Sequential Measurements

Sequential measurements are measurements performed in a specified temporal order, with each outcome operation determining the state presented to the next stage. For ideal projective measurements, the ordered record $a$ then $b$ has probability

$$p_{A\to B}(a,b) = \operatorname{Tr} \left( Q_bP_a\rho P_aQ_b \right),$$

where $\{P_a\}$ is measured first and $\{Q_b\}$ second.

Order is part of the experimental procedure. Interchanging the two apparatuses generally changes the branches and their probabilities. When all outcome projectors commute, the sequence reduces to an ordinary sharp joint measurement; when they do not, the first measurement can alter the statistics of the second.

Purpose and Scope

This page is the canonical home for ordered measurement probabilities in the core formalism. It develops:

• measuring $A$ and then $B$;
• conditional, joint, and marginal probabilities;
• reversal of measurement order;
• compatible and incompatible projective sequences;
• immediate repetition and repetition after dynamics;
• unread intermediate measurements;
• arbitrary finite sequences;
• generalized instrument composition;
• fixed and adaptive second measurements;
• spin-$1/2$ examples.

The State Update Rule owns branch normalization and selective versus nonselective conditioning. Compatible Observables owns the general equivalence among commuting projectors, simultaneous diagonalization, and sharp joint measurability. Here those ingredients are used to calculate temporal records.

Ordered Notation

Let the first projective measurement have outcome projectors

$$\{P_a\}, \qquad \sum_aP_a=I,$$

and the second have

$$\{Q_b\}, \qquad \sum_bQ_b=I.$$

Unless an intervening evolution is written explicitly, the second measurement follows immediately after the first. The notation

$$p_{A\to B}(a,b)$$

means the probability for the ordered classical record “$a$ from $A$, then $b$ from $B$.” It is an ordinary nonnegative probability distribution for that one experimental procedure. It should not be interpreted as a symmetric joint distribution of pre-existing values for incompatible observables.

At a Glance

The unnormalized two-outcome branch is

$$\widetilde\rho_{a,b} = Q_bP_a\rho P_aQ_b.$$

Its trace is the ordered joint probability:

$$p_{A\to B}(a,b) = \operatorname{Tr}\widetilde\rho_{a,b}.$$

If

$$p_A(a) = \operatorname{Tr}(\rho P_a) > 0,$$

then the conditional probability of $b$ after $a$ is

$$p_{B\mid A}(b\mid a) = \frac{ p_{A\to B}(a,b) }{p_A(a)}.$$

If the full ordered record has nonzero probability, the final conditional state is

$$\rho_{a,b} = \frac{ \widetilde\rho_{a,b} }{ p_{A\to B}(a,b) }.$$

For a pure input,

$$p_{A\to B}(a,b) = \left\lVert Q_bP_a\lvert\psi\rangle \right\rVert^2.$$

The operator order reads from right to left: first $P_a$, then $Q_b$.

Measuring A Then B

For a normalized pure input $\lvert\psi\rangle$, the probability of the first outcome is

$$p_A(a) = \langle\psi\rvert P_a\lvert\psi\rangle.$$

If $p_A(a)>0$, the Lüders state after recording $a$ is

$$\lvert\psi_a\rangle = \frac{P_a\lvert\psi\rangle} {\sqrt{p_A(a)}}.$$

The conditional probability for the second outcome is

$$\begin{aligned} p_{B\mid A}(b\mid a) &= \langle\psi_a\rvert Q_b\lvert\psi_a\rangle \\ &= \frac{ \langle\psi\rvert P_aQ_bP_a \lvert\psi\rangle }{p_A(a)}. \end{aligned}$$

Multiplying by the first probability gives

$$\begin{aligned} p_{A\to B}(a,b) &= p_A(a) p_{B\mid A}(b\mid a) \\ &= \langle\psi\rvert P_aQ_bP_a \lvert\psi\rangle. \end{aligned}$$

Because $Q_b^2=Q_b$,

$$\begin{aligned} \left\lVert Q_bP_a\lvert\psi\rangle \right\rVert^2 &= \langle\psi\rvert P_aQ_b^2P_a \lvert\psi\rangle \\ &= p_{A\to B}(a,b). \end{aligned}$$

The final pure state, when this probability is nonzero, is

$$\lvert\psi_{a,b}\rangle = \frac{ Q_bP_a\lvert\psi\rangle }{ \sqrt{ p_{A\to B}(a,b) } }.$$

