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For a pure state with two descending squared Schmidt coefficients ${\lambda }_0\ge {\lambda }_1$, this quantum instrument acts in one party's Schmidt basis. The first outcome balances the two Schmidt amplitudes when $\tan \theta =\sqrt{{\lambda }_1/{\lambda }_0}$, with probability $2{\lambda }_0{\lambda }_1$. This is not the optimal unrestricted success probability $2{\lambda }_1$ in single-copy entanglement concentration. An embedding in a larger subsystem requires a POVM on the occupied support or a full-space completion.

A generalized quantum measurement can be described by Kraus operators $M_{i\alpha }$, where $i$ labels the recorded outcome and $\alpha$ labels alternatives unresolved by that record. Completeness requiresFor an input density operator $\rho$,when $p_i>0$. The one-operator-per-outcome form is the special case with one $\alpha$ for each $i$.

The POVM effect $E_i={\sum }_{\alpha }{M}_{i\alpha }^{†}{M}_{i\alpha }$ determines the outcome probability, while the whole outcome map determines the post-measurement state. Keeping only $i$ can mix states that a refined record $(i,\alpha )$ would distinguish. Forgetting every outcome gives the trace-preserving quantum channel $\rho \mapsto {\sum }_{i,\alpha }{M}_{i\alpha }\rho M_{i\alpha }^{†}$. This is the distinction between a positive operator-valued measure and a quantum instrument. ^[ref]

For example, two unresolved operators $M_{00}=|0⟩⟨0|$ and $M_{01}=|0⟩⟨1|$ give the sole recorded outcome with certainty and reset every input qubit to $|0⟩$. A single operator with effect $I$ would instead act as an linear isometry; it cannot describe that reset on the same two-dimensional output.

The Lüders rule specifies the ideal quantum instrument for a projective measurement. Conditional on an outcome with nonzero probability, it givesfor a density operator and a pure state, respectively. The second formula applies only to a pure state input. If the result is discarded, the output is the nonselective projective measurement ${\sum }_i{P}_i\rho P_i$.

The crucial feature for a degenerate eigenvalue is preservation of quantum coherence within its eigenspace. For a qutrit with orthonormal basis $|0⟩,|1⟩,|2⟩$, considerOn $|\chi ⟩=(|0⟩+|1⟩)/\sqrt{2}$, outcome zero has probability one and the Lüders rule leaves $|\chi ⟩$ unchanged. Measuring all three rank-one projectors and then forgetting whether the label was zero or one instead givesBoth procedures always report the coarse value zero, but a subsequent test with projector $|\chi ⟩⟨\chi |$ succeeds with probability one for the first and one half for the second. Resolving extra information destroyed coherence. This explains why a quantum nondemolition measurement of a degenerate quantity must specify what information the apparatus actually records.

Gerhart Lüders proposed this state-change rule in 1951, explicitly discussing degenerate quantities and distinguishing selected subensembles from the aggregate measured ensemble. Kirkpatrick's translation was first posted in 2004; equations (7) and (8) give the unnormalized selective update and the nonselective update. ^[ref].

Repeatability follows briefly from $P_j{P}_i={\delta }_{ij}{P}_i$: after outcome $i$, $\mathrm{Tr}(P_j{\rho }_i)={\delta }_{ij}$. This assumes a second application of the same ideal instrument without intervening dynamics. An instrument with the same effects can rotate or reset the output and need not satisfy this conclusion; see POVM does not determine the post-measurement state.

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.

If a convex roof extension is non-increasing on average for every pure input to a quantum instrument, then it is non-increasing on average for mixed inputs too. For an input ensemble $\{p_j,{\pi }_j\}$, let $q_{k|j}$ and ${\sigma }_{jk}$ be its conditional outcome probabilities and states. Linearity of each unnormalized outcome map gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. Convexity gives ${\sum }_k{P}_{k}F({\rho }_k)\le {\sum }_{j,k}{p}_j{q}_{k|j}F({\sigma }_{jk})\le {\sum }_j{p}_j\mu ({\pi }_j)$. Taking the infimum over input ensembles proves the result, even when conditional outputs of pure inputs are mixed, provided the averages are well defined.

