Kraus Representation of Quantum Channels

Every physically allowed transformation of a quantum state — a noisy gate, a decohering wire, a measurement whose outcome you forget, a system leaking into its surroundings — is one mathematical object: a quantum channel. Abstractly a channel obeys two demands. It must be completely positive: it sends density operators to density operators, even when acting on only part of a larger entangled system. And it must be trace-preserving: it conserves total probability. Together these are the CPTP conditions — a pair of abstract inequalities on a superoperator, hard to check and hard to build with. The Kraus, or operator-sum, representation turns that abstraction into arithmetic. Every CPTP map $\mathcal{E}$ can be written as a finite sum $\mathcal{E}(\rho)=\sum_i K_i\rho K_i^\dagger$ of one-sided conjugations by ordinary linear operators $K_i$ , subject to the single constraint $\sum_i K_i^\dagger K_i=I$ . Each term is manifestly positive; the constraint is exactly conservation of probability; and complete positivity is automatic, because the same operators tensored with the identity act correctly on any extension. The whole content of "physically allowed" is compressed into a short list of matrices. That list is not unique — and its non-uniqueness is not a defect but a physical fact, the mathematical shadow of the freedom to read out a hidden environment in any basis. This article builds the operator-sum form from complete positivity, proves exactly how much freedom the $K_i$ carry, and places the Kraus picture at the centre of a triangle with its equivalent siblings, the Stinespring dilation and the Choi matrix.

Key takeaways

The Kraus (operator-sum) representation writes any quantum channel as $\mathcal{E}(\rho)=\sum_i K_i\rho K_i^\dagger$ , with the Kraus operators $\{K_i\}$ linear maps from input to output Hilbert space.
Trace preservation is exactly the completeness relation $\sum_i K_i^\dagger K_i=I$ ; relaxing it to $\le I$ gives a trace non-increasing operation attached to one measurement outcome.
Having a Kraus form is equivalent to being CPTP — the concrete face of the abstract condition, dual to the Choi matrix and the Stinespring dilation.
The representation is not unique: $\{K_i\}$ and $\{L_j\}$ give the same channel iff $L_j=\sum_i u_{ji}K_i$ for an isometry $u$ ; only basis-independent quantities are physical.
A channel on a $d$ -dimensional system needs at most $d^2$ Kraus operators — the Kraus rank equals the rank of the Choi matrix and bounds the minimal environment in any dilation.
Reading $K_i=\langle i|_E\,U\,|0\rangle_E$ exposes the physics: a channel is a unitary on system-plus-environment, then discard the environment.
Key idea: a CPTP map is exactly a sum of conjugations $\rho\mapsto\sum_i K_i\rho K_i^\dagger$ with $\sum_i K_i^\dagger K_i=I$ ; the Kraus operators are the matrix elements of a system–environment unitary in an environment basis, unique only up to an isometric mixing of the index.

The core insight

A CPTP map is exactly a sum of one-sided conjugations $\rho\mapsto\sum_i K_i\rho K_i^\dagger$ with $\sum_i K_i^\dagger K_i=I$ ; the Kraus operators are the matrix elements of a system–environment unitary in an environment basis, and they are unique only up to an isometric mixing of the index.

Where this lives today

Kraus operators are the working currency of quantum error correction. Every QEC noise model is a Kraus set — bit flip $\{\sqrt{1-p}\,I,\sqrt{p}\,X\}$ , depolarising, amplitude and phase damping — and the Knill–Laflamme conditions $P K_i^\dagger K_j P=\alpha_{ij}P$ are stated directly on those operators. Digitising continuous noise into a discrete (often Pauli) Kraus basis is what makes fault tolerance possible: a code that corrects a finite Kraus basis automatically corrects everything in its linear span. To characterise real hardware, process and gate-set tomography reconstruct a channel's Choi matrix from measured data; its eigendecomposition yields a minimal orthogonal Kraus set, and the number of significant eigenvalues — the effective Kraus rank — measures how strongly a superconducting or trapped-ion qubit decoheres, while comparing the dominant Kraus operator to the intended unitary isolates coherent gate error. Open-system simulators either compile a Lindblad master equation into a CPTP map with explicit Kraus operators or unravel it into stochastic trajectories where each jump applies one $K_i$ — and the isometric freedom in the Kraus set is exactly the freedom to choose the unravelling, all reproducing the same averaged density matrix. At the frontier, the complementary channel built from the same Kraus set defines the coherent information $I_c=S(\mathcal{E}(\rho))-S(\mathcal{E}^c(\rho))$ governing quantum capacity, and whether that capacity has a single-letter formula (degradability, antidegradability, the open question of superadditivity) is read off from the relation between $\mathcal{E}$ and $\mathcal{E}^c$ .

