x)\$，动量算符是\$\\hat{p}\\psi(x)=-i\\hbar\\frac{\\partial}{\\partial x}\\psi(x)\$.\n可以用其线性组合定义湮灭算符\$\\hat{a} = \\frac{1}{\\sqrt{2\\hbar\\omega}} (\\omega \\hat{x} + i\\hat{p})\$和产生算符\$\\hat{a}^\\dagger = \\frac{1}{\\sqrt{2\\hbar\\omega}} (\\omega \\hat{x} - i\\hat{p})\$. 注意这两个算符都不是Hermitian算符，因此不对应可观测量，它们互为Hermitian共轭：\$\\hat{a}^\\dagger = (\\hat{a})^\\dagger\$. \n\n所有复数\$\\alpha\$都是湮灭算符\$\\hat{a}\$的本征值，对应的本征态为相干态\$|\\alpha\\rangle\$，即\$\\hat{a} |\\alpha\\rangle = \\alpha |\\alpha\\rangle\$. 其中0对应的本征态为真空态\$|0\\rangle\$，即\$\\hat{a} |0\\rangle = 0\$，在物理上对应无粒子的状态.\n产生算符作用在真空态可以得到粒子数为1的状态，即\${\\hat{a}^\\dagger\\,|0\\rangle = |1\\rangle}\$，作用\$n\$次可以产生粒子数为\$n\$的状态\$|n\\rangle\$，即Fock态：\${(\\hat{a}^\\dagger)^n\\,|0\\rangle = \\sqrt{n!}\\;|n\\rangle}\$. 相反，湮灭算符作用在Fock态\$|n\\rangle\$上可以得到\$|n-1\\rangle\$，即\$\\hat{a}|n\\rangle = \\sqrt{n}\\,|n-1\\rangle\$，这也是产生算符和湮灭算符名字的由来。\n\n用湮灭算符和产生算符可以定义粒子数算符\$\\hat{N} = \\hat{a}^\\dagger \\hat{a}\$，这是一个Hermitian算符，其本征值为量子系统中的粒子数，本征态为Fock态，即\$\\hat{N}|n\\rangle = n |n\\rangle\$\n\n相干态\$|\\alpha\\rangle\$用Fock态\$|n\\rangle\$展开是\$|\\alpha\\rangle = e^{-\\frac{|\\alpha|^2}{2}} \\sum_{n=0}^{\\infty} \\frac{\\alpha^n}{\\sqrt{n!}} |n\\rangle\$，证明留作习题（划掉）\n\nWeyl位移算符定义为\$\\hat{D}(\\alpha) = e^{\\alpha \\hat{a}^\\dagger - \\alpha^* \\hat{a}}\$，其作用效果是把湮灭和产生算符分别平移\$\\alpha\$和\$ \\alpha^*\$，即\$\\hat{D}^\\dagger(\\alpha) \\hat{a} \\hat{D}(\\alpha) = \\hat{a} + \\alpha\$, \$\\hat{D}^\\dagger(\\alpha) \\hat{a}^\\dagger \\hat{D}(\\alpha) = \\hat{a}^\\dagger + \\alpha^*\$，所以作用在真空态\$|0\\rangle\$上可以得到相干态：\$|\\alpha\\rangle = \\hat{D}(\\alpha) |0\\rangle\$.","cooked":"

量子光学简介(1)

众所周知，量子力学中，位置算符是(\\hat{x}\\psi(x)=x\\psi(x))，动量算符是(\\hat{p}\\psi(x)=-i\\hbar\\frac{\\partial}{\\partial x}\\psi(x)).
\n可以用其线性组合定义湮灭算符(\\hat{a} = \\frac{1}{\\sqrt{2\\hbar\\omega}} (\\omega \\hat{x} + i\\hat{p}))和产生算符(\\hat{a}^\\dagger = \\frac{1}{\\sqrt{2\\hbar\\omega}} (\\omega \\hat{x} - i\\hat{p})). 注意这两个算符都不是Hermitian算符，因此不对应可观测量，它们互为Hermitian共轭：(\\hat{a}^\\dagger = (\\hat{a})^\\dagger).

