In short
A quantum channel is a CPTP map — a completely positive, trace-preserving linear map on density matrices. Three conditions, each doing real work: linear means the channel respects classical mixtures, \mathcal E(p\rho_1 + (1-p)\rho_2) = p\,\mathcal E(\rho_1) + (1-p)\,\mathcal E(\rho_2); trace-preserving means probabilities stay normalised, \text{tr}(\mathcal E(\rho)) = \text{tr}(\rho) = 1; completely positive means that even when \mathcal E acts on only half of a larger entangled system, the joint output is still a valid density matrix. The transpose map T(\rho) = \rho^\top is positive but not completely positive — that single counter-example is why plain "positive" is not enough. CPTP is the axiomatic definition; Kraus representation and Stinespring dilation are equivalent concrete forms of the same thing.
The Kraus representation you met last chapter lets you write any noise process, any measurement, any unitary gate in one template: \mathcal E(\rho) = \sum_k K_k\rho K_k^\dagger with \sum_k K_k^\dagger K_k = I. The formula is clean. But where does it come from? Why that formula and not another? If someone showed you a brand-new linear map on density matrices and asked "is this physically realisable?", how would you check — without writing down Kraus operators first?
You need the axiomatic definition. Three words, each a physical requirement the universe imposes on any honest operation: the map should be linear (because classical mixtures must mix linearly), trace-preserving (because total probability must stay 1), and completely positive (because applying the operation to half of an entangled pair must not produce a state with negative probabilities). Any map with all three properties is called a CPTP map, and the Choi-Kraus theorem promises: every CPTP map has a Kraus representation, and every Kraus representation gives a CPTP map. So CPTP is not a new object — it is the clean mathematical characterisation of the things you already know how to write down.
This chapter walks you through those three properties one at a time, shows you why each is physically necessary, and then uses the most surprising one — complete positivity — to kill the transpose map. The transpose is positive on a single system. On two systems entangled together, it produces a matrix with a negative eigenvalue. That is why you never see a "transpose gate" in any quantum circuit: the universe does not permit it.
The three properties
Write \mathcal D(\mathcal H) for the set of density matrices on a finite-dimensional Hilbert space \mathcal H. A quantum channel is a function
that takes a density matrix in and produces a density matrix out. (Input and output spaces can differ — for instance, measurement with the outcome ignored keeps the dimension, while partial trace reduces it. For most of this chapter we take \mathcal H' = \mathcal H.) The three properties below are the conditions on \mathcal E.
Linear
\mathcal E is linear means: for any two density matrices \rho_1, \rho_2 and any probabilities p, 1 - p,
More generally, \mathcal E extends to a linear map on the whole space of linear operators, not just density matrices: \mathcal E(\alpha A + \beta B) = \alpha\mathcal E(A) + \beta\mathcal E(B) for any operators A, B and any complex numbers \alpha, \beta.
Why linearity is forced: suppose you prepare \rho_1 with probability p and \rho_2 with probability 1 - p. The resulting ensemble is described by the mixture p\rho_1 + (1 - p)\rho_2 (see density operator). Whether you apply \mathcal E before or after labelling the ensemble must give the same answer. That is the linearity condition, and it is a consequence of how classical probabilities combine with quantum states. Any "operation" that was non-linear on density matrices would let you distinguish between ensembles that are supposed to be indistinguishable — a contradiction.
Linearity sounds innocent but it is the condition that forbids quantum state cloning (the no-cloning theorem) and faster-than-light signalling through entanglement. Both would require operations that respond non-linearly to the input state, and quantum mechanics does not have those.
Trace-preserving (TP)
\mathcal E is trace-preserving means: for every density matrix \rho,
The trace of a density matrix is the total probability of all outcomes. Trace-preservation says the channel does not lose or create probability — what goes in comes out, summing to 1.
For an operator A that is not a density matrix (not unit trace), the condition extends by linearity: \text{tr}(\mathcal E(A)) = \text{tr}(A).
Trace-preservation is the condition you get if the channel has no post-selection — every branch of the Kraus sum is kept, not conditioned on. A post-selected measurement (where you throw away the runs with the "wrong" outcome) gives a trace-non-increasing map, not trace-preserving. When the quantum-computing literature says "channel" without qualification, it almost always means trace-preserving.
