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I see somewhere that this happens:

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But I wonder if this is just identity.

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Yes, it is. If the bottom qubit is 0, neither gate does anything to the top qubit. If the bottom qubit is 1, both gates apply $X$. But since $X^2=\mathbb{I}$, the net effect is that nothing happens. Hence, overall, nothing happens.

Another way to see this is to look at the unitary matrix of controlled-not. $$ \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right) $$ It’s reasonably easy to see that the eigenvalues are 1,1,1,-1 (evidently, $|00\rangle$ and $|01\rangle$ are +1 eigenvectors, leaving behind a $2\times 2$ matrix like Pauli $X$, which we know has $\pm 1$ eigenvalues), so the square obviously has eigenvalues 1,1,1,1 and the only 4x4 unitary matrix with all ones eigenvalues is the identity matrix.

Equally, direct calculation: $$ \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right)\cdot \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) $$

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Any Hermitian gate is "self-canceling". Proof: since any gate $U$ is unitary $$ UU^{\dagger}=U^{\dagger}U=I $$ If $U$ is also Hermitian, $U=U^{\dagger}$ and $$UU=I$$

CNOT gate $$ \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right) $$ is Hermitian by inspection.

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