POVMs

A positive-operator-valued measurement, or POVM, is the general operator language for the probabilities of measurement outcomes. It extends projective measurement by allowing effects that are positive but not necessarily projectors.

The core message is:

$$\text{POVMs determine probabilities, not post-measurement states.}$$

State updates require a measurement operator model or, more generally, a quantum instrument.

Definition

For a finite outcome set, a POVM on a Hilbert space $\mathcal H$ is a collection of positive operators

$$\{F_m\}$$

such that

$$F_m\ge0, \qquad \sum_m F_m=I.$$

The operators $F_m$ are called effects. For a density operator $\rho$, the probability of outcome $m$ is

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

These three equations are the finite-outcome POVM formalism. Positivity gives nonnegative probabilities, and completeness gives normalization:

$$\sum_m p(m) = \operatorname{Tr} \left( \rho\sum_mF_m \right) = \operatorname{Tr}\rho =1.$$

Effects Are Not Projectors in General

Each effect is positive, so it is Hermitian and has nonnegative eigenvalues in finite dimensions. Completeness also implies

$$0\le F_m\le I.$$

The upper bound follows because

$$I-F_m = \sum_{n\ne m}F_n \ge0.$$

An effect need not be idempotent:

$$F_m^2 \ne F_m$$

in general. This is the algebraic difference between a general POVM and a projection-valued measurement.

Projective Measurements as a Special Case

An ideal projective measurement with projectors $\{P_a\}$ is a POVM with

$$F_a=P_a,$$

where

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

Every projective measurement is a POVM. The converse is false: most POVMs cannot be represented as a set of mutually orthogonal projectors acting only on the original system Hilbert space.

The reason POVMs are needed is not that projective measurement is wrong. It is that many real or effective measurements are not sharp projective measurements on the system alone.

Why POVMs Appear

POVMs appear whenever the observed outcome statistics are more general than a sharp projective readout:

• a detector has finite efficiency;
• a readout has false positives or false negatives;
• a continuous pointer is binned into finite windows;
• several microscopic detector histories are coarse grained into one reported label;
• the system is measured indirectly through an ancilla or field mode;
• the measurement is intentionally weak or unsharp;
• one uses an effective description after ignoring inaccessible degrees of freedom.

In many of these cases, a larger system-plus-apparatus model can still be projective at the end. The POVM is what remains on the original system after the apparatus degrees of freedom are eliminated from the probability calculation.

Indirect Measurement Origin

A useful construction starts with a system $S$, an ancilla or apparatus $A$, a fixed apparatus state $\eta_A$, a joint unitary $U$, and a projective pointer measurement $\{Q_m\}$ on the apparatus. The outcome probability is

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

For each $m$, this can be written as

$$p(m)=\operatorname{Tr}_S(\rho_S F_m)$$

for a positive system effect $F_m$. Thus an indirect projective measurement on a larger Hilbert space induces a POVM on the system.

This is the conceptual content behind dilation theorems: generalized measurements on a system can often be represented as ordinary projective measurements on a larger space.

POVM Versus Instrument

A POVM gives the probability rule. It does not specify the output state.

If an instrument $\{\mathcal I_m\}$ realizes the measurement, then

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

The associated POVM effect is the operator $F_m$ satisfying

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

for every input state $\rho$.

If the outcome has Kraus operators $M_{m\alpha}$,

$$\mathcal I_m(\rho) = \sum_\alpha M_{m\alpha}\rho M_{m\alpha}^\dagger,$$

then

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

Different instruments can have the same $F_m$ and therefore the same outcome probabilities while giving different post-measurement states.

Example: Inefficient Click Detector

Consider a qubit in basis $\{|0\rangle,|1\rangle\}$. A detector is intended to click on $|1\rangle$ but has efficiency $0\le\eta\le1$. A simple two-outcome POVM is

$$F_{\mathrm{click}} = \eta |1\rangle\langle1|,$$ $$F_{\mathrm{no}} = |0\rangle\langle0| + (1-\eta)|1\rangle\langle1|.$$

Both effects are positive and

$$F_{\mathrm{click}}+F_{\mathrm{no}}=I.$$

For

$$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle,$$

the click probability is

$$p(\mathrm{click}) = \eta|\beta|^2.$$

When $0\lt\eta\lt1$, $F_{\mathrm{click}}$ is not a projector:

$$F_{\mathrm{click}}^2 = \eta^2 |1\rangle\langle1| \ne F_{\mathrm{click}}.$$

The POVM describes the click statistics. A separate instrument is needed to say what state remains after a click or no-click outcome.

Example: Qubit Trine POVM

POVMs can have more outcomes than the Hilbert-space dimension. A standard qubit example is the trine POVM. Let $\mathbf n_0,\mathbf n_1,\mathbf n_2$ be three unit Bloch vectors in the equatorial plane separated by $120^\circ$, so

$$\mathbf n_0+\mathbf n_1+\mathbf n_2=0.$$

Define rank-one projectors

$$\Pi_k = \frac{1}{2} \left( I+\mathbf n_k\cdot\boldsymbol\sigma \right), \qquad k=0,1,2.$$

The trine effects are

$$F_k = \frac{2}{3}\Pi_k.$$

They are positive and satisfy

$$\sum_{k=0}^2F_k = \frac{2}{3} \sum_{k=0}^2 \frac{1}{2} \left( I+\mathbf n_k\cdot\boldsymbol\sigma \right) =I.$$

For a qubit state

$$\rho = \frac{1}{2} \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

the probabilities are

$$p(k) = \operatorname{Tr}(\rho F_k) = \frac{1}{3} \left( 1+\mathbf r\cdot\mathbf n_k \right).$$

This is not a projective measurement on the qubit: there are three outcomes in a two-dimensional Hilbert space, and the effects are not orthogonal projectors.

