Quantum Instruments

A quantum instrument is the operational object that describes a measurement when both the classical outcome and the resulting quantum state matter.

A POVM answers one question:

$$\text{What are the probabilities of the reported outcomes?}$$

An instrument answers two questions:

$$\text{What is the probability of outcome }m, \qquad \text{and what state is left after }m?$$

This distinction is essential for sequential measurements, detector backaction, postselection, continuous monitoring, and realistic readout models. Different instruments can have the same POVM but make different predictions for later measurements.

Helpful background. Kraus operators give concrete state transformations; POVMs supply the probability rule. Bayesian quantum measurement connects this rule to conditioning.

Outcome operations and conditional states

For finite-dimensional input and output spaces and finitely many outcomes, a quantum instrument is a family of completely positive, trace-nonincreasing linear maps whose sum is trace preserving:

$$\{\mathcal I_m\}$$

The input and output Hilbert spaces may differ. The map $\mathcal I_m$ is the unnormalized state transformation associated with outcome $m$; it extends linearly to all input operators. Dividing by its trace is a separate, generally nonlinear conditioning step.

For an input state $\rho$, the probability of outcome $m$ is

$$p(m) = \operatorname{Tr}\mathcal I_m(\rho).$$

If $p(m)\ne0$, the conditional output state is

$$\rho_m = \frac{\mathcal I_m(\rho)} {\operatorname{Tr}\mathcal I_m(\rho)}.$$

If the measurement is performed but the outcome is ignored, the nonselective output is

$$\Phi(\rho) = \sum_m \mathcal I_m(\rho).$$

For a complete measurement with no discarded failure branch, the sum

$$\Phi = \sum_m\mathcal I_m$$

is trace preserving. Equivalently,

$$\sum_m \operatorname{Tr}\mathcal I_m(\rho) = \operatorname{Tr}\rho$$

for every input operator $\rho$. In particular, normalized density operators give total probability one.

Kraus Form

In finite dimensions, each outcome operation can be written using Kraus operators:

$$\mathcal I_m(\rho) = \sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger.$$

Trace nonincrease for each outcome means

$$\sum_\alpha K_{m\alpha}^\dagger K_{m\alpha} \le I.$$

Completeness of the whole instrument means

$$\sum_{m,\alpha} K_{m\alpha}^\dagger K_{m\alpha} =I.$$

The label $\alpha$ indexes a representation of the operation. In a specified apparatus model it may describe unresolved detector modes or unobserved environment states, but it is not uniquely determined by the instrument and need not label literal physical histories.

Associated POVM

Every instrument has an associated POVM. In Kraus form, the effect for outcome $m$ is

$$F_m = \sum_\alpha K_{m\alpha}^\dagger K_{m\alpha}.$$

Then

$$p(m) = \operatorname{Tr}\mathcal I_m(\rho) = \operatorname{Tr}(\rho F_m).$$

More invariantly, let $\mathcal I_m^*$ be the adjoint map in the trace pairing:

$$\operatorname{Tr} \left[ A\,\mathcal I_m(\rho) \right] = \operatorname{Tr} \left[ \mathcal I_m^*(A)\rho \right].$$

The associated effect is

$$F_m = \mathcal I_m^*(I).$$

This formulation makes the hierarchy clear:

instrument -> POVM
POVM       -> probabilities only

The reverse arrow is not unique. A POVM does not determine the post-measurement state.

Why POVMs Are Not Enough

Consider a sharp two-outcome measurement with projectors

$$P_0=|0\rangle\langle0|, \qquad P_1=|1\rangle\langle1|.$$

The Lüders instrument is

$$\mathcal I_0^{\mathrm L}(\rho)=P_0\rho P_0, \qquad \mathcal I_1^{\mathrm L}(\rho)=P_1\rho P_1.$$

Its associated POVM is $\{P_0,P_1\}$. It is repeatable: immediately measuring the same sharp observable again gives the same outcome with probability one.

Now compare a measure-and-prepare instrument:

$$\mathcal I_0^{\mathrm{prep}}(\rho) = \operatorname{Tr}(P_0\rho)\sigma_0, \qquad \mathcal I_1^{\mathrm{prep}}(\rho) = \operatorname{Tr}(P_1\rho)\sigma_1,$$

where $\sigma_0$ and $\sigma_1$ are fixed output states. This instrument has the same POVM $\{P_0,P_1\}$, because the same probabilities are reported. But the post-measurement states are $\sigma_0$ and $\sigma_1$, not the projected input components.

Thus the phrase “measure $\sigma_z$” is incomplete unless the later state does not matter or the instrument is specified by context.

