State Update Rule

The state update rule specifies the quantum state to use for later predictions after a measurement record has been obtained, ignored, or only partially retained. For an ideal projective outcome $a$, the unnormalized branch is $P_a\rho P_a$; its trace is the probability of $a$, and division by that probability gives the state conditional on the recorded outcome.

This rule is also called state reduction, the projection postulate, or, especially for pure states, wavefunction collapse. Those names do not settle what the update represents physically or ontologically. Here the rule is treated operationally: it maps an input state and available measurement record to the state used for subsequent probability assignments.

Purpose and Scope

The Projective Measurement page owns the definition of a PVM and its sharp outcome subspaces. This page owns the logic of state conditioning:

• how unnormalized outcome branches encode both probability and state;
• why the conditional branch must be normalized;
• how pure-state and density-operator formulas agree;
• how selective and nonselective updates answer different questions;
• why an unread measurement is not the same as no measurement;
• in what sense quantum conditioning resembles Bayesian conditioning;
• why the analogy has important limits;
• what the formal rule does and does not claim about measurement.

The main formulas use the ideal Lüders instrument for a discrete PVM $\{P_a\}$. The later instrument section shows the corresponding generalized structure without replacing the detailed treatment in Generalized Measurements Overview.

Three Questions Before Updating

Before applying any state-update formula, identify:

1. Which measurement was implemented? Outcome probabilities alone may not determine the state change.
2. Which record is available? A known outcome, a partially coarse-grained record, and an ignored record lead to different state assignments.
3. What prediction comes next? The updated state is the input for later evolution and measurement.

The phrase “after the measurement” is incomplete until these questions are answered.

At a Glance

For a discrete projective measurement $\{P_a\}$ and input density operator $\rho$, define the unnormalized branch

$$\widetilde\rho_a = P_a\rho P_a.$$

Its trace is the outcome probability:

$$p(a) = \operatorname{Tr}\widetilde\rho_a = \operatorname{Tr}(\rho P_a).$$

If outcome $a$ is recorded and $p(a)>0$, the selective Lüders update is

$$\rho_a = \frac{\widetilde\rho_a}{p(a)} = \frac{P_a\rho P_a} {\operatorname{Tr}(\rho P_a)}.$$

If the measurement occurs but the outcome is ignored, the nonselective update is

$$\rho' = \sum_a\widetilde\rho_a = \sum_aP_a\rho P_a.$$

For a normalized pure input $\lvert\psi\rangle$, the conditional vector is

$$\lvert\psi_a\rangle = \frac{P_a\lvert\psi\rangle} {\sqrt{ \langle\psi\rvert P_a\lvert\psi\rangle }}, \qquad p(a)>0.$$

The Branch Before Normalization

The unnormalized branch is the most useful starting point because one object carries both parts of the prediction:

• its trace or squared norm gives the branch probability;
• its normalized direction or density operator gives the conditional state.

Pure-state branch

For a pure input, define

$$\lvert\widetilde\psi_a\rangle = P_a\lvert\psi\rangle.$$

Its squared norm is

$$\begin{aligned} \left\lVert \lvert\widetilde\psi_a\rangle \right\rVert^2 &= \langle\psi\rvert P_a^\dagger P_a \lvert\psi\rangle \\ &= \langle\psi\rvert P_a\lvert\psi\rangle \\ &= p(a). \end{aligned}$$

The projector removes components outside the outcome subspace. The remaining vector has norm $\sqrt{p(a)}$, not generally one.

Density-operator branch

For a general state, define

$$\widetilde\rho_a = P_a\rho P_a.$$

This operator is positive because, for every $\lvert\phi\rangle$,

$$\begin{aligned} \langle\phi\rvert \widetilde\rho_a \lvert\phi\rangle &= \langle\phi\rvert P_a\rho P_a \lvert\phi\rangle \\ &= \langle P_a\phi\rvert \rho \lvert P_a\phi\rangle \\ &\geq 0. \end{aligned}$$

Its trace is

$$\begin{aligned} \operatorname{Tr}\widetilde\rho_a &= \operatorname{Tr}(P_a\rho P_a) \\ &= \operatorname{Tr}(\rho P_a^2) \\ &= \operatorname{Tr}(\rho P_a) \\ &= p(a). \end{aligned}$$

Keeping branches unnormalized makes their weighted recombination immediate:

$$\sum_a\widetilde\rho_a = \sum_aP_a\rho P_a.$$

No probability factors have to be inserted by hand because each branch already has trace $p(a)$.

