6.3.2 POVM

6.3.2 POVM#

Prompts

  • What is a POVM, and how does it generalize projective measurement? Why can a POVM have more outcomes than the Hilbert space dimension?

  • How does a measurement with a POVM work — what determines the probability of each outcome, and how does the state update afterward? Why does the post-measurement state involve the square root \(\sqrt{\hat{M}_i}\)?

  • Explain unambiguous state discrimination: how can a three-outcome POVM distinguish non-orthogonal states with certainty (when it gives a definite answer)?

  • State Naimark’s theorem. What does it mean physically that every POVM can be realized as a projective measurement on a larger system?

  • Give an example of a POVM that is not a projective measurement and explain what advantage it provides.

Lecture Notes#

Overview#

A key limitation of projective measurements is that they cannot reliably distinguish non-orthogonal quantum states. If we relax the requirement that measurement operators be projectors and allow general positive operators—as long as they sum to the identity—we gain flexibility to improve state discrimination. This generalization defines positive operator-valued measures (POVMs).

POVM Definition#

Where a projective measurement assigns each outcome a mutually orthogonal eigenspace, a POVM relaxes that orthogonality constraint, keeping only the conditions needed for a valid probability rule.

A positive operator-valued measure (POVM) is a set of operators \(\{\hat{M}_1, \hat{M}_2, \ldots, \hat{M}_n\}\) on Hilbert space \(\mathcal{H}\) satisfying two conditions:

1. Positive semidefiniteness:

(239)#\[ \hat{M}_i \geq 0 \quad \text{for all } i \quad (\text{all eigenvalues} \geq 0) \]

2. Completeness:

(240)#\[ \sum_i \hat{M}_i = \hat{I} \]

These conditions ensure that measurement probabilities are non-negative and sum to one.

Measurement with a POVM#

If a system in state \(\hat{\rho}\) is measured using POVM \(\{\hat{M}_i\}\):

Outcome probability:

(241)#\[ p_i = \operatorname{Tr}(\hat{M}_i \hat{\rho}) \]

Post-measurement state:

(242)#\[ \hat{\rho}'_i = \frac{\sqrt{\hat{M}_i} \hat{\rho} \sqrt{\hat{M}_i}}{\operatorname{Tr}(\hat{M}_i \hat{\rho})} \]

where \(\sqrt{\hat{M}_i}\) is the unique positive square root (defined via spectral decomposition: \(\hat{M}_i = \sum_j \lambda_{i,j} \vert v_{i,j}\rangle\langle v_{i,j}\vert \Rightarrow \sqrt{\hat{M}_i} = \sum_j \sqrt{\lambda_{i,j}} \vert v_{i,j}\rangle\langle v_{i,j}\vert\)).

This post-measurement-state formula is the Lüders update, the canonical state-update rule for a POVM. Unlike the outcome probability, however, the post-measurement state is not fixed by the POVM alone: the same element \(\hat{M}_i\) can be implemented by different physical measurements that reproduce its outcome statistics yet leave the system in different states. The Lüders form is the standard choice taken when no further detail of the measuring apparatus is specified.

Normalization check: \(\sum_i p_i = \sum_i \operatorname{Tr}(\hat{M}_i \hat{\rho}) = \operatorname{Tr}(\hat{I} \hat{\rho}) = 1\) ✓

Discussion: More Outcomes Than Dimensions

A POVM can have more outcomes than the Hilbert space dimension (\(n > \dim(\mathcal{H})\)). Why is this impossible for projective measurements? What does this overcomplete structure enable?

Poll: POVM vs. projective measurement

A POVM is a set of operators \(\{\hat{M}_i\}\) satisfying \(\sum_i \hat{M}_i = \hat{I}\) and \(\hat{M}_i \geq 0\), but the \(\hat{M}_i\) need not be projectors. What is a key advantage of POVMs over projective measurements?

