# Naimark's dilation theorem

In operator theory, Naimark's dilation theorem is a result that characterizes positive operator valued measures. It can be viewed as a consequence of Stinespring's dilation theorem.

## Some preliminary notions

Let X be a compact Hausdorff space, H be a Hilbert space, and L(H) the Banach space of bounded operators on H. A mapping E from the Borel σ-algebra on X to $L(H)$ is called an operator-valued measure if it is weakly countably additive, that is, for any disjoint sequence of Borel sets $\{B_{i}\}$, we have
    $\langle E(\cup _{i}B_{i})x,y\rangle =\sum _{i}\langle E(B_{i})x,y\rangle$
for all x and y. Some terminology for describing such measures are:
* E is called regular if the scalar valued measure
    $B\rightarrow \langle E(B)x,y\rangle$
is a regular Borel measure, meaning all compact sets have finite total variation and the measure of a set can be approximated by those of open sets.
* E is called bounded if $|E|=\sup _{B}\|E(B)\|<\infty$.
* E is called positive if E(B) is a positive operator for all B.
* E is called self-adjoint if E(B) is self-adjoint for all B.
* E is called spectral if it is self-adjoint and $E(B_{1}\cap B_{2})=E(B_{1})E(B_{2})$ for all $B_{1},B_{2}$.
We will assume throughout that E is regular.
Let C(X) denote the abelian C*-algebra of continuous functions on X. If E is regular and bounded, it induces a map $\Phi _{E}:C(X)\rightarrow L(H)$ in the obvious way:
    $\langle \Phi _{E}(f)h_{1},h_{2}\rangle =\int _{X}f(x)\langle E(dx)h_{1},h_{2}\rangle$
The boundedness of E implies, for all h of unit norm
    $\langle \Phi _{E}(f)h,h\rangle =\int _{X}f(x)\langle E(dx)h,h\rangle \leq \|f\|_{\infty }\cdot |E|.$
This shows $\;\Phi _{E}(f)$ is a bounded operator for all f, and $\Phi _{E}$ itself is a bounded linear map as well.
The properties of $\Phi _{E}$ are directly related to those of E:
* If E is positive, then $\Phi _{E}$, viewed as a map between C*-algebras, is also positive.
* $\Phi _{E}$ is a homomorphism if, by definition, for all continuous f on X and $h_{1},h_{2}\in H$,
    $\langle \Phi _{E}(fg)h_{1},h_{2}\rangle =\int _{X}f(x)\cdot g(x)\;\langle E(dx)h_{1},h_{2}\rangle =\langle \Phi _{E}(f)\Phi _{E}(g)h_{1},h_{2}\rangle .$
Take f and g to be indicator functions of Borel sets and we see that $\Phi _{E}$ is a homomorphism if and only if E is spectral.
* Similarly, to say $\Phi _{E}$ respects the * operation means
    $\langle \Phi _{E}({\bar {f}})h_{1},h_{2}\rangle =\langle \Phi _{E}(f)^{*}h_{1},h_{2}\rangle .$
The LHS is
    $\int _{X}{\bar {f}}\;\langle E(dx)h_{1},h_{2}\rangle ,$
and the RHS is
    $\langle h_{1},\Phi _{E}(f)h_{2}\rangle ={\overline {\langle \Phi _{E}(f)h_{2},h_{1}\rangle }}=\int _{X}{\bar {f}}(x)\;{\overline {\langle E(dx)h_{2},h_{1}\rangle }}=\int _{X}{\bar {f}}(x)\;\langle h_{1},E(dx)h_{2}\rangle$
So, taking f a sequence of continuous functions increasing to the indicator function of B, we get $\langle E(B)h_{1},h_{2}\rangle =\langle h_{1},E(B)h_{2}\rangle$, i.e. E(B) is self adjoint.
* Combining the previous two facts gives the conclusion that $\Phi _{E}$ is a *-homomorphism if and only if E is spectral and self adjoint. (When E is spectral and self adjoint, E is said to be a projection-valued measure or PVM.)

## Naimark's theorem

The theorem reads as follows: Let E be a positive L(H)-valued measure on X. There exists a Hilbert space K, a bounded operator $V:K\rightarrow H$, and a self-adjoint, spectral L(K)-valued measure F on X, such that
    $\;E(B)=VF(B)V^{*}.$

### Proof

We now sketch the proof. The argument passes E to the induced map $\Phi _{E}$ and uses Stinespring's dilation theorem. Since E is positive, so is $\Phi _{E}$ as a map between C*-algebras, as explained above. Furthermore, because the domain of $\Phi _{E}$, C(X), is an abelian C*-algebra, we have that $\Phi _{E}$ is completely positive. By Stinespring's result, there exists a Hilbert space K, a *-homomorphism $\pi :C(X)\rightarrow L(K)$, and operator $V:K\rightarrow H$ such that
    $\;\Phi _{E}(f)=V\pi (f)V^{*}.$
Since π is a *-homomorphism, its corresponding operator-valued measure F is spectral and self adjoint. It is easily seen that F has the desired properties.

## Finite-dimensional case

In the finite-dimensional case, there is a somewhat more explicit formulation.
Suppose now $X=\{1,\dotsc ,n\}$, therefore C(X) is the finite-dimensional algebra $\mathbb {C} ^{n}$, and H has finite dimension m. A positive operator-valued measure E then assigns each i a positive semidefinite m × m matrix $E_{i}$. Naimark's theorem now states that there is a projection-valued measure on X whose restriction is E.
Of particular interest is the special case when $\sum _{i}E_{i}=I$ where I is the identity operator. (See the article on POVM for relevant applications.) In this case, the induced map $\Phi _{E}$ is unital. It can be assumed with no loss of generality that each $E_{i}$ takes the form $x_{i}x_{i}^{*}$ for some potentially subnorrmalized vector $x_{i}\in \mathbb {C} ^{m}$. Under such assumptions, the case $n<m$ is excluded and we must have either
* $n=m$ and E is already a projection-valued measure (because $\sum _{i=1}^{n}x_{i}x_{i}^{*}=I$ if and only if $\{x_{i}\}$ is an orthonormal basis),
* $n>m$ and $\{E_{i}\}$ does not consist of mutually orthogonal projections.
For the second possibility, the problem of finding a suitable projection-valued measure now becomes the following problem. By assumption, the non-square matrix
    $M={\begin{bmatrix}x_{1}\cdots x_{n}\end{bmatrix}}$
is a co-isometry, that is $MM^{*}=I$. If we can find a $(n-m)\times n$ matrix N where
    $U={\begin{bmatrix}M\\N\end{bmatrix}}$
is a n × n unitary matrix, the projection-valued measure whose elements are projections onto the column vectors of U will then have the desired properties. In principle, such a N can always be found.

## Spelling

In the physics literature, it is common to see the spelling “Neumark” instead of “Naimark.” The latter variant is according to the romanization of Russian used in translation of Soviet journals, with diacritics omitted (originally Naĭmark). The former is according to the etymology of the surname of Mark Naimark.