所有复数(\\alpha)都是湮灭算符(\\hat{a})的本征值，对应的本征态为相干态(|\\alpha\\rangle)，即(\\hat{a} |\\alpha\\rangle = \\alpha |\\alpha\\rangle). 其中0对应的本征态为真空态(|0\\rangle)，即(\\hat{a} |0\\rangle = 0)，在物理上对应无粒子的状态.
\n产生算符作用在真空态可以得到粒子数为1的状态，即({\\hat{a}^\\dagger,|0\\rangle = |1\\rangle})，作用(n)次可以产生粒子数为(n)的状态(|n\\rangle)，即Fock态：({(\\hat{a}^\\dagger)^n,|0\\rangle = \\sqrt{n!};|n\\rangle}). 相反，湮灭算符作用在Fock态(|n\\rangle)上可以得到(|n-1\\rangle)，即(\\hat{a}|n\\rangle = \\sqrt{n},|n-1\\rangle)，这也是产生算符和湮灭算符名字的由来。

用湮灭算符和产生算符可以定义粒子数算符(\\hat{N} = \\hat{a}^\\dagger \\hat{a})，这是一个Hermitian算符，其本征值为量子系统中的粒子数，本征态为Fock态，即(\\hat{N}|n\\rangle = n |n\\rangle)

相干态(|\\alpha\\rangle)用Fock态(|n\\rangle)展开是(|\\alpha\\rangle = e^{-\\frac{|\\alpha|^2}{2}} \\sum_{n=0}^{\\infty} \\frac{\\alpha^n}{\\sqrt{n!}} |n\\rangle)，证明留作习题（划掉）

Weyl位移算符定义为(\\hat{D}(\\alpha) = e^{\\alpha \\hat{a}^\\dagger - \\alpha^* \\hat{a}})，其作用效果是把湮灭和产生算符分别平移(\\alpha)和( \\alpha^)，即(\\hat{D}^\\dagger(\\alpha) \\hat{a} \\hat{D}(\\alpha) = \\hat{a} + \\alpha), (\\hat{D}^\\dagger(\\alpha) \\hat{a}^\\dagger \\hat{D}(\\alpha) = \\hat{a}^\\dagger + \\alpha^)，所以作用在真空态(|0\\rangle)上可以得到相干态：(|\\alpha\\rangle = \\hat{D}(\\alpha) |0\\rangle).

\n","createdAt":"2026-05-14 01:22:43.858395","updatedAt":"2026-05-14 01:22:43.858395","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":414,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":1411,"username":"lwe","name":null}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":72097,"postNumber":41,"raw":"# 量子光学简介(2)\n量子态\$\\hat\\rho\$的特征函数是\$\\chi(\\alpha) = \\operatorname{Tr}\\big[\\hat{\\rho}\\,\\hat{D}(\\alpha)\\big]\$，这与经典随机变量的特征函数\$\\phi(k) = \\mathbb{E}\\left[e^{ikX}\\right] = \\int_{-\\infty}^{\\infty} e^{ikx} P(x) dx\$形式类似。\n\n特征函数\$\\chi(\\alpha) \$的Fourier变换是相空间中的Wigner函数\$W(\\beta) \$，即\$\\chi(\\alpha) = \\int e^{\\alpha^* \\beta - \\alpha \\beta^*} W(\\beta) d^2\\beta\$. Wigner函数还可以用位置和动量来定义：对于一维量子态的密度矩阵\$\\rho(x,x') = \\langle x|\\rho|x'\\rangle\$，有\$W(x,p) = \\frac{1}{2\\pi\\hbar} \\int_{-\\infty}^{+\\infty} \\rho\\left(x+\\frac{y}{2}, x-\\frac{y}{2}\\right) e^{-\\frac{i}{\\hbar} p y} dy\$. 两种定义的换算关系为\$W(\\beta) = 2W\\left(\\sqrt{2}\\text{Re}\\beta, \\sqrt{2}\\text{Im}\\beta\\right)\$.\n\nWigner函数是一种准概率分布函数，虽然不满足非负性，也无法直接对应相空间的概率密度（有不确定性关系限制），但满足归一性，且边缘分布为位置或动量的概率密度函数：\$\\int_{-\\infty}^{+\\infty} W(x,p) dp = |\\psi(x)|^2, \\int_{-\\infty}^{+\\infty} W(x,p) dx = |\\tilde{\\psi}(p)|^2\$.\n\nWigner函数在信号处理领域通常称为Wigner-Ville分布，其定义为\${W_x(t, f) = \\int_{-\\infty}^{\\infty} x\\left(t+\\frac{\\tau}{2}\\right) x^*\\left(t-\\frac{\\tau}{2}\\right) e^{-i2\\pi f \\tau} d\\tau}\$.\n\n信号处理领域也需要用到Wigner函数，是因为时域信号无法看出频率信息，频域信号无法看出时间信息；含时Fourier变换、小波变换可以同时看出时间和频率信息，但都不精确，这是因为信号的时间和频率也满足不确定性关系：\${\\Delta t \\cdot \\Delta f \\geq \\frac{1}{4\\pi}}\$. 事实上，量子态的位置表象和动量表象互为Fourier变换，正如时域信号与频域信号，这才是不确定性关系更本质的数学原理。\n\n没想到同最初几条呼应上了\nhttps://xjtu.app/t/topic/6869/22","cooked":"