Pure-State Selective Update

Suppose outcome $a$ is recorded and $p(a)>0$. Normalizing the branch gives

$$\lvert\psi_a\rangle = \frac{\lvert\widetilde\psi_a\rangle} {\left\lVert \lvert\widetilde\psi_a\rangle \right\rVert} = \frac{P_a\lvert\psi\rangle} {\sqrt{p(a)}}.$$

The normalization check is

$$\begin{aligned} \langle\psi_a\vert\psi_a\rangle &= \frac{ \langle\psi\rvert P_a^2\lvert\psi\rangle }{p(a)} \\ &= \frac{p(a)}{p(a)} \\ &= 1. \end{aligned}$$

The output lies in the selected subspace:

$$P_a\lvert\psi_a\rangle = \lvert\psi_a\rangle.$$

If $P_a$ has rank one, the conditional ray is the corresponding eigenstate ray. If $P_a$ has higher rank, the update retains the normalized component of the input state within the entire eigenspace. The latter case is developed in Degenerate Measurements and Lüders Rule.

An overall phase acquired during normalization has no effect on the pure-state ray. Relative phases among components inside a degenerate outcome subspace can remain physically significant.

Density-Operator Selective Update

For a mixed or otherwise general state, the conditional Lüders update is

$$\rho \longmapsto \rho_a = \frac{P_a\rho P_a}{p(a)}, \qquad p(a)>0.$$

The result is a valid density operator:

$$\rho_a\geq0, \qquad \operatorname{Tr}\rho_a=1.$$

It is supported inside the outcome subspace:

$$P_a\rho_a = \rho_aP_a = \rho_a.$$

For a pure input

$$\rho = \lvert\psi\rangle\langle\psi\rvert,$$

the density-operator update becomes

$$\begin{aligned} \rho_a &= \frac{ P_a \lvert\psi\rangle\langle\psi\rvert P_a }{p(a)} \\ &= \lvert\psi_a\rangle \langle\psi_a\rvert. \end{aligned}$$

Thus the pure-vector and density-operator formulations are equivalent whenever the conditional branch remains pure. The density-operator formula is more general and avoids introducing arbitrary phases.

Zero-probability records

If $p(a)=0$, then positivity implies

$$P_a\rho P_a = 0.$$

The normalized expression $P_a\rho P_a/p(a)$ is undefined. Standard conditioning does not assign a state to an impossible record. A real detector model may replace the ideal zero by a small nonzero error probability, but that changes the instrument and must be stated explicitly.

Selective and Nonselective Updates

A selective update conditions on a known record. A nonselective update averages over records that were produced but are unavailable or deliberately ignored.

If $p(a)>0$, then

$$\widetilde\rho_a = p(a)\rho_a.$$

Summing over outcomes gives

$$\begin{aligned} \rho' &= \sum_ap(a)\rho_a \\ &= \sum_aP_a\rho P_a. \end{aligned}$$

The first line displays the classical averaging over conditional states. The second line is the quantum channel written directly in terms of the input state.

Trace preservation

The nonselective Lüders map

$$\mathcal L(\rho) = \sum_aP_a\rho P_a$$

is trace preserving:

$$\begin{aligned} \operatorname{Tr}\mathcal L(\rho) &= \sum_a \operatorname{Tr}(P_a\rho P_a) \\ &= \operatorname{Tr} \left( \rho\sum_aP_a \right) \\ &= \operatorname{Tr}\rho. \end{aligned}$$

It is also unital:

$$\mathcal L(I) = \sum_aP_aIP_a = I.$$

Block structure

Insert the resolution of identity on both sides of $\rho$:

$$\rho = \sum_{a,b} P_a\rho P_b.$$

The nonselective update keeps only the diagonal blocks:

$$\mathcal L(\rho) = \sum_aP_a\rho P_a.$$

Terms $P_a\rho P_b$ with $a\neq b$ encode coherence between distinct measured outcome subspaces. The ideal unread measurement removes those terms while preserving structure within each block.

The map is idempotent:

$$\mathcal L^2 = \mathcal L.$$

Indeed,

$$\begin{aligned} \mathcal L^2(\rho) &= \sum_{a,b} P_aP_b\rho P_bP_a \\ &= \sum_aP_a\rho P_a. \end{aligned}$$

Once the state is block diagonal in the measured decomposition, repeating the unread Lüders measurement produces no further change. Equivalently,

$$\mathcal L(\rho) = \rho$$

if and only if

$$[\rho,P_a] = 0 \qquad \text{for every }a.$$

An unread measurement is not no measurement

If no measurement occurs and no other dynamics acts, then

$$\rho \longmapsto \rho.$$

If the projective measurement occurs but its result is ignored, then

$$\rho \longmapsto \sum_aP_a\rho P_a.$$

These are equal only when the input is already block diagonal in the measurement decomposition. Forgetting a classical record does not reverse the physical interaction that correlated the system with the apparatus.

