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Amplitude amplification increases the probability of obtaining an answer accepted by a specified Boolean function. It requires a coherent preparation $A$, its inverse, and a marked-state phase oracle that negates accepted components. Alternating that sign change with a reflection operator about the prepared state directs interference toward the accepted subspace. Brassard, Høyer, Mosca, and Tapp developed the general framework, extending Grover's algorithm beyond a uniform quantum superposition. ^[ref].

If a normalized quantum state has good amplitude $\sin \theta$, the product of the reflection in the bad axis and the reflection in the state line rotates its good-bad plane by $2\theta$. After $k$ iterations the good probability isso $O(1/\sin \theta )$ iterations raise a small good amplitude close to one. Success subsequently oscillates, making the stopping time part of the algorithm.

Compute the exact Grover search on four entries directly. Order the computational basis as $00,01,10,11$ and mark $11$. The Hadamard transform prepares amplitudes $(1/2,1/2,1/2,1/2)$. The marked-state phase oracle changes these to $(1/2,1/2,1/2,-1/2)$, whose arithmetic mean is $\overline{a}=1/4$. The Grover diffusion operator performs inversion about the mean:The three unmarked amplitudes become $1/2-1/2=0$, while the marked amplitude becomes $1/2-(-1/2)=1$. All three stages have squared norm one. Here $\sin \theta =1/2$, so $\theta =\pi /6$ and one iteration gives ${\sin }^2(3\theta )=1$, agreeing with the explicit calculation.

Grover's original search paper was submitted on 29 May 1996; its algorithm combines selective sign changes with inversion about the mean. The example exposes the beauty of that construction: a change initially invisible to computational-basis measurement becomes complete cancellation of every wrong answer. ^[ref].

The marked-state phase oracle is an assumed operation, whose implementation cost must be counted. Computational-basis measurement yields a candidate label, here $11$; using it to retrieve a record requires the relevant data access. Preparing a quantum superposition does not make all its entries available as classical outputs. The improvement concerns the number of oracle queries needed to obtain an accepted label.

Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

The $m$-qubit bit-flip repetition code stores one logical qubit in $|0_L⟩=|0{⟩}^{\otimes m}$ and $|1_L⟩=|1{⟩}^{\otimes m}$. Its stabilizer generators are $Z_j{Z}_{j+1}$ for $j=1,\dots ,m-1$, and its logical operators are $\overline{X}=X_1\cdots X_m$ and $\overline{Z}=Z_1$. Interchanging the Pauli X gate and Pauli Z gate gives the phase-flip repetition code. For odd $m$, majority-vote decoding of a repetition code corrects up to $(m-1)/2$ physical Pauli X gate errors, assuming ideal operations and a noise family containing only these bit flips.

For $m=3$, prepare two quantum ancillas in $|00⟩$ and apply controlled-NOT gates from the input qubit to each of them. By linearity, encoding sendsThis does not produce $(\alpha |0⟩+\beta |1⟩{)}^{\otimes 3}$, which would contain cross terms. For example, encoding $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ produces $(|000⟩+|111⟩)/\sqrt{2}$, whereas three independent copies would have eight equal computational basis amplitudes. The encoding creates quantum entanglement when both coefficients are nonzero and respects the no-cloning theorem.

The three-qubit bit-flip syndrome table corrects the family $\{I,X_1,X_2,X_3\}$ and its entire linear span, as specified by the Knill--Laflamme condition. It cannot correct arbitrary errors on one physical qubit: the phase flipcommutes with both checks and acts as logical $\overline{Z}$. In particular, it takes the encoded $|+⟩$ to the orthogonal encoded $|-⟩$ with error syndrome $00$. Thus its full quantum code distance is one, despite correcting one error of the restricted Pauli X gate type. The Shor code, Steane code and five-qubit error correcting code provide protection against arbitrary one-qubit errors.

Suppose a known list of $2^m$ distinct $n$-bit strings $a_j$ is available and computational basis transpositions on $n+m$ qubits have polynomial-size implementations. Start with the uniform superposition state of index labels $j$, leaving the data zero. First swap $|0^n,j⟩$ with $|a_j,j⟩$ for each nonzero $a_j$. Next, for each $j\ne 0$, swap $|a_j,j⟩$ with $|a_j,0^m⟩$. Distinct index labels separate the first-stage pairs, and distinct data strings separate the second-stage pairs. Thus no later transposition disturbs an earlier term. At most $2^{m+1}-1$ transpositions prepare the uniform data state and return the quantum ancilla to zero. For $m=O(\log n)$ the quantum circuit has polynomial size. This assumes the list is available; a membership predicate alone does not supply it.

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

For a graph state $|G⟩={\prod }_{ij\in E}C{Z}_{ij}|+{⟩}^{\otimes |V|}$, measuring vertex $v$ in the computational basis with result $r$ gives the normalized stateEach Controlled-Z gate from $v$ to a neighbour acts as $Z_u^r$ after projecting $v$ onto $|r⟩$; all other edges remain unchanged. The projection contributes $1/\sqrt{2}$, so either result has probability $1/2$. For a four-cycle, deleting one vertex leaves a path and adds byproducts on its two endpoints.

A dephasing channel suppresses off-diagonal entries of a density matrix in a preferred basis while preserving quantum state populations. Complete dephasing is identical to a projective measurement in that basis with its outcome discarded.

For the Bell correlations versus a classical mixture example, completely dephasing Alice's qubit in the computational basis maps ${\rho }_{\mathrm{Bell}}$ to ${\rho }_{\mathrm{mix}}$. It preserves their perfect $Z$ agreement and each reduced density matrix, yet destroys the $X$ correlation. This is why testing only populations can miss damage to a quantum memory: the stored quantum coherence may have disappeared while every preferred-basis measurement still looks correct. The joint Von Neumann entropy rises from zero to one bit, although each local entropy stays fixed.