量子光学简介(2)

量子态(\\hat\\rho)的特征函数是(\\chi(\\alpha) = \\operatorname{Tr}\\big[\\hat{\\rho},\\hat{D}(\\alpha)\\big])，这与经典随机变量的特征函数(\\phi(k) = \\mathbb{E}\\left[e^{ikX}\\right] = \\int_{-\\infty}^{\\infty} e^{ikx} P(x) dx)形式类似。

特征函数(\\chi(\\alpha) )的Fourier变换是相空间中的Wigner函数(W(\\beta) )，即(\\chi(\\alpha) = \\int e^{\\alpha^* \\beta - \\alpha \\beta^*} W(\\beta) d^2\\beta). Wigner函数还可以用位置和动量来定义：对于一维量子态的密度矩阵(\\rho(x,x') = \\langle x|\\rho|x'\\rangle)，有(W(x,p) = \\frac{1}{2\\pi\\hbar} \\int_{-\\infty}^{+\\infty} \\rho\\left(x+\\frac{y}{2}, x-\\frac{y}{2}\\right) e^{-\\frac{i}{\\hbar} p y} dy). 两种定义的换算关系为(W(\\beta) = 2W\\left(\\sqrt{2}\\text{Re}\\beta, \\sqrt{2}\\text{Im}\\beta\\right)).

Wigner函数是一种准概率分布函数，虽然不满足非负性，也无法直接对应相空间的概率密度（有不确定性关系限制），但满足归一性，且边缘分布为位置或动量的概率密度函数：(\\int_{-\\infty}^{+\\infty} W(x,p) dp = |\\psi(x)|^2, \\int_{-\\infty}^{+\\infty} W(x,p) dx = |\\tilde{\\psi}(p)|^2).

Wigner函数在信号处理领域通常称为Wigner-Ville分布，其定义为({W_x(t, f) = \\int_{-\\infty}^{\\infty} x\\left(t+\\frac{\\tau}{2}\\right) x^*\\left(t-\\frac{\\tau}{2}\\right) e^{-i2\\pi f \\tau} d\\tau}).

信号处理领域也需要用到Wigner函数，是因为时域信号无法看出频率信息，频域信号无法看出时间信息；含时Fourier变换、小波变换可以同时看出时间和频率信息，但都不精确，这是因为信号的时间和频率也满足不确定性关系：({\\Delta t \\cdot \\Delta f \\geq \\frac{1}{4\\pi}}). 事实上，量子态的位置表象和动量表象互为Fourier变换，正如时域信号与频域信号，这才是不确定性关系更本质的数学原理。

没想到同最初几条呼应上了

\n $\"anon1678451\"$ \n anon1678451 — #22 in 建一个理科主题水楼！\n

Fourier 变换前后形式不变的信号——Gauss 型函数
\n

\ng_{\\alpha}\\left( t \\right) =\\frac{1}{2\\sqrt{\\pi \\alpha}}\\exp \\left\\{ -\\frac{t^2}{4\\alpha} \\right\\}\n

\n不难发现
\n

\n\\begin{align}\n\\mathscr{F} \\left\\{ g_{\\alpha}\\left( t \\right) \\right\\} &=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{+\\infty}{\\frac{1}{2\\sqrt{\\pi \\alpha}}\\exp \\left\\{ -\\frac{t^2}{4\\alpha} \\right\\} \\exp \\left\\{ -\\mathrm{i}\\omega t \\right\\} \\mathrm{d}t}\n\\\\\n&=\\frac{1}{\\sqrt{2}\\pi}\\int_{-\\infty}^{+\\infty}{\\exp \\left\\{ -\\left( \\frac{t}{2\\sqrt{\\alpha}}+\\sqrt{\\alpha}\\mathrm{i}\\omega \\right) ^2-\\alpha \\omega ^2 \\right\\} \\mathrm{d}\\frac{t}{2\\sqrt{\\alpha}}}\n\\\\\n&=\\frac{1}{\\sqrt{2\\pi}}\\exp \\left\\{ -\\alpha \\omega ^2 \\right\\} \n\\end{align}\n

