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Measurement backaction is the effect of a measurement in quantum mechanics on later predictions for the measured system. A quantum instrument with one Kraus operator $K_m$ per recorded outcome hasThe conditional density operator ${\rho }_m$ exists only when $p_m>0$. Its nonlinear normalization expresses conditioning; the unnormalized map $\rho \mapsto K_m\rho K_m^{†}$ is linear. More generally, an unresolved outcome can contain several Kraus operators: replace the numerator by ${\sum }_a{K}_{ma}\rho K_{ma}^{†}$ and take its trace for $p_m$. ^[ref].

Keeping the outcome gives a selective quantum measurement. Discarding it gives a nonselective quantum measurement with quantum channel $\rho \mapsto {\sum }_m{p}_m{\rho }_m$. The latter describes the whole measured population without its labels, whereas each ${\rho }_m$ describes a selected subpopulation. Neither description requires the measured quantity to have possessed its eventual value before the interaction.

For a concrete test, prepare the Pauli X eigenstate $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. Its $X=+1$ probability is one. An ideal $Z$ projective measurement gives zero or one with equal probability, leaving $|0⟩$ or $|1⟩$ under the Lüders rule. Discarding the result leaves the maximally mixed state $I/2$. A subsequent $X$ measurement now gives either sign with probability one half, even if the $Z$ result was kept. This observable change in statistics is backaction. Measuring $X$ on the original $|+⟩$ instead leaves it unchanged: disturbance depends on both the preparation and the instrument. Measurement-based quantum feedback uses the recorded outcome to choose a subsequent control operation, so both parts of the instrument matter.

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.

Initialize one qubit in $|0⟩$, then apply a Hadamard gate, a T gate, and a Hadamard gate, in that temporal order. Defining the Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, the successive quantum states are

Let $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ be the Pauli X eigenstates. EncodeBoth have zero total $X$ eigenvalue, so $R_x(\varphi )\otimes R_x(\varphi )$ is the identity on their span. This is the decoherence-free subspace obtained from the previous code by applying a Hadamard gate to each physical qubit.

Since $Z|+⟩=|-⟩$ and $Z|-⟩=|+⟩$, a correlated $Z\otimes Z$ error exchanges the codewords. It acts as logical $X_L$, not as harmless identity. The Pauli Y gate obeys $Y|+⟩=-i|-⟩$ and $Y|-⟩=i|+⟩$, soThe phases cancel within each codeword. Thus either unknown correlated gate gives a logical bit error on a general encoded state, though a logical $X_L$ eigenstate is unchanged as a physical ray. The code prevents collective $x$ rotations, not every possible correlated error.

The quantum bit-flip channel has Kraus operatorsThusThe orthonormal diagonal linear polarizations $|D⟩=(|H⟩+|V⟩)/\sqrt{2}$ and $|A⟩=(|H⟩-|V⟩)/\sqrt{2}$ are the Pauli X eigenstates. Each is transmitted perfectly as a density operator for every $p$; the possible minus sign on $|A⟩$ is global. Encode a classical zero and one as $D$ and $A$, then measure in that basis at the receiver. This sends one classical bit per transmitted qubit without error. General superpositions of $D$ and $A$ need not be preserved, because their relative phase is randomized.

Use the same protocol: distribute one resource photon through $\mathcal{C}$, and send Alice's two classical bits error-free in the $D,A$ alphabet from part a. At $p=1/2$ the distributed resource isFor the second component Bob has an extra $X$ before his conditional correction. Conjugation by that correction changes $X$ at most by a sign, which cancels in the density operator. This bit-flipped Bell resource teleportation therefore induces the bit-flip map,Using the squared pure-target quantum fidelity convention,The maximum is $1$ precisely for the Pauli X eigenstates $D,A$. The minimum is $1/2$ whenever $\mathrm{Re}({\alpha }^{\ast }\beta )=0$, the great circle with zero $x$ component of the Bloch vector, including $H,V$ and both circular polarizations. Uniform averaging over pure qubits gives $\overline{F}=2/3$.

The resource-distribution and classical-encoding assumptions are essential. If a perfect shared pair had already been supplied independently and this channel were used only for the robustly encoded classical bits, teleportation would remain perfect even at $p=1/2$. Conversely, unprotected $H,V$ transmission of Alice's correction bits would add classical errors. The stated expression applies to the resource sent through the given channel and the error-free classical encoding explicitly available in part a.

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.

A projective measurement has self-adjoint orthogonal projections $P_i$ satisfying $P_i{P}_j={\delta }_{ij}{P}_i$ and ${\sum }_i{P}_i=I$. Measuring an observable with discrete spectral decomposition $A={\sum }_i{a}_i{P}_i$ reports eigenvalue $a_i$ with Born rule probability $p_i=\mathrm{Tr}(P_i\rho )$. A projector can have rank greater than one: the outcome then identifies an eigenspace, rather than a particular eigenvector inside it. Projectors specify the outcome statistics; the usual ideal conditional update is the additional Lüders rule.

In the qubit example from measurement in quantum mechanics, $Z=|0⟩⟨0|-|1⟩⟨1|$. At $\theta =\pi /3$, the zero and one probabilities are $3/4$ and $1/4$, regardless of the relative quantum phase $\varphi$. Under the Lüders rule, obtaining zero prepares $|0⟩$; obtaining one prepares $|1⟩$, up to global phase. Repeating the same measurement returns the same label with certainty, provided no intervening evolution or operation changes the measured sector. Repeatability concerns the conditional output, and does not mean the first outcome was predetermined.

Change to $X$ and its Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Adding the two probability amplitudes before squaring givesAt $\theta =\pi /3$ and $\varphi =0$, $p_{+}=(2+\sqrt{3})/4\simeq 0.9330$ and $p_{-}\simeq 0.0670$. At $\varphi =\pi /2$, both are $1/2$; at $\varphi =\pi$, the first pair is exchanged. The $Z$ statistics remain $3/4,1/4$ throughout. This is the same addition-and-cancellation principle as wave interference: relative quantum phase changes how amplitudes reinforce or cancel in another basis.

The beauty of this basis dependence is that information invisible to one measurement becomes accessible to another. A $Z$ measurement of an ensemble estimates its quantum state populations, while $X$ probes one component of its quantum coherence. These are experiments on separate identically prepared copies. Sequentially measuring $Z$ and then $X$ on one copy instead probes the disturbed post-measurement state. Quantum state tomography combines suitable ensemble statistics to reconstruct the original state.