If a quantum instrument leaves every input state unchanged even after conditioning on its classical output, then its output distribution is independent of the input state. Thus an ordinary quantum process cannot learn arbitrary state information while preserving every state.

Use deterministic two-qubit entanglement dilution, retaining the specified complex probability amplitudes. Alice implements a quantum instrument on $A$ with Kraus operatorsThe Kraus representation completeness condition holds even for complex $\alpha ,\beta$:To realize this using the permitted quantum gates and a basic measurement, Alice initializes an ancilla qubit $C$ to $|0⟩$. Conditional on her computational-basis state being $j$, she applies to $C$ the unitary matrixEach has orthonormal columns. Therefore $W=|0⟩⟨0{|}_A\otimes V_0+|1⟩⟨1{|}_A\otimes V_1$ is a unitary gate, withA computational-basis measurement of $C$ implements the desired quantum instrument. Its two unnormalized branches on $AB$ areEach has probability $1/2$. Alice sends her outcome $m$ as one classical bit. For $m=0$, both parties keep their qubits unchanged. For $m=1$, both apply a Pauli X gate, givingThus both outcomes produce the exact target using one classical bit and local operations. The construction includes $\alpha =0$ or $\beta =0$ and arbitrary relative quantum phases; it requires no quantum channel between the parties.

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.

The generalized measurement postulate specifies Kraus operators $M_{i\alpha }$, where $i$ is the recorded outcome and $\alpha$ permits unobserved alternatives for that outcome, satisfyingFor a density operator $\rho$, outcome $i$ has probability and conditional stateWith one operator per outcome, this is the usual formula $p_i=\mathrm{Tr}(M_i\rho M_i^{†})$, ${\rho }_i=M_i\rho M_i^{†}/p_i$. The positive operators $E_i={\sum }_{\alpha }{M}_{i\alpha }^{†}{M}_{i\alpha }$ form a POVM, while the state-update maps form a quantum instrument.

The full postulate reduces to the usual projective measurement with the Lüders update when there is one operator per outcome and $M_i=P_i$, with $P_i=P_i^{†}=P_i^2$, $P_i{P}_j=0$ for $i\ne j$, and ${\sum }_i{P}_i=I$. Then $p_i=\mathrm{Tr}(P_i\rho )$ and ${\rho }_i=P_i\rho P_i/p_i$. Projective effects $E_i=P_i$ alone ensure these probabilities but do not fix the state update: an additional outcome-dependent unitary could still be applied. The choice $M_i=P_i$ makes the required projective state-update condition explicit.

The construction is a convex roof extension. First establish its convexity, without assuming that the minimizing quantum state ensemble exists. If $\tau ={\sum }_j{w}_j{\tau }_j$, choose a pure state ensemble for each ${\tau }_j$ whose average value of $\mu$ is within $\epsilon$ of $F({\tau }_j)$. Combining these ensembles, with their weights multiplied by $w_j$, gives a pure state ensemble for $\tau$. Taking arbitrarily small $\epsilon >0$ provesAlso, $F(\pi )=\mu (\pi )$ for a pure state $\pi$. Indeed, if $\pi ={\sum }_j{p}_j{\pi }_j$ and $v$ is orthogonal to its one-dimensional support, then $0={\sum }_j{p}_j⟨v|{\pi }_j|v⟩$. All terms are nonnegative, so every positive-weight ${\pi }_j$ has that same support and equals $\pi$.

Represent outcome $k$ of the operation by a linear unnormalized map ${\mathcal{I}}_k$, as for a quantum instrument. Normalization is performed only after applying this map: ${\mathcal{I}}_k(\rho )=P_k{\rho }_k$. For any preparation ensemble $\rho ={\sum }_j{p}_j{\pi }_j$, putLinearity gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. For $P_k>0$, Bayes' theorem identifies the conditional preparation weightsOutcomes with zero probability contribute nothing; normalized states on such branches need not be defined. The conditional states ${\sigma }_{jk}$ may be mixed, so convexity of the convex roof is essential here.

