# How to say “apply CNOT on q-bit 1 controlled by q-bit 2”?

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?

• Just as you say it! Everyone will understand. – DaftWullie Sep 27 '18 at 5:02

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 $C_{20}|abc\rangle$.
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.