Completely Positive Maps

Completely positive maps are the mathematical language for physically allowed quantum state transformations. They appear in measurement theory, noise models, reduced dynamics, quantum information, and master equations.

For the broader operation/channel vocabulary, including trace-preserving and selected trace-nonincreasing branches, see Quantum Operations.

The key point is not just that a map should send density operators to density operators. It must remain physically valid when the system is part of a larger entangled system:

$$\rho_{SR}\ge0 \quad\Longrightarrow\quad (\Phi_S\otimes\mathrm{id}_R)(\rho_{SR})\ge0.$$

Here $R$ is a reference system that is not touched. Complete positivity is the requirement that this condition hold for every possible reference system.

Positivity

Let $\Phi$ be a linear map from operators on $\mathcal H_S$ to operators on $\mathcal H_{S'}$. The map is positive if

$$X\ge0 \quad\Longrightarrow\quad \Phi(X)\ge0.$$

Positivity is necessary because a density operator is positive. If $\rho\ge0$, then a valid output state should also be positive.

But positivity alone is not enough. A system can be entangled or classically correlated with degrees of freedom that are outside the map’s direct action. A transformation that looks positive on isolated states of $S$ may fail on joint states of $SR$.

Complete Positivity

For a positive integer $n$, the map $\Phi$ is $n$-positive if

$$\Phi\otimes\mathrm{id}_n$$

is positive on operators acting on $\mathcal H_S\otimes\mathbb C^n$.

The map is completely positive if it is $n$-positive for every $n$:

$$X_{SR}\ge0 \quad\Longrightarrow\quad (\Phi\otimes\mathrm{id}_R)(X_{SR})\ge0$$

for every finite-dimensional reference system $R$.

In finite dimensions, it is enough to test a reference system whose dimension equals the input dimension of $S$. This is the content behind the Choi criterion.

Physical Meaning

Complete positivity is a locality consistency condition. If an operation is applied only to $S$, it should not produce an unphysical operator merely because $S$ is entangled with an untouched reference $R$.

For example, a laboratory operation may act on a qubit in a device. The same qubit might be entangled with another qubit in a memory, a photon mode, an inaccessible environment, or a mathematical purifying system used in a calculation. The local operation must make sense in all of these situations.

This is why complete positivity is stronger than positivity:

positive map             valid on isolated positive operators
completely positive map  valid when tensored with any idle reference

The reference system need not be a real laboratory object in every application. It is a consistency test for compatibility with composite quantum systems.

Transpose Map Example

The transpose map

$$T(\rho)=\rho^{\mathsf T}$$

is positive. If $\rho\ge0$, then $\rho^{\mathsf T}\ge0$ because transposition preserves the spectrum of a Hermitian matrix.

However, $T$ is not completely positive. To see this, take the Bell state

$$|\Omega\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2}.$$

Its density operator is

$$|\Omega\rangle\langle\Omega| = \frac12 \left( |00\rangle\langle00| +|00\rangle\langle11| +|11\rangle\langle00| +|11\rangle\langle11| \right).$$

Apply transposition to the first qubit only:

$$(T\otimes\mathrm{id}) \left( |\Omega\rangle\langle\Omega| \right) = \frac12 \left( |00\rangle\langle00| +|10\rangle\langle01| +|01\rangle\langle10| +|11\rangle\langle11| \right).$$

In the ordered basis

$$|00\rangle,\ |01\rangle,\ |10\rangle,\ |11\rangle,$$

this operator is

$$\frac12 \begin{pmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 \end{pmatrix}.$$

The middle $2\times2$ block has eigenvalues $+1/2$ and $-1/2$. Therefore the output is not positive.

The transpose map is a perfectly useful mathematical operation, and partial transpose is an important entanglement diagnostic. But transposition is not a physical quantum channel acting locally on an unknown system.

Choi Positivity Test

For a map $\Phi:M_d\to M_{d'}$, define the unnormalized maximally entangled vector

$$|\Omega_d\rangle = \sum_{j=1}^d |j\rangle\otimes |j\rangle.$$

The Choi matrix of $\Phi$ is

$$J(\Phi) = (\Phi\otimes\mathrm{id}_d) \left( |\Omega_d\rangle\langle\Omega_d| \right).$$

Choi’s theorem says that, in finite dimensions,

$$\Phi\ \text{is completely positive} \quad\Longleftrightarrow\quad J(\Phi)\ge0.$$

This is often the most practical test for complete positivity. It is also the bridge to process tomography, channel-state duality, semidefinite optimization, and minimal Kraus representations.

The transpose example above is exactly the Choi test in dimension $2$: the Choi matrix of transposition has a negative eigenvalue.

Kraus Maps Are Completely Positive

Any map with a Kraus form

$$\Phi(\rho) = \sum_\alpha K_\alpha\rho K_\alpha^\dagger$$

is completely positive. To see why, attach a reference $R$:

$$(\Phi\otimes\mathrm{id}_R)(X) = \sum_\alpha (K_\alpha\otimes I_R) X (K_\alpha^\dagger\otimes I_R).$$

If $X\ge0$, then each term

$$(K_\alpha\otimes I_R) X (K_\alpha^\dagger\otimes I_R)$$

is positive, and a sum of positive operators is positive. Therefore the extended map is positive for every $R$.

This simple argument is one reason Kraus representations are so useful: complete positivity is automatic.

Trace Conditions Are Separate

Complete positivity controls positivity of the output. It does not by itself control normalization.