Partial Records and Coarse Outcomes

Suppose the fine outcome is $a$, but the retained record only says that $a$ belongs to a set $S$. The correct update depends on what physical measurement occurred.

If the fine PVM $\{P_a\}$ was measured and the fine record was later discarded within $S$, the conditional state is

$$\rho_{S,\mathrm{fine}} = \frac{ \sum_{a\in S}P_a\rho P_a }{ \sum_{a\in S}p(a) }.$$

If the apparatus instead directly implements the coarse Lüders measurement with

$$P(S) = \sum_{a\in S}P_a,$$

then the conditional state is

$$\rho_{S,\mathrm{coarse}} = \frac{ P(S)\rho P(S) }{ \operatorname{Tr}\!\left(\rho P(S)\right) }.$$

The two expressions need not agree. Expanding the coarse numerator gives

$$P(S)\rho P(S) = \sum_{a,b\in S} P_a\rho P_b.$$

The direct coarse Lüders update retains coherence between fine alternatives inside $S$, whereas fine measurement followed by forgetting removes the terms with $a\neq b$. A classical label alone does not determine the update; the instrument matters.

This subtlety is treated systematically in Degenerate Measurements and Lüders Rule.

Linear Branches and Nonlinear Conditioning

For each outcome, the unnormalized Lüders operation

$$\mathcal I_a(\rho) = P_a\rho P_a$$

is linear:

$$\begin{aligned} \mathcal I_a \left( t\rho_1+(1-t)\rho_2 \right) &= t\mathcal I_a(\rho_1) \\ &\qquad + (1-t)\mathcal I_a(\rho_2). \end{aligned}$$

The normalized conditional map

$$\mathcal N_a(\rho) = \frac{\mathcal I_a(\rho)} {\operatorname{Tr}\mathcal I_a(\rho)}$$

is generally nonlinear because its denominator depends on the input state.

Let

$$p_i = \operatorname{Tr}\mathcal I_a(\rho_i), \qquad \rho_{i\mid a} = \frac{\mathcal I_a(\rho_i)}{p_i}.$$

For the mixture

$$\rho = t\rho_1+(1-t)\rho_2, \qquad 0\leq t\leq1,$$

the conditional state is

$$\mathcal N_a(\rho) = \frac{ tp_1\rho_{1\mid a} + (1-t)p_2\rho_{2\mid a} }{ tp_1+(1-t)p_2 }.$$

The posterior weights are proportional to the prior weights times the likelihoods $p_i$. They are not generally $t$ and $1-t$.

This nonlinearity is the ordinary nonlinearity of normalization after conditioning. The physical outcome operations remain linear on unnormalized density operators, and the summed nonselective channel

$$\sum_a\mathcal I_a$$

is linear and trace preserving.

General Instrument Form

The projective Lüders rule is one special quantum instrument. A general discrete instrument can be written using measurement operators $M_{a\mu}$:

$$\mathcal I_a(\rho) = \sum_\mu M_{a\mu}\rho M_{a\mu}^\dagger.$$

Its outcome effect is

$$E_a = \sum_\mu M_{a\mu}^\dagger M_{a\mu},$$

and the outcome probability is

$$\begin{aligned} p(a) &= \operatorname{Tr}\mathcal I_a(\rho) \\ &= \operatorname{Tr}(\rho E_a). \end{aligned}$$

The conditional state is

$$\rho_a = \frac{\mathcal I_a(\rho)}{p(a)}, \qquad p(a)>0.$$

Normalization of the total probability requires

$$\sum_{a,\mu} M_{a\mu}^\dagger M_{a\mu} = I.$$

The Lüders instrument for a PVM is recovered with one operator per outcome:

$$M_a = P_a.$$

Different sets of measurement operators can produce the same effects $\{E_a\}$ and therefore the same outcome probabilities while producing different conditional states. A POVM specifies the effects; a quantum instrument specifies both probabilities and state changes.

Relation to Bayesian Conditioning

There is a precise structural analogy between classical Bayesian conditioning and quantum state update.

