CX Gate and Entanglement

CX Gate (CNOT) and Entanglement

qc/gates qc/entanglement

Two-qubit gate: flips the target qubit iff the control qubit is |1⟩. When the control is in superposition, the two qubits become entangled — their measurement outcomes become perfectly correlated with no classical explanation.

$CX=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right)$

This is the mechanism behind Bell States and GHZ States.

CZ — the symmetric sibling

$CZ=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right)$

CZ is the native two-qubit gate on IBM’s Heron devices (see Backend Properties) — it applies a $-1$ phase only when both qubits are $|1⟩$, which makes it symmetric under swapping the two qubits (unlike CX, which clearly distinguishes control from target). CX and CZ are interconvertible with single-qubit gates ($CZ=(I\otimes H)CX(I\otimes H)$), which is why the Error Mitigation section’s Ising circuits can build everything from CZ alone.

Related

• H Gate
• Bridge Gate Identity — a non-local CX rewritten as nearest-neighbor CX gates
• Circuit Depth — CX gates are usually the bottleneck for depth
• Backend Properties — CZ as the hardware-native gate
• Dressed Gates and Pauli Propagation

Self-Check

• Could you explain CNOT to someone who’s never heard of it, including why it needs a “control” and a “target”?
• Why does entanglement only happen when the control qubit is in superposition, not when it’s in a definite state?
• Why are CX gates usually the bottleneck for Circuit Depth?
• Why is CZ symmetric under swapping its two qubits, but CX isn’t?