Density-Operator Branch Calculus

For a general input state, the unnormalized first branch is

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

Applying the second Lüders operation gives

$$\widetilde\rho_{a,b} = Q_b\widetilde\rho_aQ_b = Q_bP_a\rho P_aQ_b.$$

The branch trace can be written in several equivalent ways:

$$\begin{aligned} p_{A\to B}(a,b) &= \operatorname{Tr} \left( Q_bP_a\rho P_aQ_b \right) \\ &= \operatorname{Tr} \left( \rho P_aQ_bP_a \right) \\ &= \operatorname{Tr} \left( Q_bP_a\rho P_a \right). \end{aligned}$$

The last two forms use cyclicity of the trace and projector idempotence. The first form keeps the physical branch operation visible and is the safest expression when deriving the final state:

$$\rho_{a,b} = \frac{ Q_bP_a\rho P_aQ_b }{ \operatorname{Tr} \left( Q_bP_a\rho P_aQ_b \right) }.$$

No normalized intermediate state is required if only the joint probability or final branch is needed.

Probability Checks and Marginals

The ordered probabilities obey the ordinary probability axioms for the recorded sequence.

Positivity

The branch operator is positive:

$$\widetilde\rho_{a,b} \geq 0,$$

so

$$p_{A\to B}(a,b) \geq 0.$$

First-outcome marginal

Summing over the second result recovers the undisturbed probability of the first:

$$\begin{aligned} \sum_b p_{A\to B}(a,b) &= \sum_b \operatorname{Tr} \left( Q_bP_a\rho P_aQ_b \right) \\ &= \operatorname{Tr}(P_a\rho P_a) \\ &= p_A(a). \end{aligned}$$

Second-outcome marginal

Summing over the first result gives the probability of $b$ after the unread first measurement:

$$\begin{aligned} \sum_a p_{A\to B}(a,b) &= \operatorname{Tr} \left[ Q_b \sum_aP_a\rho P_a \right] \\ &= \operatorname{Tr} \left[ Q_b\mathcal L_A(\rho) \right]. \end{aligned}$$

This need not equal the probability for measuring $B$ directly on $\rho$.

Total normalization

Summing both records gives

$$\begin{aligned} \sum_{a,b} p_{A\to B}(a,b) &= \sum_a p_A(a) \\ &= 1. \end{aligned}$$

Unread Intermediate Measurement and Interference

If $B$ is measured directly, its outcome probability is

$$p_B^{\mathrm{direct}}(b) = \operatorname{Tr}(Q_b\rho).$$

Insert the first PVM on both sides of $\rho$:

$$\rho = \sum_{a,a'} P_a\rho P_{a'}.$$

Then

$$p_B^{\mathrm{direct}}(b) = \sum_{a,a'} \operatorname{Tr} \left( Q_bP_a\rho P_{a'} \right).$$

The second marginal of the sequential experiment contains only the diagonal terms:

$$p_B^{A\text{-first}}(b) = \sum_a \operatorname{Tr} \left( Q_bP_a\rho P_a \right).$$

Their difference is the interference contribution

$$\begin{aligned} & p_B^{\mathrm{direct}}(b) - p_B^{A\text{-first}}(b) \\ &\qquad = \sum_{\substack{a,a'\\a\neq a'}} \operatorname{Tr} \left( Q_bP_a\rho P_{a'} \right). \end{aligned}$$

Thus a naive classical law of total probability,

$$p_B(b) \stackrel{?}{=} \sum_a p_A(a) p_{B\mid A}(b\mid a),$$

computes the statistics of a different procedure: one in which $A$ was actually measured and its record ignored. It equals the direct $B$ probability only when the relevant interference terms vanish.

Reversing the Order

If $B$ is measured first and $A$ second, the ordered probability is

$$p_{B\to A}(b,a) = \operatorname{Tr} \left( P_aQ_b\rho Q_bP_a \right).$$

Equivalently,

$$p_{B\to A}(b,a) = \operatorname{Tr} \left( \rho Q_bP_aQ_b \right).$$

The difference between the two procedures for the same pair of labels is

$$\begin{aligned} \Delta_{ab}(\rho) &= p_{A\to B}(a,b) - p_{B\to A}(b,a) \\ &= \operatorname{Tr} \left[ \rho \left( P_aQ_bP_a - Q_bP_aQ_b \right) \right]. \end{aligned}$$

This order effect depends on the input state as well as the projectors. A vanishing difference for one special $\rho$ does not establish compatibility. Compatibility is a state-independent structural statement about every pair $P_a,Q_b$.

