Generalized Measurements Overview

Projective measurement is indispensable, but it is not the most general way a quantum system can produce a classical record. A detector can be inefficient, an outcome can combine unresolved microscopic records, a probe can interact weakly with the system, or an ancilla can be measured after an indirect interaction. In such cases the outcome probabilities and the conditional state changes need not be described by orthogonal projectors on the system.

The generalized formalism separates three objects:

• a quantum instrument specifies the state-changing operation associated with each recorded outcome;
• a POVM effect specifies the probability of that outcome;
• the sum of all outcome operations is the unread channel.

A single calculation with an unnormalized branch gives its probability and, after normalization, its conditional state. We use finite-dimensional input and output systems and finitely many recorded outcomes; the input and output spaces need not have the same dimension.

Helpful background. Projective measurement provides the sharp special case, and density operators describe the mixed states that can arise when microscopic records are unresolved.

Why Generalize Projective Measurement?

A projective measurement has projectors $\{P_a\}$ satisfying

$$P_aP_b = \delta_{ab}P_a, \qquad \sum_a P_a = I.$$

This structure is appropriate when each ideal outcome identifies an orthogonal subspace. Several common situations have a different effective description:

• Indirect readout: the system interacts with a probe, and the probe rather than the system is measured.
• Unsharp information: one outcome only changes the relative plausibility of alternatives instead of selecting an eigenspace.
• Readout error: a sharp microscopic result passes through a noisy classical reporting channel.
• Inefficiency: “no click” is itself an outcome with state-dependent or state-independent probability.
• Coarse graining: several detector records are reported as one outcome.
• Monitored dynamics: records such as “jump” and “no jump” identify branches of an evolving open system.
• More outcomes than dimension: a qubit measurement can have three, four, or more nonorthogonal outcomes.

Generalized measurement is not synonymous with weak measurement or experimental imperfection. It is the broader probability-and-update framework that includes those cases as well as projective measurement.

The Instrument First

Consider a finite set of recorded outcomes $i$. A quantum instrument assigns to each outcome a linear map

$$\mathcal I_i : \rho \longmapsto \mathcal I_i(\rho).$$

Each $\mathcal I_i$ is completely positive and trace nonincreasing. Complete positivity ensures that the operation remains positive when it acts on one part of an entangled system. Trace nonincrease means

$$0 \leq \operatorname{Tr}\mathcal I_i(\rho) \leq 1$$

for every normalized input state.

The total map

$$\mathcal E = \sum_i\mathcal I_i$$

must preserve trace. It is the channel describing the quantum output when the measurement is performed but the classical outcome is ignored.

The probability of outcome $i$ is

$$p(i) = \operatorname{Tr}\mathcal I_i(\rho).$$

If $p(i)>0$, the normalized conditional state is

$$\rho_i = \frac{\mathcal I_i(\rho)}{p(i)}.$$

If the outcome is unread, the output is

$$\rho' = \sum_i\mathcal I_i(\rho) = \mathcal E(\rho).$$

The unnormalized branch $\mathcal I_i(\rho)$ therefore stores two things at once:

$$\operatorname{Tr}\mathcal I_i(\rho) = p(i), \qquad \frac{\mathcal I_i(\rho)} {\operatorname{Tr}\mathcal I_i(\rho)} = \rho_i.$$

This is the same branch-first logic used in State Update Rule, now without assuming projectors.

Zero-probability outcomes

If $p(i)=0$, outcome $i$ never occurs for the specified input. The numerator $\mathcal I_i(\rho)$ is then the zero operator, and the normalized expression is undefined. No conditional state needs to be assigned to an impossible record.