Positive (P)
\mathcal E is positive means: for every positive-semi-definite (PSD) operator A, the output \mathcal E(A) is also PSD.
Recall: a PSD operator has non-negative eigenvalues, or equivalently \langle\psi|A|\psi\rangle \geq 0 for every vector |\psi\rangle. Density matrices are PSD (this is one of their three axioms). So a positive map sends density matrices to matrices whose eigenvalues are \geq 0 — which is part of what you need for the output to be a valid density matrix.
Combined with trace-preservation, positivity gives you: density matrix in, density matrix out. All three density-matrix axioms (Hermiticity, PSD, unit trace) are preserved.
This sounds like enough. The density matrix is still a density matrix; no eigenvalues went negative; probabilities still sum to 1. What more could you want?
The answer is the key insight of this chapter — and the reason the "complete" in "completely positive" is there.
Completely positive — and why positive alone is not enough
A positive map on system A alone is not the same as a positive map when extended to act on A together with an unrelated spectator system B.
Definition. \mathcal E (a map on operators of \mathcal H_A) is completely positive if, for every positive integer n, the extended map
is still positive — i.e. takes PSD operators to PSD operators on the bigger system \mathcal H_A \otimes \mathbb C^n. Here \text{id}_n is the identity map on the n-dimensional spectator \mathcal B.
Why the auxiliary system B is the whole point: nothing in the laws of physics says you can only apply a channel to an isolated qubit. You can apply it to one qubit of a Bell pair, leaving the other qubit alone. You can apply it to one qubit of a ten-qubit entangled register. If the channel is physically real, doing nothing to the rest of the register must still give a valid quantum state for the combined system. If it doesn't — if applying \mathcal E to part of an entangled state produces a matrix with a negative eigenvalue — then the map is not physical, even if it looked fine on isolated states.
Complete positivity is strictly stronger than positivity. A map can be positive on \mathcal H_A and yet fail to be positive on \mathcal H_A \otimes \mathcal H_B when applied to one side of an entangled state.
The transpose map is the counter-example
The matrix transpose is the textbook example of a map that is positive but not completely positive. It is worth working through this example in full — it is the single sharpest justification for why "complete" has to be in the definition.
Define T : \mathcal B(\mathbb C^2) \to \mathcal B(\mathbb C^2) by
Concretely, T\begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}a & c \\ b & d\end{pmatrix}.
T is linear. Obvious from the definition.
T is trace-preserving. \text{tr}(A^\top) = \text{tr}(A). Always.
T is positive. A PSD matrix A has all eigenvalues \geq 0. Transposing preserves the eigenvalues (because \det(A^\top - \lambda I) = \det(A - \lambda I)^\top = \det(A - \lambda I), so the characteristic polynomial is unchanged). So A^\top is also PSD. On a single system, the transpose map is a perfectly sensible positive map.
T is not completely positive. And here is where it breaks.
Take n = 2 (a second qubit as the spectator). The extended map T \otimes \text{id}_2 acts on 2 \times 2 blocks of a 4 \times 4 matrix. Apply it to the maximally entangled Bell state projector
In the computational basis \{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}, this is the 4\times 4 matrix
This is PSD (eigenvalues 1, 0, 0, 0). The non-zero eigenvector is \frac{1}{\sqrt 2}(|00\rangle + |11\rangle) — the Bell state itself.
Apply T \otimes \text{id}_2. The first register is transposed while the second is left alone. Explicitly, the action on basis projectors is
swapping the first-register bra and ket. So
Why the middle two terms got swapped: |00\rangle\langle 11| = |0\rangle\langle 1| \otimes |0\rangle\langle 1|. Transpose on the first register sends |0\rangle\langle 1| to |1\rangle\langle 0|. Identity on the second register leaves |0\rangle\langle 1| alone. Product: |1\rangle\langle 0| \otimes |0\rangle\langle 1| = |10\rangle\langle 01|. The other cross term works the same way.
