矩阵输入测试

Chern posted @ 2014年9月28日 23:19 in 数学物理 , 2798 阅读

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

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

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

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

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

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