<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Quantum Entanglement on Haifei's Home</title><link>https://haifei.pro/en/tags/quantum-entanglement/</link><description>Recent content in Quantum Entanglement on Haifei's Home</description><generator>Hugo</generator><language>en</language><managingEditor>hfwang132@gmail.com (hfwang132)</managingEditor><webMaster>hfwang132@gmail.com (hfwang132)</webMaster><copyright>This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.</copyright><lastBuildDate>Sun, 30 Jul 2023 02:51:44 +0800</lastBuildDate><atom:link href="https://haifei.pro/en/tags/quantum-entanglement/index.xml" rel="self" type="application/rss+xml"/><item><title>Distinguishing Mixed States from Entangled States</title><link>https://haifei.pro/en/post_20230730_%E8%BE%A8%E6%9E%90%E6%B7%B7%E5%90%88%E6%80%81%E4%B8%8E%E7%BA%A0%E7%BC%A0%E6%80%81/</link><pubDate>Sun, 30 Jul 2023 02:51:44 +0800</pubDate><author>hfwang132@gmail.com (hfwang132)</author><guid>https://haifei.pro/en/post_20230730_%E8%BE%A8%E6%9E%90%E6%B7%B7%E5%90%88%E6%80%81%E4%B8%8E%E7%BA%A0%E7%BC%A0%E6%80%81/</guid><description>&lt;h2 id="i-pure-states-and-mixed-states"&gt;I. Pure States and Mixed States&lt;/h2&gt;
&lt;p&gt;To understand pure states and mixed states, one must first understand the &lt;strong&gt;density operator&lt;/strong&gt;, or &lt;strong&gt;density matrix&lt;/strong&gt;.&lt;/p&gt;
&lt;h3 id="11-what-is-a-density-operator"&gt;1.1 What Is a Density Operator?&lt;/h3&gt;
&lt;p&gt;When I first encountered the density operator, I found it quite remarkable. This is because it takes the &lt;strong&gt;tensor product of a state with its own dual&lt;/strong&gt;:&lt;/p&gt;
\[v \mapsto v\otimes v^*\]&lt;p&gt;Or, written in a more physics-style manner:&lt;/p&gt;
\[| \psi \rangle \mapsto |\psi \rangle \langle \psi |\]&lt;p&gt;We all know that a quantum state \(v\) is merely a vector in a Hilbert space, namely \(v \in H\) . But why insist on making the density operator \(v\otimes v^*\in H\otimes H^* \cong \operatorname{End}(H)\) ? What is the point of this?&lt;/p&gt;</description></item></channel></rss>