The physical picture

Think of a channel as a black box that secretly couples your system to a fresh, hidden environment, lets the two interact through one unitary $U$ , then throws the environment away — a single story that reproduces every channel there is. If you could recover the discarded environment and measure it in some basis $\{|i\rangle_E\}$ , you would learn "outcome $i$ " occurred and your system would jump to $K_i\rho K_i^\dagger/p_i$ with probability $p_i=\mathrm{Tr}(K_i^\dagger K_i\rho)$ . Because you did not look, you average over the branches: $\sum_i K_i\rho K_i^\dagger$ . The Kraus operators are precisely those branches — the matrix elements $K_i=\langle i|_E\,U\,|0\rangle_E$ of the joint unitary. The first figure is that story drawn as a circuit.

Figure

A channel couples the system to a fresh environment through a joint unitary U and then discards it, so E of rho equals the partial trace over the environment of U acting on rho tensored with the environment ground state; reading the discarded environment in a basis peels off one Kraus operator per outcome

Each $K_i$ is a partial, non-unitary evolution — it can shrink, rotate, and entangle — but the completeness relation $\sum_i K_i^\dagger K_i=I$ laces the branches together so probabilities still add to one: it is the quantum generalisation of a stochastic matrix whose columns sum to one. The number of independent branches — the Kraus rank — measures how many distinguishable ways the environment carries information off: rank one is reversible unitary evolution, higher rank is genuine decoherence. Crucially the branches are not observer-independent: measuring the environment in a different basis mixes the $K_i$ by a unitary and changes nothing about the channel, so no single Kraus operator is "what really happened." The second figure makes that non-uniqueness concrete on a real channel, with real numbers.

Figure

Two different Kraus sets describe the very same phase-damping channel, the one that shrinks off-diagonal coherence to a factor 0.6 of its input value; the two sets are linked by a 2 by 2 isometry u, showing that no individual Kraus operator is physical while the channel and its Kraus rank are

The core equations

The definition is one line: a channel is a sum of one-sided conjugations tied together by a completeness relation that enforces probability conservation. Everything below is a consequence or repackaging of it.

\mathcal{E}(\rho)=\sum_i K_i\,\rho\,K_i^\dagger,\qquad \sum_i K_i^\dagger K_i = I_A \qquad (1)

where: $\mathcal{E}$ — the quantum channel, a CPTP linear map from operators on $\mathcal{H}_A$ to operators on $\mathcal{H}_B$ ; $K_i$ — Kraus operators, linear maps $\mathcal{H}_A\to\mathcal{H}_B$ ; $\rho$ — input density operator, $\rho\ge 0$ , $\mathrm{Tr}\,\rho=1$ ; $I_A$ — identity on the input space. Read it as: apply each operator, add the results, and completeness keeps the output a valid state. Trace preservation is not an extra axiom — it is precisely the completeness relation, as cyclicity of the trace shows. This is the single most-used calculation in the subject.

\mathrm{Tr}\,\mathcal{E}(\rho)=\sum_i\mathrm{Tr}\!\left(K_i\rho K_i^\dagger\right)=\sum_i\mathrm{Tr}\!\left(K_i^\dagger K_i\,\rho\right)=\mathrm{Tr}\!\left(\Big(\textstyle\sum_i K_i^\dagger K_i\Big)\rho\right)=\mathrm{Tr}\,\rho \qquad (2)

where: the second equality is cyclicity, moving $K_i^\dagger$ to the front; the last uses $\sum_i K_i^\dagger K_i=I$ . If instead $\sum_i K_i^\dagger K_i\le I$ , the trace can only decrease, and $\mathrm{Tr}\,\mathcal{E}(\rho)$ is the probability of the associated measurement outcome — a trace non-increasing operation, not a channel. The operator-sum form is not unique: the freedom is exactly an isometry on the Kraus index, and two sets describe the same channel if and only if they are related this way.

L_j=\sum_i u_{ji}\,K_i,\qquad \sum_j u_{ji}^{*}u_{jk}=\delta_{ik}\ \ (u^\dagger u=I) \qquad (3)

where: $L_j$ — an alternative Kraus set for the same channel; $u_{ji}$ — entries of an isometry (a unitary once both sets are padded to equal size); $\delta_{ik}$ — Kronecker delta. Mixing the branches by any $u$ with orthonormal columns leaves the channel untouched — the algebraic image of remeasuring the environment in a rotated basis. The Kraus form is the operator-sum incarnation of the Stinespring dilation: dilate the channel to a unitary on system-plus-environment, fix an environment basis, and each Kraus operator is a block of that unitary.