Kraus Operators for Outcome Branches

In finite dimensions, an outcome operation can be written

$$\mathcal I_i(\rho) = \sum_{\alpha} M_{i\alpha}\rho M_{i\alpha}^\dagger.$$

The index $i$ is the visible outcome. In a particular apparatus model, $\alpha$ can label unobserved environmental or pointer alternatives. A Kraus list is not unique, so its index is not an intrinsic list of physical events. The normalization condition for the whole instrument is

$$\sum_{i,\alpha} M_{i\alpha}^\dagger M_{i\alpha} = I.$$

It implies

$$\begin{aligned} \sum_i p(i) &= \sum_{i,\alpha} \operatorname{Tr} \left( M_{i\alpha}\rho M_{i\alpha}^\dagger \right)\\ &= \operatorname{Tr} \left[ \rho \sum_{i,\alpha} M_{i\alpha}^\dagger M_{i\alpha} \right]\\ &= \operatorname{Tr}\rho = 1. \end{aligned}$$

The operators $M_{i\alpha}$ are Kraus operators for the outcome branch. A Kraus representation is not unique; different operator lists can represent the same map $\mathcal I_i$.

The one-operator-per-outcome case

Many introductory formulas use one operator $M_i$ for each outcome:

$$\mathcal I_i(\rho) = M_i\rho M_i^\dagger, \qquad \sum_iM_i^\dagger M_i = I.$$

Then

$$p(i) = \operatorname{Tr} \left( M_i\rho M_i^\dagger \right),$$

and

$$\rho_i = \frac{ M_i\rho M_i^\dagger }{ p(i) }.$$

For a pure input, the selected state remains pure:

$$\lvert\psi_i\rangle = \frac{ M_i\lvert\psi\rangle }{ \sqrt{p(i)} }.$$

This is a special case, not the definition of every generalized measurement. If several unresolved $\alpha$ contribute to one outcome, a pure input can have a mixed conditional output.

Why the hidden index matters

Suppose a device has only one displayed outcome, but its internal operation completely dephases a qubit:

$$\mathcal I(\rho) = P_0\rho P_0 + P_1\rho P_1.$$

The two Kraus operators are $P_0$ and $P_1$. The associated outcome occurs with probability one, so the record reveals no information. Nevertheless, the system is disturbed. For the pure input $\lvert+\rangle$,

$$\mathcal I \left( \lvert+\rangle\langle+\rvert \right) = \frac12P_0+\frac12P_1,$$

which is mixed. No single Kraus operator can reproduce this branch for all inputs. Visible outcome count and Kraus rank are different notions.

Effects and POVMs

The effect associated with outcome $i$ is

$$E_i = \sum_\alpha M_{i\alpha}^\dagger M_{i\alpha}.$$

It is positive because, for every vector $\lvert\phi\rangle$,

$$\langle\phi\rvert E_i\lvert\phi\rangle = \sum_\alpha \left\lVert M_{i\alpha}\lvert\phi\rangle \right\rVert^2 \geq 0.$$

The instrument normalization gives

$$\sum_iE_i = I.$$

The collection $\{E_i\}$ is a finite-outcome positive-operator-valued measure, or POVM. Its probabilities are

$$p(i) = \operatorname{Tr}(\rho E_i).$$

Thus the POVM is the probability layer of the instrument.

Effects do not determine the update

Knowing $\{E_i\}$ is not enough to predict a later measurement. Different instruments can satisfy

$$\sum_\alpha M_{i\alpha}^\dagger M_{i\alpha} = E_i$$

while producing different conditional states.

Even in the one-operator case, if

$$N_i = U_iM_i$$

with $U_i$ unitary, then

$$N_i^\dagger N_i = M_i^\dagger M_i = E_i,$$

but the selected output is rotated by $U_i$. Outcome probabilities agree for every input; sequential statistics generally do not.

POVMs: First Encounter focuses on effects and their probability interpretation. Quantum Instruments develops the detailed theory of outcome operations.

Projective Measurements as a Special Case

For a PVM $\{P_a\}$, the Lüders instrument has

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

It follows that

$$E_a = P_a^\dagger P_a = P_a,$$

and the generalized formulas reduce to

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

But the PVM alone does not force the Lüders instrument. For example,

$$\mathcal J_a(\rho) = U_aP_a\rho P_aU_a^\dagger$$

has the same effect $P_a$ for any unitary $U_a$. If $U_a$ carries the output outside the range of $P_a$, an immediate repetition need not return $a$. Sharp probabilities and repeatable backaction are separate properties.

