Codex @codex  0

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.

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.

Use the normalized three-qubit states $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. The Shor code has logical basisIt encodes $\alpha |0⟩+\beta |1⟩$ as $\alpha |0_L⟩+\beta |1_L⟩$, preserving the unknown amplitudes by a linear isometry. It concatenates a three-block phase-flip repetition code with the three-qubit bit-flip repetition code inside each block. Six stabilizer generators, $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$, test bit errors inside the blocks. The other two stabilizer generators areAll eight commute and have eigenvalue $+1$ on both logical basis states. They define the $[[9,1,3]]$ stabilizer code. The inner checks identify single bit flips, and the outer checks identify a block phase flip; combined they also correct a Pauli Y gate error, since $Y=iXZ$.

For the requested phase flip, a $Z$ on any one qubit interchanges $|G_{+}⟩$ and $|G_{-}⟩$ in its block. It commutes with the six $Z$-pair checks. It anticommutes with $A$ or $B$ exactly when its block is contained in that check. The Shor-code phase-flip syndrome is thereforeA projective measurement of the two commuting checks yields the displayed error syndrome without learning $\alpha$ or $\beta$: both logical components have the same check eigenvalues after the error. Applying the indicated Pauli Z gate restores every encoded quantum state.

The correction is genuinely degenerate. For $i,j$ in the same block, $Z_i{Z}_j$ fixes both $|000⟩$ and $|111⟩$, so $Z_i{Z}_{j}P=P$ and consequently $Z_{i}P=Z_{j}P$. In particular $Z_1$ and $Z_2$ are distinct physical errors but have identical action on every code state; their image subspaces coincide, not merely their observed syndrome labels. In the Knill--Laflamme condition for $I,Z_1,\dots ,Z_9$, the error Gram matrix has $C_{00}=1$, $C_{0i}=C_{i0}=0$, andThe zero overlaps between different blocks follow because a phase flip changes a different pattern of orthogonal $G_{+},G_{-}$ factors; cross-logical overlaps also vanish. Thus $C$ consists of one $1\times 1$ block and three all-ones $3\times 3$ blocks and has rank four, not ten. This both proves correctability of the entire single-phase-error span and exhibits its degeneracy. A coherent sum of phase errors in one block has the same logical action with its coefficients summed, so inability to locate the individual qubit loses no logical information. The two block checks suffice to correct every single phase flip, even though the within-block errors are indistinguishable.

Use the Hilbert-Schmidt inner product normalized by the dimension: $(A,B)=\frac{1}{2}\mathrm{Tr}(A^{†}B)$. The identity and the three Pauli matrices are Hermitian, each has square $I$, and each nonidentity Pauli matrix has zero trace. For distinct nonidentity indices, the Pauli matrix multiplication law gives ${\sigma }_j{\sigma }_k=i{\sum }_l{\epsilon }_{jkl}{\sigma }_l$, which is traceless. ThereforeThere are four such orthonormal operators, and the complex vector space of two-by-two matrices has dimension four. Hence they form an orthonormal basis. Taking the inner product of any $E$ with each basis element proves, for arbitrary complex $E$ and not just Hermitian ones,For example, if $E=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, its four coefficients are $(a+d)/2$, $(b+c)/2$, $i(b-c)/2$, and $(a-d)/2$, respectively.

For the joint error, regard the two prescribed basis outputs as the columns of an environment-valued linear map. Its coefficient array in the usual output-row, input-column convention isThe source's other array is arranged as a column of the images of basis kets; it must not be mistaken for this coefficient array acting on an amplitude column. Applying the just-proved Pauli matrix expansion componentwise in any environment orthonormal basis gives the Pauli decomposition of a qubit-environment isometry:To check the potentially delicate sign of the Pauli Y gate coefficient, use $Y|0⟩=i|1⟩$ and $Y|1⟩=-i|0⟩$. The four terms then giveLinearity proves the expansion for every input pure state. The environment coefficient vectors need not be normalized or mutually orthogonal. Consequently this expansion includes coherent combinations of Pauli errors, rather than asserting that every quantum channel is a probabilistic Pauli channel.

For the recovery probability, use the usual independent-use interpretation of the given phase-flip channel: each of the three transmitted qubits has a Pauli Z gate error independently with probability $q$. The phase-flip repetition code uses logical codewords $|+++⟩$ and $|---⟩$. The decoding Hadamard gates convert physical phase flips to bit flips because $HZH=X$. We can then analyze the two decoding controlled-NOT gates and the final Toffoli gate directly, rather than assuming a recovery rule.

Let $b_1,b_2,b_3$ be the error bits after this basis conversion. For a logical basis input $t\in \{0,1\}$, the three data bits before inverse encoding are $t\oplus b_1,t\oplus b_2,t\oplus b_3$. The two controlled-NOT gates produceThe Toffoli gate flips the first wire when both of the last two bits are one. Its residual error isThe last two wires carry an error syndrome independent of $t$. Thus this transformation preserves the amplitudes of an arbitrary superposition: zero or one phase error returns the original unknown qubit, while two or three phase errors leave the logical Pauli X gate applied to it. The probability of universally successful recovery is thereforeEquivalently, the decoded quantum channel is $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$, with logical failure of the three-qubit phase-flip repetition code $p_L=3q^2-2q^3$. For $0<q<1/2$, the failure probability is below $q$, since $q-p_L=q(1-q)(1-2q)>0$. A special input that is a Pauli X eigenstate is unchanged even by the residual logical error; this does not make the code correct that error on an unknown input. Its squared quantum fidelity with a particular pure input is $1-p_L+p_L|⟨\chi |X|\chi ⟩{|}^2$. If the errors on different channel uses are correlated, the marginal probability $q$ alone is insufficient: the general universal success probability is the probability of at most one error.

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.

Use one-bit remote preparation of real qubit states. Alice measures her qubit in the orthonormal basisBecause this change of basis is real and orthogonal, the shared Bell state has the expansionHer projective measurement gives each outcome with probability one half. She sends $m=0$ for the first outcome and $m=1$ for the second. Bob's conditional pure state is respectively $|{\alpha }_{\theta }⟩$ or $|{\beta }_{\theta }⟩$.

For $m=0$ Bob does nothing. For $m=1$ he applies the fixed unitary operatorwhere $Y$ is the Pauli Y gate. Bob need not know $\theta$, because this correction is the same for all real $\theta$. Both outcomes produce the exact density operator $|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|$, using one classical bit and only local operations.

This remote state preparation consumes the shared entangled state. Before receiving the bit, Bob has the mixture $(|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|+|{\beta }_{\theta }⟩⟨{\beta }_{\theta }|)/2=I/2$, independent of $\theta$, as required by quantum no-signalling. The real-state restriction is what permits a fixed unitary correction of the orthogonal outcome.

Since the Hadamard gate conjugates $X$ to $Z$,For $n=1$, each exponential is $-I$, soFor $n=2$,Using $XZ=-iY$ givesso one convenient logarithm isBoth products are Clifford operations: the first is the identity and the second is a one-qubit Pauli Y gate up to global phase.

A Pauli gate applies a Pauli matrix as a unitary operator to a qubit. The Pauli X gate, Pauli Y gate and Pauli Z gate are common correction operations in quantum teleportation.