Exercise 4.18#
The control-\(Z\) operation on the left can be expressed as follow,
(1)#\[
{\rm control-}Z_1: |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes Z \tag{1} \label{1}
\]
The control-\(Z\) operation on the right can be expressed as follow,
(2)#\[
{\rm control-}Z_2: I\otimes |0\rangle\langle 0| + Z\otimes |1\rangle\langle 1|\tag{2}\label{2}
\]
If we write eq. (1) in the matrix form, we have
(3)#\[\begin{split}
\begin{align}
{\rm control-}Z_1&: \begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix} \otimes \begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix} + \begin{pmatrix}
0 & 0 \\ 0 & 1
\end{pmatrix} \otimes \begin{pmatrix}
1 & 0 \\ 0 & -1
\end{pmatrix} \\
&= \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix} + \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix} \\
&= \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix}
\end{align}\tag{3}\label{3}
\end{split}\]
If we write eq. (2) in the matrix form, we have
(4)#\[\begin{split}
\begin{align}
{\rm control-}Z_2&: \begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix} \otimes \begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix} + \begin{pmatrix}
1 & 0 \\ 0 & -1
\end{pmatrix}\otimes\begin{pmatrix}
0 & 0 \\ 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix} + \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix} \\
&= \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix}
\end{align}\tag{4}\label{4}
\end{split}\]
Compare eq. (3) and eq. (4), we could conclude that the \({\rm control-}Z_1 = {\rm control-}Z_2\).