Projective measurement is therefore a special case at the effect level, while the ideal Lüders rule is a particular instrument associated with that PVM.

Indirect Measurements Through an Ancilla

Generalized measurements arise naturally from ordinary unitary dynamics on a larger system. Let the system $S$ and ancilla $A$ start in the product state $\rho\otimes|0\rangle\langle0|_A$, with the same fixed ancilla state for every input. Initial correlations would require a separate preparation model. Apply a unitary $U$ and measure the ancilla in orthogonal pointer subspaces.

Choose an orthonormal ancilla basis $\{\lvert i,\alpha\rangle_A\}$, where $i$ is the displayed record and $\alpha$ is hidden. Define

$$M_{i\alpha} = {}_A\langle i,\alpha\rvert U \lvert0\rangle_A.$$

The bracket over the ancilla leaves an operator on the system. The resulting branch is

$$\mathcal I_i(\rho) = \sum_\alpha M_{i\alpha}\rho M_{i\alpha}^\dagger.$$

The complete pointer basis resolves the ancilla identity:

$$R_A = \sum_{i,\alpha} \lvert i,\alpha\rangle\langle i,\alpha\rvert = I_A.$$

Unitarity then guarantees completeness:

$$\begin{aligned} \sum_{i,\alpha} M_{i\alpha}^\dagger M_{i\alpha} &= {}_A\langle0\rvert U^\dagger R_A U \lvert0\rangle_A\\ &= {}_A\langle0\rvert U^\dagger U \lvert0\rangle_A\\ &= I_S. \end{aligned}$$

This construction explains why nonprojective effects on $S$ are compatible with projective pointer readout on a larger Hilbert space. The enlarged-space representation of the effects is developed in Naimark Dilation.

Example: An Unsharp Qubit Measurement

Let

$$0 \leq \eta \leq 1$$

and define two effects

$$E_\pm = \frac12 \left( I\pm\eta\sigma_z \right).$$

The parameter $\eta$ is the sharpness. For

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol{\sigma} \right),$$

the probabilities are

$$p(\pm) = \frac12 \left( 1\pm\eta r_z \right).$$

At $\eta=1$, the effects are the projectors $P_0$ and $P_1$. At $\eta=0$, both effects equal $I/2$, so the outcome is a fair random label independent of the state.

One possible implementation is the square-root instrument

$$M_\pm = \sqrt{E_\pm}.$$

In the computational basis,

$$\begin{aligned} M_+ &= \sqrt{\frac{1+\eta}{2}}\,P_0 + \sqrt{\frac{1-\eta}{2}}\,P_1,\\ M_- &= \sqrt{\frac{1-\eta}{2}}\,P_0 + \sqrt{\frac{1+\eta}{2}}\,P_1. \end{aligned}$$

For this particular instrument, ignoring the outcome gives

$$\rho' = \begin{pmatrix} \rho_{00} & \sqrt{1-\eta^2}\,\rho_{01}\\ \sqrt{1-\eta^2}\,\rho_{10} & \rho_{11} \end{pmatrix}.$$

The populations are unchanged, while the coherence is reduced. The endpoints are informative:

• $\eta=0$ gives $M_\pm=I/\sqrt2$, an undisturbed state and a random record;
• $\eta=1$ gives unread projective dephasing in the computational basis.

This information–disturbance relation describes the chosen square-root instrument. The effects alone permit other backaction, for example outcome-dependent unitaries after the square-root operation.

