Naimark Dilation

Naimark dilation says that a generalized measurement can be viewed as an ordinary projective measurement after enlarging the Hilbert space. In finite dimensions, every POVM on a system can be obtained by coupling the system to an ancilla and then projectively measuring the larger system.

The slogan is:

$$\text{POVM on }S \quad=\quad \text{PVM on a larger space, compressed back to }S.$$

This result explains why POVMs are not an ad hoc replacement for projective measurement. They are the effective probability rules seen by a subsystem when the apparatus, ancilla, or environment is not kept as part of the final description.

Helpful background. POVMs define the effects being represented; tensor products let us add an apparatus.

Naimark’s theorem for a finite POVM

Let $\mathcal H_S$ be finite dimensional and let $\{F_m\}$ be a finite-outcome POVM. Watrous, theorem 2.42, gives a finite-dimensional version of the dilation theorem:

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

A Naimark dilation consists of a larger Hilbert space $\mathcal K$, an isometry

$$V:\mathcal H_S\to\mathcal K, \qquad V^\dagger V=I_S,$$

and a projective measurement $\{\Pi_m\}$ on $\mathcal K$ such that

$$F_m = V^\dagger\Pi_m V.$$

The projectors obey

$$\Pi_m\Pi_n=\delta_{mn}\Pi_m, \qquad \sum_m\Pi_m=I_{\mathcal K}.$$

For any system state $\rho$, the POVM probability is reproduced by the projective measurement on the dilated state:

$$\operatorname{Tr}_S(F_m\rho) = \operatorname{Tr}_{\mathcal K} \!\left[ \Pi_m\,V\rho V^\dagger \right].$$

The theorem is about probabilities. A particular dilation can also define a state update, but a POVM alone does not uniquely determine that update.

Canonical Construction

For a finite POVM, take the unique positive square root of each effect:

$$M_m=\sqrt{F_m}.$$

Let the ancilla have an orthonormal outcome basis $\{\lvert m\rangle_A\}$. Define

$$V\lvert\psi\rangle = \sum_m M_m\lvert\psi\rangle \otimes \lvert m\rangle_A.$$

This map is an isometry because

$$\begin{aligned} \langle\phi\rvert V^\dagger V\lvert\psi\rangle &= \sum_m \langle\phi\rvert M_m^\dagger M_m\lvert\psi\rangle\\ &= \langle\phi\rvert \left(\sum_m F_m\right) \lvert\psi\rangle\\ &= \langle\phi|\psi\rangle. \end{aligned}$$

Equivalently,

$$V^\dagger V=I_S.$$

Now measure the ancilla projectively with

$$\Pi_m = I_S\otimes \lvert m\rangle\langle m\rvert_A.$$

Then

$$V^\dagger\Pi_m V = M_m^\dagger M_m = F_m.$$

Thus the POVM has been realized as a projective measurement on $S\otimes A$ after the isometric embedding $V$.

From Isometry to Unitary Apparatus

In this finite-dimensional construction, the isometry maps a subspace to another subspace of the same dimension. Completing orthonormal bases for their complements extends that map to a unitary on $\mathcal H_S\otimes\mathcal H_A$. Operationally, prepare the ancilla in a fixed ready state $\lvert0\rangle_A$ and choose a unitary $U$ such that

$$U(\lvert\psi\rangle_S\otimes\lvert0\rangle_A) = \sum_m M_m\lvert\psi\rangle_S \otimes \lvert m\rangle_A$$

for all $\lvert\psi\rangle_S$.

Then a projective measurement of the ancilla in the $\{\lvert m\rangle_A\}$ basis gives

$$p(m) = \operatorname{Tr} \!\left[ (I_S\otimes\lvert m\rangle\langle m\rvert) U(\rho_S\otimes\lvert0\rangle\langle0\rvert) U^\dagger \right].$$

The result reduces to

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

This is the standard indirect-measurement picture: couple system to apparatus, then perform a sharp pointer readout on the apparatus.

Associated Instrument

The construction above does more than reproduce probabilities. If the ancilla outcome is $m$ and the ancilla is discarded afterward, the system operation is

$$\mathcal I_m(\rho) = M_m\rho M_m^\dagger.$$

For the square-root choice $M_m=\sqrt{F_m}$, this is the square-root instrument:

$$\mathcal I_m(\rho) = \sqrt{F_m}\,\rho\,\sqrt{F_m}.$$

But the same POVM can be realized by other measurement operators $K_m$ satisfying

$$K_m^\dagger K_m=F_m.$$

For example, $K_m=U_m\sqrt{F_m}$ has the same effect for any unitary $U_m$:

$$K_m^\dagger K_m=F_m.$$

It generally has different backaction. Therefore:

$$\text{Naimark dilation explains POVM probabilities;} \qquad \text{the instrument specifies the state update.}$$

See Measurement Backaction for this distinction in physical terms.

Example: Unsharp Qubit Measurement

Consider the two-outcome qubit POVM

$$F_\pm = \frac{1}{2} \left( I\pm \eta\sigma_z \right), \qquad 0\le\eta\le1.$$

For $\eta=1$, the effects are the projectors onto $\sigma_z$ eigenstates. For $0\lt\eta\lt1$, they are positive but not projectors:

$$F_\pm^2\ne F_\pm.$$

The square-root Naimark construction uses

$$M_\pm=\sqrt{F_\pm}$$

and an ancilla basis $\{\lvert+\rangle_A,\lvert-\rangle_A\}$:

$$V\lvert\psi\rangle = M_+\lvert\psi\rangle\lvert+\rangle_A + M_-\lvert\psi\rangle\lvert-\rangle_A.$$

