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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.

Independent successive applications of the phase-flip channel $D_{ϵ}(\rho )=(1-ϵ)\rho +ϵZ\rho Z$ preserve diagonal entries and multiply off-diagonal entries by $(1-2ϵ{)}^n$. Hence $D_{ϵ}^n=D_{{ϵ}_n}$ with the displayed parameter, for nonnegative integers $n$. If $0<ϵ<1/2$, the limit is the completely dephasing channel, equivalently a nonselective projective measurement in the computational basis. A time-interval law can be iterated this way only under the memoryless composition assumption.

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

For $h:{\mathbb{Z}}_N\to {\mathbb{Z}}_M$, its modular-addition oracle is the unitary operatorFor each fixed $x$, addition by $h(x)$ permutes the answer register's computational basis. This preserves inner products, even when $h$ is not injective.

A path graph state simulates a sequence of Clifford gates $J({\alpha }_j)$, ${\alpha }_j\in \{0,\pi /2\}$, using fixed equatorial qubit measurement bases on successive path vertices. Let $s_j$ be raw outcomes, $t_j=2{\alpha }_j/\pi \in \{0,1\}$, and start the Pauli frame at $a_0=b_0=0$. Matrix commutation and $J(-\pi /2)=XJ(\pi /2)$ giveInduction shows the final unmeasured state differs from the ideal circuit output by $X^{a_m}{Z}^{b_m}$, up to global phase. A final computational basis result $d$ is corrected to $d\oplus a_m$. All bases are fixed, and projectors on distinct vertices commute, so all measurements, including the final one, can occur in a single layer. The frame recurrence is classical outcome processing, not quantum feed-forward.

A measurement in quantum mechanics with outcomes $m$ is specified, in the single-operator-per-outcome description, by linear operators $M_m$ satisfying ${\sum }_m{M}_m^{†}{M}_m=I$. Its effects $F_m=M_m^{†}{M}_m$ form a positive operator-valued measure. The two measurement postulates are the Born rule for outcomes and the conditional state update:For a density operator the corresponding formulas are $p_m=\mathrm{Tr}(M_m\rho M_m^{†})$ and ${\rho }_m=M_m\rho M_m^{†}/p_m$. More general quantum instruments have multiple operators for each outcome and sum these terms. For a projective measurement of an observable $A={\sum }_m{a}_m{P}_m$, take $M_m=P_m$; the outcome value is $a_m$ and the postmeasurement state lies in its eigenspace. A POVM alone specifies probabilities, whereas the measurement operators or instrument also specify the backaction.

For quantum teleportation, Alice and Bob initially share the Bell pair $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice also holds the input $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. The shared entanglement is a resource; the two transmitted classical bits alone could not transfer an arbitrary unknown quantum state.

Alice applies a controlled-NOT gate with $C$ as control and $A$ as target, then a Hadamard gate to $C$. Starting from $|\psi {⟩}_C|{\Phi }^{+}{⟩}_{AB}$, direct expansion givesFor example, the four receiver vectors are $\alpha |0⟩+\beta |1⟩$, $\beta |0⟩+\alpha |1⟩$, $\alpha |0⟩-\beta |1⟩$, and $-\beta |0⟩+\alpha |1⟩$. Thus the gate calculation fixes the ordering of the Pauli corrections, including the last branch's sign.

Alice measures $C,A$ in the computational basis, obtaining $(a,b)$. The Born rule gives each outcome probability $1/4$, independent of the input amplitudes. Bob's normalized conditional state is $X^b{Z}^a|\psi ⟩$. Alice sends the two outcome bits. Bob applies the inverse, $Z^a{X}^b$, and henceThe same circuit is a Bell-basis measurement on Alice's two qubits, expressed using ordinary gates followed by computational-basis measurements. It restores the state without Alice or Bob knowing $\alpha ,\beta$.

By linearity, the calculation also holds when the input is entangled with a reference: each outcome has branch map $|\psi ⟩\mapsto X^b{Z}^a|\psi ⟩/2$, and after correction every branch is one-half of the identity map. Summing the four corrected density operators gives exactly the original input density operator, with all reference correlations preserved. This is teleportation as an identity channel on a reference. Without the two classical bits Bob must average the four uncorrected states, giving $\frac{1}{4}{\sum }_{a,b}{X}^b{Z}^a\rho Z^a{X}^b=I/2$, so the process cannot signal before the message arrives. The Bell pair is consumed and Alice retains no copy of the unknown state, in accordance with the no-cloning theorem.

