现代微分几何的基本概念

矩阵输入测试

Chern posted @ 2014年9月28日 23:19 in 数学物理 , 2425 阅读

\[
\sigma _z  =
\left(
\begin{array}{*{20}c}
   1 & 0  \\
   0 & { - 1}  \\
\end{array} 
\right)
\]


\[
\left| {\psi (t)} \right\rangle  =
\left(
\begin{array}{*{20}c}
   \cos\theta  \\
   \sin\theta  \\
\end{array}
\right)
\]

\[
\left| {\psi (t)} \right\rangle  =
\left(
\begin{array}{*{20}c}
   {\cos \frac{\theta }{2}}  \\
   {\sin \frac{\theta }{2}}  \\
\end{array} 
\right)
\]
 

\[
\left| {\psi (t)} \right\rangle  =
\left(
\begin{array}{*{20}c}
   {\cos \frac{\theta }{2}}  \\   {\sin \frac{\theta }{2}} 
\end{array} 
\right)
\]


\[
\left| {\psi (t)} \right\rangle  = \exp (\mu Bt\sigma _z \frac{i}
{\hbar })
\]
\[
=
\left(
\begin{array}{*{20}c}
   \exp (\mu Bt\frac{i}{\hbar })\cos \frac{\theta }{2}  \\    \exp (\mu Bt\frac{i}{\hbar })\sin \frac{\theta }{2} 
\end{array}
\right)
\]

\[
\left| {\psi (t)} \right\rangle  = \left( {\begin{array}{*{20}c}
   {\cos \frac{1}{2}} & {\sin \frac{1}{2}}  \\
   {\sin \frac{1}{2}} & {\cos \frac{1}{2}}  \\
 \end{array} } \right)
\]

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happy wheels 说:
2019年7月29日 10:21

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