For example, circuit readout followed by resetting the measured qubit can preserve the recorded bit while replacing the quantum output. See measurement in circuits.

Nonselective Channel

The nonselective channel

$$\Phi(\rho) = \sum_m\mathcal I_m(\rho)$$

is what remains when the measurement interaction happens but the classical record is ignored. It is a completely positive, trace-preserving map for a complete instrument.

For the Lüders instrument above,

$$\Phi_{\mathrm L}(\rho) = P_0\rho P_0+P_1\rho P_1.$$

This channel removes coherence between the two eigenspaces.

For the measure-and-prepare instrument,

$$\Phi_{\mathrm{prep}}(\rho) = \operatorname{Tr}(P_0\rho)\sigma_0 + \operatorname{Tr}(P_1\rho)\sigma_1.$$

This channel discards the input state after extracting the classical distribution over outcomes. Both instruments can share the same POVM, but their nonselective channels and future predictions differ.

Sequential Measurements

Instruments are indispensable when one measurement is followed by another. Suppose the first measurement has instrument $\{\mathcal I_m\}$ and the second measurement has POVM effects $\{G_n\}$ on its output space. With no intervening evolution and no additional apparatus memory, the joint probability for first outcome $m$ and second outcome $n$ is

$$p(m,n) = \operatorname{Tr} \left[ G_n\,\mathcal I_m(\rho) \right].$$

The conditional probability for the second outcome, after learning $m$, is

$$p(n|m) = \operatorname{Tr}(G_n\rho_m).$$

The first POVM alone is not sufficient to compute these probabilities. One needs the actual state update, and that information is precisely the instrument.

This is the operational reason instruments belong between state-update rules and open-system dynamics: they tell us how classical records and quantum state transformations fit together.

An intervening channel $\Lambda_m$ changes the formula to $p(m,n)=\operatorname{Tr}[G_n\Lambda_m(\mathcal I_m(\rho))]$. The channel may depend on the recorded first outcome, as in mid-circuit measurement and feedforward.

Indirect Measurement Origin

An instrument often arises from an explicit system-apparatus model. Let $S$ be the system, $A$ an apparatus, $\eta_A$ a fixed initial apparatus state uncorrelated with the input, $U$ a joint unitary, and $\{Q_m\}$ a projective pointer measurement on the apparatus. If the apparatus is discarded after outcome $m$, the system operation can be written as

$$\mathcal I_m(\rho_S) = \operatorname{Tr}_A \left[ (I_S\otimes Q_m) U(\rho_S\otimes\eta_A)U^\dagger (I_S\otimes Q_m) \right].$$

The corresponding probability is

$$p(m) = \operatorname{Tr}_S\mathcal I_m(\rho_S).$$

This construction explains why instruments are completely positive. They are what a closed unitary model becomes after the apparatus degrees of freedom are measured, partly retained as a classical record, and otherwise ignored.

For the ideal correlation model behind projective measurement, see the von Neumann measurement model.

Coarse Graining and Hidden Records

Suppose a detector has fine-grained outcomes $(m,\alpha)$ but reports only $m$. If the fine-grained operations are $\mathcal J_{m\alpha}$, the coarse-grained instrument is

$$\mathcal I_m(\rho) = \sum_\alpha \mathcal J_{m\alpha}(\rho).$$

The associated effect is likewise coarse grained:

$$F_m = \sum_\alpha F_{m\alpha}.$$

Forgetting a label does not apply an additional physical disturbance to the system. It changes the state used for prediction from a particular conditional state to a probability-weighted mixture. The fine measurement may already have disturbed coherence. Performing that fine measurement and forgetting its label is therefore generally different from building an apparatus that never resolves the finer alternatives; the degenerate-measurement example makes the difference explicit.

Postselection and Failure Outcomes

Many experiments condition on successful runs. The operations for the retained outcomes may sum to a trace-nonincreasing map:

$$\sum_{m\in\mathrm{kept}} \operatorname{Tr}\mathcal I_m(\rho) \le \operatorname{Tr}\rho.$$

To describe the full experiment as a complete instrument, include the discarded outcomes explicitly. A common notation is

$$\mathcal I_{\mathrm{fail}} = \Phi - \sum_{m\in\mathrm{kept}}\mathcal I_m,$$

only when the difference is completely positive. An arbitrary preferred channel $\Phi$ need not allow this subtraction.

For input and output spaces of equal dimension, a simple completion of any trace-nonincreasing Kraus map uses

$$R=I-\sum_{m\in\mathrm{kept},\alpha}K_{m\alpha}^\dagger K_{m\alpha}\ge0, \qquad K_{\mathrm{fail}}=\sqrt R.$$

Then adding $K_{\mathrm{fail}}\rho K_{\mathrm{fail}}^\dagger$ gives a trace-preserving total map. This mathematical completion does not by itself identify the actual failed-run state of a particular apparatus.