(A) POVMs are simpler to implement experimentally.

(B) POVMs can distinguish between non-orthogonal states, whereas projective measurements cannot.

(C) POVMs require fewer measurement outcomes than projective measurements.

(D) POVMs provide perfect information about the state, whereas projective measurements do not.

POVM vs. Projective Measurement#

A projective measurement is a special case of a POVM where elements are orthogonal projectors \(\hat{M}_i = \hat{P}_i\) with:

  • \(\hat{P}_i^2 = \hat{P}_i\) (idempotent)

  • \(\hat{P}_i \hat{P}_j = \delta_{ij} \hat{P}_i\) (orthogonal)

  • \(\sum_i \hat{P}_i = \hat{I}\) (complete)

Key differences:

Property

Projective Measurement

General POVM

Outcomes

Limited to \(d\) (Hilbert space dimension)

Can have \(n > d\) (overcomplete)

Orthogonality

\(\hat{M}_i \hat{M}_j = 0\) for \(i \neq j\)

Elements can overlap: \(\hat{M}_i \hat{M}_j \neq 0\)

Information

Binary outcome per qubit

More outcomes = more information per measurement

Rank

Rank-1 projectors

Can be rank-1 or higher

Physical meaning of outcome

Eigenvalue of measured observable

Index number (no direct physical meaning)

Information trade-off: General POVMs extract more information about the system by allowing non-orthogonal measurement elements, at the cost of less definite outcome meaning.

Unambiguous State Discrimination#

One of the most striking applications of POVMs: distinguishing non-orthogonal quantum states with certainty when a definite answer is given.

Problem: Given a qubit in one of two non-orthogonal states \(\vert \psi_1\rangle\) or \(\vert \psi_2\rangle\) with overlap \(\langle\psi_1\vert \psi_2\rangle = c\) (where \(0 < \vert c\vert < 1\)), can you determine which state with certainty (never make an error)?

Projective measurement: No. Non-orthogonal states have overlapping eigenspaces, so any projective measurement will sometimes give wrong answers.

POVM solution: Yes, with a trade-off. Use a three-outcome POVM:

  • Outcome 1: “State is \(\vert \psi_1\rangle\)” (definite, no error)

  • Outcome 2: “State is \(\vert \psi_2\rangle\)” (definite, no error)

  • Outcome 3: “Inconclusive” (try again or use different strategy)

Construction: Unambiguous State Discrimination POVM

Work in the two-dimensional qubit space. Given two states \(\vert\psi_1\rangle, \vert\psi_2\rangle\) with overlap \(c = \langle\psi_1\vert\psi_2\rangle\), \(0 < \vert c\vert < 1\), write \(\vert\psi_1^\perp\rangle\) and \(\vert\psi_2^\perp\rangle\) for the unit vectors orthogonal to \(\vert\psi_1\rangle\) and \(\vert\psi_2\rangle\).

Step 1 — the unambiguous elements. Define the first two elements with a normalization constant \(a > 0\) left to be fixed:

\[\begin{split} \begin{split} \hat{M}_1 &= a\,(\hat{I} - \vert\psi_2\rangle\langle\psi_2\vert), \\ \hat{M}_2 &= a\,(\hat{I} - \vert\psi_1\rangle\langle\psi_1\vert). \end{split} \end{split}\]

In two dimensions \(\{\vert\psi_2\rangle, \vert\psi_2^\perp\rangle\}\) is an orthonormal basis, so its completeness relation \(\vert\psi_2\rangle\langle\psi_2\vert + \vert\psi_2^\perp\rangle\langle\psi_2^\perp\vert = \hat{I}\) gives the identity

\[ \hat{I} - \vert\psi_2\rangle\langle\psi_2\vert = \vert\psi_2^\perp\rangle\langle\psi_2^\perp\vert, \]

