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.

Helpful background. Projective measurement supplies the outcome projectors; density operators let us describe both pure and mixed inputs. The derivations below use finite-dimensional systems and finitely many recorded outcomes.

Measurement Branches and Retained Records

To choose an update, specify the physical measurement, the record that remains available, and the prediction to be made next. A known outcome calls for a conditional state; an unread outcome calls for an average. The main example is the ideal Lüders instrument for a PVM $\{P_a\}$. Later we extend the same reasoning to general instruments, because outcome probabilities alone do not specify the state change.

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.

Normalization divides by a positive real norm. Any overall phase already present in the projected vector 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 sum includes only outcomes with positive probability; zero-probability branches contribute zero in the second sum. 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 $S$. Assume this retained event has positive probability. 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:

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

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

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

These expressions assume $p_1,p_2>0$. A zero-probability component is omitted, and the total probability must be positive. 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. Its outcome operations are linear, completely positive, and trace nonincreasing; their sum preserves trace. Complete positivity means positivity also holds when the measured system is part of a larger entangled system. In finite dimensions a general 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 probability of the record

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

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 sharp observables have no joint sharp measurement valid for every input. One can still construct classical joint distributions with given marginals, but these do not reproduce all quantum measurement sequences. Ordered predictions require the instruments and their order.
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

For real $\phi$ and $0\leq\theta\leq\pi$, 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.

Common Pitfalls

Dividing a ket by a probability. Divide a vector by $\sqrt{p(a)}$, and a density operator by $p(a)$. Record the branch probability before normalizing; normalization alone discards that information.

Averaging branches with equal weights. Use the actual probabilities, $\sum_a p(a)\rho_a$. If some outcomes are grouped together, first determine whether a fine measurement occurred or a coarse Lüders measurement was performed.

Confusing conditioning with physical dynamics. Normalized conditioning is generally nonlinear; the unnormalized outcome operations and the unread channel are linear. The update rule predicts later statistics without specifying a detector’s amplification mechanism or deciding how to interpret collapse; see what measurement formalism does not settle.

Using a remote conditional state without the record. A local trace-preserving operation on $A$ leaves the reduced state of $B$ unchanged when no intervening interaction couples them. The selected subensemble can change, as in the Bell-pair example, but its selection requires the classical record.

References

• 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.
• 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

$$\lvert+\rangle\langle+\rvert = \frac12 \bigl( \lvert0\rangle\langle0\rvert + \lvert0\rangle\langle1\rvert + \lvert1\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \bigr).$$

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 $a$.

For a concrete example, take $t=1/2$, $P=\operatorname{diag}(1,1,0)$, $\rho_1=\operatorname{diag}(0.8,0,0.2)$, and $\rho_2=\operatorname{diag}(0,0.2,0.8)$. The yes probabilities are $p_1=0.8$ and $p_2=0.2$, with conditional outputs $|0\rangle\langle0|$ and $|1\rangle\langle1|$. The selected mixture is therefore $\operatorname{diag}(0.8,0.2,0)$, whereas the equal mixture of the separately normalized outputs is $\operatorname{diag}(0.5,0.5,0)$. This explicitly exhibits the nonlinearity.

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

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

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

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

Multiplying this equation on the left by $P_1$ and on the right by $P_2$ shows that $P_1\rho P_2=0$ separately. Its adjoint vanishes too: the two states agree exactly when there is no coherence between these sectors.

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.