Two-qubit gates: CNOT, CZ, SWAP

Controls and targets, the matrices, and how CZ is a CNOT in disguise.

Beginner12 min read5 sections · 2 questions+45

By the end you will be able to

01Apply CNOT to each of the four basis states.
02Convert a CZ into a CNOT with two Hadamards.
03Decompose SWAP into three CNOTs.

CNOT = 1000010000010010; (∣ c t ⟩ \to ∣ c,; t \oplus c ⟩), CZ = diag (1, 1, 1, - 1)

CNOT flips the target when the control is 1. CZ adds a −1 phase only to |11⟩ and is symmetric: either qubit can be called the control.

In the composer, the filled dot is the control and the ⊕ is the target. On a superposed control the gate acts on both branches at once: that is how CNOT turns a product state into an entangled one, the subject of the next lesson.

01Useful identities

CNOT = (I ⊗ H) · CZ · (I ⊗ H): sandwich the target with Hadamards. Hardware with a native CZ (IBM Heron) compiles every CNOT this way.
SWAP = CNOT(a,b) · CNOT(b,a) · CNOT(a,b): three two-qubit gates, which is why the transpiler avoids SWAPs when it can.
Controlled-U for any U: the composer offers CU, CP, CRX, CRY, CRZ, CH.

Hands-on

Prove CZ ≡ H·CNOT·H

Compile this circuit (any fake backend) and compare the unitary: the Levels tab shows the transpiler folding the three gates into a single CZ on a CZ-native target.

OpenQASM 3

OPENQASM 3.0;
include "stdgates.inc";
qubit[2] q;
h q[1];
cx q[0], q[1];
h q[1];

Open in the IDE Open in the Playground

Knowledge check

2 questions · one try each counts

Question 1

CNOT acts on |10⟩ (control = qubit 1 written first, target = qubit 0). What comes out?

Question 2

How many CNOTs does a SWAP cost?

01Useful identities

CNOT = (I ⊗ H) · CZ · (I ⊗ H): sandwich the target with Hadamards. Hardware with a native CZ (IBM Heron) compiles every CNOT this way.
SWAP = CNOT(a,b) · CNOT(b,a) · CNOT(a,b): three two-qubit gates, which is why the transpiler avoids SWAPs when it can.
Controlled-U for any U: the composer offers CU, CP, CRX, CRY, CRZ, CH.