Applying that convexity separately for each recorded outcome, and then the assumed average monotonicity for each pure preparation, yieldsThe left side depends only on the initial density operator and the operation, not on the chosen preparation ensemble. Taking the infimum of the last expression over all its pure state ensembles provesEquivalently, choose an ensemble within $\epsilon$ of the infimum and then let $\epsilon \downarrow 0$. This is the monotonicity of a convex roof under a quantum instrument. It uses linearity of the unnormalized branches, not a false assumption that normalized conditional states depend linearly on the input. As usual, the averages are assumed well defined; finite-valued roofs and finitely many outcomes suffice. If an input roof is $-\infty$, taking ensembles with averages tending to $-\infty$ gives the same conclusion in the extended-real interpretation whenever the output average is defined.

Use the thin singular value decomposition $a=Ud{V}^{†}$, with $U^{†}U=I_N$, $V$ unitary and $d=\mathrm{diag}(d_k)$ nonnegative. Expanding its entries givesThe columns of $U$ and of $V^{\ast }$ are orthonormal, establishing the Schmidt decomposition. Bob's basis uses the complex-conjugated columns of $V$. Omitting that conjugation would not reproduce the coefficient matrix. The squared Schmidt coefficients are the eigenvalues of $a^{†}a$; equivalently $d^2=V^{†}{a}^{†}aV$ and $d=U^{†}aV$. The two reduced density matrices are $a{a}^{†}$ and $(a^{†}a{)}^T$, which have the same nonzero eigenvalues.

The printed coefficient $0.41231$ slightly rounds a normalized state: each squared norm is $0.9999995361$. Normalize before calculating an entropy. For the two states, the resulting Gram matrices are approximatelyFor a trace-one two-dimensional Hermitian matrix with diagonal entry $u$ and off-diagonal entry $c$, its eigenvalues areSort them decreasingly and take square roots:The Von Neumann entropy is $S=-{\lambda }_0{\log }_2{\lambda }_0-{\lambda }_1{\log }_2{\lambda }_1$. $\Psi$ is more entangled: its Schmidt weights are closer to equal and its reduced-state entropy is larger. Neither state is a product state or maximally entangled on its two-dimensional support.

By Nielsen's pure-state conversion theorem, deterministic LOCC conversion requires the input vector of squared Schmidt coefficients to be majorized by the output vector. Here $0.699941<0.894238$ and both sums are one, so $\lambda (\Psi )\prec \lambda (\Phi )$. Therefore $\Psi \to \Phi$ is possible deterministically; $\Phi \to \Psi$ is not. Entanglement can be discarded by local operations but cannot increase deterministically for these pure inputs. Probabilistic concentration is a separate question and is allowed below.^[ref]

There is a dimensional qualification in the proposed measurement. Write $P=|0_A^{\prime }⟩⟨0_A^{\prime }|+|1_A^{\prime }⟩⟨1_A^{\prime }|$. The two supplied measurement operators obeyThus their effects form a POVM on the occupied support, where $P$ is the identity. Alice's full space is three-dimensional, so these two operators alone do not sum to $I_3$. A full-space completion is $M_3=|2_A^{\prime }⟩⟨2_A^{\prime }|$. It has zero probability on either input, so it changes none of the requested outcomes. Strictly, the POVM consists of the effects $E_r=M_r^{†}{M}_r$; the specified $M_r$ additionally fix a quantum instrument and therefore the conditional states.

In the appropriate Schmidt basis for either input, write $|\psi ⟩=\sqrt{{\lambda }_0}|00⟩+\sqrt{{\lambda }_1}|11⟩$. The two outcomes have probabilitiesand normalized conditional statesThese formulas give both outcomes for each state by substituting its row of the table. For example,The two inputs have different Schmidt bases; the primed measurement basis is adapted separately to the state, as the question specifies, not one common unexamined laboratory basis.

