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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.

Keep the actual odd-$n$ GHZ state from the preceding calculation. Resolving its first qubit in the Hadamard basis givesThus the $+$ outcome leaves $|{\psi }^{-}{⟩}_{23}$ and the $-$ outcome leaves $|{\psi }^{+}{⟩}_{23}$, each with probability $1/2$. Conditional on a known result the remaining qubits share a maximally entangled state.

When the result is not revealed, the appropriate density matrix is the probability-weighted mixture from a nonselective quantum measurement:The cross terms cancel. The displayed convex combination of product states proves directly that this is a separable quantum state. It still has perfect computational-basis measurement anticorrelation, so it is not itself a product state. It also equals the partial trace ${\mathrm{Tr}}_1|G⟩⟨G|$, as required by the no-communication theorem: a local quantum measurement with an unobserved outcome cannot change the remote reduced density matrix.

Assume $|\alpha {|}^2+|\beta {|}^2=1$. Let Alice's first recorded quantum measurement outcome be $a\in \{0,1\}$. Immediately afterward she assigns the conditional pure quantum state $|a⟩$, with probabilities $p_0=|\alpha {|}^2$, $p_1=|\beta {|}^2$. Bob, who has no record, assigns the density operatorThis is a nonselective quantum measurement, which removes the initial computational basis coherence. It is not generally the original pure quantum state.

After the Hadamard gate, Alice assigns $H|a⟩=|+⟩$ for $a=0$, or $|-⟩$ for $a=1$. Bob assignsBoth $|+⟩$ and $|-⟩$ give either final computational basis outcome with probability $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b⟩$, whereas Bob assignsAlice's conditional quantum states are pure; Bob's are averages over her unobserved records. They agree on all probabilities when supplied the same information. Their first descriptions can coincide if the initial quantum state is a computational basis state; after the second quantum measurement, Bob's maximally mixed state differs from each conditional pure quantum state.

The Kraus operators of the generalized measurement postulate must obeyFor a normalized pure state, the Born rule gives the outcome probability and the normalized conditional state:Here the dagger denotes the Hermitian conjugate. Completeness ensures ${\sum }_i{p}_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^{†}{A}_i$ form a POVM, but the actual operators $A_i$ also specify the post-measurement states.

Let ${\rho }_{12}=|\Psi ⟩⟨\Psi |$ be the initial joint density operator, and ${\rho }_2={\mathrm{Tr}}_1{\rho }_{12}$ its reduced density matrix on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I){\rho }_{12}(A_i^{†}\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant nonselective quantum measurement state is therefore the probability-weighted averageTo compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the trace and the fact that operators on different tensor factors commute giveBy the defining property of the partial trace, this says $\mathrm{Tr}(B{\rho }_2^{\prime })=\mathrm{Tr}(B{\rho }_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different density matrices, soThis proves remote state invariance under a local trace-preserving operation directly from completeness.

In particular, if the accessible experiment has POVM elements $F_k$, its probabilities are $\mathrm{Tr}(F_k{\rho }_2^{\prime })=\mathrm{Tr}(F_k{\rho }_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input density operator and therefore give identical outcome distributions. No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred. This is a derivation of the no-communication theorem, not an appeal to a signalling postulate.

The qualification is that the outcome is not communicated. A selective quantum measurement can give different conditional remote states, as happens when measuring one half of a Bell state; their probability-weighted average is still ${\rho }_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.

Include Alice's detector and environment before taking a partial trace. Ideal measurement produces the global stateOrthogonality of the two record states removes their cross terms from the reduced state of the particles:This nonselective quantum measurement has changed the initially entangled singlet into a separable, classically anticorrelated reduced state. Its entropy is one bit, although the full particles-plus-records state is still pure under the assumed unitary dynamics. The reduced density matrix is an improper mixture; tracing the environment does not by itself assert a literal collapse of that global state.

The Born rule gives Alice outcomes zero and one with equal probability. Conditional on a recorded result, the selective quantum measurement assigns $|01⟩$ or $|10⟩$ to the particles. The unitary measurement coupling and orthogonal records explain decoherence in this basis; they do not supply an additional rule selecting a unique outcome. That is part of the measurement problem. In a collapse postulate the individual outcome is stochastic; other interpretations account for outcome experience differently while retaining these operational probabilities.

