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.

A POVM determines outcome probabilities. It does not determine post-measurement states, which require an instrument. The discrete examples here use finite-dimensional Hilbert spaces; a final section explains continuous outcome measures.

Helpful background. Density operators and projectors supply the trace rule and sharp-measurement special case.

Positive effects and normalized probabilities

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.

Detector responses and coarse outcomes

Inefficiency, dark counts, overlapping response functions, and indirect probe readout can give nonprojective effects. The introductory examples explain why. Coarse graining alone need not do so: sums of disjoint orthogonal projectors remain projectors, although a finer measurement with discarded labels can have extra backaction.

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

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

with

$$F_m=\operatorname{Tr}_A\!\left[ (I\otimes\sqrt{\eta_A})U^\dagger(I\otimes Q_m)U (I\otimes\sqrt{\eta_A})\right].$$

The expression inside the partial trace is positive, and summing $m$ gives $I_S\otimes\eta_A$, whose partial trace is $I_S$. This proves positivity and completeness directly. The fixed, initially uncorrelated apparatus state is part of the construction.

This is the conceptual content behind dilation theorems: every finite-dimensional POVM can be represented as a projective measurement on a larger space; see Naimark dilation.

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

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

with $|\alpha|^2+|\beta|^2=1$, 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

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

For $\mathbf r=\mathbf n_0$, the dot products are $1,-1/2,-1/2$, giving $(2/3,1/6,1/6)$. For a state perpendicular to the trine plane, all probabilities are $1/3$. Thus the trine distinguishes directions within its plane but misses the perpendicular Bloch component. It is not a projective measurement on the qubit: all three effects are nonzero, with eigenvalues $0$ and $2/3$.

Example: Unambiguous State Discrimination

Suppose the two possible preparations are $|0\rangle$ and $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. Write $|-\rangle=(|0\rangle-|1\rangle)/\sqrt2$ and choose

$$F_0=c|-\rangle\langle-|,\qquad F_+=c|1\rangle\langle1|, \qquad F_?=I-F_0-F_+,\quad c=2-\sqrt2.$$

Outcome 0 can never occur for $|+\rangle$, and outcome $+$ can never occur for $|0\rangle$. The sum of the first two effects has eigenvalues $c(1\pm1/\sqrt2)$, whose larger value is 1, so $F_?\geq0$.

For either preparation, the correct conclusive outcome has probability $c/2=1-1/\sqrt2$, while the inconclusive result has probability $1/\sqrt2$. Thus no conclusive answer is wrong, but some runs supply no identification. For equal priors this attains the optimum failure probability established for two pure states (see Chefles, 1998): $|\langle0|+\rangle|=1/\sqrt2$; this optimum is a state-discrimination result, not a property of arbitrary detector noise. The broader optimization of measurement probabilities is developed in Watrous, chapter 3.

Continuous Outcome POVMs

For continuous outcomes, the finite sum is replaced by an operator-valued measure. When an operator density with respect to $dx$ exists, one writes $F(x)\,dx$ satisfying

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

with probability density

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

Unlike a discrete effect, a density $F(x)$ need not obey $F(x)\leq I$ pointwise. It has inverse-outcome units, and the bounded effect for an interval is its integral over that interval. 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. An $m$-outcome distribution has at most $m-1$ independent probabilities, so full-state informational completeness requires $m\geq d^2$, together with the spanning condition. The number of outcomes alone is not sufficient: repeated or coplanar effects can remain redundant.

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.

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.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer (2016). Publisher.
• A. Chefles, “Unambiguous discrimination between linearly independent quantum states,” Physics Letters A 239, 339–347 (1998). Article; author’s version.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd English ed., Edizioni della Normale (2011).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapters 2–3. Author’s text.