For a classical latent variable $x$ and observed record $a$, define the unnormalized posterior weight

$$\widetilde q_a(x) = q(a\mid x)q(x).$$

Its total weight is the evidence

$$q(a) = \sum_x\widetilde q_a(x),$$

and normalization gives

$$q(x\mid a) = \frac{\widetilde q_a(x)}{q(a)}.$$

The quantum instrument has the same branch-normalize pattern:

$$\begin{aligned} \widetilde\rho_a &= \mathcal I_a(\rho), \\ p(a) &= \operatorname{Tr}\widetilde\rho_a, \\ \rho_a &= \frac{\widetilde\rho_a}{p(a)}. \end{aligned}$$

This analogy is useful because it clarifies why conditional states depend on the retained record and why posterior mixture weights are reweighted by outcome likelihoods.

Where the analogy stops

The analogy does not imply that quantum update is merely classical conditioning over unknown pre-existing values.

1. Measurement can physically disturb the system. The operation $\mathcal I_a$ contains more than a classical likelihood.
2. Effects do not determine instruments. The same $p(a)$ can accompany different output states.
3. Noncommuting alternatives lack one universal joint sample space. Ordered quantum measurements require an instrument and sequence, not only a joint probability table.
4. Unread measurements can change later statistics. The nonselective channel can remove coherence even when no outcome is selected.
5. The interpretation of the quantum state is not fixed by the formula. Calling the rule “Bayesian” does not decide whether the state is epistemic, ontic, relational, operational, or part of some other interpretation.

It is therefore safest to say that quantum state update has a Bayesian conditioning structure, with an additional instrument-dependent quantum transformation.

Later Evolution and Measurements

Once outcome $a$ is known, $\rho_a$ is the input for later dynamics. If the system evolves unitarily from the first measurement to a later time,

$$\rho_a(t) = U(t)\rho_aU^\dagger(t).$$

For a later outcome $b$ represented by an effect $Q_b$, the conditional probability is

$$p(b\mid a) = \operatorname{Tr} \left[ Q_bU(t)\rho_aU^\dagger(t) \right].$$

Substituting the unnormalized first branch gives the joint probability

$$p(a,b) = \operatorname{Tr} \left[ Q_bU(t) P_a\rho P_a U^\dagger(t) \right].$$

The normalized intermediate state can therefore be avoided when only the joint probability is needed. Sequential Measurements develops ordering, compatibility, and conditional probabilities in detail.

Worked Example: Qubit Measurement Along z

Prepare

$$\lvert\psi\rangle = \cos\frac{\theta}{2}\lvert0\rangle + e^{i\phi} \sin\frac{\theta}{2}\lvert1\rangle.$$

The projectors for a computational-basis measurement are

$$P_0 = \lvert0\rangle\langle0\rvert, \qquad P_1 = \lvert1\rangle\langle1\rvert.$$

The unnormalized branches are

$$\begin{aligned} P_0\lvert\psi\rangle &= \cos\frac{\theta}{2}\lvert0\rangle, \\ P_1\lvert\psi\rangle &= e^{i\phi} \sin\frac{\theta}{2}\lvert1\rangle. \end{aligned}$$

Their squared norms give

$$\begin{aligned} p(0) &= \cos^2\frac{\theta}{2}, \\ p(1) &= \sin^2\frac{\theta}{2}. \end{aligned}$$

Conditional on the corresponding nonzero-probability result,

$$\lvert\psi_0\rangle \sim \lvert0\rangle, \qquad \lvert\psi_1\rangle \sim \lvert1\rangle.$$

The initial density operator is

$$\rho = \begin{pmatrix} \cos^2\frac{\theta}{2} & e^{-i\phi} \cos\frac{\theta}{2} \sin\frac{\theta}{2} \\ e^{i\phi} \cos\frac{\theta}{2} \sin\frac{\theta}{2} & \sin^2\frac{\theta}{2} \end{pmatrix}.$$

If the outcome is ignored, the state becomes

$$\rho' = \begin{pmatrix} \cos^2\frac{\theta}{2} & 0 \\ 0 & \sin^2\frac{\theta}{2} \end{pmatrix}.$$

The populations are unchanged, while the coherence between the two measured alternatives is removed.