\mathcal{E}(\rho)=\mathrm{Tr}_E\!\left[U\,(\rho\otimes|0\rangle\langle 0|_E)\,U^\dagger\right],\qquad K_i=\langle i|_E\,U\,|0\rangle_E \qquad (4)

where: $U$ — unitary on $\mathcal{H}_A\otimes\mathcal{H}_E$ ; $|0\rangle_E$ — fixed initial environment state; $\{|i\rangle_E\}$ — orthonormal environment basis extracting each Kraus operator; $\mathrm{Tr}_E$ — partial trace. Completeness $\sum_i K_i^\dagger K_i=I$ is exactly the statement $\langle 0|U^\dagger U|0\rangle_E=I_A$ that $U$ is unitary. The Kraus form is also dual to the Choi matrix: diagonalising it hands you a canonical, mutually orthogonal Kraus set whose size is the intrinsic Kraus rank.

J(\mathcal{E})=(\mathcal{E}\otimes\mathrm{id})\big(|\Omega\rangle\langle\Omega|\big)=\sum_{i}|v_i\rangle\langle v_i|,\quad |v_i\rangle=(K_i\otimes I)|\Omega\rangle,\quad r_{\mathrm{Kraus}}=\mathrm{rank}\,J\le d_Ad_B \qquad (5)

where: $|\Omega\rangle=\sum_n|n\rangle_A|n\rangle_{A'}$ — unnormalised maximally entangled vector; $J(\mathcal{E})$ — Choi matrix, $J\ge 0\Leftrightarrow$ complete positivity; $|v_i\rangle$ — sub-normalised eigenvectors of $J$ , the vectorised canonical Kraus operators; $r_{\mathrm{Kraus}}$ — Kraus rank, the minimal operator count, bounded by $d_Ad_B$ (so by $d^2$ when $d_A=d_B=d$ ). The canonical set is Hilbert–Schmidt orthogonal: $\mathrm{Tr}(K_i^\dagger K_j)=\lambda_i\delta_{ij}$ . The same dilating unitary defines the complementary channel — keep the environment instead of discarding it — whose matrix elements are built from the same Kraus operators, the door to entropy exchange and quantum capacity.

\mathcal{E}^{c}(\rho)=\mathrm{Tr}_B\!\left[U(\rho\otimes|0\rangle\langle 0|_E)U^\dagger\right],\qquad \big[\mathcal{E}^{c}(\rho)\big]_{ij}=\mathrm{Tr}\!\left(K_j^\dagger K_i\,\rho\right) \qquad (6)

where: $\mathcal{E}^{c}$ — complementary channel, the map from system to environment; $[\cdot]_{ij}$ — matrix element in the environment basis. Tracing out $B$ instead of $E$ swaps "kept" and "lost" information; the environment-basis choice mixes the index by the usual isometry $u$ .

Derivation AFrom complete positivity to the operator-sum form

The goal is the existence half of the theory: to prove that every completely positive map admits a Kraus representation, and that trace preservation forces the completeness relation, by constructing the operators from the spectral decomposition of the Choi matrix. This turns the abstract inequality "complete positivity" into a finite, computable list of matrices that exists whenever the map is a channel. Let $\mathcal{E}$ be linear on operators of $\mathcal{H}_A$ with $\dim\mathcal{H}_A=d$ , and set $|\Omega\rangle=\sum_{n=1}^{d}|n\rangle_A|n\rangle_{A'}$ and $J=(\mathcal{E}\otimes\mathrm{id})(|\Omega\rangle\langle\Omega|)$ .

Step 1. Complete positivity means $\mathcal{E}\otimes\mathrm{id}$ preserves positivity; since $|\Omega\rangle\langle\Omega|\ge 0$ , the Choi matrix $J\ge 0$ . Why: this converts a statement about a superoperator into positivity of a single ordinary $d^2\times d^2$ matrix, which we can diagonalise.