and likewise \(\hat{I} - \vert\psi_1\rangle\langle\psi_1\vert = \vert\psi_1^\perp\rangle\langle\psi_1^\perp\vert\). Hence \(\hat{M}_1 = a\,\vert\psi_2^\perp\rangle\langle\psi_2^\perp\vert\) and \(\hat{M}_2 = a\,\vert\psi_1^\perp\rangle\langle\psi_1^\perp\vert\) are positive rank-one operators. They are designed so that

\[ \langle\psi_2\vert\hat{M}_1\vert\psi_2\rangle = a\,\vert\langle\psi_2^\perp\vert\psi_2\rangle\vert^2 = 0, \]

and similarly \(\langle\psi_1\vert\hat{M}_2\vert\psi_1\rangle = 0\): outcome 1 can never be triggered by \(\vert\psi_2\rangle\), and outcome 2 never by \(\vert\psi_1\rangle\). This is what makes the two identifications error-free.

Step 2 — fix the normalization from positivity of \(\hat{M}_3\). The third element is the remainder,

\[ \hat{M}_3 = \hat{I} - \hat{M}_1 - \hat{M}_2, \]

assigned to the inconclusive outcome. Completeness \(\hat{M}_1 + \hat{M}_2 + \hat{M}_3 = \hat{I}\) then holds automatically; the binding constraint is positivity, \(\hat{M}_3 \geq 0\), and this is what fixes \(a\). Since \(\hat{M}_3 = \hat{I} - (\hat{M}_1 + \hat{M}_2)\), it is positive semidefinite exactly when every eigenvalue of \(\hat{M}_1 + \hat{M}_2\) is at most \(1\).

Evaluate that spectrum in the orthonormal basis \(\{\vert 0\rangle, \vert 1\rangle\}\) in which the two states take the symmetric form

\[\begin{split} \begin{split} \vert\psi_1\rangle &= \cos\tfrac{\theta}{2}\,\vert 0\rangle + \sin\tfrac{\theta}{2}\,\vert 1\rangle, \\ \vert\psi_2\rangle &= \cos\tfrac{\theta}{2}\,\vert 0\rangle - \sin\tfrac{\theta}{2}\,\vert 1\rangle, \end{split} \end{split}\]

with \(\langle\psi_1\vert\psi_2\rangle = \cos\theta = \vert c\vert\). (Any pair of states with overlap modulus \(\vert c\vert\) can be brought to this form by a unitary together with phase redefinitions, none of which changes the eigenvalues of \(\hat{M}_3\).) The orthogonal partners are \(\vert\psi_2^\perp\rangle = \sin\tfrac{\theta}{2}\vert 0\rangle + \cos\tfrac{\theta}{2}\vert 1\rangle\) and \(\vert\psi_1^\perp\rangle = \sin\tfrac{\theta}{2}\vert 0\rangle - \cos\tfrac{\theta}{2}\vert 1\rangle\). The two rank-one projectors carry off-diagonal entries \(\pm a\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}\) of opposite sign, so the cross terms cancel:

\[\begin{split} \hat{M}_1 + \hat{M}_2 = a\begin{pmatrix} 2\sin^2\tfrac{\theta}{2} & 0 \\ 0 & 2\cos^2\tfrac{\theta}{2} \end{pmatrix} = a\begin{pmatrix} 1 - \vert c\vert & 0 \\ 0 & 1 + \vert c\vert \end{pmatrix}. \end{split}\]

