Exercise 4.20

Consider a \(\rm CNOT\) gate with first qubit as control and second qubit as target, we could write the \(\rm CNOT\) gate in the following form,

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

Similarly, we should have \({\rm CNOT} = I\otimes |0\rangle\langle 0| +X\otimes |1\rangle\langle 1| \) for \(\rm CNOT\) gate with second qubit as control and first qubit as target. For Hadamard gate, we have the following properties,

(2)\[ H|0\rangle = \frac{|0\rangle+|1\rangle}{\sqrt{2}} = |+\rangle, H|1\rangle = \frac{|0\rangle-|1\rangle}{\sqrt{2}} = |-\rangle, HXH = Z \]

With eq. (1) and eq. (2), we could write the circuit shown in the question as unitary operator \(U\), and

(3)\[\begin{split} \begin{align} U &= (H\otimes H)(|0\rangle\langle 0|\otimes I + |1\rangle\langle 1| \otimes X)(H\otimes H) \\ &= H|0\rangle\langle 0|H\otimes I + H|1\rangle\langle 1|H \otimes HXH \\ &= |+\rangle\langle +|\otimes I + |-\rangle\langle -| \otimes Z \end{align} \end{split}\]

Note that for identity, \(X\) gate and \(Z\) gate, we could write them in the spectral decomposition form,

(4)\[\begin{split} \begin{align} I &= |0\rangle\langle 0 | + |1\rangle \langle 1| \\ &= |+\rangle \langle + | + |-\rangle\langle -| \\ X &= |+\rangle\langle + | - |-\rangle \langle -| \\ Z &= |0\rangle\langle 0 | - |1\rangle \langle 1| \\ \end{align} \end{split}\]

Then we could use eq. (4) and re-write eq. (3) as

(5)\[\begin{split} \begin{align} U =& |+\rangle\langle +|\otimes I + |-\rangle\langle -| \otimes Z \\ =&|+\rangle\langle +|\otimes |0\rangle\langle 0 | +|+\rangle\langle +|\otimes |1\rangle\langle 1 | \\ &+ |-\rangle\langle -|\otimes |0\rangle\langle 0 | -|-\rangle\langle -|\otimes |1\rangle\langle 1 | \\ =& I\otimes |0\rangle\langle 0| + X \otimes |1\rangle\langle 1| \end{align} \end{split}\]

We could find that the circuit shown in the problem is equivalent to \(\rm CNOT\) gate with second qubit as control and first qubit as target. Then the circuit identity is proved.

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In order to verify equation \((4.24)-(4.27)\) in the book for \(\rm CNOT\) gate with first qubit as control and second qubit as target, we could make use of the circuit identity and the properties of Hadamard gate in eq. (2).

For \(\rm CNOT(|+\rangle\otimes |+\rangle)\), according to eq. (2) and eq. (3), we could have

(6)\[\begin{split} \begin{align} {\rm CNOT}(|+\rangle\otimes |+\rangle) &= {\rm CNOT}(H\otimes H)(|0\rangle\otimes |0\rangle) \\ &= (H\otimes H)^{\dagger}(H\otimes H){\rm CNOT}(H\otimes H)(|0\rangle\otimes |0\rangle) \\ &= (H\otimes H)(H\otimes H){\rm CNOT}(H\otimes H)(|0\rangle\otimes |0\rangle) \\ &= (H\otimes H)U(|0\rangle\otimes |0\rangle) \end{align} \end{split}\]

where we have \((H\otimes H)^{\dagger} = H^{\dagger}\otimes H^{\dagger} = H\otimes H\). It means that we could use the result of \(U\) to calculate \(\rm CNOT(|+\rangle\otimes |+\rangle)\), that is,

(7)\[\begin{split} \begin{align} {\rm CNOT}(|+\rangle\otimes |+\rangle) &=(H\otimes H)U(|0\rangle\otimes |0\rangle) \\ &=(H\otimes H)(I\otimes |0\rangle\langle 0| + X \otimes |1\rangle\langle 1|)(|0\rangle\otimes |0\rangle)\\ &= (H\otimes H)(I|0\rangle\otimes |0\rangle\langle 0|0\rangle + X|0\rangle \otimes |1\rangle\langle 1|0\rangle) \\ &= (H\otimes H)(|0\rangle\otimes |0\rangle) = |+\rangle\otimes |+\rangle \end{align} \end{split}\]

For \(\rm CNOT(|-\rangle\otimes |+\rangle)\), we have

(8)\[\begin{split} begin{align} {\rm CNOT}(|-\rangle\otimes |+\rangle) &=(H\otimes H)U(|1\rangle\otimes |0\rangle) \\ &= (H\otimes H)(I\otimes |0\rangle\langle 0| + X \otimes |1\rangle\langle 1|)(|1\rangle\otimes |0\rangle) \\ &= (H\otimes H)(I|1\rangle\otimes |0\rangle\langle 0|0\rangle + X|1\rangle \otimes |1\rangle\langle 1|0\rangle) \\ &= (H\otimes H)(|1\rangle\otimes |0\rangle) = |-\rangle\otimes |+\rangle \end{align} \end{split}\]

For \(\rm CNOT(|+\rangle\otimes |-\rangle)\), we have

(9)\[\begin{split} \begin{align} {\rm CNOT}(|+\rangle\otimes |-\rangle) &=(H\otimes H)U(|0\rangle\otimes |1\rangle)\\ &=(H\otimes H)(I\otimes |0\rangle\langle 0| + X \otimes |1\rangle\langle 1|)(|0\rangle\otimes |1\rangle) \\ &= (H\otimes H)(I|0\rangle\otimes |0\rangle\langle 0|1\rangle + X|0\rangle \otimes |1\rangle\langle 1|1\rangle) \\ &= (H\otimes H)(|1\rangle\otimes |1\rangle) = |-\rangle\otimes |-\rangle \end{align} \end{split}\]

For \(\rm CNOT(|-\rangle\otimes |-\rangle)\), we have

(10)\[\begin{split} \begin{align} {\rm CNOT}(|-\rangle\otimes |-\rangle) &=(H\otimes H)U(|1\rangle\otimes |1\rangle)\\ &= (H\otimes H)(I\otimes |0\rangle\langle 0| + X \otimes |1\rangle\langle 1|)(|1\rangle\otimes |1\rangle) \\ &= (H\otimes H)(I|1\rangle\otimes |0\rangle\langle 0|1\rangle + X|1\rangle \otimes |1\rangle\langle 1|1\rangle) \\ &= (H\otimes H)(|0\rangle\otimes |1\rangle) = |+\rangle\otimes |-\rangle \end{align} \end{split}\]