\n以此为窗进行的短时 Fourier 变换称为 Gabor 变换：
\n

\n\\begin{align}\n\\mathscr{F} \\left\\{ f\\left( t \\right) g_{\\alpha}\\left( t-b \\right) \\right\\} &=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{+\\infty}{f\\left( t \\right) g_{\\alpha}\\left( t-b \\right) \\mathrm{e}^{-\\mathrm{i}\\omega t}\\mathrm{d}t}\n\\\\\n&=\\frac{1}{2\\pi}\\int_{-\\infty}^{+\\infty}{F\\left( \\xi \\right) \\mathrm{e}^{-\\mathrm{i}\\left( \\omega -\\xi \\right) b}\\mathrm{e}^{-\\alpha \\left( \\omega -\\xi \\right) ^2}\\mathrm{d}\\xi}\n\\\\\n&=\\frac{1}{2\\sqrt{\\pi \\alpha}}\\mathrm{e}^{-\\mathrm{i}\\omega b}\\int_{-\\infty}^{+\\infty}{F\\left( \\xi \\right) g_{\\frac{1}{4\\alpha}}\\left( \\xi -\\omega \\right) \\mathrm{e}^{\\mathrm{i}\\xi b}\\mathrm{d}\\xi}\n\\end{align}\n

","createdAt":"2026-05-14 01:23:26.02252","updatedAt":"2026-05-14 01:23:26.02252","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":420,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"👍️","count":1,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"}],"hasReacted":false},{"reaction":"❤️","count":1,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":72179,"postNumber":42,"raw":"# 量子光学简介(3)\n*（内容顺序可能有些混乱，因为我的学习顺序本就混乱，学完之前都不知道应该怎么学）*\n\n量子光学中一直有两种量子化方式，一种是一次量子化（first quantization），使用动量算符 \$\\hat{p}\$ 和位置算符 \$\\hat{x}\$ 描述粒子性质；另一种是二次量子化（second quantization），使用湮灭算符 \$\\hat{a}\$ 和产生算符 \$\\hat{a}^\\dagger\$ 描述场的性质，如前文所述，这两种量子化可以相互转化。\n\nWeyl位移算符的复合：\n记\$\\hat{D}_{r}=e^{i r^{\\top} \\Omega \\hat{r}}\$，则有\$\\hat{D}_{r_{1}+r_{2}}=\\hat{D}_{r_{1}} \\hat{D}_{r_{2}} e^{\\frac{i}{2} r_{1}^{\\top} \\Omega r_{2}} \\tag{3.11}\$，其中\n$$ \n\\hat{\\boldsymbol{r}}=(\\hat{x}_{1},\\hat{p}_{1},\\dots,\\hat{x}_{n},\\hat{p}_{n})^\\mathrm{T},\n\\Omega=\\bigoplus_{j=1}^{n} \\left(\\begin{matrix}\n0 & 1 \\\\\n-1 & 0\n\\end{matrix} \\right) \n$$\n\n[details=\"证明在这里\"]\n\\[\n\\begin{aligned}\ne^{i(r_{1}+r_{2})^{\\top }\\Omega \\hat {r}}\n&=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}}\\,e^{\\frac{1}{2}\\big[r_{1}^{\\top }\\Omega \\hat {r},\\,r_{2}^{\\top }\\Omega \\hat {r}\\big]}\n=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}}\\,e^{-\\frac{1}{2}\\big[r_{1}^{\\top }\\Omega \\hat {r},\\,\\hat {r}^{\\top }\\Omega r_{2}\\big]} \\\\\n&=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}}\\,e^{-\\frac{1}{2}r_{1}^{\\top }\\Omega \\big[\\hat {r},\\hat {r}^{\\top }\\big]\\Omega r_{2}}\n=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}}\\,e^{-\\frac{i}{2}r_{1}^{\\top }\\Omega ^{3}r_{2}} \\\\\n&=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}}\\,e^{\\frac{i}{2}r_{1}^{\\top }\\Omega r_{2}}\n=e^{ir_{2}^{\\top }\\Omega \\hat {r}}e^{ir_{1}^{\\top }\\Omega \\hat {r}}\\,e^{-\\frac{i}{2}r_{1}^{\\top }\\Omega r_{2}}\n\\end{aligned}\n\\]\n其中第一步用到了Baker–Campbell–Hausdorff公式\n\\[\ne^{\\hat{A}+\\hat{B}}\n= e^{\\hat{A}} e^{\\hat{B}}\n\\exp\\left(-\\frac{1}{2}\\big[\\hat{A},\\hat{B}\\big]\\right)\n\\exp\\left(\\frac{1}{6}\\Big(2\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big] + \\big[\\hat{A},\\big[\\hat{A},\\hat{B}\\big]\\big]\\Big)\\right) \n\\cdot \\exp\\left(-\\frac{1}{24}\\Big(3\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big] + 5\\big[\\hat{A},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big] + \\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\Big)\\right) \n\\cdot \\exp\\left(\\frac{1}{720}\\Big(12\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big] + 28\\big[\\hat{A},\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big] + 19\\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big] + 3\\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big]\\Big)\\right)\n\\cdots\n\\]\n[/details]\n\n二阶哈密顿（Hamiltonian）算符：\$\\hat{H}=\\frac{1}{2} \\hat{\\boldsymbol{r}}^{\\top} H \\hat{\\boldsymbol{r}}+\\hat{\\boldsymbol{r}}^{\\top} \\boldsymbol{r} \\tag{3.12}\$，其中式中 \$\\boldsymbol{r}\$ 为 \$2n\$ 维实矢量，\$H\$ 为对称矩阵，称作哈密顿矩阵。该定义包含二次项或一次项为0的退化情况。\n\n量子高斯态（quantum Gaussian states）：任意高斯态 \$\\varrho_{G}\$ 均可表示为\n\\[\\varrho_{G}=\\frac{e^{-\\beta \\hat{H}}}{\\mathrm{Tr}\\left[e^{-\\beta \\hat{H}}\\right]} \\]\n其中 \$\\beta \\in \\mathbb{R}^+\$，\$\\hat{H}\$ 为上述二阶哈密顿算符。该定义也包含如下极限情形：\n\\[\\varrho_{G}=\\lim _{\\beta \\to \\infty} \\frac{e^{-\\beta \\hat{H}}}{\\mathrm{Tr}\\left[e^{-\\beta \\hat{H}}\\right]} \\]\n当 \$\\beta\$ 取有限值时，上述定义描述的态均为**混合态**，极限情形为纯态。\n\n前文提到的真空态对应 \$H = \\omega I\$， \$\\boldsymbol{r}\$ 为零向量且 \$\\beta\$ 取无穷大的情况，相干态则对应 \$\\boldsymbol{r}\$ 不一定为零向量的情况。","cooked":"