Compatible Projective Sequences

Suppose

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

Then

$$R_{ab} = P_aQ_b$$

is a projector:

$$\begin{aligned} R_{ab}^2 &= P_aQ_bP_aQ_b \\ &= P_a^2Q_b^2 \\ &= R_{ab}. \end{aligned}$$

The nonzero $R_{ab}$ form a joint PVM because

$$\sum_{a,b}R_{ab} = I.$$

The ordered probability reduces to the Born probability for the joint event:

$$\begin{aligned} p_{A\to B}(a,b) &= \operatorname{Tr} \left( \rho P_aQ_bP_a \right) \\ &= \operatorname{Tr}(\rho P_aQ_b) \\ &= \operatorname{Tr}(\rho R_{ab}). \end{aligned}$$

Reversing the order gives the same result:

$$p_{A\to B}(a,b) = p_{B\to A}(b,a).$$

Compatibility does not imply statistical independence. In general,

$$\operatorname{Tr}(\rho P_aQ_b) \neq \operatorname{Tr}(\rho P_a) \operatorname{Tr}(\rho Q_b).$$

It means that the alternatives possess a common sharp refinement and that the ideal Lüders sequence does not create an order effect.

Incompatible Projective Sequences

If some $P_a$ and $Q_b$ fail to commute, their product need not be a projector, and no PVM with joint projectors $P_aQ_b$ exists. The ordered branch remains perfectly well defined:

$$Q_bP_a\rho P_aQ_b.$$

What fails is the interpretation of the two outcomes as one order-independent sharp joint event.

Noncommutativity enters through the state update. The first operation removes or rearranges components relevant to the second. The resulting temporal probabilities depend on:

• the initial state;
• the order of the instruments;
• whether intermediate outcomes are retained;
• the state transformation associated with each outcome;
• any evolution or feedback between stages.

The commutator is therefore a structural diagnostic, not a substitute for calculating the actual ordered branches.

Repeating the Same Measurement

For an immediate repetition of the same Lüders PVM, set

$$Q_b = P_b.$$

Then

$$\begin{aligned} p_{A\to A}(a,b) &= \operatorname{Tr} \left( \rho P_aP_bP_a \right) \\ &= \delta_{ab} \operatorname{Tr}(\rho P_a). \end{aligned}$$

Conditional on the first outcome,

$$p(b\mid a) = \delta_{ab}.$$

The first result is not necessarily predictable, but the second repeats it with certainty.

For a degenerate $P_a$, this fixes only the coarse outcome. Internal labels can remain uncertain. A refined or outcome-dependent instrument can alter those internal degrees of freedom while still returning the same coarse value; see Degenerate Measurements and Lüders Rule.

Evolution Between Measurements

Suppose the first measurement is followed by unitary time evolution $U$ before the second. The branch becomes

$$\widetilde\rho_{a,b} = Q_bUP_a\rho P_aU^\dagger Q_b.$$

The joint probability is

$$p_{A\to U\to B}(a,b) = \operatorname{Tr} \left( Q_bUP_a\rho P_aU^\dagger \right).$$

For a repeated measurement of the same PVM,

$$p(b\mid a;U) = \operatorname{Tr} \left( P_bU\rho_aU^\dagger \right).$$

This need not equal $\delta_{ab}$. Repeatability applies only when no intervening dynamics moves the state between outcome subspaces.

For a nondegenerate pure outcome $\lvert a\rangle$, the survival probability is

$$p(a\mid a;U) = \left\lvert \langle a\rvert U\lvert a\rangle \right\rvert^2.$$

Repeated measurements separated by short evolutions are the starting point for the quantum Zeno effect, but that dynamical phenomenon requires its own limiting analysis.