Apply the Born rule to the two amplitudes in the introductory calculation. In the computational basis, the probabilities areThe other basis printed in this question is $(|0⟩\pm i|1⟩)/\sqrt{2}$, the Pauli Y gate eigenbasis. It is not the real Hadamard basis. Write the output amplitudes as $a=(1+e^{i\varphi })/2$ and $b=(1-e^{i\varphi })/2$. The plus-outcome amplitude is $(a-ib)/\sqrt{2}$ and the minus-outcome amplitude is $(a+ib)/\sqrt{2}$. Since $\mathrm{Im}(\overline{a}b)=-\sin \varphi /2$, the Born rule givesIn particular, at $\varphi =\pi /2$ the output is the minus eigenstate, which checks the sign of the imaginary quadrature.

Use separate fresh runs for each measurement basis. If ${\hat{P}}_j$ denotes the empirical outcome frequency, thenThe two differences estimate $\cos \varphi$ and $\sin \varphi$, resolving the ambiguity that would remain from the first basis alone. Each Bernoulli frequency converges by the strong law of large numbers, with standard error of order the inverse square root of the number of runs. One may project the noisy estimate onto the unit circle if a phase estimate is desired, but that projection is not necessary for consistency.

Apply one Hadamard gate independently to each qubit. These gates act in parallel, so the quantum circuit implements $H^{\otimes n}$ with $n$ gates and depth one.

For a computational basis state, factorization givesThe exponent is the dot product over the finite field ${\mathbb{F}}_2$, because multiplying the separate signs adds their exponents modulo two. This is the Walsh-Hadamard transform, equivalently the quantum Fourier transform on the finite Abelian group $({\mathbb{Z}}_2{)}^n$.

In particular, the all-zero input becomes the uniform coherent superposition $2^{-n/2}{\sum }_y|y⟩$. It is a useful initial operation because an oracle can then act on all amplitudes in one call, and subsequent interference can reveal a global property. Measurement alone does not read all function values: the later interference step is what turns the preparation into an algorithmic advantage.

Subtract the two density operators. The difference is Hermitian with zero diagonal and off-diagonal entry $\alpha \overline{\beta }(1-\overline{c})$. A matrix $\left(\begin{array}{cc}0& a\\ \overline{a}& 0\end{array}\right)$ has eigenvalues $\pm |a|$, because its square is $|a{|}^{2}I$. Its two singular values are therefore both $|a|$. By the definition of the trace norm,For the intended real-overlap phase-flip channel, $1-c=2p$, givingThis vanishes for a computational basis state, since a phase flip then changes only an unobservable global phase. Its largest value at fixed $p$ is $p$, attained when $|\alpha |=|\beta |=1/\sqrt{2}$. The general formula remains correct even when the probabilistic premise of part 1 fails.

Let ${\Pi }_g$ project onto the good-output subspace and let $|g⟩=|{\Psi }_g⟩$, $|b⟩=|{\Psi }_b⟩$. They are orthonormal because their supporting computational basis strings are disjoint. The verification unitary operator is $V=I-2{\Pi }_g$, so $V|g⟩=-|g⟩$ and $V|b⟩=|b⟩$.

Also $V_0=I-2|0⟩⟨0|$. Since $A|0⟩=|\Psi ⟩$, conjugating gives $A{V}_0{A}^{†}=I-2|\Psi ⟩⟨\Psi |$. ThusBoth reflections preserve the span of $|g⟩$ and $|b⟩$: $V$ is diagonal there, and the other reflection only adds a multiple of $|\Psi ⟩$, already in that span. Direct substitution of $|\Psi ⟩=\sqrt{p}|g⟩+\sqrt{1-p}|b⟩$ yieldsThe minus sign in the first image follows from applying $V$ before the state reflection. In the ordered basis $(|b⟩,|g⟩)$ the matrix isThis exhibits the invariant plane without assumptions about the action of $A$ on the other input basis vectors.