Expand a quantum circuit matrix element by inserting computational basis resolutions between gates, then enumerate the intermediate strings depth first. The active stack stores one basis label and accumulator per quantum circuit layer, using polynomial space for polynomially many qubits and gates. Recompute entries rather than storing an exponential matrix or state vector. The output probability requires the Born rule sum of squared amplitude moduli.

The $n$-qubit Dicke state with $m$ excitations is the normalized equal superpositionwhere $|x|$ is the Hamming weight in the computational basis. The binomial coefficient counts the mutually orthogonal summands, proving normalization. Every permutation of the qubits leaves the vector unchanged. Splitting the sum according to the first bit and using $\left(\genfrac{}{}{0ex}{}{n-1}{m}\right)/\left(\genfrac{}{}{0ex}{}{n}{m}\right)=(n-m)/n$ givesTerms with zero coefficient are omitted at the endpoints. This recursion is useful for partial traces and for distributing a logical qubit among many symmetric physical qubits.

Entanglement distillation uses local operations and classical communication on many noisy entangled states to produce fewer high-fidelity Bell pairs. It is also called entanglement purification. The protocol may fail or abort; naming a purification step does not guarantee successful output for arbitrary inputs. If the output is an exact pure state of Bell pairs, every purification of a density operator factors between those pairs and the environment, so measuring each pair in the computational basis gives a shared uniform bit independent of an eavesdropper. Approximate privacy follows when the entire retained output is sufficiently close to that pure target. A separable quantum state cannot be distilled into a Bell pair by LOCC, because each local branch preserves separability.

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

Label a four-cycle $0-1-2-3-0$. Measure vertex $0$ in the computational basis with result $r$, then vertex $1$ in the equatorial qubit measurement basis at angle $\alpha$ with result $s$. Graph-state vertex deletion and one-bit teleportation leaveFinal computational basis measurements with raw results $d_2,d_3$ therefore simulate the ideal circuit $E_{23}J(\alpha {)}_2$ on $|++⟩$ after the classical correction $k=d_2\oplus r\oplus s$, $l=d_3$. The output probability distribution is correct in every prior branch; no physical Pauli frame correction is needed.

Let $|s⟩=H^{\otimes n}|0^n⟩$, let $U_f$ be the marked-state phase oracle, and replace reflection about $|0^n⟩$ by reflection about a known computational basis vector $|y⟩$. Define $T_y={⨂}_{j=0}^{n-1}{Z}_j^{y_j}$, so $T_y|x⟩=(-1{)}^{x\cdot y}|x⟩$ and $H^{\otimes n}|y⟩=T_y|s⟩$. The resulting Grover iterate obeysbecause the two diagonal operators $T_y,U_f$ commute. Thus the good and bad uniform vectors are replaced by their signed vectors $T_y|g⟩,T_y|b⟩$, with exactly the same Grover rotation angle. Initializing in $H^{\otimes n}|y⟩$ gives $G_y^r|s_y⟩=T_y{G}^r|s⟩$. Computational-basis measurement probabilities are unchanged because $T_y$ only multiplies amplitudes by signs.

The Hadamard gate is the unitary operatorwhere $I$ is the identity operator. It maps the computational basis to the Hadamard basis:

These outputs have equal computational-basis measurement probabilities, but their relative quantum phase differs. Applying another Hadamard gate sends them to different definite outputs. Calling this gate a classical coin toss would discard precisely the information that makes this possible. ^[ref].

Starting from $|0⟩$, compare two quantum circuits, reading matrix products from right to left:Here $Z$ is the Pauli Z gate. After the first Hadamard gate, the probability amplitudes are $(1/\sqrt{2},1/\sqrt{2})$. Inserting $Z$ changes them to $(1/\sqrt{2},-1/\sqrt{2})$ without changing their squared magnitudes. At the final gate, the contributions to output zero cancel and those to output one add. Without $Z$, cancellation occurs at output one instead. This is controlled interference, with the quantum phase determining which answer emerges.

The same calculation works for any phase gate $P_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$:For example, $\varphi =\pi /3$ gives $P(1)=1/4$, while $\varphi =\pi /2$ gives $P(1)=1/2$. The output-one amplitude at $\varphi =\pi /2$ is $(1-i)/2$: its two complex contributions add before the Born rule is applied. A phase gate alone leaves the populations unchanged; recombination converts its relative quantum phase into observable probabilities.

If quantum decoherence completely removes the computational basis coherence between the gates, the intermediate density matrix becomes $I/2$. Both the phase gate and final Hadamard gate leave $I/2$ unchanged, so $P(1)=1/2$ for every $\varphi$. The disappearing fringe explains why preserving quantum coherence matters: it enables the gate sequence to carry phase information into the final measurement.

The normalized Hadamard transform maps a vector indexed by binary strings through entries $2^{-n/2}(-1{)}^{x\cdot y}$, where the dot product is evaluated modulo two. It is its own inverse and an orthogonal transformation, hence also a unitary operator on complex input vectors. The Hadamard gate is the $n=1$ quantum case, and the Hadamard basis is the image of the computational basis. The Walsh-Hadamard transform gives the same tensor product construction.

For two qubits, applying one Hadamard gate to each prepares the uniform quantum superpositionEach probability amplitude is $1/2$, so each computational-basis measurement outcome has probability $1/4$. Applying the Hadamard transform again reconstructs $|00⟩$: signed contributions cancel at the other three outputs. Its usefulness lies in combining simple local quantum gates into an operation on exponentially many amplitudes. Extracting useful information still requires an appropriate subsequent transformation and measurement.

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.