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A two-qubit gate in which an interaction shifts the energy of a selected joint computational basis state and thereby accumulates a relative quantum phase. With Hamiltonian operator $H=\Omega |01⟩⟨01|$, where $\Omega$ has units of energy, an interaction time $\tau$ gives $U_{\varphi }=\mathrm{diag}(1,e^{-i\varphi },1,1)$ with $\varphi =\Omega \tau /\hslash$. The selected basis state is part of the convention: this gate differs from the usual Controlled-Z gate at $\varphi =\pi$. Conjugation by a Pauli X gate on the first qubit gives $(X\otimes I)U_{\pi }(X\otimes I)=\mathrm{CZ}$, because the conjugation interchanges $|01⟩$ and $|11⟩$.

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.

The contrast of a phase-sensitive pulse sequence depends on the state during phase accumulation. Coherent preparation can create two populated paths whose relative quantum phase is converted to a population by readout; preparation into one path gives no such signal. For the Hadamard-phase-Hadamard qubit rotation, a pure $z$-polarized input yields full contrast, a pure $x$-polarized input yields none, and an unpolarized input also yields none. The last two cases have different purity despite identical measured populations. Purity alone is therefore not a phase-sensitivity criterion.

Following the pure state through the first two gates gives $(|1⟩+e^{i\alpha }|2⟩)/\sqrt{2}$. After the last Hadamard gate it isUsing the qubit polarization convention defined above,The output remains pure and has unit Bloch vector length. The population oscillation is the Ramsey interferometry fringe: the first gate supplies quantum coherence, and the last turns the acquired relative quantum phase into a measurable population difference.

Fix the qubit polarization convention by $\rho =(I+\vec{P}\cdot \vec{\sigma })/2$, where the Pauli matrices are ordered in the basis $(|1⟩,|2⟩)$ and ${\sigma }_z=|1⟩⟨1|-|2⟩⟨2|$. Thus the ground state is the north pole of the Bloch sphere. In a rotating frame and within the rotating-wave approximation, choose the drive phase and detuning sign so thatFor example, with this ground-state convention one may take $\Delta ={\omega }_L-{\omega }_0$ and choose the transverse coupling positive. The Bloch vector precesses around the effective field $(\Omega ,0,\Delta )$ at angular frequency ${\Omega }_R=\sqrt{{\Omega }^2+{\Delta }^2}$, tracing a circle at fixed projection on that axis. Its length is conserved under coherent unitary time evolution. On resonance the axis is transverse and the Rabi oscillation can fully invert an initially ground-state atom. Away from resonance the axis tilts, andThere is no relaxation towards an equilibrium Bloch vector in this ideal model. An inhomogeneous ensemble can nevertheless lose its averaged qubit polarization through dephasing even without spontaneous emission; the circle description assumes a common drive and detuning.

For $|a{|}^2+|b{|}^2=1$, the stated quantum state givesThe diagonal elements are the two state populations. The off-diagonal elements are quantum coherences: they retain the relative quantum phase and determine interference under a subsequent pulse. This is a pure state, even if both populations are nonzero: ${\rho }^2=\rho$, $\mathrm{Tr}{\rho }^2=1$, and $|\vec{P}|=1$. A mixed state averages projectors over preparation probabilities and generally has $\mathrm{Tr}{\rho }^2<1$. Equal populations alone do not distinguish an equal coherent superposition from an unpolarized mixture.

For use in all three cases, multiplication of the gates gives the Hadamard-phase-Hadamard qubit rotationThe global phase cancels in ${\rho }^{\prime }=U\rho U^{†}$. Thus the Bloch vector map is

The final unlabelled Ramsey spectroscopy comparison applies to the three cases as follows. In an atomic clock, an initial coherent pulse prepares a superposition, a free interval $T$ accumulates a relative quantum phase proportional to detuning times $T$, and a final coherent pulse converts that phase to a population. Here $\alpha$ plays the role of that accumulated phase, with sign and fringe offset set by pulse conventions. In (a), the first Hadamard gate creates the needed superposition and the final gate reveals full-contrast Ramsey interferometry. In (b), the first Hadamard gate instead puts the state in $|1⟩$, so the intervening phase gate has no relative quantum phase to imprint: the output is independent of $\alpha$. In (c), the unpolarized density matrix remains unchanged under every gate, so the ensemble has no Ramsey interferometry contrast. A coherent unitary gate alone cannot initialize a maximally mixed ensemble into a pure clock state.

The Ramsey interferometry analogy concerns preparation, phase accumulation, and readout; a Hadamard gate is not literally a transverse $\pi /2$ rotation on every input. In this basis $H=R_y(\pi /2){\sigma }_z$, and its Bloch sphere action is a $\pi$ rotation about $(\hat{x}+\hat{z})/\sqrt{2}$, up to a global phase. The phase conventions of physical Ramsey spectroscopy pulses must be specified to compare their outputs for arbitrary initial states.

Use the Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ and $\cos \theta +i\sin \theta =e^{i\theta }$. Their probability amplitudes areThe Born rule givesThe two probabilities sum to one. Their dependence on the relative quantum phase is the quantum interference between the two components in the computational basis.

Use deterministic two-qubit entanglement dilution, retaining the specified complex probability amplitudes. Alice implements a quantum instrument on $A$ with Kraus operatorsThe Kraus representation completeness condition holds even for complex $\alpha ,\beta$:To realize this using the permitted quantum gates and a basic measurement, Alice initializes an ancilla qubit $C$ to $|0⟩$. Conditional on her computational-basis state being $j$, she applies to $C$ the unitary matrixEach has orthonormal columns. Therefore $W=|0⟩⟨0{|}_A\otimes V_0+|1⟩⟨1{|}_A\otimes V_1$ is a unitary gate, withA computational-basis measurement of $C$ implements the desired quantum instrument. Its two unnormalized branches on $AB$ areEach has probability $1/2$. Alice sends her outcome $m$ as one classical bit. For $m=0$, both parties keep their qubits unchanged. For $m=1$, both apply a Pauli X gate, givingThus both outcomes produce the exact target using one classical bit and local operations. The construction includes $\alpha =0$ or $\beta =0$ and arbitrary relative quantum phases; it requires no quantum channel between the parties.

Initialize three qubits in $|000⟩$. Apply a Hadamard gate on the first, then controlled-NOT gates $1\to 2$ and $1\to 3$:Apply the T gate twice to the first qubit. Because $T^2=\mathrm{diag}(1,i)$, the resulting GHZ state with the specified relative quantum phase is

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

For $r$ brackets the control-zero branch still does nothing, and the control-one branch applies $G^r$ to the target. Immediately before the final two Hadamard gates, the quantum state isFor the constant function with $k=0$, the phase oracle is $U_f=-I$, so $G=(-X)(-I)=X$. Since $|+⟩$ is an eigenstate of the Pauli X gate with eigenvalue $+1$, $G^r|+⟩=|+⟩$ for every integer $r$. The final two Hadamard gates therefore giveIn particular, the first qubit always ends in $|0⟩$.

For the constant function with $k=2$, $U_f=I$ and $G=-X$, so $G^r|+⟩=(-1{)}^r|+⟩$. Thus the target remains in $|+⟩$, and the top qubit before its final Hadamard gate is $(|0⟩+(-1{)}^r|1⟩)/\sqrt{2}$. The sign is a relative quantum phase between the control branches. The Hadamard gate converts it into the integer parity of $r$:The first qubit is therefore $|0⟩$ for even $r$ and $|1⟩$ for odd $r$, including all requested $r>1$.