Controlled unitary gate

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

For a qubit control and finite-dimensional target with unitary operator V, the controlled unitary gate $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$ is entangling if and only if V is not scalar. If $V=e^{i\theta }I$, the controlled operation is a phase gate on the control tensored with the target identity and preserves every product state.

Otherwise the spectral theorem supplies two orthonormal eigenvectors with distinct eigenvalues ${\lambda }_0,{\lambda }_1$. Set $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$ and use product input $|+⟩|b⟩$. The two conditional target vectors $|b⟩$ and $V|b⟩$ are linearly independent because their eigenbasis coefficients would otherwise require ${\lambda }_0={\lambda }_1$. The conditional-vector product-state criterion shows that their coherent controlled superposition is entangled. Thus even a target gate arbitrarily close to scalar can be entangling, although the resulting entanglement need not be large.

Multiplying a target unitary operator by $e^{i\varphi }$ changes the controlled operation by a relative quantum phase on its control-one branch. Substitution in $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$ proves the displayed identity. In particular, $C(-I)=Z\otimes I$ rather than the identity operator. Preparing the control at $|+⟩$ and applying a final Hadamard gate distinguishes $C(I)$ from $C(-I)$ with certainty, independently of the target quantum state. The global phase of an uncontrolled quantum gate is unobservable, but a specified controlled unitary gate provides a branch relative to which that phase can be measured.

For a controlled unitary gate $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$, Hadamard gates on the target before and after the gate produce $C(HVH)$. Proof: the control-zero branch applies $HIH=I$, while the control-one branch applies $HVH$. Linearity extends the identity to arbitrary quantum states. Since $C(H)C(V)C(H)=C(HVH)$ too, a pair of controlled Hadamard gates surrounding the same target operation can be replaced by ordinary target Hadamard gates. No single controlled Hadamard gate is thereby identified with an ordinary 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.

Every Fredkin gate preserves the Hamming weight of a computational-basis string, so any network of these gates does too. A Toffoli gate maps $110$ to $111$ and hence cannot equal such a network on the same three wires. Fixed ancillas returned unchanged do not remove the obstruction. Extra encoded outputs or input-dependent garbage change the resource model and do not by themselves implement the same clean coherent three-qubit gate.

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 $|y⟩|x⟩$.

The SWAP gate and local unitary operators preserve product states. The displayed identity moves every exchange past a pair of local gates, reducing any such two-qubit quantum circuit to a local operation, optionally followed by SWAP gate. On more wires the same argument leaves a permutation of the tensor factors. By induction, a fully factored input remains factored. Since universal quantum computation includes operations that create quantum entanglement from a product state, these gates cannot form a universal quantum gate set.

For two spin-one-half qubits, the SWAP gate acts as one on the triplet sector and minus one on the singlet sector. The displayed unitary squares to SWAP. An isotropic exchange interaction with integrated angle $\pi /2$ implements it up to a common phase. There are other square-root branches; a pulse decomposition must use the declared branch consistently.

Let $P_{-}$ project onto $(|01⟩-|10⟩)/\sqrt{2}$ and let $P_{+}=I-P_{-}$ project onto the symmetric sector. An isotropic exchange interaction, with a chosen common energy offset, gives the continuous family displayed above. A full exchange has $\theta =\pi$ and equals SWAP gate; the conventional half exchange has $\theta =\pi /2$ and equals $P_{+}+i{P}_{-}$. Reversing the sign of the exchange gives the same full exchange at $\theta =-\pi$ but the inverse half exchange at $\theta =-\pi /2$. Thus even continuous time-independent Hamiltonian evolution and unchanged $|00⟩,|11⟩$ do not let the full-time truth table determine a unique half-time gate. A chosen generator or spectral branch is necessary. The half-exchange eigenvalues $1,i$ are the conventional principal scalar square-root choices for $1,-1$; no analytic matrix-function branch across the negative real axis is being assumed.

Two square roots of the SWAP gate, separated and followed by local $z$ rotations, implement a controlled phase gate. Local $y$ rotations of its target convert it into a CNOT gate. The construction assumes isotropic exchange and independent one-qubit control; pulse errors and extra Hamiltonian terms change the implemented operation.

With $S=P_{+}+i{P}_{-}$ the conventional square root of the SWAP gate, $P=\mathrm{diag}(1,i)$ the S gate, and operators acting rightmost first, $S{Z}_{1}S=\mathrm{diag}(1,i,-i,-1)$. Therefore $H_{2}S{Z}_{1}S{P}_2{H}_2=|0⟩⟨0|\otimes X-i|1⟩⟨1|\otimes I$. Appending $X_2$ changes the zero-controlled flip into a one-controlled flip, and $P_1$ removes the control-dependent phase. This proves the displayed exact CNOT gate identity. Both the exchange branch and the phase gate square-root convention matter; changing either requires recalculating the correction.

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

A postselection scheme in which two polarization-entangled photon pairs pass through polarizing beam splitters, a half-wave plate and a nonpolarizing beam splitter. One photon at each of a trigger and three output detectors selects a GHZ state in the three output polarizations, provided optical mode matching removes distinguishable pair labels. All four detections are needed for the selected event; the trigger alone does not guarantee the three other output occupations.