Example: Unambiguous State Discrimination

Suppose a source prepares one of two nonorthogonal states. No projective measurement can perfectly identify the state in every run without error because nonorthogonal states cannot be perfectly distinguished. A POVM can introduce an inconclusive outcome:

$$\{F_1,F_2,F_?\}, \qquad F_1+F_2+F_?=I.$$

The effects $F_1$ and $F_2$ are designed so that, when they click, they identify the corresponding state without error. The price is that $F_?$ sometimes occurs. This is a typical generalized-measurement tradeoff: avoid wrong answers by allowing a third outcome.

The details depend on the states and priors, but the conceptual point is robust. POVMs allow outcome structures unavailable to ordinary projective measurements on the original system.

Continuous Outcome POVMs

For continuous outcomes, the finite sum is replaced by an operator-valued measure. In informal density notation one writes effects $F(x)\,dx$ satisfying

$$F(x)\ge0, \qquad \int dx\,F(x)=I,$$

with probability density

$$p(x)=\operatorname{Tr}(\rho F(x)).$$

A finite-resolution position detector is often described this way: exact position projectors are smeared by the detector response function. The resulting measurement may be more realistic than a sharp position PVM.

Informational Completeness

A POVM is informationally complete if its outcome probabilities determine the unknown state $\rho$ uniquely. This requires enough linearly independent effects to span the operator space of interest.

For a $d$-dimensional Hilbert space, a generic density operator has $d^2-1$ real parameters. Informationally complete POVMs therefore need enough independent outcome statistics to reconstruct those parameters, subject to positivity and trace constraints.

Informational completeness is important for tomography, but it is not required for ordinary measurement modeling. Many useful POVMs answer a narrower question.

For detector calibration, where the effects themselves are unknown, see Measurement Tomography.

Common Mistakes

• Calling a POVM element a projector when $F_m^2\ne F_m$.
• Assuming every POVM outcome has a unique post-measurement state.
• Forgetting that all effects must sum to $I$ over the complete outcome set.
• Omitting no-click, loss, or inconclusive outcomes and accidentally making probabilities sum to less than one.
• Treating a POVM as a detector dynamics model rather than a probability model.
• Assuming a POVM with more outcomes than the Hilbert-space dimension is impossible.
• Forgetting that the same POVM can be realized by different instruments.

Cross-Links

• POVMs: First Encounter
• Generalized Measurements Overview
• Why Generalized Measurements Are Needed
• State Tomography
• Unsharp Measurements
• Projective Measurements
• State Update Rules
• Selective and Nonselective Measurements
• von Neumann Measurement Model
• Common Misconceptions

References

• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer, 1983.
• P. Busch, P. J. Lahti, and P. Mittelstaedt, The Quantum Theory of Measurement, Springer, 1996.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, Edizioni della Normale, 2011.
• J. Preskill, Lecture Notes for Physics 229: Quantum Information and Computation, California Institute of Technology.

Exercises

1. Show that every effect in a finite POVM satisfies $0\le F_m\le I$.

Solution

By definition $F_m\ge0$. Completeness gives

$$I-F_m = \sum_{n\ne m}F_n.$$

The right-hand side is a sum of positive operators, hence positive. Therefore $I-F_m\ge0$, which means $F_m\le I$.

2. For the inefficient detector effects

$$F_{\mathrm{click}}=\eta |1\rangle\langle1|, \qquad F_{\mathrm{no}}=|0\rangle\langle0|+(1-\eta)|1\rangle\langle1|,$$

compute the click and no-click probabilities for $\rho=\operatorname{diag}(p,1-p)$.

Solution

The click probability is

$$p(\mathrm{click}) = \operatorname{Tr}(\rho F_{\mathrm{click}}) = \eta(1-p).$$

The no-click probability is

$$p(\mathrm{no}) = \operatorname{Tr}(\rho F_{\mathrm{no}}) = p+(1-\eta)(1-p).$$

They sum to $1$.

3. Verify that the trine effects $F_k=(2/3)\Pi_k$ sum to $I$ when $\sum_k\mathbf n_k=0$.

Solution

Using $\Pi_k=(I+\mathbf n_k\cdot\boldsymbol\sigma)/2$,

$$\sum_{k=0}^2F_k = \frac{2}{3} \sum_{k=0}^2 \frac{1}{2} \left( I+\mathbf n_k\cdot\boldsymbol\sigma \right).$$

The identity terms give $(2/3)(3/2)I=I$. The vector terms vanish because $\sum_k\mathbf n_k=0$. Hence $\sum_kF_k=I$.

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

Solution

One instrument ignores the system and produces a random outcome while leaving the state unchanged:

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

Another first measures in the $\{|0\rangle,|1\rangle\}$ basis and then reports a random label independent of the result:

$$\mathcal J_0(\rho)=\frac{1}{2} \sum_{j=0}^1 |j\rangle\langle j|\rho|j\rangle\langle j|,$$ $$\mathcal J_1(\rho)=\frac{1}{2} \sum_{j=0}^1 |j\rangle\langle j|\rho|j\rangle\langle j|.$$

Both have effects $I/2$ for each reported outcome, but the second instrument dephases the state while the first does not.