Quantum Computing 3.5 -- Transformed CNOT

We first calculate the tensor product of the Hadamard operators $\begin{aligned} H \otimes H &= \left(H \otimes I \right) \cdot \left(I \otimes H \right) \\ &= \frac{1}{\sqrt{2}} \left(\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{array} \right) \cdot \frac{1}{\sqrt{2}} \left(\begin{array}{cccc} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{array} \right) \\ &= \frac{1}{2} \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{array} \right) \end{aligned}$ We consider the whole quantum circuit as an operator $U$ which results as matrix muliplication of the different operators $\begin{aligned} U &= (H \otimes H) \cdot \text{CNOT} \cdot (H \otimes H) \\ &= \frac{1}{4} \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{array} \right) \cdot \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) \cdot \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{array} \right) \\ &= \frac{1}{4} \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{array} \right) \cdot \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \\ 1 & 1 & -1 & -1 \end{array} \right) \\ &= \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \\ &= \text{CNOT}_{21} \end{aligned}$ The whole circuit can be represented by an CNOT gate $\text{CNOT}_{21}$ in which control and target qubit are interchanged. For distinction we label the ordenary CNOT gate with $\text {CNOT}_{12} \equiv \text{CNOT}$ . The effect of these gates on the basis states is $\begin{aligned} \text{CNOT}_{12} |ij\rangle &= |i,i \oplus j\rangle \\ \text{CNOT}_{21} |ij\rangle &= |i \oplus j,j\rangle \end{aligned}$ where $i,j \in \{0,1\}$ . By appling the CNOT $_{21}$ gate on an general state vector, we get the result $\text{CNOT}_{21} \left( \begin{array}{cc} \alpha \\ \beta \\ \gamma \\ \delta \end{array}\right) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \left( \begin{array}{cc} \alpha \\ \beta \\ \gamma \\ \delta \end{array}\right) = \left( \begin{array}{cc} \alpha \\ \delta \\ \gamma \\ \beta \end{array}\right)$