Naimark Dilation for Measurable Outcomes

Naimark’s theorem represents every normalized POVM as the compression of a PVM on a larger Hilbert space. The measurable-outcome theorem requires neither finitely many effects nor a pointwise operator density. The canonical Naimark Dilation article gives finite-dimensional implementations; this page constructs the general dilation and states its uniqueness precisely.

Helpful background. Generalized observables as POVMs supplies operator-measure additivity; Hilbert-space completion supplies the quotient-and-completion idea.

The dilation theorem

Let $E:\Sigma\to\mathcal B(\mathcal H)$ be a normalized POVM on a measurable space $(\Omega,\Sigma)$. There exist a Hilbert space $\mathcal K$, an isometry $V:\mathcal H\to\mathcal K$, and a PVM $P$ on the same outcome space such that

$$E(\Delta)=V^*P(\Delta)V \qquad(\Delta\in\Sigma).$$

The dilation is minimal if

$$\mathcal K=\overline{\operatorname{span}} \{P(\Delta)V\psi:\Delta\in\Sigma,\ \psi\in\mathcal H\}.$$

Any two minimal dilations are related by a unitary intertwining both the embedded system and every event projection. Arbitrary larger dilations can contain unused subspaces, so uniqueness is asserted only after imposing minimality.

Since $V^*V=I$, normalized system vectors remain normalized. The probability identity $\operatorname{Tr}(\rho E(\Delta))=\operatorname{Tr}(V\rho V^*P(\Delta))$ then gives the operational content.

Building the larger Hilbert space

Start with formal finite sums of vector-valued indicator functions, $s=\sum_i\mathbf1_{\Delta_i}\otimes\psi_i$. Define a sesquilinear form by

$$\langle s,t\rangle_E =\sum_{i,j} \langle\psi_i,E(\Delta_i\cap\Gamma_j)\phi_j\rangle, \qquad t=\sum_j\mathbf1_{\Gamma_j}\otimes\phi_j.$$

To see positivity, refine all sets in $s$ into a finite disjoint measurable partition. On each cell, the vector coefficient is a fixed sum of the $\psi_i$. The quadratic form becomes a sum of positive terms of the form $\langle\eta,E(C)\eta\rangle$.

Quotient by the zero-norm vectors and complete. This produces $\mathcal K$. Multiplication by $\mathbf1_\Lambda$ defines an orthogonal projection $P(\Lambda)$: it is idempotent, self-adjoint for the form, and contractive. Countable additivity follows from countable additivity of the scalar measures induced by $E$, first on the dense simple-function subspace and then by continuity.

Set $V\psi=\mathbf1_\Omega\otimes\psi$. Normalization gives $\|V\psi\|^2=\langle\psi,E(\Omega)\psi\rangle=\|\psi\|^2$. Finally,

$$\langle V\phi,P(\Delta)V\psi\rangle =\langle\phi,E(\Delta)\psi\rangle,$$

which proves the compression formula. The construction is minimal because the simple functions are spanned by vectors $P(\Delta)V\psi$.

Why minimal dilations are unitarily equivalent

Suppose $(\mathcal K_1,P_1,V_1)$ and $(\mathcal K_2,P_2,V_2)$ are minimal. Send

$$\sum_iP_1(\Delta_i)V_1\psi_i \longmapsto \sum_iP_2(\Delta_i)V_2\psi_i.$$

The inner product of two such sums depends only on $E(\Delta_i\cap\Gamma_j)$, so the map preserves inner products and sends null relations to null relations. It therefore extends by completion to an isometry. Minimality makes its range dense, hence the extension is unitary. Its definition gives $UV_1=V_2$ and $UP_1(\Delta)=P_2(\Delta)U$.

This is a uniqueness theorem for the compressed observable. It is not uniqueness of the apparatus Hamiltonian, the interaction time, or the instrument left on the original system.

Countable outcomes and strong convergence

For a countable POVM $(E_j)$ with $\sum_jE_j=I$ strongly, a convenient dilation is

$$\mathcal K=\bigoplus_j\mathcal H,\qquad V\psi=(\sqrt{E_j}\psi)_j.$$

The norm identity $\sum_j\|\sqrt{E_j}\psi\|^2=\|\psi\|^2$ shows that the vector lies in the Hilbert direct sum and $V$ is an isometry. Let $P_j$ select the $j$th component. Then $V^*P_jV=E_j$. This dilation need not be minimal; restrict to the closed span specified in the theorem when minimality matters.

For a binary effect $E$, the map is $V\psi=(\sqrt E\psi,\sqrt{I-E}\psi)$. If $E=\lambda I$, outcome one has probability $\lambda$ for every state: a projective measurement on the enlarged space can induce a completely uninformative system POVM.

What the theorem leaves unspecified

Naimark dilation guarantees an enlarged PVM and an embedding. It does not by itself choose a system-preserving measurement interaction or a conditional system output. Different instruments can have the same POVM and hence the same probability dilation.

It also does not justify generalized eigenkets for individual continuous outcomes. The enlarged PVM is an event measure; it may have zero singleton projections. Probability and conditioning must still be formulated with measurable events and the relevant instrument.

Exercises

1. For a binary POVM, prove the direct-sum dilation formula by evaluating $V^*P_1V$ on arbitrary vectors.

Solution

The first component of $V\psi$ is $\sqrt E\psi$. Therefore $\langle V\phi,P_1V\psi\rangle=\langle\sqrt E\phi,\sqrt E\psi\rangle=\langle\phi,E\psi\rangle$. This identifies the compressed operator as $E$, exactly as required by the operational dilation.

2. Why does adding an orthogonal unused Hilbert space to a minimal dilation destroy minimality without changing the POVM?

Solution

Embed $V\mathcal H$ only in the original summand and extend the PVM arbitrarily on the new summand. Compression still sees only the original part. The vectors $P(\Delta)V\psi$ never reach the added summand, so their closed span is no longer the full enlarged space.

References

• Busch, Paul, Pekka Lahti, Juha-Pekka Pellonpää, and Kari Ylinen, Quantum Measurement, Springer, 2016 — Naimark dilations of operator measures and measurement models. Publisher.
• Heinosaari, Teiko, and Mário Ziman, The Mathematical Language of Quantum Theory: From Uncertainty to Entanglement, Cambridge University Press, 2012 — dilation and observable structure. Publisher.