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?
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1 Answer
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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.
answered Aug 4, 2021 at 23:44
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$\begingroup$ Hi! Would you mind explaing why $1 \oplus b = \neg b$? I totally understand everything else in your answer! Thank you! $\endgroup$– ClaireAug 5, 2021 at 0:04
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1$\begingroup$ Just check all (i.e. both) possibilities: $1\oplus 0 = 1 = \neg 0$ and $1\oplus 1 = 0 = \neg 1$. $\endgroup$ Aug 5, 2021 at 0:06