Completely Positive Maps and Causal Quantum Channels

Complete positivity and relativistic locality answer different questions. Complete positivity asks whether a map remains physically positive when an arbitrary ancilla is included. Locality asks which spacetime algebras the map can change. A channel is a causal QFT intervention only after it passes both tests. This page makes the distinction quantitative with localized Gaussian channels whose spacelike compositions commute and whose timelike compositions are order dependent.

Required background. Field master equations and Markov generators provide the continuous-time comparison. Local measurement instruments supplies CP operations and their effects.

Helpful background. Positivity, monotonicity, and data processing explains the information inequalities preserved by channels.

Chapter map. Use the route from a local coupling to a field instrument, the chapter claim-validity table, and the three independent validity questions to keep algebraic CP, spacetime support, and model approximation separate.

Channel positivity does not specify spacetime support

A Schrödinger channel $\mathcal E$ is completely positive and trace preserving. Its Heisenberg dual $\mathcal E^*$ is completely positive and unital:

$$\operatorname{tr}[\mathcal E(\rho)A] =\operatorname{tr}[\rho\mathcal E^*(A)], \qquad \mathcal E^*(\mathbf1)=\mathbf1.$$

Complete positivity means that $\mathcal E\otimes\operatorname{id}_R$ is positive for every finite ancillary system $R$. In AQFT this condition is not merely borrowed from finite-dimensional quantum mechanics: under Schlieder independence of the two commuting algebras and proper infiniteness of the spectator local algebra, a positive operation that extends while fixing that spectator algebra must be CP Kitajima 2018, PDF, Definition 7 and Theorem 9.

Now let $\mathcal A(O)$ denote the observable algebra of a region $O$. A nonselective intervention supported in a compact region $K$ must satisfy the independent action criterion

$$\mathcal E_K^*(B)=B, \qquad B\in\mathcal A(O),\quad O\subset K^\perp.$$

Complete positivity alone does not imply this identity. Conversely, checking the identity only on positive states without ancillary extensions does not establish complete positivity. Both properties must be established separately for a claimed causal channel.

Gaussian channels and their CP matrix

For a regulated bosonic field, collect canonical variables into $R=(q_1,p_1,\ldots,q_n,p_n)^T$ with

$$[R_j,R_k]=i\Omega_{jk}.$$

Define the mean $m=\langle R\rangle$ and covariance

$$V_{jk}=\frac12\langle\{R_j-m_j,R_k-m_k\}\rangle.$$

A Gaussian channel acts as

$$m\longmapsto Xm+d, \qquad V\longmapsto XVX^T+Y,$$

where $Y=Y^T$ is real. With the present $[q,p]=i$ convention, complete positivity is equivalent to

$$Y+\frac{i}{2}\left(\Omega-X\Omega X^T\right)\ge0.$$

This is the covariance-matrix form of the twisted positive-definiteness criterion for linear bosonic channels in Holevo and Werner 2001, PDF, Eqs. (4.35), (4.42)–(4.45). It is an algebraic test; the matrices carry no spacetime meaning until the canonical variables and a supported dilation have been specified.

Composition follows directly:

$$\begin{aligned} X_{2\circ1}&=X_2X_1,\\ d_{2\circ1}&=X_2d_1+d_2,\\ Y_{2\circ1}&=X_2Y_1X_2^T+Y_2. \end{aligned}$$

First QFT benchmark: spacelike composition

Use a finite lattice regulator for a free scalar field and select two cells, $A$ and $B$, whose coupling regions are spacelike separated. The phase-space ordering is

$$R=(q_A,p_A,q_B,p_B)^T, \qquad \Omega=\Omega_A\oplus\Omega_B, \qquad \Omega_j=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$

The regulator is part of the benchmark: it supplies finite canonical pairs, while causal localization is still checked against the cell algebras and the underlying supported couplings.

Apply a vacuum-loss channel of transmissivity $\eta=1/4$ in $A$,

$$X_L=\operatorname{diag}\!\left(\frac12,\frac12,1,1\right), \qquad Y_L=\operatorname{diag}\!\left(\frac38,\frac38,0,0\right), \qquad d_L=0,$$

and a unit displacement by one in $q_B$,

$$X_D=\mathbf1_4, \qquad Y_D=0, \qquad d_D=(0,0,1,0)^T.$$

The displacement is a Gaussian unitary, so its CP matrix vanishes. For the loss channel, the nonzero $A$ block of the CP matrix is

$$\frac38\left(\mathbf1_2+i\Omega_A\right),$$

whose eigenvalues are $0$ and $3/4$. Both maps are therefore CP.

Because $X_Ld_D=d_D$, the composition rule gives

$$(X,d,Y)_{L\circ D}=(X,d,Y)_{D\circ L} =\left(X_L,d_D,Y_L\right).$$

For the two-cell vacuum $m=0$, $V=\mathbf1_4/2$, either order yields

$$m_{\mathrm{out}}=(0,0,1,0)^T, \qquad V_{\mathrm{out}}=\frac12\mathbf1_4.$$

The equality is not an accident of one expectation value: the full Gaussian triples agree. At the QFT level it is licensed by spacelike support and causal factorization, not by the choice of a convenient mode basis.

All matrices in this benchmark are exact. A numerical implementation must state its positivity and commutator tolerances; in particular, the exact zero CP eigenvalue should not be promoted to a physical negative eigenvalue because of roundoff.

