Controlled unitary gate

A controlled-$U$ gate applies $U$ to its target exactly when its control qubit is one.

A quantum variable rotation uses an angle stored in a quantum register to rotate a target qubit coherently. For a binary angle, controlled fixed-angle rotations from each bit multiply to the desired total rotation.

Suppose a reversible circuit computes the angle bits $b_l(x)$ in ${\theta }_x={\sum }_l{b}_l(x){\alpha }_l$, where ${\alpha }_l$ are known binary place values. Apply a controlled unitary gate $R_y(2{\alpha }_l)$ to the target for each set angle bit. The rotations about the y-axis commute and their angles add, giving $R_y(2{\theta }_x)$. Finally perform uncomputation of the angle and arithmetic workspace. An $L$-bit angle needs $L$ controlled rotations and twice the reversible angle-computation cost. This preserves coherent superpositions of inputs and underlies the HHL controlled reciprocal rotation.

The Fredkin gate, or controlled-SWAP gate, exchanges two target qubits exactly when its control qubit is one.

The controlled-NOT gate maps $|a,b⟩$ to $|a,b\oplus a⟩$.

Writing a CNOT gate as $P_0\otimes I+P_1\otimes X$ gives $I\otimes Q_{+}+Z\otimes Q_{-}$, where $Q_{\pm }=(I\pm X)/2$. In the Hadamard basis, $Z$ flips the sign label and $X$ is diagonal. Thus the same gate matrix reverses its control and target roles under the two local basis changes.

A CNOT gate can correlate either the original control's computational label with a computational pointer, or the original target's sign-basis label with a sign-basis pointer. For the second interpretation prepare the original control in $|+⟩$ and read it in the Hadamard basis. Its two pointer states become $|+⟩$ and $|-⟩$, measuring $X$ on the original target. An apparatus role therefore depends on the ready state and readout as well as the measurement interaction.

The SWAP gate maps $|x⟩|y⟩$ to $|y⟩|x⟩$.

Applying controlled-NOT gates with directions $1\to 2$, $2\to 1$, and $1\to 2$ mapsand therefore implements the SWAP gate.

The three-qubit GHZ state is

For $n\ge 2$, apply a Hadamard gate to one qubit, then controlled-NOTs from it to all others. Replacing each controlled-NOT gate by a target Hadamard, a Controlled-Z gate, and another target Hadamard uses $1+3(n-1)$ allowed gates. Each one-qubit reduced state is $I/2$, proving entanglement across each qubit-versus-rest bipartition.

For $|\alpha ⟩=a|0⟩+b|1⟩$, a Bell measurement on $|\alpha ⟩$ and the first leg of a GHZ state leaves the other two legs in $a|00⟩\pm b|11⟩$, after applying $X$ to both legs for the two $\Psi$ outcomes.

Starting from $|000⟩$, apply a Hadamard gate to the first qubit and then controlled-NOT gates from the first qubit to each of the other two qubits.

If two senders share a GHZ state with a receiver, one sender can encode two bits with $X^x{Z}^z$ and the other one bit with $X^b$. Sending both qubits to the receiver produces one of eight orthogonal GHZ-basis states, so a joint measurement recovers all three bits.

The GHZ theorem uses perfect multipartite quantum correlations to contradict local predetermined values without a statistical inequality. For the three-qubit GHZ state, three certain Pauli-product relations force a fourth product to have the opposite sign from its quantum-mechanical value.

The Hadamard test estimates the real part of $⟨\psi |U|\psi ⟩$ from one ancilla measurement.

If $U|\psi ⟩=e^{2\pi i\theta }|\psi ⟩$, the ordinary Hadamard test returns ancilla zero with probability

Adding a phase gate to the Hadamard test rotates the interference signal and measures the imaginary part of $⟨\psi |U|\psi ⟩$.

The swap test on pure states $|a⟩,|b⟩$ returns ancilla zero with probability