The Kraus Representation

Choi's theorem tells us which maps are completely positive, but it does not give a working formula. The Kraus representation (also called the operator-sum representation) does: it writes any CP map as a short sum of "sandwich" terms, and it is the form you actually compute with.

The operator-sum form

A linear map $\mathcal{E}$ on density operators is completely positive if and only if there exist operators $\{K_k\}$ , the Kraus operators, such that

\mathcal{E}(\rho) = \sum_{k} K_k\,\rho\,K_k^\dagger.

Why this is automatically completely positive: for any vector $|\psi\rangle$ on system-plus-reference,

\langle\psi|\,(\mathcal{E}\otimes\mathbb{1})(\sigma)\,|\psi\rangle = \sum_k \langle\psi|(K_k\otimes\mathbb{1})\,\sigma\,(K_k\otimes\mathbb{1})^\dagger|\psi\rangle = \sum_k \langle\phi_k|\sigma|\phi_k\rangle \ge 0

whenever $\sigma \succeq 0$ , where $|\phi_k\rangle = (K_k^\dagger\otimes\mathbb{1})|\psi\rangle$ . Each term is manifestly positive, and the ancilla played no special role, so positivity holds for every reference — that is exactly complete positivity. The converse (every CP map has such a decomposition) follows by spectrally decomposing the Choi operator: if $J(\mathcal{E}) = \sum_k |v_k\rangle\langle v_k|$ , then "unvectorizing" each $|v_k\rangle$ into a matrix yields a Kraus operator $K_k$ .

Trace preservation as a sum rule

The map is trace preserving exactly when the Kraus operators satisfy the completeness relation

\sum_k K_k^\dagger K_k = \mathbb{1}.

This is the operator-sum statement of "probabilities sum to one," derived in full in the next lesson. A CP map with $\sum_k K_k^\dagger K_k \preceq \mathbb{1}$ is trace non-increasing and models a process conditioned on an outcome.

Kraus operators are not unique

Two sets $\{K_k\}$ and $\{L_j\}$ describe the same channel if and only if they are related by an isometry $u$ with $\sum_k u_{jk} K_k = L_j$ (pad the smaller set with zeros). So a channel has many Kraus representations; only the channel — the map itself — is physical. The minimum number of Kraus operators needed equals the rank of the Choi operator, called the Kraus rank.

Example: the bit-flip channel

The bit-flip channel applies $X$ with probability $p$ and leaves the state alone otherwise:

\mathcal{E}(\rho) = (1-p)\,\rho + p\,X\rho X, \qquad K_0 = \sqrt{1-p}\,\mathbb{1}, \quad K_1 = \sqrt{p}\,X.

Check completeness: $K_0^\dagger K_0 + K_1^\dagger K_1 = (1-p)\mathbb{1} + p\,X^2 = (1-p)\mathbb{1} + p\,\mathbb{1} = \mathbb{1}$ , since $X^2 = \mathbb{1}$ . The channel is therefore CPTP. Acting on $\rho = |0\rangle\langle 0|$ it returns $(1-p)|0\rangle\langle 0| + p|1\rangle\langle 1|$ : with probability $p$ the bit has flipped to $|1\rangle$ .

Try it

Take the bit-flip channel with $p = \tfrac14$ and input $\rho = |+\rangle\langle +|$ , where $|+\rangle = \tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)$ . Apply the operator-sum formula $\mathcal{E}(\rho) = (1-p)\,\rho + p\,X\rho X$ by representing $2\times 2$ complex matrices in code, and return $\langle 0|\,\mathcal{E}(\rho)\,|0\rangle$ , the probability of outcome $0$ . Because $|+\rangle$ is an eigenstate of $X$ ( $X|+\rangle = |+\rangle$ ), the bit flip does nothing here, so the answer is $\tfrac12$ — confirm this by computing, not assuming.

Run your code to see the quantum state.