For the $M_1$ outcome to be a maximally entangled state, its two amplitudes must be equal. HenceThe requested numerical results areThus $\Psi$ has the larger success probability, consistent with its greater initial entanglement. The $M_2$ branch cannot be balanced in the allowed angle range for either nonmaximal input: it would require $\tan \theta =\sqrt{{\lambda }_0/{\lambda }_1}>1$.

The success probabilities belong to this particular binary Schmidt filter. They are not the optimum over every possible quantum instrument. The diagonal success operator $\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ instead gives the standard single-copy entanglement concentration probability $2{\lambda }_1$. Postselection can increase entanglement in the retained branch, while the outcome-averaged entanglement cannot increase under LOCC.

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.

A positive operator-valued measure is a family of positive operators $E_i$ satisfying ${\sum }_i{E}_i=I$. Each $E_i$ is called an effect and determines a probability $p_i=\mathrm{Tr}(E_i\rho )$ for input density operator $\rho$. The effects need not be orthogonal projections or commute. This permits measurements with more outcomes than the dimension of the Hilbert space, such as the trine qubit POVM. Coupling to a quantum ancilla and measuring the larger system supplies a physical route to generalized measurements. ^[ref].

For an elementary unsharp $Z$ example on a qubit, takeOn the eigenstate $|0⟩$, $p_{\pm }=(1\pm \eta )/2$; at $\eta =1/2$ they are $3/4$ and $1/4$. At $\eta =0$ the outcome is a fair coin independent of the input, and at $\eta =1$ these effects specify a sharp $Z$ projective measurement. Choosing $K_{\pm }=\sqrt{E_{\pm }}$ specifies one quantum instrument, but $U_{\pm }\sqrt{E_{\pm }}$ for outcome-dependent unitary operators has the same effects and different state updates. Thus POVM does not determine the post-measurement state.

A POVM fixes outcome probabilities, but a quantum instrument additionally fixes the conditional output state. For computational-basis effects $E_y=|y⟩⟨y|$, the operators $M_y=|y⟩⟨y|$ leave the measured eigenstate as output, whereas $N_y=|0⟩⟨y|$ reset every outcome to $|0⟩$. Both have $M_y^{†}{M}_y=N_y^{†}{N}_y=E_y$. Thus a specification of effects alone cannot select a unique joint state that retains the measured quantum system.

A quantum instrument associates each classical outcome with a completely positive trace-nonincreasing map; summing the outcome maps gives a quantum channel. It specifies both outcome probabilities and conditional output states.

A quantum measurement turns a quantum state preparation into a classical outcome, such as a detector label. Two questions must be distinguished: how likely is each outcome, and what quantum state remains conditional on that outcome? A positive operator-valued measure answers the first; a quantum instrument answers both. This distinction is useful whenever a measured system participates in a later experiment, including quantum state tomography and measurement-based quantum feedback.

For a density operator $\rho$, the Born rule gives $p_m=\mathrm{Tr}(E_m\rho )$, where the $E_m$ are positive operators satisfying ${\sum }_m{E}_m=I$. Here $I$ is the identity operator, and $\mathrm{Tr}$ denotes the trace. Positivity makes every $p_m$ nonnegative; completeness makes their sum one. For a pure state, $\rho =|\psi ⟩⟨\psi |$ and $p_m=⟨\psi |E_m|\psi ⟩$. A projective measurement is the special case in which the $E_m$ are orthogonal projections onto mutually orthogonal subspaces.

For a qubit, choose the computational basis and write a normalized pure state, up to global phase, asThe angle $\varphi$ is a relative quantum phase; at the poles it does not distinguish states. Measuring the Pauli matrix $Z$ reports $+1$ for label zero and $-1$ for label one, withThese probabilities describe repeated identical preparations. They do not assign a hidden, pre-existing value to every possible measurement basis.