In matrix form this is
This is \tfrac{1}{2} times the SWAP operator on two qubits — the matrix that exchanges them. Its eigenvalues are +1 (on symmetric states) and -1 (on antisymmetric states). Dividing by 2, the eigenvalues of the output are +\tfrac{1}{2}, +\tfrac{1}{2}, +\tfrac{1}{2}, -\tfrac{1}{2}.
A negative eigenvalue. The output has an eigenvalue of -\tfrac{1}{2}. That means it is not PSD — there is a vector |\psi\rangle with \langle\psi|(T\otimes\text{id})|\Phi^+\rangle\langle\Phi^+||\psi\rangle = -\tfrac{1}{2}. A negative "probability". This is not a density matrix, and no physical operation could have produced it.
So the transpose, positive on a single qubit, fails on two qubits when the input is entangled. T is positive but not completely positive. And that is why the transpose is not a quantum channel — the universe does not let you do it.
Physical reading
What is the "negative probability" actually telling you? It says: if the transpose were a gate, then applying it to one qubit of an entangled pair and measuring the two qubits in the right basis, you would predict a detection probability of -\tfrac{1}{2} for some outcome. Detection probabilities are never negative. So either (a) your prediction formula is wrong, or (b) the gate does not exist.
Option (a) would wreck quantum mechanics at its foundation — the Born rule is one of the postulates. So option (b): the transpose is not a physically realisable operation. The transpose is a mathematical operation on matrices, it is not a channel on states.
This is also why entanglement is so closely tied to complete positivity: the whole story only breaks because the input state is entangled. On a product state \rho_A \otimes \rho_B, the transpose works fine: (T\otimes\text{id})(\rho_A\otimes\rho_B) = \rho_A^\top \otimes \rho_B, which is PSD. Entanglement is what exposes the non-positivity of the transpose. In that sense, "positive but not CP" maps are entanglement witnesses — they detect entanglement by producing a non-PSD output when applied to half of an entangled state. A whole theory of entanglement is built around them (going-deeper section).
The equivalence theorem
The punchline of this chapter: there are three equivalent ways to say "\mathcal E is a quantum channel."
Equivalence of channel descriptions
For a linear map \mathcal E : \mathcal B(\mathcal H) \to \mathcal B(\mathcal H), the following are equivalent.
- CPTP: \mathcal E is completely positive and trace-preserving.
- Kraus representation: there exist operators \{K_k\} with \sum_k K_k^\dagger K_k = I such that \mathcal E(\rho) = \sum_k K_k\rho K_k^\dagger for all \rho.
- Stinespring dilation: there exist a Hilbert space \mathcal H_E, a state |0\rangle_E, and a unitary U on \mathcal H \otimes \mathcal H_E such that \mathcal E(\rho) = \text{tr}_E\bigl(U(\rho\otimes|0\rangle\langle 0|_E)U^\dagger\bigr).
The implication Kraus \Rightarrow CPTP was proved in the previous chapter: if \mathcal E has Kraus form, the trace-preserving and PSD-in-PSD-out properties follow directly. Complete positivity requires only noting that applying \mathcal E \otimes \text{id}_n to a PSD operator on \mathcal H\otimes\mathbb C^n still has a Kraus form (with operators K_k \otimes I_n), so the result is PSD.
The implication CPTP \Rightarrow Kraus is the harder direction. It is the Choi-Kraus theorem (1975), and the proof goes through the Choi matrix, which is sketched in the "going deeper" section below.
The implication Stinespring \Rightarrow Kraus is also from the previous chapter: define K_k = \langle k|_E\,U\,|0\rangle_E, where \{|k\rangle_E\} is a basis of the environment, and verify that the completeness relation holds. The reverse direction — Kraus \Rightarrow Stinespring — is the content of next chapter, which constructs the dilating unitary explicitly.
So all three descriptions — axiomatic CPTP, operational Kraus, physical Stinespring — describe the same objects. Pick whichever is easiest for the problem at hand.
Examples and non-examples
CPTP (all of these are quantum channels)
- Identity channel: \mathcal E(\rho) = \rho. Kraus K_0 = I.
- Unitary channel: \mathcal E(\rho) = U\rho U^\dagger. Kraus K_0 = U.
