Exercise 4.17

Suppose that we have control-\(X\) operation as

(1)\[ {\rm CNOT}: |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes X \]

control-\(Z\)​ operation can be expressed as follow,

(2)\[ {\rm control-}Z: |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes Z \]

Then we could simply add one Hadamard gate before the target operation, and one Hadamard gate after the target operation. That is,

(3)\[\begin{split} \begin{align} (I\otimes H){\rm control-}Z(I\otimes H) &= (I\otimes H)(|0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes Z )(I\otimes H) \\ &= (|0\rangle\langle 0| \otimes H + |1\rangle\langle 1| \otimes HZ )(I\otimes H)\\ &= |0\rangle\langle 0| \otimes HH + |1\rangle\langle 1| \otimes HZH \end{align} \end{split}\]

Note that \(H = H^{\dagger}\) and \(H^{\dagger}H = I\), and \(HZH = X\), we could conclude that

(4)\[ {\rm CNOT} = (I\otimes H){\rm control-}Z(I\otimes H) \]

Similarly, if the control-\(Z\) gate is in the following form,

(5)\[ {\rm control-}Z_2: I\otimes |0\rangle\langle 0| + Z\otimes |1\rangle\langle 1| \]

we could also write a \(\rm CNOT\) gate as

(6)\[ (H\otimes I){\rm control-}Z_2(H\otimes I) = I\otimes |0\rangle\langle 0| + X\otimes |1\rangle\langle 1| \]