When Bob makes the same kind of measurement, the global state becomesTracing both laboratories gives the same reduced matrix, ${\rho }_{AB}^{(2)}={\rho }_{AB}^{(1)}$. Bob has added a correlated record, not restored the lost reduced coherence. If Alice's outcome is known, Bob's opposite outcome is certain in this ideal common basis. Without that record, his marginal is $I/2$, so his outcome remains equiprobable. Alice cannot choose her random result or use this correlation for faster-than-light signalling, consistently with quantum no-signalling.

These common-basis measurements alone do not distinguish quantum predictions from a local source sending opposite preassigned bits. To test that explanation, prepare fresh singlets and let the observers independently choose among incompatible measurement axes before recording each result. For singlet spin measurements along unit vectors $\mathbf{a},\mathbf{b}$, quantum mechanics predicts $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. The CHSH inequality bounds $|S|\le 2$ for Bell-local models with measurement-independent settings. Choosinggives $S=E(\mathbf{a},\mathbf{b})+E(\mathbf{a},{\mathbf{b}}^{\prime })+E({\mathbf{a}}^{\prime },\mathbf{b})-E({\mathbf{a}}^{\prime },{\mathbf{b}}^{\prime })=-2\sqrt{2}$. A suitable Bell test constrains that class of explanations. Independently chosen settings, control of detection bias and separation of the measurement events matter when interpreting an actual experiment.

If Alice first performs the fixed-axis irreversible measurement described above, the resulting mixed state has $E(\mathbf{a},\mathbf{b})=-a_z{b}_z$ and cannot supply that singlet violation. The incompatible settings must belong to the Bell measurement itself. Moreover, Bell violations do not choose among all interpretations of quantum mechanics: interpretations sharing the Born predictions remain observationally equivalent for this experiment. Objective-collapse modifications could instead require additional interference tests with specified departures from unitary dynamics.

An explicit original ideal meter model explains how photon number survives while field quantum coherence changes. Let $\hat{N}$ be the number operator and prepare an atomic qubit in $|+⟩=(|e⟩+|g⟩)/\sqrt{2}$. The dispersive unitary operatorcommutes with $\hat{N}\otimes I$. For a field quantum superposition ${\sum }_n{c}_n|n⟩$, it produces ${\sum }_n{c}_n|n⟩|a_n⟩$, where $|a_n⟩=(e^{-in\chi /2}|e⟩+e^{in\chi /2}|g⟩)/\sqrt{2}$. The atomic meter states encode $n$ through relative quantum phase and can become entangled with the field. The second Ramsey interferometry pulse and detector turn those correlations into the outcome probabilities above when $C=1$.

Tracing out the atom multiplies a field density matrix entry ${\rho }_{nm}$ by $⟨a_m|a_n⟩=\cos ((n-m)\chi /2)$. Hence the nonselective quantum measurement preserves every photon number population ${\rho }_{nn}$ while changing off-diagonal quantum coherence. For $\chi =\pi /2$, coherence between $|0⟩$ and $|1⟩$ is multiplied by $1/\sqrt{2}$ per fresh meter, even with no photons absorbed. Field quantum phase information can therefore be disturbed by an energy-preserving probe. No universal self-adjoint phase operator or naive canonical number–phase commutator is needed for this conclusion.

One possible finite-contrast quantum instrument makes the distinction sharper. Define diagonal Kraus operators $K_e=\cos ((\varphi +\chi \hat{N})/2)$ and $K_g=\sin ((\varphi +\chi \hat{N})/2)$. After their ideal readout, flip the reported label with probability $r=(1-C)/2$. The unnormalized selective maps areTheir traces reproduce $L_e,L_g$ on a Fock state, and their sum reproduces the displayed dephasing quantum channel, independently of $r$. At $C=0$ the recorded labels carry no information although the physical meter interaction can still cause measurement backaction. This is an illustrative instrument, not a reconstruction of all laboratory imperfections; equal likelihoods alone do not determine the disturbance.