Exercise 2.11#
The Pauli matrices are given by
The eigenvectors, eigenvalues and diagonal representation of Pauli matrices are shown below.
For \(X\), the eigenvalue \(\lambda\) is the solution of the following equation,
(2)#\[\begin{split} |X - \lambda I|= \begin{vmatrix} -\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = \lambda^2 - 1 = 0 \iff \lambda = \pm 1 \end{split}\]The corresponding normalized eigenvectors are given by
(3)#\[\begin{split} \begin{align} X|v_1\rangle &= \begin{pmatrix} 0 & 1\\ 1 &0 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix} \iff |v_1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \\ X|v_2\rangle &= \begin{pmatrix} 0 & 1\\ 1 &0 \end{pmatrix}\begin{pmatrix} c \\ d \end{pmatrix} = -\begin{pmatrix} c \\ d \end{pmatrix} \iff |v_2\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} \end{align} \end{split}\]The coefficient \(1/\sqrt{2}\) is to normalized the eigenvector. Usually, we define \(|+\rangle = |v_1\rangle \) and \(|-\rangle = |v_2\rangle\). From the eigenvalues and eigenvectors, we could write the diagonal representation for \(X\) as
(4)#\[ X = |v_1\rangle\langle v_1| - |v_2\rangle\langle v_2| = |+\rangle\langle +| - |-\rangle\langle -| \]For \(Y\), the eigenvalue \(\lambda\) is the solution of the following equation,
(5)#\[\begin{split} |Y - \lambda I|= \begin{vmatrix} -\lambda & -i \\ i & -\lambda \end{vmatrix} = \lambda^2 - 1 = 0 \iff \lambda = \pm 1 \end{split}\]The corresponding normalized eigenvectors are given by
(6)#\[\begin{split} \begin{align} Y|w_1\rangle &= \begin{pmatrix} 0 & -i\\ i &0 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix} \iff |w_1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix} \\ Y|w_2\rangle &= \begin{pmatrix} 0 & -i\\ i &0 \end{pmatrix}\begin{pmatrix} c \\ d \end{pmatrix} = -\begin{pmatrix} c \\ d \end{pmatrix} \iff |w_2\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix} \end{align} \end{split}\]The coefficient \(1/\sqrt{2}\) is to normalized the eigenvector. From the eigenvalues and eigenvectors, we could write the diagonal representation for \(Y\) as
(7)#\[ Y = |w_1\rangle\langle w_1| - |w_2\rangle\langle w_2| \]For \(Z\), the eigenvalue \(\lambda\) is the solution of the following equation,
(8)#\[\begin{split} |Z - \lambda I|= \begin{vmatrix} 1-\lambda & 0 \\ 0 & -1-\lambda \end{vmatrix} = \lambda^2 - 1 = 0 \iff \lambda = \pm 1 \end{split}\]The corresponding normalized eigenvectors are given by
(9)#\[\begin{split} \begin{align} Z|u_1\rangle &= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix} \iff |u_1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = |0\rangle \\ Z|u_2\rangle &= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} c \\ d \end{pmatrix} = -\begin{pmatrix} c \\ d \end{pmatrix} \iff |u_2\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = |1\rangle \end{align} \end{split}\]From the eigenvalues and eigenvectors, we could write the diagonal representation for \(Z\) as
(10)#\[ Z = |0\rangle\langle 0| - |1\rangle\langle 1| \]