Why Generalized Measurements Are Needed

A projective measurement assigns orthogonal subspaces to distinct outcomes. Generalized measurements are needed when the observed probabilities cannot be described that way on the system alone—for example, an inefficient detector or a qubit measurement with three nonzero outcomes. Even when the probabilities are projective, predicting the output state still requires an apparatus model.

Helpful background. Projective measurements defines sharp outcomes; measurement as an operation separates probabilities from state changes.

Which measurement object answers the question?

Object	Prediction supplied	Additional information needed
PVM	Sharp outcome probabilities	An instrument for the state afterward; Lüders is one choice.
POVM	General outcome probabilities	An instrument for later experiments.
Instrument	Each outcome’s probability and output operation	An apparatus model if its microscopic mechanism is wanted.
Channel	Unconditional output state	An instrument to resolve observed outcomes.

Kraus operators are a representation of an operation or channel, not a unique microscopic explanation. An instrument groups its Kraus operators by the reported result. The sections below give three reasons to use this broader description.

An inefficient detector has nonprojective effects

Suppose a detector clicks on a qubit state $|1\rangle$ with efficiency $\eta$ and has no dark counts. Its effects are

$$F_{\rm click}=\eta|1\rangle\langle1|,\qquad F_{\rm no}=|0\rangle\langle0|+(1-\eta)|1\rangle\langle1|, \quad 0\leq\eta\leq1.$$

They sum to $I$, but $F_{\rm click}^2=\eta^2|1\rangle\langle1|$ differs from $F_{\rm click}$ for $0<\eta<1$. This is a POVM, not a PVM.

For $\eta=0.6$ and $\rho=\operatorname{diag}(1/4,3/4)$, the click probability is $0.6(3/4)=0.45$ and the no-click probability is $0.55$. A no-click result does not establish that the qubit was in $|0\rangle$. To predict the state after that result, use an instrument: inefficiency alone says neither how strongly the qubit was disturbed nor whether it was absorbed.

A qubit can have more than two reported outcomes

A qubit PVM has at most two nonzero orthogonal projectors. A POVM can have three or more nonzero effects. For three equatorial unit vectors $\mathbf n_k$ separated by $120^\circ$, the trine effects are

$$F_k=\frac13(I+\mathbf n_k\cdot\boldsymbol\sigma),\qquad k=0,1,2.$$

Their eigenvalues are $0$ and $2/3$, and they sum to $I$ because $\sum_k\mathbf n_k=0$. An input aligned with $\mathbf n_0$ produces probabilities $(2/3,1/6,1/6)$. These are three distinct possible results on a two-dimensional system.

This example also shows why “more outcomes” does not automatically mean full tomography: all three effects lie in one Bloch plane, so they cannot determine the perpendicular Bloch component. POVMs explains the spanning condition.

Another useful extra outcome is “inconclusive.” Two distinct nonorthogonal pure states cannot be distinguished without error in every run by any quantum measurement. A suitable three-outcome POVM can identify either state without error when it returns a conclusive result, while allowing a nonzero inconclusive probability. The explicit discrimination example calculates this tradeoff.

Measuring a probe induces a measurement on the system

Prepare an auxiliary system in a fixed state $\eta_A$, initially uncorrelated with $\rho_S$. Couple them by $U$ and measure apparatus projectors $Q_m$. Then

$$p(m)=\operatorname{Tr}_{SA}\!\left[ U(\rho_S\otimes\eta_A)U^\dagger(I_S\otimes Q_m)\right] =\operatorname{Tr}_S(\rho_SF_m).$$

Although the pointer readout is sharp, its induced system effects $F_m$ need not be projectors. Naimark dilation proves that every finite-dimensional POVM admits a projective realization on a larger space. Quantum instruments additionally describes what the coupling leaves behind.

Coarse reporting sums both effects and unnormalized outcome operations. If the fine effects are orthogonal projectors, summing disjoint groups still gives projectors. Therefore coarse graining, indirect coupling, or finite resolution does not by itself prove that the system POVM is nonprojective. The actual effects decide. In contrast, reporting a coarse result after physically resolving finer states can change the output instrument even when the coarse effects remain sharp.

Weak information need not imply weak backaction

The unsharp effects $F_\pm=(I\pm\epsilon\sigma_z)/2$ approach a random coin as $\epsilon\to0$. The square-root choice $K_\pm=\sqrt{F_\pm}$ then causes a small state change. Another instrument with the same effects could apply a large unitary kick or erase coherence. The term weak measurement must therefore specify the coupling or update, not merely the small contrast of the outcome probabilities.

When an environment carries an observed record, the instrument can be conditioned on it. Averaging over that record gives a channel. Neither the choice to ignore the record nor its unreliability removes the physical backaction. Weak measurements and continuous monitoring develop the small-step limit.

Exercises

Inefficient Click Is Not a Projector

For

$$F_{\mathrm{click}} = \eta\lvert1\rangle\langle1\rvert,$$

show that $F_{\mathrm{click}}$ is an effect for $0\le\eta\le1$ but not a projector unless $\eta=0$ or $\eta=1$.

Solution

For every vector $\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle$,

$$\langle\psi\rvert F_{\mathrm{click}}\lvert\psi\rangle = \eta|\beta|^2\ge0,$$

so the operator is positive when $\eta\ge0$. Also $F_{\mathrm{click}}\le I$ when $\eta\le1$, so it can be part of a complete two-outcome POVM.

It is a projector only if $F_{\mathrm{click}}^2=F_{\mathrm{click}}$. But

$$F_{\mathrm{click}}^2 = \eta^2\lvert1\rangle\langle1\rvert.$$

Thus $F_{\mathrm{click}}^2=F_{\mathrm{click}}$ requires $\eta^2=\eta$, so $\eta=0$ or $\eta=1$.

Same POVM, Different Backaction

Let $F_0=F_1=I/2$. Give two different instruments with this same POVM.

Solution

One instrument flips a fair classical coin and leaves the system unchanged:

$$\mathcal I_0(\rho)=\frac{1}{2}\rho, \qquad \mathcal I_1(\rho)=\frac{1}{2}\rho.$$

Another flips the same fair label but also fully dephases the qubit in the $\sigma_z$ basis:

$$\mathcal J_0(\rho) = \frac{1}{2} \sum_{k=0}^1 \lvert k\rangle\langle k\rvert \rho \lvert k\rangle\langle k\rvert,$$

and

$$\mathcal J_1(\rho) = \frac{1}{2} \sum_{k=0}^1 \lvert k\rangle\langle k\rvert \rho \lvert k\rangle\langle k\rvert.$$

Both instruments have effects $I/2$ for outcomes $0$ and $1$, but the second destroys $\sigma_z$-basis coherence while the first does not.

Choosing the Object

For each task, name the minimal object needed: outcome probabilities only; conditional post-measurement states; state after ignoring the record; continuous conditioned record.

Solution

Outcome probabilities only require a POVM. Conditional post-measurement states require an instrument, or at least a specified outcome operation. The state after ignoring the record is the nonselective channel $\sum_m\mathcal I_m$. A continuous conditioned record requires a continuous-time instrument limit, usually written as a stochastic master equation or quantum trajectory.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer (2016). Publisher.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapters 2–3. Author’s text.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). Publisher.