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Say you have $2$ q-bits, namely $q_1, q_2$. What's the right language for saying apply CNOT on $q_1$ and $q_2$ where $q_1$ is the control bit and $q_2$ is the target?

For instance, can I say "apply CNOT on $q_2$ controlled by $q_1$"? What's the standard way of saying that?

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    $\begingroup$ Just as you say it! Everyone will understand. $\endgroup$ – DaftWullie Sep 27 '18 at 5:02
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The most common convention is to refer to qbits by the index of their significance, with the least-significant qbit having index $0$. This is cribbed from binary, where the significance index is the same as the exponent in the sum of powers of two:

$1011=1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0$

So for the system $|abc\rangle = a \otimes b \otimes c$, you'd say $a$ is qbit $2$, $b$ qbit $1$, and $c$ qbit $0$.

CNOT is usually denoted $C_{c,t}$ (or $CX_{c,t}$ or even $CNOT_{c,t}$) where $c$ is the index of the control qbit and $t$ is the index of the target qbit. So a CNOT gate with the most-signifigant qbit of a three-qbit system as control and least-significant qbit as target is denoted $C_{2,0}$, or just $C_{20}$ when there are ten or fewer qbits so the comma is unnecessary. Applying this operator to your system is written as $C_{20}|abc\rangle$.

This convention is outlined in detail in section 1.4, pages 10-11 of Quantum Computer Science: An Introduction by N. David Mermin.

For plain-language phrasing, something like "apply CNOT with qbit $x$ as control, $y$ as target" or even just "apply CNOT-$x$-$y$" (if spoken) works fine.

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