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 Lüders projective measurements without intervening evolution, 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.

Helpful background. State update supplies the branch operations; compatible observables explains commuting outcome projectors. We use finite-dimensional systems with finitely many outcomes, and Lüders operations unless another instrument is specified.

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.

Measuring A Then B

The operator product $Q_bP_a$ acts from right to left: first $P_a$, then $Q_b$. 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

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

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 exactly when the sum of the relevant interference contributions vanishes. Individual terms can cancel; the classical law itself remains valid within either specified experiment.

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

Composing system operations assumes that each stage has the stated instrument on the system passed to it, with no additional unmodeled apparatus memory. A fresh or reset probe is a common way to ensure this. If a reused probe retains correlations with the system, include that probe in the propagated state: independently calibrated single-stage instruments on the system alone need not predict the sequence. The instrument formulation uses the same assumption.

The branch-composition rule extends beyond projectors. Each outcome operation is completely positive and trace nonincreasing, and each full instrument sums to a trace-preserving map. 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

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

For each fixed $a$, the second-stage operations must form a complete instrument: $\sum_b\mathcal I^B_{b\mid a}$ preserves trace. This ensures $\sum_b p(a,b)=p_A(a)$ and total probability one even when the apparatus settings depend on the first result. Examples include adaptive tomography and conditional recovery. The general theory is developed 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 the unit vector $\boldsymbol m$, with result $t=\pm1$:

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

The first probability is

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

When its outcome has positive probability, the first rank-one projector gives the conditional state $P_s(\boldsymbol n)$. 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

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

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. Outcome histograms for a sufficiently rich set of known input states can determine the effects; a single input generally cannot. Later conditional probabilities can reveal how the first apparatus transformed the quantum output.

The Sequence as One POVM

There is a useful way to regard the complete pair of recorded outcomes. Define

$$G_{ab}=P_aQ_bP_a.$$

These operators are positive because $G_{ab}=(Q_bP_a)^\dagger(Q_bP_a)$, and $\sum_{a,b}G_{ab}=I$. Thus the whole experiment is a POVM on the initial state, with $p(a,b)=\operatorname{Tr}(\rho G_{ab})$.

Its marginals explain the disturbance:

$$\sum_bG_{ab}=P_a,\qquad \sum_aG_{ab}=\sum_aP_aQ_bP_a.$$

The first marginal is the original $A$ PVM, but the second is generally a modified measurement of $B$. There is no contradiction with the impossibility of a joint sharp measurement for noncommuting observables: the second marginal need not be $Q_b$.

Common Pitfalls

Multiplying two initial Born probabilities. The second measurement acts on the first branch, so its probability is conditional. Compatibility permits correlations and does not imply statistical independence.

Dropping a projector from the output state. Idempotence and cyclicity can shorten a trace, but the unnormalized state still has the full form $Q_bP_a\rho P_aQ_b$.

Testing compatibility with one special state. A maximally mixed qubit gives the same two-axis record probabilities in both orders even when the projectors do not commute. Compatibility must hold for every input state.

Expecting repetition despite intervening evolution. Immediate Lüders repetition is certain. A unitary, a different measurement, or feedback can move the state into a different outcome subspace before the next readout.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016 — general instruments and sequential measurement theory.
• 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.
• 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.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — spin measurements and state update.

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

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

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: An adaptive second measurement

Prepare $|+z\rangle$ and measure $S_x$. After result $+x$, measure $S_z$; after result $-x$, measure $S_x$ again. Find every nonzero complete record probability and check normalization.

Solution

The first outcomes have probability $1/2$. The $+x$ branch gives either $z$ result with conditional probability $1/2$, whereas the $-x$ branch repeats $-x$ with certainty. Thus

$$p(+x,+z)=\frac14,\qquad p(+x,-z)=\frac14,\qquad p(-x,-x)=\frac12.$$

They sum to one. The second record must include the measurement setting: the last label is an $x$ result, while the first two are $z$ results. Summing over the permitted second records recovers probability $1/2$ for each first outcome.

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.