- Unread projective measurement: \mathcal E(\rho) = \sum_m P_m\rho P_m. Kraus K_m = P_m.
- Partial trace: \mathcal E(\rho_{AB}) = \text{tr}_B(\rho_{AB}). This maps from \mathcal B(\mathcal H_A\otimes\mathcal H_B) to \mathcal B(\mathcal H_A); Kraus operators \{I_A \otimes \langle k|_B\} indexed by a basis of B.
- Reset channel ("prepare and measure"): \mathcal E(\rho) = |0\rangle\langle 0| for all \rho. Kraus K_0 = |0\rangle\langle 0|, K_1 = |0\rangle\langle 1|. Check: K_0^\dagger K_0 + K_1^\dagger K_1 = |0\rangle\langle 0|\cdot|0\rangle\langle 0| + |1\rangle\langle 0|\cdot|0\rangle\langle 1| = |0\rangle\langle 0| + |1\rangle\langle 1| = I.
- Bit-flip, phase-flip, depolarizing, amplitude damping: the four standard noise channels from the Kraus chapter.
Not CPTP
- Transpose T(\rho) = \rho^\top: positive but not CP. Worked through above.
- Matrix conjugation alone \rho \mapsto \rho^*: same as transpose (since \rho^\dagger = \rho^{*\top} = \rho, so \rho^* = \rho^\top for Hermitian \rho). Anti-unitary operations are not CPTP.
- Non-linear maps: e.g. \rho \mapsto \rho^2 / \text{tr}(\rho^2). The "square then renormalise" operation is not linear, hence not a channel.
- Measurement with post-selection: \rho \mapsto P_m\rho P_m / p_m. This is a state-update rule, but it is trace-preserving only when normalised by p_m — and that normalisation is non-linear. It is not a CPTP map; it is a trace-non-increasing CP map that you renormalise after the fact. The full treatment is via instruments (CP-instruments), a generalisation that includes classical outcomes alongside quantum state updates.
Worked examples
Example 1 — Transpose fails CP on a Bell state, concrete numbers
Verify explicitly that the transpose map sends the Bell state |\Phi^+\rangle\langle\Phi^+| to a matrix with a negative eigenvalue.
Step 1. Write the Bell state in the computational basis.
As a 4\times 4 matrix in the ordering |00\rangle, |01\rangle, |10\rangle, |11\rangle:
Why only the corners are non-zero: |\Phi^+\rangle\langle\Phi^+| has exactly four non-zero entries — the ones that couple |00\rangle and |11\rangle. All other entries are 0 because the state has no |01\rangle or |10\rangle component.
Step 2. Identify the four 2\times 2 blocks. Writing the 4\times 4 matrix as a 2\times 2 block matrix with 2\times 2 blocks (first register = block index, second register = inside the block):
Step 3. Apply T \otimes \text{id}. Transpose between blocks (first register), identity inside each block (second register). Swapping the top-right and bottom-left blocks:
Why blocks swap while inside stays fixed: transpose on the first register is exactly "swap the block row-index with the block column-index." The inside of each block is the second register, which identity leaves untouched. So the (i,j) block moves to the (j,i) position.
Step 4. Recognise the result. The matrix \tfrac{1}{2} times
is \tfrac{1}{2} times the SWAP operator — the gate that exchanges two qubits. (Check: it acts as identity on |00\rangle, |11\rangle and swaps |01\rangle \leftrightarrow |10\rangle.) SWAP is a symmetric permutation whose eigenvalues are +1 on symmetric states \{|00\rangle, |11\rangle, \tfrac{1}{\sqrt 2}(|01\rangle+|10\rangle)\} and -1 on the antisymmetric state \tfrac{1}{\sqrt 2}(|01\rangle-|10\rangle).
Step 5. Write down the eigenvalues of the output.
Result. A negative eigenvalue -\tfrac{1}{2} appears. The output is not PSD. It is not a valid density matrix. So T is not a CP map, and hence not a quantum channel. Applying the transpose to one half of a Bell state would produce an object that predicts a negative probability for some measurement outcome — impossible physically, forbidden mathematically.