Compute the honest one-qubit density operators in the computational basis. Writing $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|u⟩=(5|0⟩+|1⟩)/\sqrt{26}$, the two quantum state ensembles giveThe equality includes the off-diagonal coherence; comparing only the basis outcome probabilities would not suffice.

The protocol is perfectly hiding against Bob when Alice is honest. Independence of the honest preparations makes his full received density operator ${\rho }_0^{\otimes N}={\rho }_1^{\otimes N}$. For any POVM $\{E_z\}$, including collective measurements and measurements using ancillas, the Born rule gives the same outcome probabilities $\mathrm{Tr}(E_z{\rho }_0^{\otimes N})=\mathrm{Tr}(E_z{\rho }_1^{\otimes N})$. Thus no measurement supplies information about the committed bit.

It is not binding against Alice. Here is an explicit ensemble-steering attack on quantum bit commitment. For each position, Alice prepares the normalized two-qubit pure stateretains $A$ and sends $B$. Its partial trace on $A$ is the honest common density operator. She sends the $B$ halves of $N$ independent copies while retaining all the $A$ halves, and chooses her bit only at unveiling.

To open one, she makes a projective measurement of each retained qubit in the computational basis. Outcome zero has probability $13/20$ and prepares Bob's state $|u⟩$; outcome one has probability $7/20$ and prepares $|1⟩$. She announces that list. It has exactly the honest bit-one distribution, and every declared-state test passes.

To open zero, she instead measures each retained qubit using a three-outcome POVM with effects $E_j=|v_j⟩⟨v_j|$, whereAll effects are positive. Their diagonal sums are $4/5+1/5=1$ and $4/70+16/70+5/7=1$, and the off-diagonal sum is zero, so $E_0+E_{+}+E_1=I$. This establishes that the proposed POVM is physically valid.

To verify the conditional states directly, the coefficient matrix of $|\Omega ⟩$, with Bob's index as the row and Alice's as the column, isA real rank-one effect $|v_j⟩⟨v_j|$ on Alice's half gives the unnormalized Bob state vector $C{v}_j$. For the three effects above,Their squared norms are therefore $1/2,1/4,1/4$, exactly the honest bit-zero probabilities, and Bob's conditional pure states are the corresponding declared states. Independent measurements on the retained halves produce the honest distribution of the entire list, so even checks on the declared frequencies do not defeat the attack. Every declared-state projector test again succeeds.

Alice can consequently unveil either bit with acceptance probability one, without having chosen it when sending the qubits. This is the Hughston–Jozsa–Wootters theorem mechanism implemented by an explicit measurement, rather than merely an appeal to the general impossibility of perfectly hiding and binding quantum bit commitment.

A central application is quantum order finding, which supplies the quantum step in the Shor algorithm for factoring an integer $M$. Choose an integer $a$ coprime to $M$, and let $r$ be its multiplicative order modulo $M$. On valid residue quantum states define the reversible modular-multiplication unitary operator $U_a|x⟩=|ax\mathrm{mod}M⟩$, extending it reversibly to any unused computational basis quantum states.

The orbit $|1⟩,|a⟩,\dots ,|a^{r-1}⟩$ has the orthonormal eigenstatesfor $s=0,\dots ,r-1$. One need not know $r$ to prepare an eigenstate: the easy input $|1⟩=r^{-1/2}{\sum }_s|u_s⟩$ makes quantum phase estimation sample one of these eigenphases. Its control register estimates $s/r$. With $N=2^t$ chosen at least of order $M^2$, the nearest quantum phase label has rational approximation error small enough for continued-fraction recovery in quantum order finding. For instance, $N>2M^2$ gives $|m/N-s/r|<1/(2r^2)$ on the nearest-label event. If $s$ and $r$ are coprime, the recovered denominator is $r$; otherwise one obtains a divisor and can repeat, combine candidates and verify them by modular exponentiation.