The 1999 experiment used an approximately $200\mathrm{fs}$, $394\mathrm{nm}$ pump pulse and a beta barium borate crystal. Two polarization-entangled photon pairs supplied four photons. A trigger and three other photodetectors selected a GHZ state through fourfold coincidence counting. Narrow interference filters lengthened the photon coherence time to approximately $500\mathrm{fs}$, exceeding the pump duration and suppressing timing information about which pair supplied which photon. The optical mode matching was essential to coherence, not merely to routing photons.^[ref]

The observed state was selected by all four detections, which also measured the photons. A trigger alone did not demonstrate a guaranteed surviving three-photon output. This distinction connects postselection to what an apparatus actually supplies, rather than identifying a fourfold coincidence with an unconditional source.^[ref]

For a source with exactly two photons in each input arm, let $q$ be the probability of one horizontal and one vertical photon in the trigger arm. This sector alone supplies exactly one trigger photon. Splitting the other arm across its two output ports has probability $1/2$; the final polarizing beam splitter selects equal polarizations with probability $1/2$. The fourfold success probability is therefore $q/4$. The two-pair component of polarization-entangled down-conversion has $q=1/3$, giving $1/12$; independent pair bookkeeping has $q=1/2$, giving $1/8$. Pair-label distinguishability can also destroy the conditional GHZ state coherence and must be checked separately.

Charlie measures his qubit of a GHZ state in the $X$ basis. The two equally likely outcomes leave Alice and Bob in the Bell states ${\Phi }^{+}$ or ${\Phi }^{-}$, respectively. Charlie sends the outcome bit $c$ to Bob, who applies the Pauli operator $X{Z}^c$, giving ${\Psi }^{+}$ in either case. This is deterministic LOCC entanglement localization. If Charlie's record is ignored, Alice and Bob instead have the separable mixture $(|00⟩⟨00|+|11⟩⟨11|)/2$. The classical outcome is necessary to turn the conditional entanglement into a known usable Bell pair.

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.

A three-party nonlocal game with binary questions of even Hamming weight. Its winning output parity is half the question weight modulo two. For the four possible input triples, a hypothetical deterministic classical response would give four parity equations whose left sides cancel under exclusive or, while their right sides have odd total parity. Thus no classical strategy succeeds on every allowed triple, even with pre-shared randomness.

On the GHZ state, measure the Pauli X gate as an observable for question zero and the Pauli Y gate for question one, and encode the measured eigenvalue as $(-1{)}^x$. The joint observables XXX, XYY, YXY and YYX have respective GHZ eigenvalues $+1,-1,-1,-1$. Their products of local outcomes enforce the required parity with certainty. The strategy uses pre-shared quantum entanglement and local measurements during the game, without communication. It realizes the GHZ theorem as an operational task. If the four questions are uniform, the best classical success is $3/4$: the parity contradiction bounds it above, and the deterministic rule $x=a$, $y=z=0$ wins three triples and loses only 011. The uniformity assumption is necessary for that numerical score, not for the impossibility of worst-case certainty.

If each of $XXY$, $XYX$ and $YXX$ has product $-1$ except with probability $ϵ$, predetermined local X/Y values imply that a positive YYY product requires at least one of those constraints to fail. The union bound gives the displayed result without assuming independent errors. For the mixture of $(|000⟩-i|111⟩)/\sqrt{2}$ and its phase-flipped partner with weights $1-ϵ,ϵ$, quantum mechanics instead gives $P(YYY=+1)=1-ϵ$. The violation survives for $ϵ<1/4$. A noisy local outcome histogram is not uniquely determined by these parity statistics alone.

For $|G_N^{-}⟩=(|0{⟩}^{\otimes N}-|1{⟩}^{\otimes N})/\sqrt{2}$, let $O_j$ have one Pauli X gate at site $j$ and Pauli Y gates elsewhere, and let $X_N=X^{\otimes N}$. The relations $Y|0⟩=i|1⟩$, $Y|1⟩=-i|0⟩$ show that $O_j|G_N^{-}⟩$ is proportional to $|G_N^{-}⟩$ exactly when $N$ is odd. In that case its eigenvalue is ${\lambda }_N=(-1{)}^{(N+1)/2}$, while $X_N$ always has eigenvalue $-1$.

Under the EPR criterion of reality and locality, each site's $X$ and $Y$ outcomes have predetermined values $x_j,y_j\in \{\pm 1\}$. Multiplying the $N$ certain equations $x_j{\prod }_{k\ne j}{y}_k={\lambda }_N$ gives ${\prod }_j{x}_j={\lambda }_N$ when $N$ is odd, since every $y_k$ appears an even number of times. This contradicts the certain quantum value $-1$ precisely when $N\equiv 3(\mathrm{mod}4)$. Thus the displayed GHZ state gives an infinite family of perfect-correlation contradictions with local hidden-variable theory, without an inequality.

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

For the Hadamard test with a density operator $\rho$ in the auxiliary register, the output control qubit has Pauli Z gate expectation $\mathrm{Re}m$ and Pauli Y gate expectation $-\mathrm{Im}m$, where $m=\mathrm{Tr}(\rho U)$. The minus sign comes from the final Hadamard gate, since $HYH=-Y$. Measuring in $(|0⟩\pm i|1⟩)/\sqrt{2}$ therefore estimates the imaginary part using the minus-outcome frequency minus the plus-outcome frequency.

Prepare the auxiliary register in the maximally mixed state $I/N$ and estimate both mixed-state Hadamard-test quadratures. Multiplying the normalized expectation by $N$ estimates the trace of $U$. A uniformly randomized preparation of any orthonormal basis produces $I/N$; the eigenvectors of $U$ need not be known.

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