Compatible, Incompatible, and Sequential Measurements

Sequential measurements are measurements performed in a specified order, with the state left by earlier measurement operations used as the input for later ones. The order is part of the experiment, not an annotation added after the fact.

The earlier instrument determines the state on which later probabilities depend. For projective measurements, this includes the familiar order dependence of noncommuting observables. For generalized measurements, compatible outcome statistics need not imply that a chosen implementation is nondisturbing.

Helpful background. State update rules explains instruments. The basic sequential measurement formula develops the projective special case.

Ordered Instruments

Let the first measurement have outcome operations $\{\mathcal I_a\}$ and the second measurement have outcome operations $\{\mathcal J_b\}$. The ordered joint probability for first $a$, then $b$, is

$$p(a,b) = \operatorname{Tr} \!\left[ \mathcal J_b(\mathcal I_a(\rho)) \right].$$

If only the second outcome probability is needed after the first outcome $a$ is known, write

$$p(a)=\operatorname{Tr}\mathcal I_a(\rho), \qquad \rho_a = \frac{\mathcal I_a(\rho)}{p(a)}$$

when $p(a)\ne0$. Then

$$p(b|a) = \operatorname{Tr}\mathcal J_b(\rho_a), \qquad p(a,b)=p(a)p(b|a).$$

For an $n$-step measurement record $r_1,\ldots,r_n$, the same idea becomes an ordered composition:

$$p(r_1,\ldots,r_n) = \operatorname{Tr} \!\left[ \mathcal I^{(n)}_{r_n} \circ \cdots \circ \mathcal I^{(1)}_{r_1} (\rho) \right].$$

The rightmost operation acts first. These finite-dimensional formulas assume specified instruments for each step. If a channel $\mathcal E$ acts between two measurements, insert it as $\mathcal J_b\circ\mathcal E\circ\mathcal I_a$. If a detector retains memory, its next operation can depend on the earlier record or require the detector to be included in the state.

Projective Special Case

Let the first sharp measurement have projectors $\{P_a\}$ and the second have projectors $\{Q_b\}$. In the ideal Lüders model,

$$\mathcal I_a(\rho)=P_a\rho P_a, \qquad \mathcal J_b(\rho)=Q_b\rho Q_b.$$

The ordered joint probability is

$$p(a\ \text{then}\ b) = \operatorname{Tr} \!\left[ Q_bP_a\rho P_aQ_b \right].$$

Using $Q_b^2=Q_b$ and trace cyclicity, this is often written as

$$p(a\ \text{then}\ b) = \operatorname{Tr}(P_aQ_bP_a\rho).$$

The reversed experiment has

$$p(b\ \text{then}\ a) = \operatorname{Tr}(Q_bP_aQ_b\rho).$$

These are generally different operators and therefore different experiments. The difference is not only that two symbols are written in another order; the first measurement changes the state used to predict the second.

Compatible Projective Measurements

For two projective measurements, compatibility is expressed by commuting projectors:

$$P_aQ_b=Q_bP_a \qquad \text{for all }a,b.$$

Then $P_aQ_b$ is itself a projector onto the joint outcome subspace, and

$$p(a\ \text{then}\ b) = \operatorname{Tr}(P_aQ_b\rho) = p(b\ \text{then}\ a).$$

With these Lüders instruments, the nonselective first measurement also leaves the later statistics of the second projective measurement unchanged:

$$\operatorname{Tr} \!\left[ Q_b \sum_aP_a\rho P_a \right] = \operatorname{Tr}(Q_b\rho).$$

Compatibility does not mean statistical independence. It means the alternatives can be refined into a common projective measurement with joint outcomes $(a,b)$.

The algebraic background is Compatible Observables and Noncommuting Observables.

Incompatible Measurements and Disturbance

If $P_a$ and $Q_b$ do not commute, an intermediate unread measurement of $\{P_a\}$ can change the later $Q_b$ statistics. Without the intermediate measurement,

$$p_0(b)=\operatorname{Tr}(Q_b\rho).$$

With the intermediate measurement performed but ignored,

$$p_{\mathrm{unread}}(b) = \sum_a \operatorname{Tr}(Q_bP_a\rho P_a).$$

The difference comes from the off-diagonal blocks of $\rho$ in the $P_a$ decomposition:

$$\rho = \sum_{a,c}P_a\rho P_c.$$

The unread measurement removes the terms with $a\ne c$. If $Q_b$ is sensitive to those coherences, the later outcome distribution changes.

This is measurement disturbance in its most operational form: the first measurement interaction changes the probabilities assigned to a later experiment.

Selective and Nonselective Branches

Sequential experiments force one to say whether an intermediate record is retained.

If the first outcome $a$ is known, the later probability is computed from the selective state:

$$p(b|a) = \operatorname{Tr} \!\left[ F_b \frac{\mathcal I_a(\rho)} {\operatorname{Tr}\mathcal I_a(\rho)} \right],$$

where $F_b$ is the later effect if the second measurement is only being used for probabilities.