Worked Example: Local Update of an Entangled Pair

Let two qubits $A$ and $B$ be prepared in

$$\lvert\Phi^+\rangle_{AB} = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}.$$

Measure qubit $A$ in the computational basis with

$$\begin{aligned} P_0^{(A)} &= \lvert0\rangle\langle0\rvert \otimes I_B, \\ P_1^{(A)} &= \lvert1\rangle\langle1\rvert \otimes I_B. \end{aligned}$$

The two outcomes have equal probability:

$$p(0) = p(1) = \frac12.$$

Conditional on outcome $0$,

$$\rho_{AB\mid0} = \lvert00\rangle\langle00\rvert.$$

Conditional on outcome $1$,

$$\rho_{AB\mid1} = \lvert11\rangle\langle11\rvert.$$

The corresponding reduced states of $B$ are

$$\rho_{B\mid0} = \lvert0\rangle\langle0\rvert, \qquad \rho_{B\mid1} = \lvert1\rangle\langle1\rvert.$$

If the outcome at $A$ is not available, the nonselective joint state is

$$\rho_{AB}' = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right).$$

Tracing out $A$ gives

$$\rho_B' = \frac I2,$$

which is the same reduced state $B$ had before the measurement. The conditional states of $B$ depend on the record at $A$, but the unread local operation does not transmit that record. Classical communication is required to sort later data by the outcome.

Worked Example: Same Probabilities, Different Updates

Consider the computational-basis PVM

$$P_0 = \lvert0\rangle\langle0\rvert, \qquad P_1 = \lvert1\rangle\langle1\rvert.$$

The Lüders instrument uses

$$M_0^{\mathrm L} = P_0, \qquad M_1^{\mathrm L} = P_1.$$

Now consider a reset instrument:

$$M_0^{\mathrm R} = \lvert0\rangle\langle0\rvert, \qquad M_1^{\mathrm R} = \lvert0\rangle\langle1\rvert.$$

Both instruments have the same effects:

$$\begin{aligned} \left(M_0^{\mathrm R}\right)^\dagger M_0^{\mathrm R} &= P_0, \\ \left(M_1^{\mathrm R}\right)^\dagger M_1^{\mathrm R} &= P_1. \end{aligned}$$

They therefore give the same probabilities

$$p(a) = \operatorname{Tr}(\rho P_a).$$

Their conditional outputs differ. The Lüders instrument outputs $\lvert0\rangle$ after outcome $0$ and $\lvert1\rangle$ after outcome $1$. The reset instrument outputs $\lvert0\rangle$ after either outcome:

$$\rho_0^{\mathrm R} = \rho_1^{\mathrm R} = \lvert0\rangle\langle0\rvert.$$

The reset device extracts the same classical computational-basis record and then prepares a fixed state. It is not repeatable for outcome $1$. This example shows why the probability effects must not be mistaken for a complete state-update rule.

What the Formal Rule Does Not Settle

The update rule provides a conditional state assignment within the operational formalism. It does not, by itself, establish:

• whether the quantum state is ontic, epistemic, relational, or something else;
• whether collapse is a physical process, an information update, an effective description, or an emergent branch-relative rule;
• why one definite macroscopic record is experienced;
• where a fundamental system–apparatus boundary lies;
• how a detector amplifies, stores, and reports an outcome;
• whether an ideal projection occurs instantaneously in spacetime;
• which interpretation of quantum mechanics is correct.

The nonselective local update also does not permit superluminal signaling. Conditional remote states depend on a local record, but an observer without that record uses the averaged state. No controllable message is available until the record is communicated through an ordinary causal channel.

These boundaries are developed in What Measurement Formalism Does Not Settle and What the Postulates Do Not Say.

A Reliable Calculation Workflow

1. Specify the instrument. For an ideal projective measurement, state that the Lüders instrument is being used.

2. Compute each unnormalized branch.

   $$\widetilde\rho_a = \mathcal I_a(\rho).$$

3. Extract the probabilities.

   $$p(a) = \operatorname{Tr}\widetilde\rho_a.$$

4. Check normalization.

   $$\sum_ap(a) = 1.$$

5. Condition on the actual record. If $a$ is known and $p(a)>0$,

   $$\rho_a = \frac{\widetilde\rho_a}{p(a)}.$$

6. Average only when the record is unavailable.

   $$\rho' = \sum_a\widetilde\rho_a.$$

7. Propagate the resulting state. Use $\rho_a$ for a selected branch or $\rho'$ for the unread ensemble in every later prediction.

8. Check the scope of the model. If the apparatus is noisy, destructive, inefficient, or followed by feedback, use its actual instrument rather than an ideal projector by habit.