What this shows. Complete positivity is not a mathematical nicety; it is the concrete condition that distinguishes physical operations from unphysical ones. Positivity on a single system can hold for a non-physical map, and only the extension to a larger entangled system exposes the failure.
Example 2 — Depolarizing channel is CPTP, verified from the Kraus operators
Show that the single-qubit depolarizing channel with parameter p is a valid CPTP map.
Step 1. Write the Kraus operators. From the Kraus chapter,
Step 2. Check the completeness relation \sum_k K_k^\dagger K_k = I. Each Kraus operator is (scalar) times a Pauli; the Paulis are Hermitian, and X^2 = Y^2 = Z^2 = I^2 = I, so K_k^\dagger K_k = |c_k|^2 I where c_k is the Kraus scalar.
Why the arithmetic works out: the "do nothing" branch contributes 1 - \tfrac{3p}{4}; each of the three Pauli branches contributes \tfrac{p}{4} because the amplitude is \tfrac{\sqrt p}{2} and the scalar coefficient is |c_k|^2 = \tfrac{p}{4}. Adding \tfrac{3p}{4} from the Pauli branches cancels the -\tfrac{3p}{4} in the identity branch, giving a clean I.
Step 3. TP follows automatically. From the Kraus theorem, any \{K_k\} with \sum_k K_k^\dagger K_k = I gives a trace-preserving map, by the argument in the previous chapter.
Step 4. CP follows automatically. By the Choi-Kraus equivalence, any Kraus representation gives a completely positive map. (A direct check: for any auxiliary dimension n, the extended map \mathcal E\otimes\text{id}_n has Kraus operators \{K_k\otimes I_n\}, and these still satisfy completeness: \sum_k (K_k\otimes I_n)^\dagger(K_k\otimes I_n) = \sum_k K_k^\dagger K_k \otimes I_n = I\otimes I_n = I. So the extended map is itself in Kraus form, hence positive.)
Step 5. Verify on a specific state — the Bell state |\Phi^+\rangle\langle\Phi^+| with depolarizing applied to qubit A. This is a single-qubit-on-one-side channel; complete positivity demands the output be a valid density matrix. Working through the algebra, (\mathcal E_{\text{dep}}\otimes\text{id})(|\Phi^+\rangle\langle\Phi^+|) = (1 - p)|\Phi^+\rangle\langle\Phi^+| + \tfrac{p}{4}I_4 — a weighted mixture of the Bell state and the maximally mixed state on both qubits. All eigenvalues of this mixture are non-negative: the largest is (1 - p) + \tfrac{p}{4} = 1 - \tfrac{3p}{4}, and the other three are all \tfrac{p}{4}. Both are \geq 0 for p \in [0, 1]. PSD. Valid.
Result. The depolarizing channel is CPTP: the Kraus representation verifies completeness (TP), the Kraus form implies CP, and direct computation on a Bell state confirms the extended map produces a valid density matrix for any p \in [0, 1].
What this shows. The Kraus form is the certificate of CPTP-ness. Once you have a valid Kraus representation, the map is automatically CPTP — no further checks needed. The equivalence theorem does all the work for you. This is the practical power of the theorem: you do not need to directly verify complete positivity on every possible entangled input; you just write down Kraus operators and confirm \sum_k K_k^\dagger K_k = I.
Common confusions
-
"CP is the same as P — any positive map is completely positive." False. The transpose is the standard counter-example. CP is strictly stronger. The extension \mathcal E\otimes\text{id}_n is what reveals the difference, and an entangled input is what triggers the failure for non-CP positive maps.
-
"The transpose is a valid quantum operation if you ignore the complex conjugation issue." No. The transpose of a Hermitian matrix is already a Hermitian matrix with the same eigenvalues — so the transpose is positive, trace-preserving, and linear on single-system density matrices. All those properties look fine. It still fails to be a channel because of complete positivity, which requires a bipartite test. This is the sharpest illustration of why CP is needed.
-
"CPTP maps are always unitary." No. Unitary channels are the special case where there is exactly one Kraus operator, K_0 = U, and the CPTP condition reduces to U^\dagger U = I. General CPTP maps have many Kraus operators and are non-unitary. Measurement (with outcome ignored), noise, partial trace — all CPTP, none unitary.