If the first measurement happens but the outcome is ignored, the later probability is

$$p(b) = \operatorname{Tr} \!\left[ F_b \sum_a\mathcal I_a(\rho) \right].$$

These are predictions for different ensembles. The selective state describes the subensemble with a known record. The nonselective state describes the unsorted ensemble after the physical measurement interaction. See Selective and Nonselective Measurements for the general distinction.

Compatible unsharp measurements can still disturb each other

Two POVMs $\{E_s\}$ and $\{F_t\}$ are jointly measurable if there is a positive joint POVM $\{G_{st}\}$ with marginals $\sum_tG_{st}=E_s$ and $\sum_sG_{st}=F_t$. Unlike PVMs, compatible POVMs need not commute.

For example, take noisy qubit $z$ and $x$ effects

$$E_s=\frac{I+s\eta\sigma_z}{2},\qquad F_t=\frac{I+t\eta\sigma_x}{2},\qquad s,t=\pm1.$$

For $0\leq\eta\leq1/\sqrt2$, the operators

$$G_{st}=\frac14[I+\eta(s\sigma_z+t\sigma_x)]$$

are positive: their eigenvalues are $(1\pm\sqrt2\eta)/4$. Summing either label yields the required marginal. For $\eta>0$ the $E$ and $F$ effects do not commute, yet this joint measurement exists.

Now implement the first POVM by its square-root instrument $K_s=\sqrt{E_s}$. Its unread channel multiplies the input $x$ component by $\sqrt{1-\eta^2}$. Starting from $|+x\rangle$, the subsequent $F_+$ probability changes from $(1+\eta)/2$ to

$$\frac{1+\eta\sqrt{1-\eta^2}}2.$$

At $\eta=1/2$, these are $3/4$ and $(4+\sqrt3)/8\simeq0.717$. Joint measurability is an existence claim about a joint apparatus; it does not make every sequential implementation nondisturbing.

Nondisturbance for a Later Test

For a general first measurement, let the nonselective channel be

$$\Phi(\rho)=\sum_a\mathcal I_a(\rho).$$

The first measurement is nondisturbing for a later effect $F_b$ if

$$\operatorname{Tr}[F_b\Phi(\rho)] = \operatorname{Tr}(F_b\rho)$$

for all input states $\rho$. In the Heisenberg-picture adjoint channel, this is

$$\Phi^\dagger(F_b)=F_b.$$

For an ideal projective measurement of $\{P_a\}$ followed by another projective question $\{Q_b\}$, this nondisturbance condition is guaranteed when all $P_a$ commute with all $Q_b$. For generalized measurements, nondisturbance can be subtler: an unsharp measurement may disturb less than a sharp one, and a noisy instrument may disturb more than its POVM alone suggests.

Spin One-Half Sequence

For a spin-$1/2$ particle, the projector for outcome $s=\pm1$ along unit vector $\boldsymbol n$ is

$$P_{\boldsymbol n,s} = \frac{1}{2} \left( I+s\,\boldsymbol n\cdot\boldsymbol\sigma \right).$$

If the system is first projected into outcome $s$ along $\boldsymbol n$, a subsequent ideal measurement along $\boldsymbol m$ gives outcome $t=\pm1$ with probability

$$p(t|s) = \operatorname{Tr} \!\left( P_{\boldsymbol m,t} P_{\boldsymbol n,s} \right) = \frac{1}{2} \left( 1+st\,\boldsymbol n\cdot\boldsymbol m \right).$$

For perpendicular axes, $\boldsymbol n\cdot\boldsymbol m=0$, so the second outcome is unbiased. Thus a $z$-definite beam sent through an $x$ analyzer and then a $z$ analyzer loses the original certainty of the final $z$ outcome. Conditioning on the intermediate result prepares an $S_x$ eigenstate; discarding it gives an equal mixture of the two $S_x$ eigenstates.

If the same axis is repeated, $\boldsymbol n\cdot\boldsymbol m=1$. Then $p(t=s|s)=1$ and $p(t=-s|s)=0$, the ideal repeatability result.

The Stern–Gerlach implementation and analyzer language are developed in Stern–Gerlach Revisited.

Degenerate outcomes in a sequence

A later observable can probe coherences inside a degenerate eigenspace. An initial Lüders measurement preserves them; a refined measurement followed by discarding its finer label can erase them. Both can have the same coarse PVM and be repeatable for its outcomes. The three-level example calculates the resulting difference in later probabilities.

Temporal Correlations

Sequential measurement data often enter through temporal correlation functions. If outcomes $q_i=\pm1$ are recorded at times $t_i$, one can define

$$C_{ij} = \sum_{q_i,q_j=\pm1} q_iq_j\,p(q_i,q_j).$$

The probabilities in this expression are sequential probabilities for a specified measurement protocol. They are not automatically the same as expectation values of operator products computed without measurement backaction.