Common Mistakes

• Dividing by the wrong quantity. A pure vector is divided by $\sqrt{p(a)}$; a density operator is divided by $p(a)$.
• Normalizing before recording the probability. The branch norm or trace is the probability information.
• Conditioning on an unspecified outcome. A selective state requires a definite retained record.
• Conditioning on a zero-probability result. The normalized branch is undefined.
• Averaging normalized states without weights. The nonselective state is $\sum_ap(a)\rho_a$, not $\sum_a\rho_a$.
• Reusing the input state for later predictions. A measurement can change the state even when its record is ignored.
• Treating no measurement and unread measurement as equivalent. Coherences between outcome sectors can distinguish them.
• Assuming the PVM fixes the update. Effects determine probabilities; an instrument determines state change.
• Using the coarse Lüders formula after a fine measurement. Fine-and-forgotten and directly coarse measurements can preserve different coherences.
• Calling the normalized update a linear dynamical map. The unnormalized branch operation is linear; normalization makes conditioning nonlinear.
• Taking the Bayesian analogy as an interpretation theorem. The shared normalization structure does not settle what the quantum state represents.
• Inferring superluminal signaling from conditional remote states. An unavailable outcome must be averaged over.
• Treating projection as a detector model. The formal rule omits coupling, amplification, noise, and readout physics.

Canonical Boundaries and Cross-Links

• Measurement in the Formalism owns the distinction among records, effects, instruments, and apparatus models.
• Projective Measurement owns PVM structure, sharp outcome subspaces, and ideal repeatability.
• Degenerate Measurements and Lüders Rule owns coherence within degenerate sectors and measurement refinements.
• Sequential Measurements owns ordered joint and conditional probabilities.
• Generalized Measurements Overview introduces measurement operators beyond projectors.
• POVMs: First Encounter owns positive effects and explains why a POVM does not fix outputs.
• Quantum Instruments gives the detailed theory of completely positive outcome maps.
• Reduced States owns partial traces and subsystem state assignments.
• What Measurement Formalism Does Not Settle owns the boundary with interpretation, decoherence, and detector dynamics.

Summary

State update begins with an unnormalized outcome operation:

$$\widetilde\rho_a = \mathcal I_a(\rho).$$

The branch trace is the outcome probability,

$$p(a) = \operatorname{Tr}\widetilde\rho_a,$$

and a retained nonzero-probability record gives the conditional state

$$\rho_a = \frac{\widetilde\rho_a}{p(a)}.$$

If the record is ignored, the state is the sum of unnormalized branches:

$$\rho' = \sum_a\widetilde\rho_a.$$

For the Lüders instrument, $\mathcal I_a(\rho)=P_a\rho P_a$. The branch maps are linear, the normalized conditional map is generally nonlinear, and the nonselective sum is a linear trace-preserving channel. This formal structure supports reliable future predictions without deciding what state reduction means ontologically.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958 — standard state-vector reduction rule.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955 — projection postulate and measurement framework.
• G. Lüders, “Über die Zustandsänderung durch den Meßprozeß,” Annalen der Physik 8, 322–328, 1951; K. A. Kirkpatrick, “Translation of Lüders’ ‘Über die Zustandsänderung durch den Messprozess’,” Annalen der Physik 15, 663–670, 2006, arXiv:quant-ph/0403007 — the minimally refining update for degenerate observables.
• E. B. Davies and J. T. Lewis, “An operational approach to quantum probability,” Communications in Mathematical Physics 17, 239–260, 1970, doi:10.1007/BF01647093 — quantum instruments and operational conditional probabilities.
• M. Ozawa, “An operational approach to quantum state reduction,” Annals of Physics 259, 121–137, 1997, doi:10.1006/aphy.1997.5706, arXiv:quant-ph/9706027 — conditional state reduction from measurement operations.
• M. Ozawa, “Quantum State Reduction and the Quantum Bayes Principle,” in Quantum Communication, Computing, and Measurement, Plenum, 1997, arXiv:quant-ph/9705030 — operational formulation of the conditioning analogy.
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer, 1983 — operation and instrument formalism.
• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016 — modern treatment of effects, operations, instruments, and state change.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010 — finite-dimensional measurement operators and conditional states.

Exercises

Exercise 1: Normalize a pure branch

Let

$$\lvert\psi\rangle = \frac{ 2\lvert0\rangle +i\lvert1\rangle +\lvert2\rangle }{\sqrt6}$$

and

$$P = \lvert0\rangle\langle0\rvert + \lvert2\rangle\langle2\rvert.$$

Find the unnormalized yes branch, its probability, and the normalized conditional state.