量子光学简介(3)

（内容顺序可能有些混乱，因为我的学习顺序本就混乱，学完之前都不知道应该怎么学）

量子光学中一直有两种量子化方式，一种是一次量子化（first quantization），使用动量算符 (\\hat{p}) 和位置算符 (\\hat{x}) 描述粒子性质；另一种是二次量子化（second quantization），使用湮灭算符 (\\hat{a}) 和产生算符 (\\hat{a}^\\dagger) 描述场的性质，如前文所述，这两种量子化可以相互转化。

Weyl位移算符的复合：
\n记(\\hat{D}{r}=e^{i r^{\\top} \\Omega \\hat{r}})，则有(\\hat{D}{r_{1}+r_{2}}=\\hat{D}{r{1}} \\hat{D}{r{2}} e^{\\frac{i}{2} r_{1}^{\\top} \\Omega r_{2}} \\tag{3.11})，其中
\n

\n\\hat{\\boldsymbol{r}}=(\\hat{x}_{1},\\hat{p}_{1},\\dots,\\hat{x}_{n},\\hat{p}_{n})^\\mathrm{T},\n\\Omega=\\bigoplus_{j=1}^{n} \\left(\\begin{matrix}\n0 & 1 \\\\\n-1 & 0\n\\end{matrix} \\right)\n

证明在这里

[
\n\\begin{aligned}
\ne^{i(r_{1}+r_{2})^{\\top }\\Omega \\hat {r}}
\n&=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}},e^{\\frac{1}{2}\\big[r_{1}^{\\top }\\Omega \\hat {r},,r_{2}^{\\top }\\Omega \\hat {r}\\big]}
\n=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}},e^{-\\frac{1}{2}\\big[r_{1}^{\\top }\\Omega \\hat {r},,\\hat {r}^{\\top }\\Omega r_{2}\\big]} \\
\n&=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}},e^{-\\frac{1}{2}r_{1}^{\\top }\\Omega \\big[\\hat {r},\\hat {r}^{\\top }\\big]\\Omega r_{2}}
\n=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}},e^{-\\frac{i}{2}r_{1}^{\\top }\\Omega ^{3}r_{2}} \\
\n&=e^{ir_{1}^{\\top }\\Omega \\hat {r}}e^{ir_{2}^{\\top }\\Omega \\hat {r}},e^{\\frac{i}{2}r_{1}^{\\top }\\Omega r_{2}}
\n=e^{ir_{2}^{\\top }\\Omega \\hat {r}}e^{ir_{1}^{\\top }\\Omega \\hat {r}},e^{-\\frac{i}{2}r_{1}^{\\top }\\Omega r_{2}}
\n\\end{aligned}
\n]
\n其中第一步用到了Baker–Campbell–Hausdorff公式
\n[
\ne^{\\hat{A}+\\hat{B}}
\n= e^{\\hat{A}} e^{\\hat{B}}
\n\\exp\\left(-\\frac{1}{2}\\big[\\hat{A},\\hat{B}\\big]\\right)
\n\\exp\\left(\\frac{1}{6}\\Big(2\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big] + \\big[\\hat{A},\\big[\\hat{A},\\hat{B}\\big]\\big]\\Big)\\right)
\n\\cdot \\exp\\left(-\\frac{1}{24}\\Big(3\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big] + 5\\big[\\hat{A},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big] + \\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\Big)\\right)
\n\\cdot \\exp\\left(\\frac{1}{720}\\Big(12\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big] + 28\\big[\\hat{A},\\big[\\hat{B},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big] + 19\\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{B},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big] + 3\\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{A},\\big[\\hat{A},\\hat{B}\\big]\\big]\\big]\\big]\\Big)\\right)
\n\\cdots
\n]