Step 2. Because $J\ge 0$ , decompose it with eigenvalues absorbed: $J=\sum_i|v_i\rangle\langle v_i|$ with $|v_i\rangle=\sqrt{\lambda_i}|w_i\rangle$ , $\lambda_i\ge 0$ and $\{|w_i\rangle\}$ orthonormal in $\mathcal{H}_B\otimes\mathcal{H}_{A'}$ . Why: positivity makes the eigenvalues non-negative, so their square roots are real and the $|v_i\rangle$ are well defined — no positivity, no Kraus operators.

Step 3. Devectorise each eigenvector: define $K_i$ by $|v_i\rangle=(K_i\otimes I)|\Omega\rangle$ , i.e. $K_i=\sum_{m,n}\langle m|_B\langle n|_{A'}|v_i\rangle\,|m\rangle\langle n|$ . Why: $|\Omega\rangle$ sets up the Choi–Jamiołkowski isomorphism between vectors in $\mathcal{H}_B\otimes\mathcal{H}_{A'}$ and operators $\mathcal{H}_A\to\mathcal{H}_B$ , so each eigenvector is an operator in disguise.

Step 4. Contract the second factor of $J$ against $\rho$ ; the structure $\sum_i|v_i\rangle\langle v_i|$ reassembles as $\mathcal{E}(\rho)=\sum_i K_i\rho K_i^\dagger$ . Why: the same isomorphism turns the eigen-expansion of $J$ back into the channel — one fact, two writings.

Step 5. Impose trace preservation. Tracing out the output of $J$ gives $\mathrm{Tr}_B J=\sum_i K_i^\dagger K_i$ (transposed on $A'$ ), so $\mathrm{Tr}_B J=I_{A'}$ is equivalent to $\sum_i K_i^\dagger K_i=I$ . Why: the partial-trace condition on the Choi matrix and the completeness relation are literally the same equation, which is why "trace-preserving" and " $\sum K_i^\dagger K_i=I$ " are interchangeable. *The equivalence $\mathcal{E}$ CP $\iff J\ge 0\iff$ operator-sum form exists is the central result of this derivation. The rank of $J$ is the minimal number of Kraus operators, bounded by $d_Ad_B$ .*

Go deeper · Derivation B · 1 minThe Kraus operators are unique only up to an isometry

Derivation BThe Kraus operators are unique only up to an isometry

The goal is to pin down exactly how much freedom the operator-sum form carries: to prove that two Kraus sets represent the same channel if and only if they are related by an isometry on the index, so the genuine non-uniqueness is neither more nor less than a change of environment basis. This licenses reshuffling error models and switching between trajectory unravellings without changing the physics — and forbids attaching meaning to any single Kraus operator. Take two Kraus sets $\{K_i\}_{i=1}^{m}$ and $\{L_j\}_{j=1}^{n}$ into $\mathcal{H}_B$ ; pad the shorter with zeros so $m=n$ .

Step 1. Sufficiency by direct substitution. If $L_j=\sum_i u_{ji}K_i$ with $\sum_j u_{ji}^{*}u_{jk}=\delta_{ik}$ , then $\sum_j L_j\rho L_j^\dagger=\sum_{j,i,k}u_{ji}u_{jk}^{*}K_i\rho K_k^\dagger=\sum_{i,k}\delta_{ik}K_i\rho K_k^\dagger=\sum_i K_i\rho K_i^\dagger$ . Why: the isometry relation collapses the double sum to a single diagonal one, so an isometric mixing provably leaves the channel invariant.

Step 2. For necessity, encode each set as one purifying vector, $|\Phi_K\rangle=\sum_i (K_i\otimes I)|\Omega\rangle\otimes|i\rangle_E$ and $|\Phi_L\rangle=\sum_j (L_j\otimes I)|\Omega\rangle\otimes|j\rangle_{E'}$ ; tracing out the ancilla returns the Choi matrix. Why: bundling the set into one state turns "same channel" into "same reduced state," which we can answer.

Step 3. Equal channels have equal Choi matrices, so $|\Phi_K\rangle$ and $|\Phi_L\rangle$ purify the same positive operator $J$ . By the HJW / GHJW theorem, any two purifications are related by an isometry on the purifying system: $|v_j^{(L)}\rangle=\sum_i u_{ji}|v_i^{(K)}\rangle$ . Why: purifications of a fixed operator form one isometry-equivalence class — the operator version of the fact that all ensembles of a density matrix are unitarily related.