When the verified order is even and $a^{r/2}\not\equiv -1(\mathrm{mod}M)$, the greatest common divisorsgive nontrivial factors: their product is divisible by $M$, while neither factor inside the gcd is zero modulo $M$. Unsuccessful choices of $a$ or quantum phase samples are handled by repetition and classical verification. Controlled powers of $U_a$ are implemented by efficient reversible modular arithmetic, rather than by naively applying $U_a$ exponentially many times.

Quantum phase estimation turns modular-multiplication eigenphases into an efficiently usable period, enabling quantum factoring in time polynomial in the bit length of the integer. Factoring is important to public-key cryptography, and this is a major example of a quantum algorithm improving on known classical methods. Another use is extracting energy eigenvalues by applying quantum phase estimation to a controlled time-evolution operator, with care to avoid energy aliasing and to prepare an input having overlap with the desired eigenstate.

Number the physical wires $1,\dots ,5$ from top to bottom and write a computational basis ket in that order. The first four are ancilla qubits and the fifth carries the input. It is enough to follow the basis input $|b⟩$, $b\in \{0,1\}$, because the encoding quantum circuit acts linearly.

The first fan-out consists of four controlled-NOT gates from wire 5 to wires 1–4. It sends $|0000b⟩$ to $|bbbbb⟩$. The three Hadamard gates then giveThe two Pauli X gates on wires 2 and 3, followed by a Controlled-Z gate and the same two X gates, restore the bits but contribute $(-1{)}^{(1-c)(1-d)}$: the sign is negative exactly when the original bits on both wires were zero. Finally, the right-hand controlled-NOT gates are $3\to 4$, $2\to 5$, $1\to 4$ and $1\to 5$. They leave the first three bits unchanged and turn the last two into $b\oplus a\oplus d$ and $b\oplus a\oplus c$. ThusExplicitly, the logical basis quantum states produced by the printed gates areEach has eight orthogonal terms of equal magnitude, and their supports are disjoint, so they are normalized and mutually orthogonal. The arbitrary input is therefore encoded asThis is a realization of the five-qubit error correcting code. The input amplitudes and their relative quantum phase are preserved in the logical subspace; the encoding is an entangled-state linear isometry, not five independent copies of the unknown input. The explicit signs and bit ordering above are fixed by this circuit, rather than by a convention for a different encoder.

The Pauli X gate interchanges the two basis quantum states. Because the bit-flip outcome is a classical unobserved error rather than a coherent superposition, the quantum bit-flip channel givesFor $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its explicit density matrix isFor a pure reference quantum state, the square of the paper's quantum fidelity is $F^2=⟨\psi |{\rho }_2|\psi ⟩$. ThereforeWith $|\psi ⟩=\cos (\theta /2)|0⟩+e^{i\varphi }\sin (\theta /2)|1⟩$, the expectation is $⟨X⟩=\sin \theta \cos \varphi$, the $x$ component of the Bloch vector. Hence“All possible quantum states” is interpreted as the rotationally invariant uniform pure-qubit average, with measure $d\mu =\sin \theta d\theta d\varphi /(4\pi )$. Thusso the average pure-qubit fidelity of a bit-flip channel isEquivalently, rotational symmetry gives $⟨r_x^2⟩=⟨r_y^2⟩=⟨r_z^2⟩=1/3$, since a pure-state Bloch vector has unit length. The $X$ eigenstates are unchanged as density matrices and have fidelity one, while a computational basis quantum state has $F^2=1-p$; these provide checks on the angular expression. A uniform distribution in $\theta$ rather than in solid angle would produce the wrong ensemble average. The requested quantity is the mean squared fidelity, not the square of the mean unsquared fidelity.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

Let $e_j$ be the bit string with one in position $j$ and zero elsewhere, and use the standard convention $Y=iXZ$. On the computational basis, the single-site Pauli operators act asThe identity leaves the basis state unchanged. A tensor-product Pauli operator applies these actions at each occupied site; operators at different sites commute. Their linear span is the full operator algebra, so correcting their span handles coherent combinations as well as random Pauli errors.

Let $P$ project onto the code $\mathcal{C}$, with orthonormal logical basis $|i_L⟩$. A set $\mathcal{E}=\{E_a\}$ is correctable exactly when the Knill--Laflamme condition holds:for a matrix $C$ independent of the logical indices. A set $\mathcal{D}$ is detectable whenThese conditions extend by linearity to the corresponding operator spans. Detection allows an error to act as a harmless scalar on the code; it need not assign a distinct syndrome to such an error.