Example: Sharp Readout Followed by Classical Error

Suppose a sharp PVM $\{P_a\}$ occurs microscopically, but the reported label $i$ is drawn from a classical conditional distribution $q(i\mid a)$. Then

$$q(i\mid a) \geq 0, \qquad \sum_i q(i\mid a) = 1.$$

The observed effects are

$$E_i = \sum_a q(i\mid a)P_a.$$

If the microscopic process is a Lüders measurement followed only by corruption of its classical record, the outcome operation is

$$\mathcal I_i(\rho) = \sum_a q(i\mid a) P_a\rho P_a.$$

A Kraus representation is

$$M_{ia} = \sqrt{q(i\mid a)}\,P_a.$$

For symmetric binary readout error $0\leq\epsilon\leq1$,

$$\begin{aligned} E_0 &= (1-\epsilon)P_0+\epsilon P_1,\\ E_1 &= \epsilon P_0+(1-\epsilon)P_1. \end{aligned}$$

If $p_{\mathrm{true}}(a)=\operatorname{Tr}(\rho P_a)$, then

$$p_{\mathrm{obs}}(0) = (1-\epsilon)p_{\mathrm{true}}(0) + \epsilon p_{\mathrm{true}}(1).$$

This model separates quantum backaction from classical reporting error. For $0\leq\epsilon\leq1/2$, its effects equal those of the preceding unsharp example with $\eta=1-2\epsilon$. Yet here the unread state is fully dephased for every $\epsilon$, because the sharp microscopic measurement has already happened. The square-root instrument instead retains the factor $\sqrt{1-\eta^2}$ of the coherence. Matching outcome frequencies does not establish matching state changes.

Example: Monitored Decay

Consider a two-level system with decay probability $0\leq\gamma\leq1$ per time interval. Assume decay from excited $|1\rangle$ to ground $|0\rangle$, no thermal excitation, and perfect monitoring of the emitted excitation. In an interaction picture with free phases removed, a simple model has

$$M_0 = P_0 + \sqrt{1-\gamma}\,P_1,$$

for “no jump,” and

$$M_1 = \sqrt{\gamma} \lvert0\rangle\langle1\rvert$$

for “jump.” These operators satisfy

$$M_0^\dagger M_0 + M_1^\dagger M_1 = I.$$

The jump probability is

$$p(1) = \gamma\rho_{11}.$$

When a jump is observed and $p(1)>0$, the selected state is

$$\rho_1 = P_0.$$

The no-jump branch is also informative: the absence of a detected jump changes the relative weight of the excited component. Ignoring the record gives the amplitude-damping channel

$$\rho' = M_0\rho M_0^\dagger + M_1\rho M_1^\dagger.$$

This example shows why generalized outcomes need not correspond to eigenvalues of a fixed observable. They can label alternative dynamical histories.

The Classical–Quantum Output

It is often useful to retain the classical record explicitly. With orthogonal register states $\{\lvert i\rangle_C\}$, the complete output can be written

$$\Omega_{CQ} = \sum_i \lvert i\rangle\langle i\rvert_C \otimes \mathcal I_i(\rho).$$

Its trace is one. Reading register $C$ gives $p(i)$, and conditioning on $i$ gives $\rho_i$. Tracing out $C$ gives the unread channel output:

$$\operatorname{Tr}_C\Omega_{CQ} = \sum_i\mathcal I_i(\rho).$$

This representation keeps selective and nonselective descriptions in one normalized state. Tracing out the register describes the predictions available when its record is inaccessible; no physical erasure is required. Ignoring a record does not undo the interaction that produced it.

Common Pitfalls

Normalizing unresolved terms separately. Sum all $M_{i\alpha}\rho M_{i\alpha}^\dagger$ for the recorded outcome first, then divide by the total branch probability. Separately normalized terms need not have equal weights.

Demanding that every branch preserve trace. A branch trace is its outcome probability, often less than one. It is the sum over all outcomes that must preserve trace.

Inferring disturbance from information alone. The one-outcome dephasing example gives no state information but can change the state. An uninformative POVM does not force an identity channel.

Choosing an update from the effects alone. The square-root instrument is one valid implementation, not a universal rule. Use the instrument specified by the physical model for sequential predictions.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016.
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer, 1983.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press, 2010.

Exercises

1. A random record that reveals nothing

Let

$$M_0 = \sqrt q\,I, \qquad M_1 = \sqrt{1-q}\,I.$$

Here $0\leq q\leq1$. Verify that these operators define an instrument. Find the probabilities, selected states, and unread output for arbitrary $\rho$.