Longer Projective Sequences

Consider $n$ projective measurements with recorded outcomes

$$\boldsymbol a = (a_1,\ldots,a_n).$$

Let $P_{a_k}^{(k)}$ denote the projector at stage $k$. Define the branch operator

$$K_{\boldsymbol a} = P_{a_n}^{(n)} \cdots P_{a_2}^{(2)} P_{a_1}^{(1)}.$$

The unnormalized final branch is

$$\widetilde\rho_{\boldsymbol a} = K_{\boldsymbol a} \rho K_{\boldsymbol a}^\dagger,$$

and the probability of the complete ordered record is

$$p(\boldsymbol a) = \operatorname{Tr} \left( K_{\boldsymbol a} \rho K_{\boldsymbol a}^\dagger \right).$$

For a pure input,

$$p(\boldsymbol a) = \left\lVert K_{\boldsymbol a} \lvert\psi\rangle \right\rVert^2.$$

Summing over every possible record gives one because the PVM completeness relation can be applied successively from the last stage backward.

If unitary operators $U_{k+1,k}$ act between stages, insert them in temporal order:

$$\begin{aligned} K_{\boldsymbol a} &= P_{a_n}^{(n)} U_{n,n-1} P_{a_{n-1}}^{(n-1)} \cdots \\ &\qquad \cdots U_{2,1} P_{a_1}^{(1)}. \end{aligned}$$

General Instruments and Adaptive Sequences

The branch-composition rule extends beyond projectors. Let

$$\mathcal I_a^A$$

be the first outcome operation and

$$\mathcal I_b^B$$

the second. The unnormalized sequential branch is

$$\widetilde\rho_{a,b} = \mathcal I_b^B \left( \mathcal I_a^A(\rho) \right).$$

The joint probability is

$$p_{A\to B}(a,b) = \operatorname{Tr} \left[ \mathcal I_b^B \left( \mathcal I_a^A(\rho) \right) \right].$$

The order of composition is essential:

$$\mathcal I_b^B \circ \mathcal I_a^A \neq \mathcal I_a^A \circ \mathcal I_b^B$$

in general.

If the second measurement is chosen using the first record, write its operation as

$$\mathcal I_{b\mid a}^{B}.$$

Then

$$\widetilde\rho_{a,b} = \mathcal I_{b\mid a}^{B} \left( \mathcal I_a^A(\rho) \right).$$

This describes feed-forward protocols, adaptive tomography, measurement-based control, and conditional recovery. Detailed completely positive instrument theory belongs in Quantum Instruments.

Spin Along Two Arbitrary Axes

Let the initial qubit state be

$$\rho = \frac12 \left( I+\boldsymbol r\cdot\boldsymbol\sigma \right), \qquad \lVert\boldsymbol r\rVert\leq1.$$

First measure spin along the unit vector $\boldsymbol n$, with result $s=\pm1$:

$$P_s(\boldsymbol n) = \frac12 \left( I+s\boldsymbol n\cdot\boldsymbol\sigma \right).$$

Then measure along $\boldsymbol m$, with result $t=\pm1$:

$$P_t(\boldsymbol m) = \frac12 \left( I+t\boldsymbol m\cdot\boldsymbol\sigma \right).$$

The first probability is

$$p_{\boldsymbol n}(s) = \frac12 \left( 1+s\boldsymbol r\cdot\boldsymbol n \right).$$

Because the first projector has rank one, the conditional state is $P_s(\boldsymbol n)$. The second conditional probability is

$$\begin{aligned} p_{\boldsymbol m\mid\boldsymbol n}(t\mid s) &= \operatorname{Tr} \left[ P_t(\boldsymbol m) P_s(\boldsymbol n) \right] \\ &= \frac12 \left( 1+st\boldsymbol n\cdot\boldsymbol m \right). \end{aligned}$$

The ordered joint distribution is

$$\begin{aligned} p_{\boldsymbol n\to\boldsymbol m}(s,t) &= \frac14 \left( 1+s\boldsymbol r\cdot\boldsymbol n \right) \\ &\qquad\times \left( 1+st\boldsymbol n\cdot\boldsymbol m \right). \end{aligned}$$

The later marginal is

$$\begin{aligned} p_{\boldsymbol m}^{\boldsymbol n\text{-first}}(t) &= \sum_s p_{\boldsymbol n\to\boldsymbol m}(s,t) \\ &= \frac12 \left[ 1+ t (\boldsymbol r\cdot\boldsymbol n) (\boldsymbol n\cdot\boldsymbol m) \right]. \end{aligned}$$

By contrast, measuring along $\boldsymbol m$ directly gives

$$p_{\boldsymbol m}^{\mathrm{direct}}(t) = \frac12 \left( 1+t\boldsymbol r\cdot\boldsymbol m \right).$$

The intermediate measurement retains only the component of the initial Bloch vector along $\boldsymbol n$.