Solution

The unnormalized branch is

$$\lvert\widetilde\psi_{\mathrm y}\rangle = P\lvert\psi\rangle = \frac{ 2\lvert0\rangle+\lvert2\rangle }{\sqrt6}.$$

Its squared norm is

$$\begin{aligned} p(\mathrm y) &= \left\lVert \lvert\widetilde\psi_{\mathrm y}\rangle \right\rVert^2 \\ &= \frac{4+1}{6} \\ &= \frac56. \end{aligned}$$

Dividing by $\sqrt{5/6}$ gives

$$\lvert\psi_{\mathrm y}\rangle = \frac{ 2\lvert0\rangle+\lvert2\rangle }{\sqrt5}.$$

Exercise 2: Validate a density-operator branch

Let $\rho$ be a density operator, $P$ a projector, and

$$p = \operatorname{Tr}(\rho P) > 0.$$

Prove that

$$\rho_P = \frac{P\rho P}{p}$$

is positive, has unit trace, and satisfies $P\rho_PP=\rho_P$.

Solution

For any $\lvert\phi\rangle$,

$$\langle\phi\rvert P\rho P \lvert\phi\rangle = \langle P\phi\rvert \rho \lvert P\phi\rangle \geq 0,$$

so $P\rho P$ and $\rho_P$ are positive. Cyclicity of the trace gives

$$\begin{aligned} \operatorname{Tr}\rho_P &= \frac{ \operatorname{Tr}(P\rho P) }{p} \\ &= \frac{ \operatorname{Tr}(\rho P^2) }{p} \\ &= 1. \end{aligned}$$

Finally,

$$P\rho_PP = \frac{P^2\rho P^2}{p} = \rho_P.$$

Thus the conditional state is normalized and supported in $\operatorname{Ran}P$.

Exercise 3: Selective versus unread qubit measurement

Let

$$\lvert+\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

A computational-basis measurement is performed.

1. Find the conditional states for outcomes $0$ and $1$.
2. Find the nonselective state when the outcome is ignored.
3. Compare this state with the state obtained if no measurement occurs.

Solution

Both outcomes have probability $1/2$. The selective states are

$$\rho_0 = \lvert0\rangle\langle0\rvert, \qquad \rho_1 = \lvert1\rangle\langle1\rvert.$$

Ignoring the result gives

$$\begin{aligned} \rho' &= \frac12\rho_0+\frac12\rho_1 \\ &= \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) \\ &= \frac I2. \end{aligned}$$

If no measurement occurs, the state remains

$$\begin{aligned} \lvert+\rangle\langle+\rvert &= \frac12 \bigl( \lvert0\rangle\langle0\rvert + \lvert0\rangle\langle1\rvert \\ &\qquad + \lvert1\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \bigr). \end{aligned}$$

The unread measurement removes the off-diagonal coherence; doing nothing does not.

Exercise 4: Conditional nonlinearity

Let an outcome operation be $\mathcal I_a$. Define

$$\rho = t\rho_1+(1-t)\rho_2$$

and

$$p_i = \operatorname{Tr}\mathcal I_a(\rho_i) > 0.$$

Derive the conditional state for $\rho$ and determine when its posterior mixture weights equal the prior weights $t$ and $1-t$.

Solution

Linearity of the unnormalized operation gives

$$\mathcal I_a(\rho) = t\mathcal I_a(\rho_1) + (1-t)\mathcal I_a(\rho_2).$$

Write

$$\rho_{i\mid a} = \frac{\mathcal I_a(\rho_i)}{p_i}.$$

The total outcome probability is

$$p(a) = tp_1+(1-t)p_2.$$

Therefore

$$\rho_a = \frac{ tp_1\rho_{1\mid a} + (1-t)p_2\rho_{2\mid a} }{ tp_1+(1-t)p_2 }.$$

The posterior weights equal $t$ and $1-t$ when $p_1=p_2$, apart from trivial cases such as $t=0$ or $t=1$. In general, conditioning reweights the alternatives according to their likelihood of producing $a$.

Exercise 5: Fixed points of the unread Lüders map

For a PVM $\{P_a\}$, let

$$\mathcal L(\rho) = \sum_aP_a\rho P_a.$$

Show that $\mathcal L(\rho)=\rho$ if $\rho$ commutes with every $P_a$. Conversely, show that $\mathcal L(\rho)=\rho$ implies $P_a\rho=\rho P_a$ for every $a$.