Solution

Completeness holds:

$$M_0^\dagger M_0 + M_1^\dagger M_1 = qI+(1-q)I = I.$$

The probabilities are

$$p(0) = q, \qquad p(1) = 1-q,$$

independent of $\rho$. For either outcome of nonzero probability,

$$\rho_i = \rho.$$

The unread output is

$$q\rho+(1-q)\rho = \rho.$$

The device produces a classical random label but gains no state information and causes no quantum disturbance.

2. Derive the effect conditions

Starting from

$$E_i = \sum_\alpha M_{i\alpha}^\dagger M_{i\alpha},$$

prove that $E_i\geq0$, that $\sum_iE_i=I$, and that $0\leq E_i\leq I$.

Solution

For every $\lvert\phi\rangle$,

$$\langle\phi\rvert E_i\lvert\phi\rangle = \sum_\alpha \left\lVert M_{i\alpha}\lvert\phi\rangle \right\rVert^2 \geq 0,$$

so $E_i\geq0$. Completeness of the instrument gives

$$\sum_iE_i = \sum_{i,\alpha} M_{i\alpha}^\dagger M_{i\alpha} = I.$$

Finally,

$$I-E_i = \sum_{j\neq i}E_j \geq 0.$$

Hence $E_i\leq I$.

3. Quantify unread disturbance in the square-root instrument

For the unsharp qubit effects

$$E_\pm = \frac12 \left( I\pm\eta\sigma_z \right),$$

use $M_\pm=\sqrt{E_\pm}$ to derive the factor multiplying $\rho_{01}$ in the unread state. What happens for $\eta=0$ and $\eta=1$?

Solution

Write

$$a = \sqrt{\frac{1+\eta}{2}}, \qquad b = \sqrt{\frac{1-\eta}{2}}.$$

Then

$$\begin{aligned} M_+ &= aP_0+bP_1, \\ M_- &= bP_0+aP_1. \end{aligned}$$

The $01$ entry of the unread output is

$$(\rho')_{01} = ab\rho_{01} + ba\rho_{01} = 2ab\rho_{01}.$$

Since

$$2ab = \sqrt{1-\eta^2},$$

the coherence factor is $\sqrt{1-\eta^2}$. At $\eta=0$ it is one, so this instrument does not disturb the state. At $\eta=1$ it is zero, giving complete dephasing in the computational basis.

4. Add symmetric readout error

A true projective qubit measurement has

$$p_{\mathrm{true}}(0) = \frac34, \qquad p_{\mathrm{true}}(1) = \frac14.$$

Each classical label is flipped with probability $\epsilon=1/10$. Find the reported distribution and verify normalization.

Solution

The reported zero probability is

$$\begin{aligned} p_{\mathrm{obs}}(0) &= \frac{9}{10}\frac34 + \frac{1}{10}\frac14\\ &= \frac{27+1}{40} = \frac{7}{10}. \end{aligned}$$

Similarly,

$$p_{\mathrm{obs}}(1) = \frac{1}{10}\frac34 + \frac{9}{10}\frac14 = \frac{3+9}{40} = \frac{3}{10}.$$

The probabilities sum to one. The classical error moves the distribution toward $(1/2,1/2)$.

5. One POVM, two one-outcome instruments

Compare the identity channel

$$\mathcal I_{\mathrm{id}}(\rho) = \rho$$

with the dephasing channel

$$\mathcal I_{\mathrm{deph}}(\rho) = P_0\rho P_0+P_1\rho P_1.$$

Each instrument displays only one outcome. Show that both have the same one-element POVM, then distinguish them using input $\lvert+\rangle$.