Step 4. Devectorise to $L_j=\sum_i u_{ji}K_i$ with $u^\dagger u=I$ , which guarantees both sets obey completeness. Why: the Choi–Jamiołkowski dictionary turns the vector relation into the operator mixing, closing the equivalence both ways. *Padding the shorter Kraus set with zero operators makes $u$ a genuine unitary on the larger index set, completing the equivalence in both directions.*

Go deeper · Derivation C · 1 minOne dephasing channel written two ways, checked with numbers

Derivation COne dephasing channel written two ways, checked with numbers

The goal is to make non-uniqueness fully concrete: to write one physical channel — phase damping that shrinks a qubit's off-diagonal coherence to a factor $\eta$ of its input — in two visibly different Kraus sets, verify by matrix multiplication that both give the identical map, and exhibit the explicit isometry linking them. Parametrising by the coherence-survival factor $\eta\in[0,1]$ (not a decay probability) makes the two forms equal at the same parameter; we take $\eta=0.6$ so every entry is checkable by hand. The target action is $\rho=\begin{pmatrix}a&b\\ b^{*}&c\end{pmatrix}\mapsto\begin{pmatrix}a&\eta b\\ \eta b^{*}&c\end{pmatrix}$ .

Step 1. Random-unitary form: $K_0=\sqrt{\tfrac{1+\eta}{2}}\,I=\sqrt{0.8}\,I$ and $K_1=\sqrt{\tfrac{1-\eta}{2}}\,Z=\sqrt{0.2}\,Z$ . Completeness: $\tfrac{1+\eta}{2}I+\tfrac{1-\eta}{2}I=I$ . Off-diagonal: $\tfrac{1+\eta}{2}b+\tfrac{1-\eta}{2}(-b)=\eta b=0.6\,b$ . Why: a random application of $I$ or $Z$ , whose $Z$ branch flips the sign of the coherence, so mixing them attenuates it by exactly $\eta$ .

Step 2. Diagonal "damping" form: $\tilde K_0=\mathrm{diag}(1,\eta)=\mathrm{diag}(1,0.6)$ and $\tilde K_1=\mathrm{diag}(0,\sqrt{1-\eta^2})=\mathrm{diag}(0,0.8)$ . Completeness: $\mathrm{diag}(1,0.36)+\mathrm{diag}(0,0.64)=I$ . The upper-right entry of $\tilde K_0\rho\tilde K_0^\dagger$ is $\eta b=0.6\,b$ , and $\tilde K_1$ contributes nothing off-diagonal. Why: the "environment reads the populations" form looks nothing like Step 1 yet must give the same map.

Step 3. Both maps send $\rho\mapsto\begin{pmatrix}a&0.6\,b\\ 0.6\,b^{*}&c\end{pmatrix}$ , fixing populations and multiplying coherences by $0.6$ ; direct $2\times2$ multiplication confirms identical output on every $\rho$ . Why: two different-looking lists implement one channel, so their difference cannot be physical.

Step 4. The isometry: $u=\begin{pmatrix}\sqrt{0.8}&\sqrt{0.2}\\ \sqrt{0.2}&-\sqrt{0.8}\end{pmatrix}$ satisfies $u^\dagger u=I$ , and $\tilde K_0=u_{00}K_0+u_{01}K_1$ , $\tilde K_1=u_{10}K_0+u_{11}K_1$ reproduce $\mathrm{diag}(1,0.6)$ and $\mathrm{diag}(0,0.8)$ exactly. Why: the previous derivation guarantees such a $u$ exists — here it is with numerical entries, non-uniqueness demonstrated rather than asserted. *Both sets satisfy $\sum_i K_i^\dagger K_i=I$ ; direct matrix multiplication confirms that both reproduce $b\mapsto 0.6\,b$ ; and the isometry $u$ maps Set A to Set B exactly.*

Limiting cases

Special cases of the operator-sum form recover the familiar transformations, each fixed by how many Kraus operators survive.

Case	Result	Meaning
Unitary (closed) evolution	$\mathcal{E}(\rho)=U\rho U^\dagger,\ \ U^\dagger U=I$	A single Kraus operator, Kraus rank $1$ . No environment is entangled; the evolution is reversible and entropy-preserving.
Measurement with recorded outcome	$\mathcal{E}(\rho)=\sum_i K_i\rho K_i^\dagger\otimes\|i\rangle\langle i\|$	A quantum instrument: the index is copied to a classical register. Each $E_i=K_i^\dagger K_i$ is a POVM element and $\sum_i E_i=I$ .
Random unitary / dephasing	$\mathcal{E}(\rho)=\sum_i p_i\,U_i\rho U_i^\dagger,\ \ \sum_i p_i=1$	$K_i=\sqrt{p_i}\,U_i$ are proportional to unitaries; unital. Qubit dephasing uses $K_0=\sqrt{1-p}\,I,\ K_1=\sqrt{p}\,Z$ .
Amplitude damping (non-unital)	$K_0=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix},\ K_1=\begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix}$	Models spontaneous emission. $K_1$ takes $\|1\rangle\to\|0\rangle$ ; $\sum_i K_i^\dagger K_i=I$ but $\sum_i K_iK_i^\dagger\ne I$ .
Trace non-increasing operation	$\Phi(\rho)=\sum_i K_i\rho K_i^\dagger,\ \ \sum_i K_i^\dagger K_i\le I$	One outcome of a selective measurement; $\mathrm{Tr}\,\Phi(\rho)\in[0,1]$ is its probability. Renormalise to get the post-selected state.

