The CNOT Gate

Every gate so far has acted on one qubit at a time — spin it, flip it, rotate its phase. But a qubit alone can never become entangled with anything; for that you need a gate that lets one qubit reach across and touch another. The simplest such gate is the controlled-NOT, or CNOT. It reads a control qubit and, only if that control is |1\rangle, flips a second target qubit — the quantum cousin of the classical rule "if this switch is on, toggle that light." It sounds almost boring. Yet run it on a qubit that is both on and off at once, and out comes the strangest object in the theory: two qubits with a single, shared identity.

The rule: control decides, target flips

CNOT takes two qubits. We call the first the control and the second the target. The rule is one sentence:

if the control is |1\rangle, apply the X (NOT) gate to the target — i.e. flip it;
if the control is |0\rangle, leave the target alone.

Either way the control itself is never changed. Taking the control as the first (left) qubit, watch what happens to each of the four basis states:

The first two are untouched (control 0, nothing happens); the last two have their second bit flipped (control 1, target toggles). Notice the output target bit is exactly control XOR target, c \oplus t — CNOT is the reversible quantum version of the classical XOR.

The circuit symbol

In a circuit diagram each qubit is a horizontal wire, read left to right in time. CNOT is drawn as a filled dot on the control wire, a ⊕ (a NOT/target circle) on the target wire, and a vertical line joining them. The control comes out unchanged; the target comes out as c \oplus t.

As a matrix

Written in the ordered basis |00\rangle, |01\rangle, |10\rangle, |11\rangle, CNOT is the 4\times 4 permutation matrix

\mathrm{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}.

The top-left 2\times 2 block is the identity (control 0: leave the target), and the bottom-right block is the X gate \begin{psmallmatrix} 0 & 1 \\ 1 & 0 \end{psmallmatrix} (control 1: flip the target). Because it merely permutes the basis states, it is unitary, and swapping the same two rows back again shows it is its own inverse: \mathrm{CNOT}^2 = I. Apply it twice and you are exactly where you started.

Worked example 1: \mathrm{CNOT}\,|10\rangle

By the rule. The control (first qubit) is 1, so we flip the target (second qubit) from 0 to 1. The control stays 1. Result: |10\rangle \to |11\rangle.

By the matrix. In the ordered basis, |10\rangle is the third basis vector (0,0,1,0)^{\mathsf T}. Multiplying picks out the third column of CNOT:

\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} = |11\rangle.

Both routes agree: \mathrm{CNOT}\,|10\rangle = |11\rangle. On any single basis state CNOT is this dull, entirely classical, look-up. The magic only appears when the control is a superposition.

Worked example 2: two gates make entanglement

Start with both qubits in |0\rangle, the state |00\rangle. First apply a Hadamard to the control (first) qubit, which sends |0\rangle \to \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle):

|00\rangle \;\xrightarrow{\;H\text{ on qubit 1}\;}\; \tfrac{1}{\sqrt2}\big(|0\rangle + |1\rangle\big)\otimes|0\rangle = \tfrac{1}{\sqrt2}\big(|00\rangle + |10\rangle\big).

Now the control is genuinely both 0 and 1. Apply CNOT. It acts on each piece of the superposition at once — leaving |00\rangle alone and turning |10\rangle into |11\rangle:

\tfrac{1}{\sqrt2}\big(|00\rangle + |10\rangle\big) \;\xrightarrow{\;\mathrm{CNOT}\;}\; \tfrac{1}{\sqrt2}\big(|00\rangle + |11\rangle\big).

That output is the Bell state |\Phi^+\rangle. Try to write it as a product |\psi\rangle \otimes |\varphi\rangle and you will fail — there is no way to split it into "the first qubit does this, the second does that." It is entangled. This "H then CNOT" circuit is the standard recipe for making a Bell state. Step through it below.

Correlate, do not copy

It is tempting to read the Bell result as "CNOT measured the control and copied it onto the target." It did neither. Nothing was measured — the state is still a clean superposition, not a definite 0 or 1. And nothing was copied — the outcome \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle) is not two independent copies of \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle) (that would be \tfrac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle), a product state). What CNOT actually did was correlate: it wove the two qubits together so their measurements must agree, without ever pinning either one down.

CNOT is a two-qubit gate with a control and a target: if the control is |1\rangle it applies X (a flip) to the target; if |0\rangle it does nothing. The control is unchanged;
on the basis (control = first qubit): |00\rangle\to|00\rangle, |01\rangle\to|01\rangle, |10\rangle\to|11\rangle, |11\rangle\to|10\rangle — the target becomes c\oplus t;
its matrix is the 4\times 4 permutation \begin{psmallmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{psmallmatrix}; it is unitary and its own inverse (\mathrm{CNOT}^2 = I);
on a superposed control it entangles: |00\rangle \xrightarrow{H,\ \mathrm{CNOT}} \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle), a Bell state. It correlates the qubits; it does not measure or clone them.

It is remarkable how little it takes. Entanglement — the resource behind teleportation, superdense coding, and the speed-ups that make quantum computers interesting — is manufactured by exactly two elementary gates. A Hadamard splits one qubit into an even blend of 0 and 1; a CNOT then ties a second qubit to that undecided coin. Neither gate is exotic, and neither on its own does anything spooky. But composed in this order they turn a plain |00\rangle into a state that Einstein could not bring himself to accept. Almost every entangling circuit you will ever meet has this same little H–CNOT motif hiding somewhere inside it.

On the four basis states CNOT looks completely classical — it just copies the control onto the target when the control is 1 (|10\rangle \to |11\rangle), exactly like a wire fanning out. That intuition is a trap. Feed it a superposed control and it does not produce two independent copies; it produces entanglement. There is no gate that clones an unknown qubit — the no-cloning theorem forbids it — and CNOT is not a secret exception. The control's superposition survives untouched only when it is a basis state; a genuine superposition gets braided into the target instead of duplicated onto it.