-
"The identity channel and 'do nothing' are the same." They are — but only on the system you are thinking about. Partial trace is not the identity channel; it is a CPTP map that reduces the dimension. "Do nothing on A" in the presence of another system B really means \text{id}_A \otimes (something on B). Keep track of which system you are working on.
-
"Non-linear operations could still be physical if they preserve trace and positivity." No. Linearity is forced by the structure of ensembles and classical mixtures. A non-linear "operation" would let you distinguish mixtures that are supposed to be physically identical — a contradiction with the predictive content of the density matrix formalism.
-
"Trace-non-increasing CP maps (\sum_k K_k^\dagger K_k \leq I, not =I) are also channels." No — the standard definition of "channel" requires equality. Trace-non-increasing CP maps are called CP subchannels or effect operations and correspond to measurement outcomes without post-selection renormalisation. They are the building blocks of quantum instruments, which combine classical outcomes with quantum state updates.
-
"Measurement with a recorded outcome is CPTP." It is CP and it is trace-preserving on average (summed over all outcomes), but for a specific recorded outcome, the state-update rule \rho \to P_m\rho P_m/p(m) is non-linear in \rho (because p(m) = \text{tr}(P_m\rho) depends on \rho). The correct way to handle recorded measurement in the CPTP framework is through instruments: the quantum instrument assigns, to each outcome m, a CP subchannel \mathcal E_m such that \sum_m \mathcal E_m is CPTP. The individual \mathcal E_m is trace-non-increasing.
Going deeper
If you are here for the axiomatic definition of a quantum channel as a CPTP map, the reason complete positivity is needed (the transpose counter-example), and the equivalence with Kraus, you have the core. The rest of this section builds the Choi matrix, proves the Kraus-CPTP equivalence through Choi, and points to positive-but-not-CP maps as entanglement witnesses.
The Choi-Jamiolkowski isomorphism
There is a bijection between linear maps on operators of \mathcal H (an n-dimensional space) and operators on \mathcal H \otimes \mathcal H (an n^2-dimensional space). Given a linear map \mathcal E : \mathcal B(\mathcal H) \to \mathcal B(\mathcal H), define the Choi matrix of \mathcal E by
where |\Omega\rangle is the un-normalised maximally entangled state on \mathcal H\otimes\mathcal H. J(\mathcal E) is an n^2\times n^2 operator. The map \mathcal E \mapsto J(\mathcal E) is a bijection (Choi-Jamiolkowski isomorphism, 1972/1975).
The key theorem. \mathcal E is completely positive if and only if J(\mathcal E) is PSD. Complete positivity of the map — which seems like a statement about every possible extension — reduces to positive-semi-definiteness of a single n^2\times n^2 matrix.
Proof sketch (only-if direction): if \mathcal E is CP, then \mathcal E\otimes\text{id} applied to |\Omega\rangle\langle\Omega| (a PSD operator) is PSD by CP. That is exactly J(\mathcal E). Done.
Proof sketch (if direction): suppose J(\mathcal E) is PSD. Diagonalise it: J(\mathcal E) = \sum_k |\phi_k\rangle\langle\phi_k|, where each |\phi_k\rangle is an (unnormalised) vector in \mathcal H\otimes\mathcal H. Reshape each |\phi_k\rangle — which has n^2 components — into an n\times n matrix K_k by writing |\phi_k\rangle = \sum_{ij}(K_k)_{ij}|i\rangle|j\rangle. Direct calculation shows \mathcal E(\rho) = \sum_k K_k\rho K_k^\dagger for all \rho. So \mathcal E has a Kraus representation, hence is CP.
This proves the Kraus-CPTP equivalence: if \mathcal E is CP, then J(\mathcal E) is PSD, from which Kraus operators can be extracted by diagonalising J(\mathcal E). And conversely, the Kraus form directly produces a CP map.
Positive-but-not-CP maps as entanglement witnesses
Positive maps that are not completely positive are not physical operations, but they are extremely useful. The failure mode — applying them to a density matrix produces something non-PSD — becomes a diagnostic test.