Common trapDo not ask which Kraus operator really happened

The operator-sum decomposition is not unique, so an individual Kraus operator $K_i$ has no observer-independent meaning — it is not "the error that occurred." Any two valid Kraus sets for the same channel are related by $L_j=\sum_i u_{ji}K_i$ with $u$ an isometry, which physically means reading the discarded environment in a different basis. Only basis-independent objects carry invariant meaning: the channel, its Kraus rank, the Choi spectrum, the entropies. A second, algebraic trap: completeness is $\sum_i K_i^\dagger K_i=I$ (daggers on the inside, acting on the input space), not $\sum_i K_iK_i^\dagger=I$ — the latter equals $\mathcal{E}(I)$ and is $I$ only for a unital channel, amplitude damping being the standard counterexample.

Worked examples

1 · The depolarising channel: completeness and Kraus rank (fully worked)

The depolarising channel mixes in white noise: $\mathcal{E}(\rho)=(1-p)\rho+\tfrac{p}{3}(X\rho X+Y\rho Y+Z\rho Z)$ . Take $p=0.3$ ; read off the Kraus operators, verify completeness, find the Bloch-vector effect, and state the Kraus rank.

Step 1. The four Kraus operators are $K_0=\sqrt{1-p}\,I=\sqrt{0.7}\,I$ and $K_{1,2,3}=\sqrt{p/3}\,\{X,Y,Z\}=\sqrt{0.1}\,\{X,Y,Z\}$ . Four Pauli-proportional operators.

Step 2. Completeness: since $X^\dagger X=Y^\dagger Y=Z^\dagger Z=I$ , $\sum_i K_i^\dagger K_i=(1-p)I+3\cdot\tfrac{p}{3}I=(1-p)I+pI=I$ . Trace-preserving for every $p$ . Σ Kᵢ†Kᵢ = I confirmed.

Step 3. Effect on $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ : each Pauli conjugation flips the sign of the two orthogonal Bloch components, so $\sum_{P}P(\vec r\cdot\vec\sigma)P=-\vec r\cdot\vec\sigma$ and the Bloch vector contracts uniformly, $\vec r\mapsto(1-\tfrac{4p}{3})\vec r=0.6\,\vec r$ . The Bloch ball shrinks isotropically by 0.6 toward the maximally mixed centre.

Step 4. Kraus rank: $\{I,X,Y,Z\}$ are Hilbert–Schmidt orthogonal, so none is redundant and the rank is $4$ — the maximum a qubit channel can have, saturating $r\le d^2=4$ . Any representation with more operators reduces to these four. Kraus rank = 4 (maximal).

2 · Amplitude damping: completeness, non-unitality, and a decayed population

The amplitude-damping channel with $\gamma=0.3$ has $K_0=\mathrm{diag}(1,\sqrt{0.7})$ and $K_1=\begin{pmatrix}0&\sqrt{0.3}\\0&0\end{pmatrix}$ . Verify completeness, show it is non-unital, and evolve the excited state $|1\rangle\langle 1|$ .

Step 1. Completeness: $K_0^\dagger K_0=\mathrm{diag}(1,0.7)$ and $K_1^\dagger K_1=\mathrm{diag}(0,0.3)$ , so $\sum_i K_i^\dagger K_i=\mathrm{diag}(1,1)=I$ . The channel is trace-preserving. Σ Kᵢ†Kᵢ = I confirmed.

Step 2. The other combination: $K_0K_0^\dagger=\mathrm{diag}(1,0.7)$ and $K_1K_1^\dagger=\mathrm{diag}(0.3,0)$ , so $\sum_i K_iK_i^\dagger=\mathrm{diag}(1.3,0.7)\ne I$ — the channel is non-unital, draining population toward the ground state rather than fixing the maximally mixed state. Σ KᵢKᵢ† = diag(1.3, 0.7) ≠ I.

