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$\mathrm{C}H$

controlled· 2 qubits

Controlled-Hadamard

Applies a Hadamard to the target qubit when the control qubit is in the $|1\rangle$ state.

Open in Quirk ↗ hermitian

The controlled-Hadamard gate performs a conditional change of basis, moving the target between the computational basis and the Hadamard basis only when the control is $|1\rangle$.

$$ \mathrm{C}H = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ 0 & 0 & \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{bmatrix} $$

Properties

  • Hermitian and self-inverse, like the Hadamard itself.
  • Real, so it lies in the orthogonal group.
  • Although $H$ is Clifford, $\mathrm{C}H$ is not. Adding a control to a Clifford gate does not preserve Clifford-ness in general (compare CNOT, which stays Clifford, versus $\mathrm{C}H$ and $\mathrm{C}S$, which do not).

Decompositions

  • A single entangling gate suffices: $\mathrm{C}H = (I \otimes R_y(\pi/4)) \, \mathrm{C}Z \, (I \otimes R_y(-\pi/4))$, which works because $R_y(\pi/4) \, Z \, R_y(-\pi/4) = H$.

Also written

  • $\Lambda(H)$
  • $\text{controlled-}H$

Related

  • Controlled version of Hadamard

Groups

  • orthogonal

Data

  • This gate as JSON
  • Edit this page on GitHub

The unitary, drawn

Every entry of the matrix as a disc. The area is the transition probability $|U_{jk}|^2$; the colour and the hand are the phase. Rows are outputs $\langle j|$, columns inputs $|k\rangle$, in the computational basis.

Phase

Disc area $= |U_{jk}|^2$. Hover an entry for its value, or take the JSON.

Weyl chamber

Where this gate sits among all two-qubit gates, up to single-qubit rotations before and after.

  • this gate
  • other gates
  • perfect entanglers

Coordinates:

not entangling entangling perfect entangler

$(a, b, c)$ means the gate equals $e^{i(a\,XX + b\,YY + c\,ZZ)}$ up to single-qubit gates, reduced to the chamber $\pi/4 \ge a \ge b \ge |c|$ (the convention of the canonical gate, Qiskit and Cirq). The $a = \pi/4$ face is a mirror: $(\pi/4, b, c)$ and $(\pi/4, b, -c)$ are the same gate. Drag the figure to rotate it, double-click to reset.

SDK support

How to call this gate in each framework. Hover a ⓘ for a note; the support matrix compares every gate at once.

SDKName
Qiskit qiskit.circuit.library.CHGate
PennyLane pennylane.CH
Cirq — ⓘ
Q# — ⓘ
PyQuil — ⓘ
Braket — ⓘ
BQSKit bqskit.ir.gates.CHGate
Qibo qibo.gates.CH
pytket pytket.circuit.OpType.CH
Stim —
OpenQASM stdgates.inc: ch

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