Solution

The identity channel has Kraus operator $I$, so its only effect is

$$E_{\mathrm{id}} = I^\dagger I = I.$$

The dephasing channel has Kraus operators $P_0$ and $P_1$, so its only effect is

$$E_{\mathrm{deph}} = P_0^\dagger P_0 + P_1^\dagger P_1 = P_0+P_1 = I.$$

Both outcomes occur with probability one for every input. Yet

$$\mathcal I_{\mathrm{id}} \left( \lvert+\rangle\langle+\rvert \right) = \lvert+\rangle\langle+\rvert,$$

whereas

$$\mathcal I_{\mathrm{deph}} \left( \lvert+\rangle\langle+\rvert \right) = \frac12P_0+\frac12P_1.$$

The same POVM can therefore accompany different unread and conditional quantum outputs.

6. Recover completeness from an ancilla model

An ancilla begins in $\lvert0\rangle_A$, a unitary $U$ acts on $S\otimes A$, and the ancilla is measured in a complete basis $\{\lvert\mu\rangle_A\}$. Define

$$M_\mu = {}_A\langle\mu\rvert U \lvert0\rangle_A.$$

Show that $\sum_\mu M_\mu^\dagger M_\mu=I_S$.

Solution

Insert the ancilla resolution of the identity:

$$\sum_\mu \lvert\mu\rangle\langle\mu\rvert_A = I_A.$$

It follows that

$$\begin{aligned} \sum_\mu M_\mu^\dagger M_\mu &= {}_A\langle0\rvert U^\dagger I_A U \lvert0\rangle_A\\ &= {}_A\langle0\rvert U^\dagger U \lvert0\rangle_A\\ &= I_S. \end{aligned}$$

The last equality uses both unitarity and normalization of the ancilla ready state.

7. Analyze a monitored jump

For

$$\begin{aligned} M_0 &= P_0+\sqrt{1-\gamma}\,P_1, \\ M_1 &= \sqrt\gamma\,\lvert0\rangle\langle1\rvert, \end{aligned}$$

take the normalized pure input

$$\lvert\psi\rangle = \alpha\lvert0\rangle+\beta\lvert1\rangle.$$

Find the jump probability and both normalized conditional states when their probabilities are nonzero.

Solution

The jump branch is

$$M_1\lvert\psi\rangle = \sqrt\gamma\,\beta\lvert0\rangle,$$

so

$$p(1) = \gamma\lvert\beta\rvert^2, \qquad \lvert\psi_1\rangle = \lvert0\rangle$$

up to an irrelevant phase.

The no-jump branch is

$$M_0\lvert\psi\rangle = \alpha\lvert0\rangle + \sqrt{1-\gamma}\,\beta\lvert1\rangle.$$

Its probability is

$$p(0) = \lvert\alpha\rvert^2 + (1-\gamma)\lvert\beta\rvert^2,$$

and the normalized state is

$$\lvert\psi_0\rangle = \frac{ \alpha\lvert0\rangle + \sqrt{1-\gamma}\,\beta\lvert1\rangle }{ \sqrt{ \lvert\alpha\rvert^2 + (1-\gamma)\lvert\beta\rvert^2 } }.$$

The absence of a jump updates the relative amplitudes whenever $\gamma>0$ and both input amplitudes are nonzero.

8. Compose two generalized measurements

The first measurement has one Kraus operator $M_i$ per outcome. The second has one Kraus operator $N_j$ per outcome. Derive the ordered joint probability $p(i,j)$ and the final selected state. Explain why the first POVM effects alone are insufficient.

Solution

The unnormalized branch for ordered record $(i,j)$ is

$$\widetilde\rho_{ij} = N_jM_i\rho M_i^\dagger N_j^\dagger.$$

Therefore

$$p(i,j) = \operatorname{Tr} \left( N_jM_i\rho M_i^\dagger N_j^\dagger \right).$$

When $p(i,j)>0$, the final selected state is

$$\rho_{ij} = \frac{ N_jM_i\rho M_i^\dagger N_j^\dagger }{ p(i,j) }.$$

The expression depends on the post-$i$ branch $M_i\rho M_i^\dagger$, not only on

$$E_i = M_i^\dagger M_i.$$

Different instruments with the same first POVM can therefore produce different conditional and joint statistics for the second measurement.