Step 3. Evolve $|1\rangle\langle 1|=\mathrm{diag}(0,1)$ : $K_0|1\rangle\langle 1|K_0^\dagger=\mathrm{diag}(0,0.7)$ and $K_1|1\rangle\langle 1|K_1^\dagger=\mathrm{diag}(0.3,0)$ sum to $\mathrm{diag}(0.3,0.7)$ — a fraction $\gamma=0.3$ of the excited population has decayed to ground, exactly as spontaneous emission over one interval should. |1⟩⟨1| ↦ 0.3 |0⟩⟨0| + 0.7 |1⟩⟨1|.

Physics of the portrait

The operator-sum representation answers a foundational question — what is the most general way a quantum state can change? — and bears the name of the physicist who posed it.

Origins · 1 minKarl Kraus and how this was discovered

Karl Kraus

1938–1988, German theoretical physicist · foundations of quantum theory, quantum operations and effects

Karl Kraus (born 21 March 1938 in Hohenelbe, today Vrchlabí; died 9 June 1988) took his doctorate in 1962 under Kurt Just, then became an assistant to Günther Ludwig at Marburg. It was Ludwig's operational programme — describing quantum theory through states, effects, and operations rather than wavefunctions alone — that shaped his central contribution. In a 1971 Annals of Physics paper, "General state changes in quantum theory," Kraus asked precisely the right question: what is the most general map that must send every valid quantum state to another valid state, even when the system is entangled with a reference it never touches? The answer is the operator-sum form $\rho\mapsto\sum_i K_i\rho K_i^\dagger$ now universally called the Kraus representation. He built on Stinespring's 1955 dilation theorem and worked alongside the matrix criterion Choi published in 1975; his 1983 Springer lecture notes, States, Effects, and Operations, unified states, effects (POVM elements), and operations into a language now standard across quantum information. Appointed professor at Würzburg in 1971, he died in 1988 at fifty — but the operators bearing his name are today the everyday vocabulary of quantum error correction, open-system simulation, and channel-capacity theory.

History

1955

— Stinespring proves every completely positive map dilates to a unitary on a larger space

1961

— Sudarshan, Mathews and Rau cast open-system evolution as a linear map on density matrices

1971

— Kraus derives the operator-sum form of general state changes in quantum theory

1975

— Choi characterizes completely positive maps by positivity of the Choi matrix

1983

— Kraus's States, Effects, and Operations systematizes operations and effects

Going further

Texts. Nielsen and Chuang, Quantum Computation and Quantum Information, §8.2–8.3 · Preskill, Lecture Notes for Physics 219, Ch. 3 · Watrous, The Theory of Quantum Information, Ch. 2 · Wilde, Quantum Information Theory, Ch. 4 · Kitaev, Shen and Vyalyi, Classical and Quantum Computation, §11 (superoperators). Key results to keep in mind: existence of the operator-sum form follows from diagonalising the Choi matrix and confirming $J\ge0\Leftrightarrow$ CP; the unitary-freedom theorem gives $L_j=\sum_i u_{ji}K_i$ ; the Kraus-rank bound is $r\le d_Ad_B$ from $\mathrm{rank}\,J$ ; trace preservation re-derives via cyclicity. For the phase-damping example at $\eta=0.6$ : both sets satisfy $\sum_i K_i^\dagger K_i=I$ , both send $b\mapsto 0.6\,b$ , and $u=[\sqrt{0.8},\sqrt{0.2};\sqrt{0.2},-\sqrt{0.8}]$ (with $u^\dagger u=I$ ) maps Set A to Set B. For amplitude damping at $\gamma=0.3$ : $\sum_i K_iK_i^\dagger=\mathrm{diag}(1.3,0.7)\ne I$ and $|1\rangle\langle1|\mapsto\mathrm{diag}(0.3,0.7)$ .

Builds on: Quantum Channels and Completely Positive Trace-Preserving Maps · Von Neumann Entropy · Stinespring Dilation Theorem. Leads to: Choi–Jamiołkowski Isomorphism · Complementary Channels and Degradability · Data-Processing Inequality · Schumacher Quantum Data Compression.

Ask the tutor: ask me to build the explicit Kraus operators for any channel you name — bit flip, depolarising, amplitude damping — and I will check completeness, find the Kraus rank from the Choi matrix, and show you a second, isometry-related set for the very same channel.

