The following circuit with a CNOT gate has the following effect on a computational basis state $|a, b\rangle$, where all additions are done modulo 2.
Why is the state of the second qubit changed to $a\oplus b$ after CNOT is applied?
Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.
Sign up to join this communityThe following circuit with a CNOT gate has the following effect on a computational basis state $|a, b\rangle$, where all additions are done modulo 2.
Why is the state of the second qubit changed to $a\oplus b$ after CNOT is applied?
This is the definition of the CNOT gate. The definition has been chosen this way, because
$$ 0 \oplus b = b \\ 1 \oplus b = \neg b $$
where $\neg$ denotes bit-flip and $\oplus$ denotes addition modulo $2$. In other words, $a\oplus b$ does nothing to $b$ if $a=0$ and it flips $b$ if $a=1$. This is what one would expect a controlled-NOT gate to do.