This distinction matters in Leggett–Garg tests, which compare temporal correlations against assumptions such as macroscopic realism and noninvasive measurability. One common Leggett–Garg form for dichotomic outcomes is

$$K=C_{12}+C_{23}-C_{13}\le1,$$

under its macrorealist assumptions. Quantum protocols can violate such inequalities, but interpreting the violation requires careful control of measurement invasiveness, detector inefficiency, and state preparation. For the assumptions and experimental interpretation, see Leggett–Garg inequalities.

Common pitfalls

Conditional is not marginal. Use $p(a,b)=p(a)p(b\mid a)$, with the second factor evaluated on the state after outcome $a$. Multiplying two probabilities evaluated on the original state usually describes a different experiment.

Compatibility is not independence. Commuting projectors admit joint probabilities, but those probabilities can be correlated. The Lüders assumption is also essential when inferring nondisturbance from commutation.

One input is not every input. A disturbed channel can accidentally leave one preparation unchanged. The adjoint condition above demands unchanged later statistics for all input states.

Exercises

Ordered Projective Probability

For projectors $P_a$ followed by projectors $Q_b$, derive

$$p(a\ \text{then}\ b) = \operatorname{Tr}(Q_bP_a\rho P_aQ_b).$$

Solution

The first operation is $\mathcal I_a(\rho)=P_a\rho P_a$. The second operation is $\mathcal J_b(\rho)=Q_b\rho Q_b$. The ordered joint probability is the trace after both trace-nonincreasing operations:

$$p(a,b) = \operatorname{Tr} \!\left[ Q_bP_a\rho P_aQ_b \right].$$

This is the desired expression. Since $Q_b^2=Q_b$, cyclicity of the trace also gives $\operatorname{Tr}(P_aQ_bP_a\rho)$.

Unread Intermediate Spin Measurement

A qubit starts in $\lvert+z\rangle$. An unread ideal $\sigma_x$ measurement is performed, followed by a $\sigma_z$ measurement. What is the probability of obtaining $+z$ at the end?

Solution

The unread $\sigma_x$ measurement maps the initial pure state to

$$\rho' = \frac{1}{2} \lvert+x\rangle\langle+x\rvert + \frac{1}{2} \lvert-x\rangle\langle-x\rvert = \frac{I}{2}.$$

Therefore

$$p(+z) = \operatorname{Tr} \!\left[ \lvert+z\rangle\langle+z\rvert \frac{I}{2} \right] = \frac{1}{2}.$$

The intermediate unread measurement erased the coherence needed to guarantee the later $+z$ outcome.

Commuting Projectors

Assume $P_aQ_b=Q_bP_a$ for all $a,b$. Show that the unread $P$ measurement does not change the later probability of outcome $b$ for the $Q$ measurement.

Solution

The unread state is $\sum_aP_a\rho P_a$. The later probability is

$$\sum_a \operatorname{Tr}(Q_bP_a\rho P_a).$$

Using commutation and cyclicity,

$$\operatorname{Tr}(Q_bP_a\rho P_a) = \operatorname{Tr}(P_aQ_bP_a\rho) = \operatorname{Tr}(Q_bP_a\rho).$$

Summing over $a$ gives

$$\sum_a\operatorname{Tr}(Q_bP_a\rho) = \operatorname{Tr} \!\left[ Q_b \sum_aP_a \rho \right] = \operatorname{Tr}(Q_b\rho).$$

Same Effect, Different Sequence

Let $N_a=U_aM_a$ with $U_a$ unitary. Show that $M_a$ and $N_a$ have the same effect but can give different later probabilities for an effect $F_b$.

Solution

The effects agree:

$$N_a^\dagger N_a = M_a^\dagger U_a^\dagger U_aM_a = M_a^\dagger M_a.$$

Thus the probability of outcome $a$ is the same. But the conditional output states are related by

$$\rho_a^N = U_a\rho_a^M U_a^\dagger.$$

The later probability for effect $F_b$ is $\operatorname{Tr}(F_b\rho_a^N)$ instead of $\operatorname{Tr}(F_b\rho_a^M)$. These are equal for all inputs only under additional conditions, such as $F_b$ being invariant under the relevant unitary action on the conditional states.

References

• P. Busch, P. J. Lahti, and P. Mittelstaedt, The Quantum Theory of Measurement, 2nd ed., Springer (1996). Publisher.
• C. Emary, N. Lambert, and F. Nori, “Leggett–Garg inequalities,” Reports on Progress in Physics 77, 016001 (2014). DOI.
• T. Heinosaari, T. Miyadera, and M. Ziman, “An invitation to quantum incompatibility,” Journal of Physics A: Mathematical and Theoretical 49, 123001 (2016). Author preprint.
• A. J. Leggett and A. Garg, “Quantum mechanics versus macroscopic realism: Is the flux there when nobody looks?”, Physical Review Letters 54, 857–860 (1985). DOI.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). Publisher.