Postselection is legitimate, but probabilities and normalized states must be reported conditionally. Forgetting the failure branch often leads to apparent violations of trace preservation or exaggerated success probabilities.

Continuous-Measurement Preview

A continuous measurement can be viewed as a limit of instruments applied over short time steps. For a small interval $dt$, the record increment $r$ has an operation $\mathcal I_r^{dt}$. A full measurement record is a sequence

$$r_1,r_2,\ldots,r_N, \qquad Ndt=t.$$

The unnormalized conditional state is obtained by composing the corresponding operations:

$$\tilde\rho_{r_1\cdots r_N} = \mathcal I_{r_N}^{dt} \circ\cdots\circ \mathcal I_{r_1}^{dt}(\rho_0).$$

Its trace is the probability weight of a discrete record, or the probability density relative to a specified product measure for continuous records, and the normalized state is the conditional state along the observed trajectory. This is the discrete skeleton behind Continuous Monitoring, quantum jumps, diffusive trajectories, and stochastic master equations.

Exercises

1. Instrument to POVM. Let $\mathcal I_m(\rho)=\sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger$ and assume $\sum_{m,\alpha}K_{m\alpha}^\dagger K_{m\alpha}=I$. Prove that $F_m=\sum_\alpha K_{m\alpha}^\dagger K_{m\alpha}$ defines a POVM.

Solution

Each $F_m$ is positive because for every vector $|\psi\rangle$,

$$\langle\psi|F_m|\psi\rangle = \sum_\alpha \|K_{m\alpha}|\psi\rangle\|^2 \ge0.$$

Completeness follows directly:

$$\sum_mF_m = \sum_{m,\alpha} K_{m\alpha}^\dagger K_{m\alpha} =I.$$

Thus $\{F_m\}$ is a collection of positive effects summing to the identity.

2. Same POVM, different future. Let $\rho=|+\rangle\langle+|$, where $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. Compare the nonselective output of the Lüders instrument for $\{P_0,P_1\}$ with a reset instrument

$$\mathcal I_0(\rho)=\operatorname{Tr}(P_0\rho)|0\rangle\langle0|, \qquad \mathcal I_1(\rho)=\operatorname{Tr}(P_1\rho)|0\rangle\langle0|.$$

Both instruments have the same first-measurement POVM. What is the probability of obtaining $0$ in a second $\sigma_z$ measurement after the first outcome is ignored?

Solution

For the Lüders instrument,

$$\Phi_{\mathrm L}(|+\rangle\langle+|) = \frac12 |0\rangle\langle0| + \frac12 |1\rangle\langle1|.$$

A second $\sigma_z$ measurement gives outcome $0$ with probability $1/2$.

For the reset instrument,

$$\Phi_{\mathrm{reset}}(|+\rangle\langle+|) = \frac12 |0\rangle\langle0| + \frac12 |0\rangle\langle0| = |0\rangle\langle0|.$$

The second $\sigma_z$ measurement gives outcome $0$ with probability $1$. The first measurement statistics were the same, but the instruments produced different later states.

3. Sequential probability formula. A first measurement has instrument $\{\mathcal I_m\}$, and a second measurement has effects $\{G_n\}$. Show that $p(m,n)=\operatorname{Tr}[G_n\mathcal I_m(\rho)]$ is normalized if the first instrument is complete and the second POVM is complete.

Solution

Sum over the second outcome first:

$$\sum_n p(m,n) = \sum_n \operatorname{Tr} \left[ G_n\mathcal I_m(\rho) \right] = \operatorname{Tr} \left[ \left(\sum_nG_n\right)\mathcal I_m(\rho) \right].$$

Since $\sum_nG_n=I$,

$$\sum_n p(m,n) = \operatorname{Tr}\mathcal I_m(\rho).$$

Now sum over $m$:

$$\sum_{m,n}p(m,n) = \operatorname{Tr} \left[ \sum_m\mathcal I_m(\rho) \right] = \operatorname{Tr}\rho =1,$$

where completeness of the first instrument was used in the second equality and $\rho$ was assumed normalized.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer (2016). Publisher.
• E. B. Davies and J. T. Lewis, “An operational approach to quantum probability,” Communications in Mathematical Physics 17, 239–260 (1970). Article.
• M. Ozawa, “Quantum measuring processes of continuous observables,” Journal of Mathematical Physics 25, 79–87 (1984). Article.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). Publisher.