# Show that conjugating $\text{CNOT}$ by $H\otimes H$ exchanges control and target qubits

I have gotten so far to show that

$$(H⊗H)\text{CNOT}(H⊗H) = (H⊗I)CZ(H⊗I)\,.$$

How do I proceed from here in first proving the identity and then using the identity to show that on the dual basis, when $$\text{CNOT}$$ is applied target and control are swapped.

• Does this answer your question? Equivalence between quantum circuit: CNOT changes control and target qubit Commented Mar 3 at 2:10
• I am trying to show in the operator format Commented Mar 3 at 2:25
• What do you mean by "show that on the dual basis"? I am not familiar with this terminology. Commented Mar 3 at 4:29
• it just means |+> and |-> state for the qubits Commented Mar 3 at 4:59

The key observation is that the CZ gate, whose action on the computational basis is \begin{align} |00\rangle&\mapsto|00\rangle\\ |01\rangle&\mapsto|01\rangle\\ |10\rangle&\mapsto|10\rangle\\ |11\rangle&\mapsto-|11\rangle\\ \end{align}\tag1 is symmetric under the exchange of qubits. In other words, $$C_1Z_2=C_2Z_1\tag2$$ where $$C_iU_j$$ denotes the controlled-$$U$$ gate with qubit $$i$$ as the control and qubit $$j$$ as the target. Therefore, \begin{align} (H\otimes H)\circ C_1X_2\circ(H\otimes H)&=(H\otimes I)\circ C_1Z_2\circ(H\otimes I)\tag3\\ &=(H\otimes I)\circ C_2Z_1\circ(H\otimes I)\tag4\\ &=C_2X_1\tag5 \end{align} where we first used $$HXH=Z$$, then $$(2)$$ and then $$HZH=X$$ again. $$\square$$