Von Neumann Algebra Channels and Positive Normal Maps

A channel between von Neumann algebras is most cleanly specified in the Heisenberg picture as a normal unital completely positive map. Complete positivity controls arbitrary finite ancillary systems, normality supplies a predual action on normal states, and unitality preserves total probability. None of these conditions follows from the other two, and none by itself says that the operation is localized in spacetime.

Required background. Tomita–Takesaki theory and standard forms supply the von Neumann-algebra topology and normal functionals used below; states, GNS representations, and folia distinguish normal state changes within a representation from singular changes. Helpful background. Modular automorphisms and standard forms clarify support projections, type-III local algebras explain why no local trace is assumed, and causal quantum channels develop spacetime localization separately.

Normal completely positive maps

Let $\mathcal M$ be the output-observable algebra and $\mathcal N$ the input-observable algebra. A linear map

$$\Phi:\mathcal M\longrightarrow\mathcal N$$

is positive when $A\geq0$ implies $\Phi(A)\geq0$, and completely positive when every amplification $\operatorname{id}_{M_n}\otimes\Phi$ is positive on $M_n(\mathcal M)$. It is normal when it preserves suprema of bounded increasing positive nets. Equivalently, it is ultraweakly continuous and has a bounded predual

$$\Phi_*:\mathcal N_*\longrightarrow\mathcal M_*, \qquad (\Phi_*\omega)(A)=\omega(\Phi(A)).$$

Thus a normal state $\omega$ remains normal after the Schrödinger-picture update $\Phi_*$. If $\Phi(1)=1$, normalized states stay normalized. A selective operation is instead normal completely positive and subunital; its probability in $\omega$ is $\omega(\Phi(1))$.

The representation theorem says that a completely positive map has

$$\Phi(A)=V^*\pi(A)V,$$

for a representation $\pi$ and bounded $V$; unitality is $V^*V=1$. For a normal map between von Neumann algebras, $\pi$ may be chosen normal. This is the precise content of the Stinespring theorem, not a universal claim that every channel has a Kraus sum inside either algebra Stinespring 1955, Theorem 1, pp. 212–214. Countable Kraus formulas require an appropriate spatial/type-I representation and separability assumptions.

Faithfulness is another independent property: $\Phi(A^*A)=0\Rightarrow A=0$. A channel can be normal and completely positive while erasing an observable subspace. Likewise, the support of $\Phi_*\omega$ need not be $1$ even when the input state is faithful unless the map satisfies an additional faithfulness condition.

A useful nondisturbance test is the multiplicative domain. If a unital completely positive map obeys

$$\Phi(A^*A)=\Phi(A)^*\Phi(A),\qquad \Phi(AA^*)=\Phi(A)\Phi(A)^*,$$

then block-matrix positivity implies $\Phi(AB)=\Phi(A)\Phi(B)$ and $\Phi(BA)=\Phi(B)\Phi(A)$ for every $B$. In the CAR construction below, every even causal-complement observable fixed by the local coupling lies in this domain. This conclusion is stronger than equality of one expectation value: it says that all mixed products with the unaffected algebra are transported homomorphically.

A local even CAR channel

Consider a quasilocal CAR net with local system algebra $\mathcal M(O)$ and one ancillary fermionic mode, whose full matrix algebra is $M_2$. Work with the even observable subalgebra and choose a parity-invariant normal ancillary state $\tau$. For an even unitary

$$U\in\mathcal M(O)_{\mathrm{even}}\,\overline\otimes\,M_{2,\mathrm{even}},$$

define

$$\Phi(A)=(\operatorname{id}\,\overline\otimes\,\tau) \bigl(U^*(A\otimes1)U\bigr).$$

Conjugation by $U$ is a normal $*$-automorphism, and the slice map $\operatorname{id}\,\overline\otimes\,\tau$ is normal, unital, and completely positive. Their composition therefore has all three channel properties. Because $U$ and $\tau$ are parity even, $\Phi$ maps even system observables to even system observables. If $A$ is localized spacelike to $O$, graded locality reduces to ordinary commutation for even observables, so $[A,U]=0$ and $\Phi(A)=A$. This is the promised local CAR realization of the abstract causal-channel construction.

The same calculation provides an independent check. Introduce a purification $|\Omega_\tau\rangle$ of $\tau$ and set $V\psi=U(\psi\otimes|\Omega_\tau\rangle)$. Then $V^*\pi(A)V=\Phi(A)$ in the enlarged representation, while $V^*V=1$. Positivity and normalization are therefore verified twice: once by closure of completely positive maps and once by an explicit dilation.

Why positivity and normality are separate

The transpose $T:M_2\to M_2$ is positive and unital, but it is not completely positive. If $|\Omega\rangle=(|00\rangle+|11\rangle)/\sqrt2$, then

$$(\operatorname{id}\otimes T)(|\Omega\rangle\langle\Omega|)=\tfrac12 F,$$

where the flip $F$ has eigenvalue $-1$ on the antisymmetric vector. The amplified output is therefore not positive. Choi’s matrix criterion makes this failure necessary and sufficient for maps on matrix algebras Choi 1975, Theorem 2, pp. 287–288.

Normality has a different failure mode. On $B(\ell^2)$ choose a singular state $s$ that vanishes on compact operators and define $\Psi(A)=s(A)1$. This map is unital and completely positive, but for finite-rank projections $P_n\uparrow1$ one has $\Psi(P_n)=0$ while $\Psi(1)=1$. Hence it does not preserve monotone suprema and has no predual map on normal states. An update built from it can leave the chosen folium even though every finite matrix amplification remains positive.

Exercises

1. Selective branches. Let $K_0,K_1$ be bounded operators with $K_0^*K_0+K_1^*K_1=1$. Show that $\Phi_i(A)=K_i^*AK_i$ are normal completely positive operations and that $\Phi_0+\Phi_1$ is a channel.

Solution

Each branch has the one-operator Stinespring form, hence is completely positive; multiplication by fixed bounded operators is ultraweakly continuous, hence normal. Moreover $\Phi_i(1)=K_i^*K_i\leq1$, while $(\Phi_0+\Phi_1)(1)=1$. For a normal state $\omega$, the branch probability is $\omega(K_i^*K_i)$ and the two probabilities sum to one.

2. Detect nonnormality. Verify directly that $\Psi(A)=s(A)1$ above cannot be the Heisenberg adjoint of a map $B(\ell^2)_*\to B(\ell^2)_*$.

Solution

Any Banach preadjoint would make $\Psi$ ultraweakly continuous. But $P_n\to1$ ultraweakly and $\Psi(P_n)=0\not\to1=\Psi(1)$. This contradiction is independent of complete positivity.

References

• Choi, Man-Duen. “Completely Positive Linear Maps on Complex Matrices.” Linear Algebra and Its Applications 10 (1975): 285–290. DOI.
• Stinespring, W. Forrest. “Positive Functions on C*-Algebras.” Proceedings of the American Mathematical Society 6 (1955): 211–216. DOI.