Entanglement witness. Given a bipartite state \rho_{AB} on \mathcal H_A\otimes\mathcal H_B, apply (T\otimes\text{id}) (transpose on A, identity on B). If the result has a negative eigenvalue, \rho_{AB} must have been entangled. Why: if \rho_{AB} had been separable (a convex combination of product states), the output would be separable too (because the transpose of a product state is another product state), and in particular PSD. A negative eigenvalue shows separability fails.
This is the Peres-Horodecki criterion (Peres 1996, Horodecki 1996): a state \rho_{AB} is NPT (Negative Partial Transpose) if (T\otimes\text{id})\rho_{AB} has a negative eigenvalue, which implies entanglement. For 2\times 2 and 2\times 3 systems, NPT is equivalent to entanglement; in higher dimensions there are PPT-entangled (Positive Partial Transpose but still entangled) states, which are trickier.
Practical use. Peres-Horodecki is a polynomial-time test for entanglement in small systems. No matter how complicated the state looks, you compute the partial transpose (reshape and transpose a 2\times 2 block) and check for a negative eigenvalue. If you find one, the state is entangled — no further analysis needed.
The fact that the transpose is "the right tool" for witnessing entanglement is not a coincidence. The broad theorem (Horodecki-Horodecki-Horodecki 1996): a state is entangled if and only if there exists some positive-but-not-CP map \Lambda such that (\Lambda\otimes\text{id})\rho has a negative eigenvalue. Non-CP positive maps are the natural language of entanglement detection.
Beyond CPTP: quantum instruments and general operations
Relaxing the conditions on \mathcal E gives richer structures used in quantum-information theory:
-
Quantum instrument: a collection of CP maps \{\mathcal E_m\} with \sum_m \mathcal E_m CPTP. The index m is a classical outcome; the instrument tells you both the probability of outcome m (via \text{tr}(\mathcal E_m(\rho))) and the post-measurement state (proportional to \mathcal E_m(\rho)). Recorded projective measurement, POVMs with state updates, and feedback-controlled operations all fit this framework.
-
Non-trace-preserving CP maps: \sum_k K_k^\dagger K_k \leq I. Used to describe single outcomes of measurements without normalisation. Their trace gives the outcome probability; the rest is absorbed into the "discarded" outcome.
-
Anti-unitary maps: complex conjugation, time-reversal, etc. These are not linear in the required sense (they are anti-linear) and are not channels, but they play a role in fundamental-physics formulations. Wigner's theorem classifies symmetries of quantum mechanics as either unitary or anti-unitary.
The CPTP framework is the workhorse for almost all practical quantum-computing questions (noise models, error correction, channel capacity). The richer structures come in when you need to track classical outcomes, post-selection, or fundamental symmetries.
Where this leads next
- Stinespring Dilation — the next chapter, constructing the dilating unitary that realises any CPTP map as a unitary on a larger system.
- Kraus Representation — the operator-sum form equivalent to CPTP, from the previous chapter.
- Standard Channels — the bit-flip, phase-flip, depolarizing, and amplitude-damping channels worked through in depth.
- Fidelity and Trace Distance — measures of how close two CPTP maps (or their outputs) are.
- Purification — the state-level analogue of Stinespring: every density matrix comes from a pure state on a larger system.
- Partial Trace — the CPTP map that takes you from \rho_{AB} down to \rho_A, the specific channel that makes Stinespring work.
References
- Wikipedia, Quantum operation — CPTP definition, equivalence with Kraus and Stinespring, standard examples.
- Nielsen and Chuang, Quantum Computation and Quantum Information, §8.2 (the operator-sum representation and CPTP axioms) — Cambridge University Press.
- John Preskill, Lecture Notes on Quantum Computation, Ch. 3 (the full derivation of CPTP from physical axioms) — theory.caltech.edu/~preskill/ph229.
- John Watrous, The Theory of Quantum Information (2018), Ch. 2 (completely positive maps, Choi representation, diamond norm) — cs.uwaterloo.ca/~watrous/TQI.
- Wikipedia, Choi's theorem on completely positive maps — the 1975 Choi-Kraus equivalence and the Choi matrix.
- Wikipedia, Peres-Horodecki criterion — the transpose map as an entanglement witness, with explicit examples.