<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:content="http://purl.org/rss/1.0/modules/content/"><channel><title>Quantum Mechanics on Martin‘s Physics Club</title><link>https://martinphysics.club/tags/quantum-mechanics/</link><description>Recent content in Quantum Mechanics on Martin‘s Physics Club</description><generator>Hugo</generator><language>en-US</language><copyright>2020-2026 martinphysics.club</copyright><lastBuildDate>Thu, 20 Aug 2026 07:51:38 +0000</lastBuildDate><atom:link href="https://martinphysics.club/tags/quantum-mechanics/index.xml" rel="self" type="application/rss+xml"/><item><title>Quantum Wave Function Visualization</title><link>https://martinphysics.club/posts/wp-100002/</link><pubDate>Thu, 04 Jun 2020 11:31:38 +0000</pubDate><guid>https://martinphysics.club/posts/wp-100002/</guid><description>&lt;div class="video-container"&gt;&lt;iframe src="https://www.youtube-nocookie.com/embed/KKr91v7yLcM?start=1" title="YouTube video" loading="lazy" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen&gt;&lt;/iframe&gt;&lt;/div&gt;</description></item><item><title>解一维谐振子</title><link>https://martinphysics.club/posts/wp-100001/</link><pubDate>Thu, 26 Mar 2020 05:28:09 +0000</pubDate><guid>https://martinphysics.club/posts/wp-100001/</guid><description>&lt;p&gt;一维谐振子，是稳态系统。对应的是定态薛定谔方程。&lt;/p&gt;
&lt;p&gt;对一维谐振子，其哈密顿量为：&lt;/p&gt;
&lt;span class="math-expression" data-display="true"&gt;\hat{H}=\frac{p_{x}^{2}}{2m}+\frac{1}{2}m\omega ^{2}x^{2}&lt;/span&gt;
&lt;p&gt;对应定态的薛定谔方程为：&lt;/p&gt;
&lt;span class="math-expression" data-display="true"&gt;\hat{H}\psi (x)=E\psi (x),p=-i\hbar\frac{d}{dx}&lt;/span&gt;
&lt;p&gt;代入可得：&lt;/p&gt;
&lt;span class="math-expression" data-display="true"&gt;-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi (x)+(\frac{1}{2}m\omega ^{2}x^{2}-E)\psi (x)=0&lt;/span&gt;
&lt;p&gt;上述是二阶变系数常微分方程，在不知道一个特解的情况下无法（我不会）降阶解方程，所以做变换。&lt;/p&gt;</description></item><item><title>通过旧量子论解一维无限深势阱</title><link>https://martinphysics.club/posts/wp-100011/</link><pubDate>Wed, 25 Mar 2020 05:25:04 +0000</pubDate><guid>https://martinphysics.club/posts/wp-100011/</guid><description>&lt;p&gt;有一个质量为m的球在一个一维的盒子里运动，盒子长度是a，盒子壁不可穿透，球和盒壁的碰撞是弹性的（即碰撞不会改变球的能量），盒子是固定的；球的大小忽略不计，不考虑摩擦力：
1、画出球在相空间的轨迹(至少两条)
2、按照旧量子论，求出这个球的量子能级**。**
3、解定态薛定谔方程。&lt;/p&gt;</description></item><item><title>解一维无限深势阱(解定态薛定谔方程)</title><link>https://martinphysics.club/posts/wp-100009/</link><pubDate>Fri, 13 Mar 2020 05:17:39 +0000</pubDate><guid>https://martinphysics.club/posts/wp-100009/</guid><description>&lt;h6 id="定态薛定谔方程求解"&gt;定态薛定谔方程求解：&lt;/h6&gt;
&lt;p&gt;假设粒子在一维势场中运动，势场满足：&lt;/p&gt;
&lt;span class="math-expression" data-display="true"&gt;U(x)=\left\{\begin{matrix}0,0\leq x\leq a\\+\infty,x&gt;a,x&lt;0\end{matrix}\right.&lt;/span&gt;
&lt;p&gt;解：&lt;/p&gt;
&lt;p&gt;设粒子的波函数为&lt;span class="math-expression" data-display="false"&gt;\psi (x)&lt;/span&gt;
, 由定态薛定谔方程得：&lt;/p&gt;
&lt;span class="math-expression" data-display="true"&gt;\hat{H}\psi (x)=E\psi (x)&lt;/span&gt;
&lt;p&gt;其中&lt;span class="math-expression" data-display="false"&gt;\hat{H}&lt;/span&gt;
为粒子的哈密顿量，为粒子动能与势能的总和，E为粒子的能量。&lt;/p&gt;</description></item></channel></rss>