A completely positive map may be trace preserving:

$$\operatorname{Tr}\Phi(\rho) = \operatorname{Tr}\rho,$$

as for a deterministic quantum channel. It may also be trace nonincreasing:

$$\operatorname{Tr}\Phi(\rho) \le \operatorname{Tr}\rho,$$

as for a selected measurement outcome.

In Kraus form, trace preservation is the condition

$$\sum_\alpha K_\alpha^\dagger K_\alpha=I,$$

while trace nonincrease is

$$\sum_\alpha K_\alpha^\dagger K_\alpha\le I.$$

Thus the usual physical classes are:

Map type	Positivity condition	Trace condition	Meaning
CP	completely positive	none specified	unnormalized operation may change trace arbitrarily
CPTNI	completely positive	trace nonincreasing	possible selected outcome or postselected branch
CPTP	completely positive	trace preserving	deterministic channel or nonselective evolution

Relation to Open Systems

Complete positivity arises naturally from a system-environment model. Suppose

$$\Phi(\rho_S) = \operatorname{Tr}_E \left[ U_{SE}(\rho_S\otimes\eta_E)U_{SE}^\dagger \right],$$

where $\eta_E$ is a fixed environment state and $U_{SE}$ is unitary. This map is completely positive and trace preserving.

It is trace preserving because unitary evolution and partial trace preserve total trace. It is completely positive because the same construction remains positive after adding an untouched reference:

$$(\Phi\otimes\mathrm{id}_R)(\rho_{SR}) = \operatorname{Tr}_E \left[ (U_{SE}\otimes I_R) (\rho_{SR}\otimes\eta_E) (U_{SE}^\dagger\otimes I_R) \right].$$

Every step on the right preserves positivity.

This is the reduced-dynamics origin of quantum channels. A closed unitary theory for a larger system becomes a completely positive map for a subsystem when inaccessible degrees of freedom are ignored. The channel-side representation is organized in Stinespring Representation.

When Complete Positivity Can Be Subtle

Complete positivity is the standard requirement for maps that act on arbitrary system states and are compatible with arbitrary reference systems. There are subtleties when a map is defined only on a restricted set of states, or when the system has fixed initial correlations with an environment.

In such cases, one must distinguish:

• a universally valid channel on all system states;
• an effective map on a restricted compatibility domain;
• a dynamical assignment that depends on initial system-environment correlations;
• a mathematical positive map used as a diagnostic rather than as a physical operation.

Most textbook noise models, measurement instruments, and Markovian master equations use completely positive maps because they are meant to apply independently of unknown entanglement with a reference.

Common Mistakes

• Thinking positivity alone is enough for a physical quantum channel.
• Treating the transpose or partial transpose as a possible local operation.
• Confusing complete positivity with trace preservation.
• Forgetting the idle reference system when testing a proposed map.
• Assuming a map derived on a restricted state family automatically extends to a valid channel on all states.
• Treating a non-CP intermediate map in an approximate derivation as harmless without checking the approximation domain.

References

• W. F. Stinespring, “Positive functions on C*-algebras,” Proceedings of the American Mathematical Society 6, 211-216 (1955).
• E. C. G. Sudarshan, P. M. Mathews, and J. Rau, “Stochastic dynamics of quantum-mechanical systems,” Physical Review 121, 920-924 (1961).
• K. Kraus, “General state changes in quantum theory,” Annals of Physics 64, 311-335 (1971).
• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285-290 (1975).
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer (1983).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).
• M. M. Wilde, Quantum Information Theory, Cambridge University Press, 2nd ed. (2017).

Exercises

1. Transpose is positive. Prove that if $\rho\ge0$, then $\rho^{\mathsf T}\ge0$.

Solution

For any vector $|v\rangle$, let $|v^*\rangle$ be the entrywise complex conjugate in the basis defining the transpose. Then

$$\langle v|\rho^{\mathsf T}|v\rangle = \langle v^*|\rho|v^*\rangle.$$

If $\rho\ge0$, the right-hand side is nonnegative for every $|v\rangle$. Hence $\rho^{\mathsf T}\ge0$.

2. A Kraus map is completely positive. Let $\Phi(\rho)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$. Show directly that $\Phi\otimes\mathrm{id}_R$ is positive for every reference $R$.

Solution

For any positive $X_{SR}$,

$$(\Phi\otimes\mathrm{id}_R)(X_{SR}) = \sum_\alpha (K_\alpha\otimes I_R) X_{SR} (K_\alpha^\dagger\otimes I_R).$$

If $X_{SR}\ge0$, then $A X_{SR} A^\dagger\ge0$ for every operator $A$. Each summand is therefore positive, and the sum is positive. Since $R$ was arbitrary, $\Phi$ is completely positive.

3. Trace preservation in Kraus form. Show that $\Phi(\rho)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$ is trace preserving for all $\rho$ if and only if $\sum_\alpha K_\alpha^\dagger K_\alpha=I$.

Solution

Using cyclicity of the trace,

$$\operatorname{Tr}\Phi(\rho) = \sum_\alpha \operatorname{Tr} \left( K_\alpha\rho K_\alpha^\dagger \right) = \operatorname{Tr} \left[ \rho \sum_\alpha K_\alpha^\dagger K_\alpha \right].$$

If $\sum_\alpha K_\alpha^\dagger K_\alpha=I$, then $\operatorname{Tr}\Phi(\rho)=\operatorname{Tr}\rho$.

Conversely, if this equality holds for all $\rho$, then

$$\operatorname{Tr} \left[ \rho \left( \sum_\alpha K_\alpha^\dagger K_\alpha-I \right) \right] =0$$

for all density operators, and hence for all operators by linearity. Therefore $\sum_\alpha K_\alpha^\dagger K_\alpha=I$.