二阶哈密顿（Hamiltonian）算符：(\\hat{H}=\\frac{1}{2} \\hat{\\boldsymbol{r}}^{\\top} H \\hat{\\boldsymbol{r}}+\\hat{\\boldsymbol{r}}^{\\top} \\boldsymbol{r} \\tag{3.12})，其中式中 (\\boldsymbol{r}) 为 (2n) 维实矢量，(H) 为对称矩阵，称作哈密顿矩阵。该定义包含二次项或一次项为0的退化情况。

量子高斯态（quantum Gaussian states）：任意高斯态 (\\varrho_{G}) 均可表示为
\n[\\varrho_{G}=\\frac{e^{-\\beta \\hat{H}}}{\\mathrm{Tr}\\left[e^{-\\beta \\hat{H}}\\right]} ]
\n其中 (\\beta \\in \\mathbb{R}^+)，(\\hat{H}) 为上述二阶哈密顿算符。该定义也包含如下极限情形：
\n[\\varrho_{G}=\\lim _{\\beta \\to \\infty} \\frac{e^{-\\beta \\hat{H}}}{\\mathrm{Tr}\\left[e^{-\\beta \\hat{H}}\\right]} ]
\n当 (\\beta) 取有限值时，上述定义描述的态均为混合态，极限情形为纯态。

前文提到的真空态对应 (H = \\omega I)， (\\boldsymbol{r}) 为零向量且 (\\beta) 取无穷大的情况，相干态则对应 (\\boldsymbol{r}) 不一定为零向量的情况。