The correlation of the two recorded signs is

$$\begin{aligned} \langle st\rangle_{\boldsymbol n\to\boldsymbol m} &= \sum_{s,t} st\, p_{\boldsymbol n\to\boldsymbol m}(s,t) \\ &= \boldsymbol n\cdot\boldsymbol m. \end{aligned}$$

For ideal rank-one spin measurements, this temporal correlation is independent of the initial Bloch vector.

Worked Spin Sequence: z, Then x, Then z

Prepare $\lvert+z\rangle$. The first $S_z$ measurement returns $+z$ with probability one.

The subsequent $S_x$ measurement has outcomes

$$p(+x\mid+z) = p(-x\mid+z) = \frac12.$$

After either $x$ outcome, the final $S_z$ measurement gives

$$p(+z\mid\pm x) = p(-z\mid\pm x) = \frac12.$$

The four nonzero complete records therefore have probabilities

$$\begin{aligned} p(+z,+x,+z) &= \frac14, \\ p(+z,+x,-z) &= \frac14, \\ p(+z,-x,+z) &= \frac14, \\ p(+z,-x,-z) &= \frac14. \end{aligned}$$

If $S_z$ were measured twice without the intermediate $S_x$ measurement, the final result would remain $+z$ with certainty. The inserted incompatible measurement changes the temporal statistics even if its $x$ record is later ignored.

Degeneracy and Instrument Dependence

For degenerate PVMs, the formulas on this page assume Lüders outcome operations

$$\mathcal I_a^{\mathrm L}(\rho) = P_a\rho P_a.$$

A refinement or outcome-dependent transformation inside the eigenspace can give the same first probabilities while changing the second measurement statistics. Thus

$$p_{A\to B}(a,b)$$

is not determined by the PVM effects $\{P_a\}$ and $\{Q_b\}$ alone unless the intermediate instruments are specified.

This is why sequential experiments can diagnose measurement disturbance. A single first-outcome histogram determines effects, while later conditional probabilities can reveal how the first apparatus transformed the quantum output.

What an Ordered Probability Means

The distribution

$$p_{A\to B}(a,b)$$

is a joint distribution of two records produced by one temporal protocol. It is not generally:

• symmetric under interchange of $A$ and $B$;
• a probability distribution for simultaneous sharp values;
• a context-free property of the input state alone;
• determined by outcome effects without state-update operations;
• evidence that the system possessed both values before either measurement.

These distinctions prevent a correct operational probability from being overinterpreted as a classical joint distribution for incompatible observables.

A Reliable Calculation Workflow

1. Write the temporal order. Name which apparatus acts first, second, and later.

2. Specify every outcome operation. For ideal projective measurements, state whether the Lüders rule is assumed.

3. Build the unnormalized branch from right to left.

   $$\widetilde\rho_{a,b} = Q_bP_a\rho P_aQ_b.$$

4. Take its trace for the ordered probability.

   $$p(a,b) = \operatorname{Tr}\widetilde\rho_{a,b}.$$

5. Normalize only when a conditional state is needed.

   $$\rho_{a,b} = \frac{\widetilde\rho_{a,b}}{p(a,b)}.$$

6. Insert dynamics in the correct position. A unitary between measurements belongs between their projectors.

7. Check marginals. The first marginal must reproduce the first measurement statistics; the second marginal refers to the disturbed or unread intermediate state.

8. Check normalization.

   $$\sum_{a,b}p(a,b) = 1.$$

9. Reverse the order explicitly if an order comparison is required. Do not rearrange noncommuting factors.

Common Mistakes

• Multiplying independent Born probabilities. The second probability is conditional on the updated first branch.
• Reading operator products from left to right. In $Q_bP_a\lvert\psi\rangle$, $P_a$ acts first.
• Dropping the second projector around a density operator. The branch is $Q_bP_a\rho P_aQ_b$.
• Using the direct $B$ probability as the second marginal. The first unread measurement may change it.
• Applying the classical law of total probability to an unperformed intermediate measurement. The missing terms are quantum coherences.
• Assuming ordered probabilities are symmetric. The reversed protocol has different branch operators.
• Treating one vanishing order effect as proof of compatibility. Special states can hide noncommutativity.
• Equating compatibility with independence. Compatible outcomes can be strongly correlated.
• Claiming repeatability after intervening evolution. The state may leave the selected eigenspace.
• Ignoring degeneracy or instrument details. The same PVM effects can accompany different intermediate disturbances.
• Averaging conditional states without probabilities. Unread branches are summed unnormalized.
• Treating an ordered record distribution as pre-existing simultaneous values. It belongs to a specified temporal experiment.