Why isn't the Kraus decomposition of a channel unique?

Because the discarded environment has no preferred basis. The Kraus operators are the blocks $K_i=\langle i|_E U|0\rangle_E$ of the dilating unitary in a chosen environment basis; choose a different basis and you mix them by a unitary $u$ , giving $L_j=\sum_i u_{ji}K_i$ — the identical channel. Padding with zeros is a further trivial freedom. The invariant content survives all such changes: the channel, its Kraus rank, the Choi spectrum, and any entropy computed from them.

What is the Kraus rank and why does it matter?

The Kraus rank is the minimal number of Kraus operators for a channel, equal to the rank of the Choi matrix. Unlike the raw count — which you can inflate by adding zeros — the rank is intrinsic. It bounds the smallest environment in any Stinespring dilation, so it measures how many distinguishable "ways to leak" the channel has: rank $1$ is reversible unitary evolution, higher rank is genuine decoherence. In tomography the number of significant Choi eigenvalues is the effective rank, a direct read-out of device noise. For a $d$ -dimensional system it never exceeds $d^2$ .

Why is the completeness relation Σ Kᵢ†Kᵢ = I and not Σ KᵢKᵢ† = I?

Trace preservation demands $\mathrm{Tr}\,\mathcal{E}(\rho)=\mathrm{Tr}\,\rho$ for every input. By cyclicity $\mathrm{Tr}\,\mathcal{E}(\rho)=\mathrm{Tr}\big((\sum_i K_i^\dagger K_i)\rho\big)$ , which equals $\mathrm{Tr}\,\rho$ for all $\rho$ only if $\sum_i K_i^\dagger K_i=I$ — daggers inside, acting on the input space. The other combination $\sum_i K_iK_i^\dagger$ equals $\mathcal{E}(I)$ and is $I$ only for a unital channel. Amplitude damping is the standard non-unital counterexample, with $\sum_i K_iK_i^\dagger=\mathrm{diag}(1+\gamma,1-\gamma)\ne I$ .

How does the Kraus form relate to the Stinespring and Choi pictures?

Three faces of one object. Stinespring writes the channel as $\mathcal{E}(\rho)=\mathrm{Tr}_E[U(\rho\otimes|0\rangle\langle0|)U^\dagger]$ ; contracting in an environment basis gives $K_i=\langle i|_E U|0\rangle_E$ . Choi packs the channel into one positive matrix $J$ , and diagonalising $J$ returns a canonical orthogonal Kraus set. Use Stinespring for complementary channels and entropy, Choi to test complete positivity ( $J\ge0$ ), and Kraus for explicit error-model algebra. The Kraus unitary freedom is the same as the environment-basis choice in Stinespring and the eigenbasis degeneracy in Choi.

Can a channel require infinitely many Kraus operators?

Not in finite dimensions. The Choi matrix of a channel between dimensions $d_A$ and $d_B$ is $d_Ad_B\times d_Ad_B$ , so its rank — the Kraus rank — is at most $d_Ad_B$ , and that many operators always suffice. You may write more (padding with zeros) but never need more, since diagonalising the Choi matrix reduces any representation to the minimal orthogonal one. Infinitely many operators arise only in infinite-dimensional (continuous-variable) settings, where the Choi operator can have infinite rank.

Is an individual Kraus operator something physical that happened?

No. Reading $\mathcal{E}(\rho)=\sum_i K_i\rho K_i^\dagger$ as "error $K_i$ occurred with some probability" is basis-dependent: a different valid set $L_j=\sum_i u_{ji}K_i$ describes the same channel and corresponds to measuring the environment in a rotated basis, a choice no experiment on the system alone can settle. Two people can disagree about "which error happened" yet agree on every prediction. Only basis-independent quantities are real. When a specific unravelling is singled out (a photodetector clicks), that fixes a basis — but then the measurement, not the Kraus form, does the work.

How do I get Kraus operators for a real, measured noise process?

Run process or gate-set tomography to reconstruct the Choi matrix $J$ , then diagonalise it: $J=\sum_i\lambda_i|w_i\rangle\langle w_i|$ . Each eigenvector, rescaled by $\sqrt{\lambda_i}$ and devectorised through the maximally entangled state, is a canonical, Hilbert–Schmidt-orthogonal Kraus operator; the eigenvalues above the noise floor give the effective Kraus rank. Comparing the dominant Kraus operator to the intended gate isolates coherent error, while the smaller eigenvalues quantify decoherence — the split that tells you whether a gate is calibration- or decoherence-limited.