The Born rule grew out of Max Born's 1926 work on quantum scattering: squared probability amplitudes determine the relative frequencies of scattering outcomes. Born explained this statistical interpretation again in his 1954 Nobel lecture. This probability law does not by itself select a post-measurement state. ^[ref]; ^[ref].

Consider an original balanced-readout model, rather than a fit to the experiment. Let $n\in \{0,\dots ,N_{\max }\}$, with a normalized prior probability $p_0(n)\ge 0$ and ${\sum }_n{p}_0(n)=1$. A detected chemical atom has outcome $s\in \{e,g\}$. Define the conditional probabilitiesHere $\chi$ is the accumulated quantum phase per photon, $\varphi$ the controlled analysis quantum phase, and $C$ the fringe contrast. The two probabilities sum to one for every $n$. Actual calibration may also require an offset and asymmetric detection errors; $C$ alone does not specify a quantum instrument.

Assume no photon number change during the record and fresh independently prepared atomic meters. For observed outcomes $s_1,\dots ,s_k$ at recorded settings ${\varphi }_1,\dots ,{\varphi }_k$, the Born rule and Bayes' theorem giveThe posterior probability is defined only for $Z_k>0$. Summing its numerator over $n$ proves normalization. This is also the update of diagonal elements of a field density operator under a measurement in quantum mechanics; it does not assume that an initial quantum superposition was an unknown pre-existing integer. The observed settings are part of the record, especially when selected adaptively.

For an exact finite example take $p_0=(1,1,1,1)/4$, $C=4/5$, and $\chi =\pi /2$. An $e$ outcome at $\varphi =0$ has likelihood $(9,5,1,5)/10$, giving $p_1=(9,5,1,5)/20$. Another $e$ at $\varphi =-\pi /2$ gives $p_2=(9,9,1,1)/20$. A subsequent $g$ at $\varphi =0$ givesThus three atomic records raise the posterior probability of $n=1$ from $1/4$ to $45/68$, while leaving finite uncertainty. No one of these outcomes identifies $n$ with certainty.

The periodic likelihood function imposes an identifiability limit. If $\chi =2\pi /q$ for an integer $q>0$, then $L_s(n+q;\varphi )=L_s(n;\varphi )$ for every setting: the posterior odds between $n$ and $n+q$ never change. Varying $\varphi$ can resolve a cosine reflection ambiguity within a period, but cannot remove this exact period. Changing $\chi$ can supply additional information. A finite cutoff is therefore a physical assumption to justify, not evidence that excluded high photon numbers were absent. At $C=0$ no outcome changes the prior probability; at $0<C<1$ neither binary outcome has an exactly zero likelihood.

The no-loss assumption also has a time limit. In an original zero-temperature loss model, let $\kappa =1/T_c$. A Fock state $|n⟩$ loses its first photon at rate $n\kappa$, so its survival probability is $\exp (-n\kappa t)$ and its mean lifetime is $T_c/n$ for $n>0$. Several records must be acquired within that lifetime to identify a plateau. When losses matter, first predict $p_k^{-}(n)={\sum }_m{T}_{nm}(\Delta t)p_{k-1}(m)$ using a normalized transition matrix, then apply the atomic Bayesian inference update to $p_k^{-}$. Thermal excitation requires upward transitions too. A delayed or broadened inference after a jump need not indicate that an atom caused the jump.

For one pure rank-two entangled state with descending squared Schmidt coefficients $({\lambda }_0,{\lambda }_1)$, the success operator $F=\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ gives a Bell pair with probability $2{\lambda }_1$. The complementary operator $\sqrt{I-F^{†}F}$ makes a valid quantum instrument. Optimality follows from the Schmidt-tail entanglement monotone: a successful Bell branch has smallest weight $1/2$, so its probability cannot exceed $2{\lambda }_1$. Failure branches have nonnegative monotone and cannot loosen this bound.