{"categories":["quantum_computing"],"version":"1.0","blog_url":"https://randommemory.hatenablog.com/","height":"190","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Frandommemory.hatenablog.com%2Fentry%2F2022%2F01%2F22%2F181849\" title=\"Qiskit (17) \u2015 Deutsch-Jozsa \u30a2\u30eb\u30b4\u30ea\u30ba\u30e0 - \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":"Deutsch-Jozsa \u30a2\u30eb\u30b4\u30ea\u30ba\u30e0\u306e\u8a71\u3002\u30c9\u30a4\u30c1-\u30b8\u30e7\u30b5\u306e\u30a2\u30eb\u30b4\u30ea\u30ba\u30e0 \u3082\u53c2\u8003\u306b\u3057\u305f\u3044\u3002$\\ket{\\psi_0} = \\ket{0}^{\\otimes n}\\ket{1}$ \u3068\u3057\u3066\\begin{align*} \\ket{\\psi_1} &= H^{\\otimes n + 1}\\ket{\\psi_0} \\\\ &= (H\\ket{0})^{\\otimes n} H\\ket{1} \\end{align*}\u3092\u8a08\u7b97\u3059\u308b\u3002\\begin{align*} \\begin{cases} H\\ket{0} &= \\frac{1}{\\sqrt{2}}(\\ket{0} + \\ket{1}) \\\\ H\\ket{1}\u2026","image_url":null,"blog_title":"\u3089\u3093\u3060\u3080\u306a\u8a18\u61b6","author_url":"https://blog.hatena.ne.jp/derwind/","author_name":"derwind","url":"https://randommemory.hatenablog.com/entry/2022/01/22/181849","published":"2022-01-22 18:18:49","type":"rich","width":"100%","provider_url":"https://hatena.blog","provider_name":"Hatena Blog","title":"Qiskit (17) \u2015 Deutsch-Jozsa \u30a2\u30eb\u30b4\u30ea\u30ba\u30e0"}