For distinct Pauli representatives modulo phase, a nondegenerate quantum error-correcting code has orthogonal error-image subspaces, giving $C_{ab}={\delta }_{ab}$ because each Pauli is unitary. In the usual nondegenerate, or pure, detection convention, every nonidentity detectable Pauli has $PDP=0$, while $PIP=P$. More generally a nonsingular error-overlap matrix can be diagonalized and the resulting error operators rescaled to obtain orthogonal syndrome spaces; a singular overlap matrix expresses degeneracy of the specified independent error family.

For an $[[n,k,d]]$ code, the quantum code distance is the smallest Pauli weight for which scalar compression fails. Hence every Pauli of weight at most $d-1$ is detectable. If $E_a,E_b$ each have weight at most $t$, their product $E_a^{†}{E}_b$ has weight at most $2t$. When $2t<d$, its compression is scalar, so the Knill--Laflamme condition proves simultaneous correction of all such errors. ThereforeThe integer part is understood when the printed half-distance is not an integer. Arbitrary operators on such subsets are covered by their Pauli expansion.

For errors confined to a known set of $m$ positions, every pairwise product is still supported on that same set and has weight at most $m$, rather than $2m$. Thus $m<d$ supplies quantum erasure correction, with maximum guaranteed numberThis guarantee cannot hold for every set of $d$ positions: a minimum-weight undetectable Pauli has support in such a set, and the pair consisting of it and the identity violates the Knill--Laflamme condition. Particular larger location sets can sometimes be correctable; $d-1$ is the maximum uniform guarantee over all possible known sets.

For the Shor code, write $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. The logical codewords are $|G_{+}{⟩}^{\otimes 3}$ and $|G_{-}{⟩}^{\otimes 3}$. A phase flip $Z_j$ in one block interchanges $G_{+}$ and $G_{-}$ in that block, independently of its position within the block. Let $B_1=X_1{X}_2{X}_3$, $B_2=X_4{X}_5{X}_6$, $B_3=X_7{X}_8{X}_9$. Measure the two commuting stabilizer generators $S_1=B_1{B}_2$ and $S_2=B_2{B}_3$. Both are $+1$ on the entire logical code space. The Shor-code phase-flip syndrome isThese parity measurements reveal the erroneous block without revealing the logical amplitudes, so they preserve arbitrary superpositions of the two codewords. Applying the listed representative phase flip restores the state.

For any two physical positions $i,j$ in the same block, $Z_i{Z}_j$ acts as $+1$ on both $|000⟩$ and $|111⟩$, and hence on the code. Therefore $Z_{i}P=Z_{j}P$ and $P{Z}_i^{†}{Z}_{j}P=P$ even when $i\ne j$. Their error spaces coincide, and the error-overlap matrix has a nonzero off-diagonal entry. The correction is degenerate because different within-block phase errors have exactly the same action on every logical state.

For the normalized vector defining the orthogonal projection $P=|\varphi ⟩⟨\varphi |$, $P^k=P$ for every $k\ge 1$. Thuswhich proves the projector phase rotation identity in this case:It changes the sign of the component along $\varphi$ and fixes the orthogonal complement, so it is a reflection operator.

For a marked computational basis vector $w$, $U_{P_w}=I-2|w⟩⟨w|$ is the marked-state phase oracle. With $|\psi ⟩=H^{\otimes N}|0^N⟩$, the other reflection isThe Hadamard gates thus implement its change of basis from the known all-zero state. The usual Grover diffusion operator is $D=2|\psi ⟩⟨\psi |-I=-U_{P_{\psi }}$. Consequently a Grover search algorithm iteration isThe minus sign is an irrelevant global phase. These two reflections rotate the search state in the span of the marked vector and the unmarked uniform vector, with the Grover rotation angle controlled by the initial marked amplitude. Iterating them approximately $\pi \sqrt{2^N}/4$ times gives high-probability search. The same pair of projectors appears additively in the continuous-time Hamiltonian, but the product of full reflections and evolution under their sum are distinct dynamics.