Canonical Boundaries and Cross-Links

• Projective Measurement owns PVM structure and immediate ideal repeatability.
• State Update Rule owns unnormalized branches, conditioning, and unread channels.
• Degenerate Measurements and Lüders Rule owns coarse versus refined intermediate instruments.
• Compatible Observables owns state-independent sharp compatibility criteria.
• Commutators owns commutator algebra.
• Probability in Different Bases owns single-measurement basis-overlap probabilities.
• Unitary Time Evolution owns propagation between measurement times.
• Quantum Instruments owns general completely positive outcome-map composition.

Summary

Sequential measurement probabilities are obtained by composing outcome operations in temporal order. For projective Lüders measurements,

$$\widetilde\rho_{a,b} = Q_bP_a\rho P_aQ_b,$$

and

$$p_{A\to B}(a,b) = \operatorname{Tr}\widetilde\rho_{a,b}.$$

The first marginal reproduces $p_A(a)$, while the second marginal describes $B$ after the unread $A$ measurement. Commuting projectors produce an order-independent joint PVM; noncommuting projectors generally produce order effects. Longer sequences, intervening dynamics, generalized instruments, and adaptive choices all follow the same branch-composition rule.

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955 — projective measurement and successive state reduction.
• G. Lüders, “Über die Zustandsänderung durch den Meßprozeß,” Annalen der Physik 8, 322–328, 1951; English translation and discussion by K. A. Kirkpatrick, Annalen der Physik 15, 663–670, 2006, arXiv:quant-ph/0403007 — ideal state update and compatibility.
• 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 — instruments, conditional probabilities, and repeated measurements.
• L. M. Johansen, “Quantum theory of successive projective measurements,” Physical Review A 76, 012119, 2007, doi:10.1103/PhysRevA.76.012119, arXiv:0705.0229 — ordered projective probabilities and disturbance terms.
• S. Gudder, “Sequential products of quantum measurements,” Reports on Mathematical Physics 60, 273–288, 2007, doi:10.1016/S0034-4877(07)80139-X — sequential products, compatibility, and conditioning.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995 — operational analysis of compatible and successive measurements.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — spin measurements and state update.
• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016 — general instruments and sequential measurement theory.

Exercises

Exercise 1: Derive the two-projector branch

For a pure input $\lvert\psi\rangle$, derive

$$p_{A\to B}(a,b) = \left\lVert Q_bP_a\lvert\psi\rangle \right\rVert^2$$

from conditional probability. Find the normalized final state when the joint probability is nonzero.

Solution

The first outcome probability is

$$p_A(a) = \langle\psi\rvert P_a\lvert\psi\rangle.$$

After recording $a$,

$$\lvert\psi_a\rangle = \frac{P_a\lvert\psi\rangle} {\sqrt{p_A(a)}}.$$

The conditional probability for $b$ is

$$\begin{aligned} p_{B\mid A}(b\mid a) &= \langle\psi_a\rvert Q_b\lvert\psi_a\rangle \\ &= \frac{ \langle\psi\rvert P_aQ_bP_a \lvert\psi\rangle }{p_A(a)}. \end{aligned}$$

Multiplication by $p_A(a)$ gives

$$\begin{aligned} p_{A\to B}(a,b) &= \langle\psi\rvert P_aQ_bP_a \lvert\psi\rangle \\ &= \left\lVert Q_bP_a\lvert\psi\rangle \right\rVert^2. \end{aligned}$$

The final state is

$$\lvert\psi_{a,b}\rangle = \frac{ Q_bP_a\lvert\psi\rangle }{ \sqrt{ p_{A\to B}(a,b) } }.$$

Exercise 2: The second marginal and missing interference

Let $\{P_a\}$ be measured and ignored before a projective measurement $\{Q_b\}$. Show that

$$p_B^{A\text{-first}}(b) = \sum_a \operatorname{Tr} \left( Q_bP_a\rho P_a \right).$$

Express the difference from the direct probability $\operatorname{Tr}(Q_b\rho)$ in terms of off-diagonal blocks $P_a\rho P_{a'}$.