Its largest eigenvalue is \(a(1 + \vert c\vert)\), so \(\hat{M}_3 \geq 0\) requires \(a(1 + \vert c\vert) \leq 1\). The sharpest discrimination takes the largest admissible value,

\[ a = \frac{1}{1 + \vert c\vert}, \]

for which \(\hat{M}_3\) has eigenvalues \(1 - a(1 + \vert c\vert) = 0\) and \(1 - a(1 - \vert c\vert) = \tfrac{2\vert c\vert}{1 + \vert c\vert}\) — positive semidefinite, with the smallest eigenvalue exactly zero. The completed POVM is

\[\begin{split} \begin{split} \hat{M}_1 &= \frac{\hat{I} - \vert\psi_2\rangle\langle\psi_2\vert}{1 + \vert c\vert}, \\ \hat{M}_2 &= \frac{\hat{I} - \vert\psi_1\rangle\langle\psi_1\vert}{1 + \vert c\vert}, \\ \hat{M}_3 &= \hat{I} - \hat{M}_1 - \hat{M}_2. \end{split} \end{split}\]

Step 3 — outcome probabilities. Suppose the input is \(\vert\psi_1\rangle\). The correct identification has probability

\[\begin{split} \begin{split} P(1\vert\psi_1) = \langle\psi_1\vert\hat{M}_1\vert\psi_1\rangle &= \frac{1}{1 + \vert c\vert}\,\langle\psi_1\vert(\hat{I} - \vert\psi_2\rangle\langle\psi_2\vert)\vert\psi_1\rangle \\ &= \frac{1 - \vert\langle\psi_1\vert\psi_2\rangle\vert^2}{1 + \vert c\vert} = \frac{1 - \vert c\vert^2}{1 + \vert c\vert} = 1 - \vert c\vert. \end{split} \end{split}\]

The wrong identification never happens, \(P(2\vert\psi_1) = \langle\psi_1\vert\hat{M}_2\vert\psi_1\rangle = 0\) (Step 1), so the inconclusive outcome carries the rest,

\[ P(3\vert\psi_1) = 1 - P(1\vert\psi_1) - P(2\vert\psi_1) = \vert c\vert. \]

By the \(\psi_1 \leftrightarrow \psi_2\) symmetry of the construction, an input \(\vert\psi_2\rangle\) gives \(P(2\vert\psi_2) = 1 - \vert c\vert\), \(P(1\vert\psi_2) = 0\), and \(P(3\vert\psi_2) = \vert c\vert\).

Key property. Each state is identified with zero error — outcomes 1 and 2 are never false — at the cost of an inconclusive outcome 3 occurring with probability \(\vert c\vert\), exactly the overlap modulus. The more similar the states (larger \(\vert c\vert\)), the less often the discrimination succeeds; orthogonal states (\(\vert c\vert \to 0\)) are never inconclusive.

Naimark’s Theorem: POVM as Extended Projective Measurement#

Every POVM on a Hilbert space can be realized as a projective measurement on an extended Hilbert space (original space plus auxiliary system).

Naimark’s Theorem

Every POVM \(\{\hat{M}_i\}\) on \(\mathcal{H}\) can be realized as a projective measurement on an extended space \(\mathcal{H} \otimes \mathcal{H}_\text{aux}\). Specifically, there exist:

  • An auxiliary (ancilla) Hilbert space \(\mathcal{H}_\text{aux}\)

  • Orthogonal projectors \(\{\hat{P}_i\}\) on \(\mathcal{H} \otimes \mathcal{H}_\text{aux}\)

such that:

(243)#\[ \hat{M}_i = \operatorname{Tr}_\text{aux}(\hat{P}_i (\hat{I}_{\mathcal{H}} \otimes \vert 0\rangle\langle 0\vert_\text{aux})) \]

Physical picture: Entangle the system with an ancilla, perform a projective measurement on the combined system, and trace out (discard) the ancilla measurement outcome.

This theorem reveals a profound truth: POVMs are not fundamentally more general than projective measurements—they are equivalent to projective measurements on a larger system. The apparent “softness” of a POVM comes from not observing (or not having access to) the ancilla measurement outcome.

