在阅读该页内容之前,我们向量子计算的开创者费曼和Deutsch致敬,同时向三位量子信息学的奠基人Charles H. Bennett, David Deutsch, Peter Shor表示敬意.
问题：
Find the eigenvectors, eigenvalues, and diagonal representations of the Pauli matrices$X$, $Y$ and $Z$.
找出泡利矩阵$X,Y,Z$的特征向量，特征值和对角表示.
解答
$X$=$\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)$,建立方程$\left|\begin{array}{cc}
0-\lambda & 1 \\
1 & 0-\lambda
\end{array}\right|=0$，
得特征值为$\lambda_{1}$=1,$\lambda_{2}$=$-1$,
对应的特征向量为$\left|v_{1}\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right),\left|v_{2}\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1 \\ -1\end{array}\right).$
对角表示为$X=\frac{1}{2}\left(\begin{array}{l}1 \\ 1\end{array}\right)\left(\begin{array}{ll}1 & 1\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}1 \\ -1\end{array}\right)\left(\begin{array}{ll}1 & -1\end{array}\right).$
$Y$=$\left(\begin{array}{ll}
0 & -i \\
i & 0
\end{array}\right)$,建立方程$\left|\begin{array}{cc}
0-\lambda & -i \\
i & 0-\lambda
\end{array}\right|=0$，
得特征值为$\lambda_{1}$=$1$,$\lambda_{2}$=$-1$,
对应的特征向量为$\left|v_{1}\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ i\end{array}\right),\left|v_{2}\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1 \\ -i\end{array}\right).$
对角表示为$Y=\frac{1}{2}\left(\begin{array}{l}1 \\ i\end{array}\right)\left(\begin{array}{ll}1 & -i\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}1 \\ -i\end{array}\right)\left(\begin{array}{ll}1 & i\end{array}\right).$
$Z$=$\left(\begin{array}{ll}
1 & 0 \\
0 & -1
\end{array}\right)$,建立方程$\left|\begin{array}{cc}
1-\lambda & 0 \\
0 & -1-\lambda
\end{array}\right|=0$，
得特征值为$\lambda_{1}$=1,$\lambda_{2}$=-1,
对应的特征向量为$\left|v_{1}\right\rangle=\left(\begin{array}{l}1 \\ 0\end{array}\right),\left|v_{2}\right\rangle=\left(\begin{array}{l}0 \\ 1\end{array}\right)$
对角表示为$Z=\left(\begin{array}{l}1 \\ 0\end{array}\right)(1 \ 0)-\left(\begin{array}{l}0 \\ 1\end{array}\right)(0 \ 1).$
参考
[1]www.qtumist.com
参与者
作者：HKL, W65
贡献者:Dingyan, Wjw,Wxw