Solution

After the unread first measurement,

$$\rho' = \sum_aP_a\rho P_a.$$

Therefore

$$\begin{aligned} p_B^{A\text{-first}}(b) &= \operatorname{Tr}(Q_b\rho') \\ &= \sum_a \operatorname{Tr} \left( Q_bP_a\rho P_a \right). \end{aligned}$$

For the direct experiment, insert identity resolutions:

$$\begin{aligned} \operatorname{Tr}(Q_b\rho) &= \sum_{a,a'} \operatorname{Tr} \left( Q_bP_a\rho P_{a'} \right). \end{aligned}$$

Subtracting the sequential marginal leaves

$$\begin{aligned} & \operatorname{Tr}(Q_b\rho) - p_B^{A\text{-first}}(b) \\ &\qquad = \sum_{\substack{a,a'\\a\neq a'}} \operatorname{Tr} \left( Q_bP_a\rho P_{a'} \right). \end{aligned}$$

The difference consists of coherences between distinct first-measurement sectors.

Exercise 3: Construct the compatible joint PVM

Suppose $[P_a,Q_b]=0$ for all $a,b$. Let

$$R_{ab} = P_aQ_b.$$

Show that the nonzero $R_{ab}$ are mutually orthogonal projectors that sum to the identity, and show that either measurement order gives $\operatorname{Tr}(\rho R_{ab})$.

Solution

Commutation and idempotence give

$$\begin{aligned} R_{ab}^2 &= P_aQ_bP_aQ_b \\ &= P_a^2Q_b^2 \\ &= R_{ab}. \end{aligned}$$

The adjoint is

$$R_{ab}^\dagger = Q_bP_a = R_{ab}.$$

For different pairs,

$$\begin{aligned} R_{ab}R_{a'b'} &= P_aP_{a'}Q_bQ_{b'} \\ &= \delta_{aa'}\delta_{bb'} R_{ab}. \end{aligned}$$

Completeness gives

$$\sum_{a,b}R_{ab} = \left( \sum_aP_a \right) \left( \sum_bQ_b \right) = I.$$

Finally,

$$\begin{aligned} p_{A\to B}(a,b) &= \operatorname{Tr}(\rho P_aQ_bP_a) \\ &= \operatorname{Tr}(\rho P_aQ_b) \\ &= \operatorname{Tr}(\rho R_{ab}), \end{aligned}$$

and the reversed order gives the same expression.

Exercise 4: Spin along arbitrary axes

A qubit with Bloch vector $\boldsymbol r$ is measured first along $\boldsymbol n$ and then along $\boldsymbol m$. Derive

$$p(s,t) = \frac14 \left( 1+s\boldsymbol r\cdot\boldsymbol n \right) \left( 1+st\boldsymbol n\cdot\boldsymbol m \right).$$

Check the limits $\boldsymbol m=\boldsymbol n$ and $\boldsymbol m\perp\boldsymbol n$.

Solution

The first probability is

$$p(s) = \operatorname{Tr} \left[ \rho P_s(\boldsymbol n) \right] = \frac12 \left( 1+s\boldsymbol r\cdot\boldsymbol n \right).$$

After a nonzero-probability rank-one outcome, the state is $P_s(\boldsymbol n)$. Therefore

$$\begin{aligned} p(t\mid s) &= \operatorname{Tr} \left[ P_t(\boldsymbol m) P_s(\boldsymbol n) \right] \\ &= \frac12 \left( 1+st\boldsymbol n\cdot\boldsymbol m \right). \end{aligned}$$

Multiplying gives the stated joint distribution.

If $\boldsymbol m=\boldsymbol n$, then

$$p(t\mid s) = \frac12(1+st) = \delta_{st},$$

so the outcome repeats. If $\boldsymbol m\perp\boldsymbol n$, then

$$p(t\mid s) = \frac12,$$

so the second result is unbiased for either first result.

Exercise 5: A z, x, z sequence

A spin-$1/2$ system begins in $\lvert+z\rangle$. Compute the complete record probabilities for measurements of $S_z$, then $S_x$, then $S_z$. Find the marginal distribution of the final $S_z$ outcome.