Examples#

Example: Six-State Octahedral IC-POVM on a Qubit

One convenient informationally complete POVM for a qubit uses the six eigenstates of \(\hat{\sigma}^{x}\), \(\hat{\sigma}^{y}\), and \(\hat{\sigma}^{z}\). These states sit at the vertices of a regular octahedron on the Bloch sphere. Define

\[ \hat{M}_{k} = \tfrac{1}{3}\vert\phi_{k}\rangle\langle\phi_{k}\vert\quad (k = 1,\ldots,6). \]

This is a valid POVM because each Pauli axis contributes a pair of opposite projectors summing to \(\hat{I}\):

\[ \sum_{k=1}^{6}\hat{M}_{k} = \frac{1}{3}\sum_{i=x,y,z}\left(\vert{+}_{i}\rangle\langle{+}_{i}\vert + \vert{-}_{i}\rangle\langle{-}_{i}\vert\right) = \hat{I}. \]

The six probabilities are informationally complete: they contain enough data to reconstruct the Bloch vector of any qubit state. This octahedral POVM is therefore useful for tomography from three mutually unbiased bases.

A different standard construction is the qubit SIC-POVM, which uses four tetrahedral states (the minimal number \(d^{2}=4\) for informational completeness in dimension \(d=2\)). Both are IC-POVMs; the octahedral version is overcomplete, while the SIC-POVM is minimal and symmetric.

Example: Heterodyne Detection (Optical POVM)

In quantum optics, measuring quadrature amplitudes (amplitude and phase of a light field) is described by a continuous POVM:

\[ \hat{M}(\alpha) = \frac{1}{\pi}\vert \alpha\rangle\langle\alpha\vert_\text{coherent} \]

where \(\vert \alpha\rangle\) are coherent states indexed by complex amplitude \(\alpha \in \mathbb{C}\).

Properties:

  • Continuous outcome space (not discrete)

  • Coherent states are highly non-orthogonal: \(\vert\langle\alpha\vert \beta\rangle\vert = \mathrm{e}^{-\vert \alpha-\beta\vert^2/2} \neq 0\) for \(\alpha \neq \beta\)

  • Enables full quantum tomography of optical field states

Summary#

  • POVM definition: Positive operator-valued measures \(\{\hat{M}_m\}\) satisfy \(\hat{M}_m \geq 0\) and \(\sum_m \hat{M}_m = \hat{I}\), but are not required to be orthogonal or satisfy \(\hat{M}_m^2 = \hat{M}_m\) (unlike projectors).

  • Measurement probabilities: \(p_m = \operatorname{Tr}(\hat{M}_m \hat{\rho})\) (same formula as projective), but post-measurement state is not determined by POVM alone—different physical realizations can give same outcome statistics.

  • Advantages over projectors: More general; one outcome can have multiple post-measurement states (depends on probe details); allows finer-grained outcome statistics with fewer outcomes than dimension of Hilbert space.

  • Unambiguous state discrimination: POVM enables non-projective measurements that sometimes identify unknown quantum states with zero ambiguity (no error-free projective equivalent exists).

  • Naimark’s theorem: Every POVM on \(\mathcal{H}\) can be realized as an ordinary projective measurement on an extended space \(\mathcal{H} \otimes \mathcal{H}_\text{aux}\)—entangle the system with an ancilla, measure the enlarged system projectively, and discard the ancilla outcome. POVMs are therefore not fundamentally more general than projective measurements; their extra flexibility is the price of not observing the ancilla.

See Also

Homework#

1. Projective Measurement and POVM. Show that any projective measurement \(\{\hat{P}_i\}\) (with \(\hat{P}_i^2 = \hat{P}_i\), \(\hat{P}_i \hat{P}_j = \delta_{ij}\hat{P}_i\), \(\sum_i \hat{P}_i = \hat{I}\)) is a special case of a POVM by setting \(\hat{M}_i = \hat{P}_i\). Verify the POVM conditions.