Solution

If $[\rho,P_a]=0$ for every $a$, then

$$\begin{aligned} \mathcal L(\rho) &= \sum_aP_a\rho P_a \\ &= \sum_a\rho P_a^2 \\ &= \rho. \end{aligned}$$

Conversely, assume

$$\rho = \sum_bP_b\rho P_b.$$

Multiplying on the left by $P_a$ gives

$$P_a\rho = P_a\rho P_a.$$

Multiplying on the right gives

$$\rho P_a = P_a\rho P_a.$$

Hence

$$P_a\rho = \rho P_a$$

for every $a$.

Exercise 6: Fine-and-forgotten versus coarse Lüders update

Let $P_1$ and $P_2$ be orthogonal projectors and define

$$Q = P_1+P_2.$$

For an input $\rho$ with $\operatorname{Tr}(\rho Q)>0$, compute the difference

$$\rho_{Q,\mathrm{coarse}} - \rho_{Q,\mathrm{fine}}.$$

Under what condition do the two conditional states agree?

Solution

The direct coarse Lüders state is

$$\rho_{Q,\mathrm{coarse}} = \frac{Q\rho Q}{\operatorname{Tr}(\rho Q)}.$$

The fine-and-forgotten state is

$$\rho_{Q,\mathrm{fine}} = \frac{ P_1\rho P_1 + P_2\rho P_2 }{ \operatorname{Tr}(\rho Q) }.$$

Expanding $Q\rho Q$ gives

$$\begin{aligned} Q\rho Q &= P_1\rho P_1 + P_1\rho P_2 \\ &\qquad + P_2\rho P_1 + P_2\rho P_2. \end{aligned}$$

Therefore

$$\rho_{Q,\mathrm{coarse}} - \rho_{Q,\mathrm{fine}} = \frac{ P_1\rho P_2+P_2\rho P_1 }{ \operatorname{Tr}(\rho Q) }.$$

The two states agree precisely when

$$P_1\rho P_2+P_2\rho P_1 = 0.$$

For a Hermitian $\rho$, this means that the relevant coherence contribution between the two fine sectors vanishes.

Exercise 7: Conditional entanglement and no signaling

For

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2},$$

qubit $A$ is measured in the computational basis.

1. Find the reduced state of $B$ before the measurement.
2. Find the two conditional reduced states of $B$.
3. Average them with their probabilities and compare with the initial reduced state.

Solution

The initial joint density operator is

$$\rho_{AB} = \lvert\Phi^+\rangle\langle\Phi^+\rvert.$$

Tracing out $A$ gives

$$\rho_B = \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) = \frac I2.$$

Outcome $0$ has probability $1/2$ and gives

$$\rho_{B\mid0} = \lvert0\rangle\langle0\rvert.$$

Outcome $1$ also has probability $1/2$ and gives

$$\rho_{B\mid1} = \lvert1\rangle\langle1\rvert.$$

Without the record at $A$, the state assigned to $B$ is

$$\begin{aligned} \rho_B' &= \frac12\rho_{B\mid0} + \frac12\rho_{B\mid1} \\ &= \frac I2 \\ &= \rho_B. \end{aligned}$$

The local record changes the conditional ensemble decomposition, but the unread state available at $B$ is unchanged.

Exercise 8: One PVM, two instruments

For

$$P_0 = \lvert0\rangle\langle0\rvert, \qquad P_1 = \lvert1\rangle\langle1\rvert,$$

compare:

$$M_0=P_0, \qquad M_1=P_1$$

with

$$N_0 = \lvert+\rangle\langle0\rvert, \qquad N_1 = \lvert+\rangle\langle1\rvert.$$

Show that both instruments have the same outcome probabilities for every input state. Find their conditional outputs and determine which instrument is repeatable.

Solution

For the first instrument,

$$M_a^\dagger M_a = P_a.$$

For the second,

$$\begin{aligned} N_0^\dagger N_0 &= \lvert0\rangle\langle+\vert+\rangle\langle0\rvert = P_0, \\ N_1^\dagger N_1 &= \lvert1\rangle\langle+\vert+\rangle\langle1\rvert = P_1. \end{aligned}$$

Both therefore assign

$$p(a) = \operatorname{Tr}(\rho P_a).$$

The first instrument has conditional outputs

$$\rho_0 = \lvert0\rangle\langle0\rvert, \qquad \rho_1 = \lvert1\rangle\langle1\rvert.$$

The second prepares the same state after either outcome:

$$\rho_0' = \rho_1' = \lvert+\rangle\langle+\rvert.$$

An immediate repetition of the same PVM after the first instrument reproduces the outcome with certainty. After the second instrument, either computational-basis outcome occurs with probability $1/2$, regardless of the first record. The first is repeatable; the second is not.