{"type":"rich","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Frandommemory.hatenablog.com%2Fentry%2F2021%2F10%2F13%2F000549\" title=\"Qiskit (2) - \u3089\u3093\u3060\u3080\u306a\u8a18\u61b6\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","description":"Hadamard \u30b2\u30fc\u30c8 $H$ \u3092 $\\ket{000}$ \u306e $q_0$ \u3060\u3051\u306b\u4f5c\u7528\u3055\u305b\u308b\u3068 3 \u91cf\u5b50\u30d3\u30c3\u30c8\u30b2\u30fc\u30c8\u3068\u3057\u3066\u306f $U = I \\otimes I \\otimes H$ \u304c\u4f5c\u7528\u3059\u308b\u306e\u3067\u3001\\begin{align*} U \\ket{000} = \\ket{0} \\otimes \\ket{0} \\otimes (H \\ket{0}) &= \\ket{0} \\otimes \\ket{0} \\otimes (\\frac{1}{\\sqrt{2}}(\\ket{0} + \\ket{1})) \\\\ & = {\\frac{1}{\\sqrt{2}}}(\\ket{000} + \\ket{001})) \u2026","author_url":"https://blog.hatena.ne.jp/derwind/","published":"2021-10-13 00:05:49","version":"1.0","provider_url":"https://hatena.blog","blog_title":"\u3089\u3093\u3060\u3080\u306a\u8a18\u61b6","categories":["quantum_computing"],"title":"Qiskit (2)","author_name":"derwind","blog_url":"https://randommemory.hatenablog.com/","height":"190","image_url":"https://cdn-ak.f.st-hatena.com/images/fotolife/d/derwind/20211013/20211013000517.png","url":"https://randommemory.hatenablog.com/entry/2021/10/13/000549","width":"100%","provider_name":"Hatena Blog"}