2. Trine POVM on Qubits. Consider a trine POVM on a qubit: \(\hat{M}_k = \frac{2}{3}\vert \phi_k\rangle\langle\phi_k\vert\) for \(k=0,1,2\), where \(\vert \phi_0\rangle = \vert 0\rangle\), \(\vert \phi_1\rangle = -\frac{1}{2}\vert 0\rangle + \frac{\sqrt{3}}{2}\vert 1\rangle\), \(\vert \phi_2\rangle = -\frac{1}{2}\vert 0\rangle - \frac{\sqrt{3}}{2}\vert 1\rangle\). Verify that \(\sum_{k=0}^2 \hat{M}_k = \hat{I}\).

3. Unambiguous State Discrimination. Consider two non-orthogonal states \(\vert \psi_1\rangle\) and \(\vert \psi_2\rangle\) with \(\langle\psi_1\vert \psi_2\rangle = c \neq 0\). A POVM for unambiguous discrimination has three outcomes: “\(\psi_1\)”, “\(\psi_2\)”, and “?” (inconclusive). Explain why this task is impossible with a projective measurement if \(c \neq 0\), but possible with a POVM. What condition must the POVM elements satisfy to make discrimination unambiguous when a definite answer is given?

4. POVM State Discrimination. Construct an explicit POVM for unambiguous discrimination of \(\vert 0\rangle\) and \(\vert +\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\). Find POVM elements \(\hat{M}_1\) (identifies \(\vert 0\rangle\)), \(\hat{M}_2\) (identifies \(\vert +\rangle\)), and \(\hat{M}_?\) (inconclusive) that satisfy all POVM conditions. What is the probability of an inconclusive result for each state?

5. Naimark Dilation. Any POVM \(\{\hat{M}_i\}\) on \(\mathcal{H}\) can be realized as a projective measurement on a larger Hilbert space \(\mathcal{H} \otimes \mathcal{H}_\text{aux}\): introduce an ancilla in state \(\vert 0\rangle_\text{aux}\), perform a unitary \(\hat{U}\) on \(\mathcal{H} \otimes \mathcal{H}_\text{aux}\), then measure the ancilla. Show that if \(\hat{U}(\vert \psi\rangle \otimes \vert 0\rangle) = \sum_i (\sqrt{\hat{M}_i}\vert \psi\rangle) \otimes \vert i\rangle_\text{aux}\), then the probability of outcome \(i\) is \(\mathrm{Tr}[\hat{M}_i \vert \psi\rangle\langle\psi\vert]\).

6. Informationally Complete POVM. An informationally complete POVM (IC-POVM) has enough elements to uniquely determine any density matrix \(\hat{\rho}\) from the measurement statistics \(\{p_i = \mathrm{Tr}[\hat{M}_i \hat{\rho}]\}\). For a \(d\)-dimensional system, what is the minimum number of POVM elements needed for an IC-POVM? Justify your answer by counting the degrees of freedom in \(\hat{\rho}\).

7. POVM Outcome Probabilities. For a POVM with elements \(\{\hat{M}_i\}\), the probability of outcome \(i\) on mixed state \(\hat{\rho}\) is \(p_i = \mathrm{Tr}[\hat{M}_i \hat{\rho}]\). Show that if \(\hat{\rho} = \sum_k q_k \hat{\rho}_k\) is a convex mixture, then \(p_i = \sum_k q_k \mathrm{Tr}[\hat{M}_i \hat{\rho}_k]\). This shows POVM statistics are linear in the state.

8. POVM vs Projective Measurement. Compare the information gain of a POVM vs. a projective measurement. For a qubit initially in an unknown state on the Bloch sphere:

(a) A projective measurement gives binary outcomes. How much classical information (in bits) does it convey?

(b) A 4-element symmetric IC-POVM (SIC-POVM) with \(\hat{M}_k = \frac{1}{2}\vert \phi_k\rangle\langle\phi_k\vert\) for 4 tetrahedral states. Explain qualitatively why this conveys more information about \(\hat{\rho}\) per measurement than a projective measurement.