Timelike order dependence

Move both operations to the same regulated local mode in causally ordered regions. Now

$$X_L=\frac12\mathbf1_2, \qquad Y_L=\frac38\mathbf1_2, \qquad d_D=(1,0)^T.$$

Loss after displacement gives

$$d_{L\circ D}=X_Ld_D=\left(\frac12,0\right)^T,$$

while displacement after loss gives

$$d_{D\circ L}=d_D=(1,0)^T.$$

Starting from the vacuum, both orders leave $V=\mathbf1_2/2$, but their output means differ by $1/2$ in $q$. The channels are individually CP and local to their respective interaction regions; timelike causality requires the physical order, not commutation. For probe-induced operations, the general statement is Fewster and Verch 2020, PDF, Eq. (3.28) and Theorem 3.5: causally ordered instruments compose in that order, while causally disjoint instruments compose in either order.

Adversarial control: a CP map that is not local

A Kraus normalization check cannot certify causal support. In a two-cell type-I regulator, select a qubit subalgebra in each cell and denote its Pauli operators by $X_j,Z_j$. For $0\le p\le1$, let $K_0^A=\sqrt{1-p}\,\mathbf1_A$ and $K_1^A=\sqrt p\,Z_A$. The channel with Kraus operators $K_j^A\otimes\mathbf1_B$ is localized in $A$. Now insert the remote unitary $X_B$ into every Kraus operator:

$$L_j=K_j^A\otimes X_B.$$

The new map remains CP and trace preserving because

$$\sum_jL_j^\dagger L_j =\sum_j(K_j^A)^\dagger K_j^A\otimes\mathbf1_B =\mathbf1_{AB}.$$

Nevertheless, its Heisenberg action on a remote observable is

$$\mathcal F^*(\mathbf1_A\otimes Z_B) =\mathbf1_A\otimes X_BZ_BX_B =-\mathbf1_A\otimes Z_B.$$

The CP test passes and the causal-support test fails maximally. This is why locality must be diagnosed from the map’s action on the observable net or from a supported dilation. Kraus decompositions are not unique, and on a proper von Neumann subalgebra a normal CP map need not even admit a Kraus representation internal to that algebra; Kitajima 2018, PDF, Definition 12, Theorems 14–15, and Corollary 16 gives the relevant distinctions and approximation result.

Scope and limitations

The numerical benchmark is a finite lattice or split-inclusion model. It does not turn an arbitrary continuum region into an oscillator tensor factor. A continuum claim must specify the local algebra, the regulator-to-continuum comparison, and either the channel action on complementary algebras or a compactly supported dilation. Finally, a finite intervention $\mathcal E_K$ need not equal $e^{t\mathcal L}$ for a time-independent Lindblad generator. Divisibility, a time parameter, generator domains, and the approximation producing Markovianity must be supplied before generator language is used.

Exercises

1. Verify the loss-channel CP condition

For one mode, set $X=\sqrt\eta\,\mathbf1_2$ and $Y=(1-\eta)\mathbf1_2/2$ with $0\le\eta\le1$. Find the eigenvalues of the CP matrix.

Solution

Because $X\Omega X^T=\eta\Omega$,

$$Y+\frac i2(\Omega-X\Omega X^T) =\frac{1-\eta}{2}(\mathbf1_2+i\Omega).$$

The Hermitian matrix $i\Omega$ has eigenvalues $\pm1$, so $\mathbf1_2+i\Omega$ has eigenvalues $0$ and $2$. The CP-matrix eigenvalues are therefore $0$ and $1-\eta$, both nonnegative in the declared range.

2. Reproduce the two orders

Use the Gaussian composition rule to calculate loss after displacement and displacement after loss for $\eta=1/4$ and $d=(1,0)^T$.

Solution

For loss after displacement,

$$d_{L\circ D}=X_Ld+d_L=\frac12d.$$

For displacement after loss,

$$d_{D\circ L}=X_Dd_L+d=d.$$

In both cases $X=\mathbf1_2/2$ and $Y=3\mathbf1_2/8$ because the displacement has $X_D=\mathbf1_2$ and $Y_D=0$. The output covariances agree, but the first moments do not.

3. Make CP pass while locality fails

For the adversarial Kraus operators $L_j=K_j^A\otimes X_B$, verify trace preservation and evaluate the action on $Z_B$. Which test identifies the failure?

Solution

The relation $\sum_j(K_j^A)^\dagger K_j^A=\mathbf1_A$ gives $\sum_jL_j^\dagger L_j=\mathbf1_{AB}$, so the map is CPTP. Its dual sends

$$\mathbf1_A\otimes Z_B \longmapsto \sum_j(K_j^A)^\dagger K_j^A\otimes X_BZ_BX_B =-\mathbf1_A\otimes Z_B.$$

Only the complementary-algebra action test detects that an intervention advertised as local to $A$ actually changes $B$.

References

• Fewster, C. J., and Verch, R. (2020). “Quantum Fields and Local Measurements.” Communications in Mathematical Physics 378, 851–889. DOI. Open PDF.
• Holevo, A. S., and Werner, R. F. (2001). “Evaluating Capacities of Bosonic Gaussian Channels.” Physical Review A 63, 032312. DOI. Open PDF.
• Kitajima, Y. (2018). “Local Operations and Completely Positive Maps in Algebraic Quantum Field Theory.” In M. Ozawa, J. Butterfield, H. Halvorson, M. Rédei, Y. Kitajima, and F. Buscemi (eds.), Reality and Measurement in Algebraic Quantum Theory, Springer Proceedings in Mathematics & Statistics 261, 83–95. DOI. Open PDF.