\n","createdAt":"2026-05-30 03:30:51.638528","updatedAt":"2026-05-30 03:37:07.040112","replyToPostNumber":null,"replyCount":0,"likeCount":2,"wordCount":500,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"👍️","count":1,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"}],"hasReacted":false},{"reaction":"❤️","count":2,"users":[{"id":30,"username":"一只玉米人","name":"玉米🌽"},{"id":1411,"username":"lwe","name":null}],"hasReacted":false}],"wiki":false,"version":2,"publicVersion":2,"deletedAt":null,"isAnonymous":false,"user":{"id":1817,"username":"anon1678451","name":null,"avatarUrl":"/uploads/image/c6/9a/c69a738e32dc09506ebfea8035a962d53e98afc1828783d42c3dac2eaa627df5.avif","admin":false,"moderator":false,"signatureRaw":null,"signatureCooked":null}},{"id":72223,"postNumber":43,"raw":"# 量子光学简介(4)\n\n## 高斯保迹完全正定映射(确定性高斯完全正定映射, Gaussian Completely Positive Trace-Preserving maps)\n\n一个映射 $ \\Phi: \\mathcal{B}(\\mathcal{H}) \\to \\mathcal{B}(\\mathcal{H}) $ 被称为高斯保迹完全正定映射，当且仅当它同时满足以下三个条件：\n\n1. 完全正定性（CP）：对任意正整数 $k$，映射 $\\Phi \\otimes \\mathbb{1}_k$ 都是正定映射\n2. 保迹性（TP）：对任意密度算符 $\\varrho$，有 $\\mathrm{Tr}[\\Phi(\\varrho)] = \\mathrm{Tr}[\\varrho]$\n3. 高斯性：对任意高斯态 $\\varrho$，$\\Phi(\\varrho)$ 仍然是高斯态\n\n### 对高斯态的作用\n\n设 $n$ 模初态为高斯态，其一阶矩矢量 $\\overline{\\boldsymbol r}$、协方差矩阵 $\\boldsymbol\\sigma$；确定性高斯完全正定映射驱动的态演化，完全由两个 $2n\\times2n$ 实矩阵 $\\boldsymbol X,\\boldsymbol Y$ 刻画，二者对统计矩的变换规则为：\n$$ \\overline{\\boldsymbol r}\\mapsto \\boldsymbol X\\overline{\\boldsymbol r}, \\boldsymbol\\sigma\\mapsto \\boldsymbol X\\boldsymbol\\sigma \\boldsymbol X^\\top+\\boldsymbol Y $$\n\n矩阵 $\\boldsymbol X,\\boldsymbol Y$ 需要满足约束：\n\n$$\\boldsymbol Y + i\\Omega \\ge i\\boldsymbol X\\Omega \\boldsymbol X^\\top .$$\n\n### 对一般态的特征函数作用\n\n设 $\\chi_\\varrho(\\boldsymbol r)$ 为量子态 $\\varrho$ 对应的特征函数。确定性高斯完全正定映射 $\\Phi$ 对特征函数的变换定义为：\n\n$$ \\chi_{\\varrho}(\\boldsymbol r) \\mapsto \\chi_{\\Phi(\\varrho)}(\\boldsymbol r)=e^{-\\frac14 \\boldsymbol r^\\top \\Omega^\\top Y \\Omega \\boldsymbol r}\\,\\chi_{\\varrho}\\big(\\Omega^\\top X^\\top \\Omega \\boldsymbol r\\big). $$\n\n### 对偶映射\n\n对任意完全正定映射 $\\Phi$，其对偶完全正定映射 $\\Phi^{*}$ 由如下迹等式定义：\n\n$$ \\mathrm{Tr}\\big[\\hat{A}\\,\\Phi^{*}(\\hat{B})\\big]=\\mathrm{Tr}\\big[\\Phi(\\hat{A})\\,\\hat{B}\\big] $$\n\n上式对所有有界算符 $\\hat{A},\\hat{B}$ 成立。\n\n设 $\\Phi$ 为由矩阵 $\\boldsymbol X,\\boldsymbol Y$ 参数化的确定性高斯完全正定映射，则其对偶映射 $\\Phi^*$ 作用于 Weyl 算符 $\\hat D_{\\Omega \\boldsymbol r}$ 的形式为：\n\n$$\\Phi^{*}\\big(\\hat{D}_{\\Omega \\boldsymbol r}\\big)=\\hat{D}_{\\Omega X^{\\top}\\boldsymbol r}\\,e^{-\\frac14 \\boldsymbol r^\\top Y \\boldsymbol r}. $$\n\n### 对 Weyl 算符的作用\n由矩阵 $\\boldsymbol X,\\boldsymbol Y$ 表征、且 $\\boldsymbol X$ 可逆的高斯完全正定映射，作用在 Weyl 算符上满足：\n\n$$ \\Phi\\left(\\hat{D}_{\\Omega \\boldsymbol r}\\right)=\\frac{e^{-\\frac14 \\boldsymbol r^\\top \\boldsymbol X^{-1}\\boldsymbol Y (\\boldsymbol X^{-1})^\\top \\boldsymbol r}}{\\left|\\det \\boldsymbol X\\right|}\\hat{D}_{\\Omega (\\boldsymbol X^{-1})^\\top \\boldsymbol r}. $$\n\n### 经典混合\n当 $\\boldsymbol Y\\succ0$ 时，完全正定映射：\n\n$$\\Phi_{Y}(\\varrho)=\\int_{\\mathbb{R}^{2 n}} d \\boldsymbol{r}' \\frac{e^{-\\boldsymbol{r}'^{\\top} \\boldsymbol{Y}^{-1} \\boldsymbol{r}'}}{\\pi^{n} \\sqrt{\\det \\boldsymbol{Y}}} \\hat{D}_{\\boldsymbol{r}'} \\varrho \\hat{D}_{\\boldsymbol{r}'}^{\\dagger}$$\n\n等价于参数取 $\\boldsymbol X=\\mathbb 1$、噪声矩阵为 $\\boldsymbol Y$ 的高斯完全正定映射。\n这个写法其实就是在做卷积。","cooked":"

量子光学简介(4)

高斯保迹完全正定映射(确定性高斯完全正定映射, Gaussian Completely Positive Trace-Preserving maps)

一个映射 $ \\Phi: \\mathcal{B}(\\mathcal{H}) \\to \\mathcal{B}(\\mathcal{H}) $ 被称为高斯保迹完全正定映射，当且仅当它同时满足以下三个条件：

