Completely Positive Maps

A map that sends density operators to density operators had better send positive operators to positive operators — otherwise it could produce "states" with negative probabilities. But ordinary positivity is not enough to be a physical channel. The correct requirement is the stronger complete positivity, and the gap between the two is one of the most instructive facts in quantum information.

Positive maps

A linear map $\mathcal{E}: \mathcal{L}(\mathcal{H}_A) \to \mathcal{L}(\mathcal{H}_B)$ on operators is positive if it maps positive semidefinite operators to positive semidefinite operators:

\rho \succeq 0 \;\Longrightarrow\; \mathcal{E}(\rho) \succeq 0.

This guarantees that if you feed $\mathcal{E}$ a valid state of system $A$ in isolation, you get a valid (unnormalized) state of $B$ . The catch is the phrase "in isolation."

Why positivity is not enough: the partial transpose

Quantum systems are not isolated — your system may be entangled with a reference $R$ that the channel does not touch. The physically meaningful map is then $\mathcal{E} \otimes \mathbb{1}_R$ , and it must also be positive. A map can be positive yet fail this test.

The canonical example is the transpose $T(\rho) = \rho^{\mathsf{T}}$ . Transposition preserves eigenvalues, so $T$ is positive: $\rho \succeq 0 \Rightarrow \rho^{\mathsf{T}} \succeq 0$ . Now apply $T \otimes \mathbb{1}$ to half of a two-qubit maximally entangled state $|\Phi^+\rangle = \tfrac{1}{\sqrt 2}(|00\rangle + |11\rangle)$ . The result is the partial transpose of $|\Phi^+\rangle\langle\Phi^+|$ , which equals $\tfrac{1}{2}$ times the SWAP operator. SWAP has eigenvalues $+1$ (symmetric subspace) and $-1$ (antisymmetric subspace), so $(T\otimes \mathbb{1})(|\Phi^+\rangle\langle\Phi^+|)$ has eigenvalues $\pm\tfrac12$ — it is not positive. Transposition is positive but not completely positive, hence not a physical channel. (This negativity is exactly the basis of the PPT entanglement criterion you may meet later.)

Complete positivity

A linear map $\mathcal{E}$ is completely positive (CP) if

\mathcal{E} \otimes \mathbb{1}_n \;\text{ is positive for every } n \ge 1,

where $\mathbb{1}_n$ is the identity on an $n$ -dimensional ancilla. Equivalently, attaching any reference system and applying $\mathcal{E}$ to one side never produces a negative operator. This is the property that makes $\mathcal{E}$ safe to use as a subroutine inside a larger quantum process.

Choi's theorem: a finite test

Checking "for every $n$ " sounds impossible, but in finite dimensions there is a single, finite criterion. Let $\{|i\rangle\}$ be a basis of $\mathcal{H}_A$ and form the (unnormalized) maximally entangled vector $|\Omega\rangle = \sum_i |i\rangle_A |i\rangle_{A'}$ . The Choi operator of $\mathcal{E}$ is

J(\mathcal{E}) = (\mathcal{E} \otimes \mathbb{1})\big(|\Omega\rangle\langle\Omega|\big) = \sum_{i,j} \mathcal{E}(|i\rangle\langle j|) \otimes |i\rangle\langle j|.

Choi's theorem. $\mathcal{E}$ is completely positive if and only if $J(\mathcal{E}) \succeq 0$ .

So an infinite family of conditions collapses to: form one operator and check that it is positive semidefinite. The transpose map fails precisely because its Choi operator is (proportional to) the SWAP, which has a negative eigenvalue. We devote a later lesson to this Choi–Jamiołkowski correspondence; here the point is that it gives complete positivity a concrete, checkable meaning.

Equivalent characterizations

Complete positivity has several faces, all equivalent in finite dimension:

$J(\mathcal{E}) \succeq 0$ (Choi's criterion).
$\mathcal{E}(\rho) = \sum_k K_k\,\rho\,K_k^\dagger$ for some operators $K_k$ (the Kraus form).
$\mathcal{E}(\rho) = \operatorname{tr}_E[\,U(\rho\otimes|0\rangle\langle 0|)U^\dagger\,]$ for some unitary $U$ (the Stinespring dilation).

Each of the next lessons makes one of these concrete. Adding trace preservation on top turns a CP map into a full quantum channel.

The takeaway

Complete positivity is positivity that survives the presence of an untouched reference system: $\mathcal{E}\otimes\mathbb{1}_n$ must be positive for all $n$ . The transpose shows the requirement has teeth — it is positive but not completely positive, and so is not a channel. Choi's theorem reduces the infinite test to a single positive-semidefinite check on the Choi operator, and CP maps are exactly those expressible in Kraus form or as a dilation.