Solution

The first result is $+z$ with probability one. Conditional on that result,

$$p(\pm x\mid+z) = \frac12.$$

For either $x$ eigenstate,

$$p(\pm z\mid\pm x) = \frac12.$$

Hence each of the four records

$$\begin{aligned} (+z,+x,+z), &\quad (+z,+x,-z), \\ (+z,-x,+z), &\quad (+z,-x,-z) \end{aligned}$$

has probability $1/4$. Summing over the intermediate $x$ record gives

$$p_{\mathrm{final}}(+z) = \frac12, \qquad p_{\mathrm{final}}(-z) = \frac12.$$

The incompatible intermediate measurement converts the initially certain final $z$ result into an unbiased distribution.

Exercise 6: Repetition after a unitary rotation

An ideal $S_z$ measurement returns $+z$. The qubit then evolves under

$$U(\theta) = \exp \left( -\frac{i\theta}{2}\sigma_y \right)$$

before $S_z$ is measured again. Find the two conditional probabilities.

Solution

The first conditional state is $\lvert+z\rangle$. A rotation about $y$ gives

$$U(\theta)\lvert+z\rangle = \cos\frac{\theta}{2} \lvert+z\rangle + \sin\frac{\theta}{2} \lvert-z\rangle,$$

up to the phase convention for $\lvert-z\rangle$. Therefore

$$\begin{aligned} p(+z\mid+z;\theta) &= \cos^2\frac{\theta}{2}, \\ p(-z\mid+z;\theta) &= \sin^2\frac{\theta}{2}. \end{aligned}$$

At $\theta=0$, immediate repeatability is recovered. At $\theta=\pi$, the outcome flips with certainty.

Exercise 7: Three projective stages

For PVMs

$$\{P_a\}, \qquad \{Q_b\}, \qquad \{R_c\},$$

derive the unnormalized branch, joint probability, and final conditional state for the record $(a,b,c)$. Show that the probabilities sum to one.

Solution

The branch operator is

$$K_{a,b,c} = R_cQ_bP_a.$$

The unnormalized density operator is

$$\widetilde\rho_{a,b,c} = R_cQ_bP_a \rho P_aQ_bR_c.$$

Its trace is

$$p(a,b,c) = \operatorname{Tr} \left( R_cQ_bP_a \rho P_aQ_bR_c \right).$$

For nonzero probability, the final state is

$$\rho_{a,b,c} = \frac{ \widetilde\rho_{a,b,c} }{p(a,b,c)}.$$

To check normalization, sum over $c$ first and use $\sum_cR_c=I$, then over $b$, then over $a$:

$$\begin{aligned} \sum_{a,b,c}p(a,b,c) &= \operatorname{Tr} \left( \rho\sum_aP_a \right) \\ &= \operatorname{Tr}\rho \\ &= 1. \end{aligned}$$

Exercise 8: Instrument composition

The first instrument measures the computational-basis effects but resets the qubit:

$$\begin{aligned} M_0 &= \lvert0\rangle\langle0\rvert, \\ M_1 &= \lvert0\rangle\langle1\rvert. \end{aligned}$$

The second measurement is the rank-one $x$-basis PVM $\{Q_+,Q_-\}$. For an arbitrary input $\rho$, find $p(a,\pm)$ and compare it with the Lüders computational-basis instrument followed by the same $x$ measurement.

Solution

The first outcome probabilities are

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

because

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

After either outcome, the reset instrument outputs $\lvert0\rangle\langle0\rvert$. Since

$$\left\lvert \langle\pm x\vert0\rangle \right\rvert^2 = \frac12,$$

the sequential probabilities are

$$p_{\mathrm R}(a,\pm) = \frac12 \operatorname{Tr}(\rho P_a).$$

The Lüders computational-basis instrument outputs $\lvert a\rangle\langle a\rvert$. Each computational-basis state is also unbiased in the $x$ basis, so

$$p_{\mathrm L}(a,\pm) = \frac12 \operatorname{Tr}(\rho P_a).$$

For this particular second measurement, the two instruments are not distinguished. A later $z$-basis measurement would distinguish them: the reset instrument always gives $0$, while the Lüders instrument repeats $a$. This illustrates that one sequential probe may be insensitive to a real difference between instruments.