Measuring the ancilla projectively gives probabilities

$$p(\pm) = \operatorname{Tr}(F_\pm\rho).$$

Here is an explicit coupling. Put $a=\sqrt{(1+\eta)/2}$ and $b=\sqrt{(1-\eta)/2}$. With the pointer initially in a fixed ready state, the required action is

$$\begin{aligned} |0\rangle|\mathrm{ready}\rangle_A&\longmapsto |0\rangle(a|+\rangle_A+b|-\rangle_A),\\ |1\rangle|\mathrm{ready}\rangle_A&\longmapsto |1\rangle(b|+\rangle_A+a|-\rangle_A). \end{aligned}$$

These two outputs are orthonormal because their system factors are orthogonal, even though their pointer factors overlap by $2ab=\sqrt{1-\eta^2}$. For $\eta=3/5$, the pointer amplitudes are $a=2/\sqrt5$, $b=1/\sqrt5$. An input $|0\rangle$ gives pointer probabilities $4/5$ and $1/5$, whereas $|1\rangle$ gives $1/5$ and $4/5$. Sharp pointer readout therefore induces unsharp system statistics. At $\eta=0$ the pointer states coincide; at $\eta=1$ they are orthogonal.

Example: More Outcomes Than Dimension

A qubit projective measurement on the qubit alone can have at most two nonzero rank-one orthogonal outcomes. A qubit POVM can have three or more nonzero outcomes.

The trine POVM has three effects

$$F_k = \frac{1}{3} \left( I+\mathbf n_k\cdot\boldsymbol\sigma \right), \qquad k=1,2,3,$$

where the unit vectors $\mathbf n_k$ lie in the equatorial plane separated by $120^\circ$ and satisfy

$$\sum_{k=1}^3\mathbf n_k=0.$$

Then

$$\sum_{k=1}^3F_k=I.$$

These three effects cannot be a three-outcome PVM on a two-dimensional Hilbert space, because three nonzero mutually orthogonal projectors cannot all fit inside a qubit space. Naimark dilation says they can be represented as a projective measurement after embedding the qubit into a larger Hilbert space.

How large must the enlarged space be?

The tensor-product construction uses $dM$ dimensions for a $d$-dimensional input and $M$ outcomes. It need not be minimal. For a finite POVM, a smaller construction uses

$$\mathcal K=\bigoplus_m\operatorname{ran}\sqrt{F_m}, \qquad V|\psi\rangle=\bigoplus_m\sqrt{F_m}|\psi\rangle.$$

Each $\Pi_m$ selects one orthogonal summand, so $V^\dagger\Pi_mV=F_m$. Its dimension is $\sum_m\operatorname{rank}F_m$. No dilation can use less: compression cannot increase rank, hence $\operatorname{rank}F_m\le\operatorname{rank}\Pi_m$, and the orthogonal projector ranks add to $\dim\mathcal K$.

Thus the rank-one trine POVM needs a three-dimensional enlarged space, although the simple qubit-times-three-state-pointer construction uses six. This minimal space need not factor as “the original system plus a separate pointer,” so a convenient physical implementation may deliberately use more dimensions.

Dilation of probabilities versus dilation of a channel

Naimark dilation reproduces effects through $F_m=V^\dagger\Pi_mV$. Stinespring dilation reproduces a state transformation through $\Phi(\rho)=\operatorname{Tr}_E(V\rho V^\dagger)$. Both can describe the same apparatus, but a POVM specifies neither a unique apparatus nor a unique surviving system state.

Exercises

Verify the isometry

Let $V\lvert\psi\rangle=\sum_m\sqrt{F_m}\lvert\psi\rangle\otimes\lvert m\rangle$. Show that $V^\dagger V=I$.

Solution

For arbitrary $\lvert\phi\rangle$ and $\lvert\psi\rangle$,

$$\langle\phi\rvert V^\dagger V\lvert\psi\rangle = \sum_m \langle\phi\rvert \sqrt{F_m}^{\,\dagger}\sqrt{F_m} \lvert\psi\rangle.$$

Since $F_m$ is positive, $\sqrt{F_m}^{\,\dagger}\sqrt{F_m}=F_m$. Therefore

$$\langle\phi\rvert V^\dagger V\lvert\psi\rangle = \langle\phi\rvert \left(\sum_mF_m\right) \lvert\psi\rangle = \langle\phi|\psi\rangle.$$

Thus $V^\dagger V=I$.

Recover the POVM effect

Using $\Pi_m=I\otimes\lvert m\rangle\langle m\rvert$, show that $V^\dagger\Pi_mV=F_m$ for the square-root construction.

Solution

The projector $\Pi_m$ selects the $m$th ancilla component:

$$\Pi_mV\lvert\psi\rangle = \sqrt{F_m}\lvert\psi\rangle\otimes\lvert m\rangle.$$

Therefore, for arbitrary $\lvert\phi\rangle$ and $\lvert\psi\rangle$,

$$\langle\phi\rvert V^\dagger\Pi_mV\lvert\psi\rangle = \langle\phi\rvert \sqrt{F_m}^{\,\dagger}\sqrt{F_m} \lvert\psi\rangle = \langle\phi\rvert F_m\lvert\psi\rangle.$$

Hence $V^\dagger\Pi_mV=F_m$.

Trine normalization

For the trine POVM, assume $\sum_k\mathbf n_k=0$. Show that $\sum_kF_k=I$.

Solution

Using

$$F_k = \frac{1}{3} \left( I+\mathbf n_k\cdot\boldsymbol\sigma \right),$$

we find

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

Since $\sum_k\mathbf n_k=0$,

$$\sum_{k=1}^3F_k=I.$$

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer (2016). Publisher.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, second English edition, Edizioni della Normale (2011). Publisher.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), theorem 2.42. Author’s text.