{"author_url":"https://blog.hatena.ne.jp/derwind/","blog_title":"\u3089\u3093\u3060\u3080\u306a\u8a18\u61b6","blog_url":"https://randommemory.hatenablog.com/","image_url":null,"published":"2022-01-21 01:30:01","title":"Qiskit (14)","width":"100%","url":"https://randommemory.hatenablog.com/entry/2022/01/21/013001","author_name":"derwind","height":"190","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Frandommemory.hatenablog.com%2Fentry%2F2022%2F01%2F21%2F013001\" title=\"Qiskit (14) - \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>","provider_url":"https://hatena.blog","description":"$H_2 \\mathrm{CNOT}_{1,2} H_2$ \u3092\u8a08\u7b97\u3057\u305f\u3044\u3002$X = \\ket{0}\\bra{1} + \\ket{1}\\bra{0}$, $Z = \\ket{0}\\bra{0} - \\ket{1}\\bra{1}$, $H = \\frac{1}{\\sqrt{2}}(X+Z)$, $\\mathrm{CNOT} = \\ket{0}\\bra{0} \\otimes I + \\ket{1}\\bra{1} \\otimes X$ \u304b\u3089\u3046\u307e\u304f\u8a08\u7b97\u3067\u304d\u308b\u3060\u308d\u3046\u304b\uff1f\\begin{align*} \\mathrm{CNOT}(\\ket{\\psi} \\otimes H\\ket{\\phi}) = \\frac{\u2026","type":"rich","categories":["quantum_computing"],"provider_name":"Hatena Blog","version":"1.0"}