完全正定性（CP）：对任意正整数 $k$ ，映射 $\\Phi \\otimes \\mathbb{1}_k$ 都是正定映射
保迹性（TP）：对任意密度算符 $\\varrho$ ，有 $\\mathrm{Tr}[\\Phi(\\varrho)] = \\mathrm{Tr}[\\varrho]$
高斯性：对任意高斯态 $\\varrho$ ， $\\Phi(\\varrho)$ 仍然是高斯态

对高斯态的作用

设 $n$ 模初态为高斯态，其一阶矩矢量 $\\overline{\\boldsymbol r}$ 、协方差矩阵 $\\boldsymbol\\sigma$ ；确定性高斯完全正定映射驱动的态演化，完全由两个 $2n\\times2n$ 实矩阵 $\\boldsymbol X,\\boldsymbol Y$ 刻画，二者对统计矩的变换规则为：
\n

\n\\overline{\\boldsymbol r}\\mapsto \\boldsymbol X\\overline{\\boldsymbol r}, \\boldsymbol\\sigma\\mapsto \\boldsymbol X\\boldsymbol\\sigma \\boldsymbol X^\\top+\\boldsymbol Y\n

矩阵 $\\boldsymbol X,\\boldsymbol Y$ 需要满足约束：

\n\\boldsymbol Y + i\\Omega \\ge i\\boldsymbol X\\Omega \\boldsymbol X^\\top .\n

对一般态的特征函数作用

设 $\\chi_\\varrho(\\boldsymbol r)$ 为量子态 $\\varrho$ 对应的特征函数。确定性高斯完全正定映射 $\\Phi$ 对特征函数的变换定义为：

\n\\chi_{\\varrho}(\\boldsymbol r) \\mapsto \\chi_{\\Phi(\\varrho)}(\\boldsymbol r)=e^{-\\frac14 \\boldsymbol r^\\top \\Omega^\\top Y \\Omega \\boldsymbol r}\\,\\chi_{\\varrho}\\big(\\Omega^\\top X^\\top \\Omega \\boldsymbol r\\big).\n

对偶映射

对任意完全正定映射 $\\Phi$ ，其对偶完全正定映射 $\\Phi^{*}$ 由如下迹等式定义：

\n\\mathrm{Tr}\\big[\\hat{A}\\,\\Phi^{*}(\\hat{B})\\big]=\\mathrm{Tr}\\big[\\Phi(\\hat{A})\\,\\hat{B}\\big]\n

上式对所有有界算符 $\\hat{A},\\hat{B}$ 成立。

设 $\\Phi$ 为由矩阵 $\\boldsymbol X,\\boldsymbol Y$ 参数化的确定性高斯完全正定映射，则其对偶映射 $\\Phi^*$ 作用于 Weyl 算符 $\\hat D_{\\Omega \\boldsymbol r}$ 的形式为：

\n\\Phi^{*}\\big(\\hat{D}_{\\Omega \\boldsymbol r}\\big)=\\hat{D}_{\\Omega X^{\\top}\\boldsymbol r}\\,e^{-\\frac14 \\boldsymbol r^\\top Y \\boldsymbol r}.\n

对 Weyl 算符的作用

由矩阵 $\\boldsymbol X,\\boldsymbol Y$ 表征、且 $\\boldsymbol X$ 可逆的高斯完全正定映射，作用在 Weyl 算符上满足：

\n\\Phi\\left(\\hat{D}_{\\Omega \\boldsymbol r}\\right)=\\frac{e^{-\\frac14 \\boldsymbol r^\\top \\boldsymbol X^{-1}\\boldsymbol Y (\\boldsymbol X^{-1})^\\top \\boldsymbol r}}{\\left|\\det \\boldsymbol X\\right|}\\hat{D}_{\\Omega (\\boldsymbol X^{-1})^\\top \\boldsymbol r}.\n

经典混合

当 $\\boldsymbol Y\\succ0$ 时，完全正定映射：

\n\\Phi_{Y}(\\varrho)=\\int_{\\mathbb{R}^{2 n}} d \\boldsymbol{r}' \\frac{e^{-\\boldsymbol{r}'^{\\top} \\boldsymbol{Y}^{-1} \\boldsymbol{r}'}}{\\pi^{n} \\sqrt{\\det \\boldsymbol{Y}}} \\hat{D}_{\\boldsymbol{r}'} \\varrho \\hat{D}_{\\boldsymbol{r}'}^{\\dagger}\n

等价于参数取 $\\boldsymbol X=\\mathbb 1$ 、噪声矩阵为 $\\boldsymbol Y$ 的高斯完全正定映射。
\n这个写法其实就是在做卷积。

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