Quantum Channels: Kraus and Stinespring

Derivation

Quantum Channels: Kraus and Stinespring

D-405 Home PU-404 Threads chance · symmetry Depends on Reduced States and the Partial Trace, spectral-theorem-hermitian-operators

Statement

Let \(\mathcal{E}:\mathcal{B}(\mathcal{H}_S)\to\mathcal{B}(\mathcal{H}_S)\) be a linear map on the operators of a finite-dimensional system Hilbert space \(\mathcal{H}_S\) with \(\dim\mathcal{H}_S=d\). If \(\mathcal{E}\) is completely positive and trace-preserving (a CPTP map, or quantum channel), then there exists a finite set of operators \(\{\hat{K}_a\}_{a=1}^{r}\) with \(\hat{K}_a:\mathcal{H}_S\to\mathcal{H}_S\), the Kraus operators, such that \(\mathcal{E}(\hat{\rho})=\sum_{a}\hat{K}_a\hat{\rho}\hat{K}_a^{\dagger}\) for every \(\hat{\rho}\), subject to the completeness relation \(\sum_a\hat{K}_a^{\dagger}\hat{K}_a=\hat{\mathbb{1}}_S\); the number of terms may be taken as \(r\le d^2\). Equivalently (Stinespring dilation) there is an environment \(\mathcal{H}_E\) of dimension \(r\), a fixed pure state \(\lvert e_0\rangle\in\mathcal{H}_E\), and a unitary \(\hat{U}\) on \(\mathcal{H}_S\otimes\mathcal{H}_E\) such that \(\mathcal{E}(\hat{\rho})=\operatorname{Tr}_E\!\big[\hat{U}\,(\hat{\rho}\otimes\lvert e_0\rangle\langle e_0\rvert)\,\hat{U}^{\dagger}\big]\).

Why it matters

The Kraus/Stinespring theorem is the structure theorem for open-system dynamics. It tells us that the most general physically admissible evolution of a quantum system — the map from an input density operator to an output density operator that could ever arise — is exactly the operator-sum form. Nothing more exotic is allowed, and nothing simpler suffices: closed-system unitary evolution is the special case \(r=1\).

Its two faces answer two different questions with one object. The Kraus form is the working representation for noise, measurement back-action, and error correction: each \(\hat{K}_a\) is a distinguishable "thing that could have happened" to the system. The Stinespring form is the ontological statement: every channel is secretly a unitary on a larger world, and irreversibility is only our ignorance of an environment we chose to trace out. The two threads meet here — chance (the stochastic sum over outcomes) and symmetry (the underlying reversible unitary).

Assumptions

Finite dimension.If \(\dim\mathcal{H}_S=\infty\) the Choi operator need not be trace class and the eigen-decomposition step fails; the theorem still holds for normal CP maps but requires the general Stinespring construction with a possibly non-separable dilation, not the elementary spectral argument used here.

Linearity of \(\mathcal{E}\).If the map depends on \(\hat{\rho}\) non-linearly (e.g. mean-field self-interaction), there is no single fixed Choi operator and no operator-sum form; superluminal signalling can appear.

Complete positivity, not mere positivity.If \(\mathcal{E}\) is positive but not completely positive (the transpose map is the canonical example), the Choi operator has negative eigenvalues, \(\hat{K}_a\hat{\rho}\hat{K}_a^{\dagger}\) cannot be assembled, and applying \(\mathcal{E}\otimes\hat{\mathbb{1}}\) to an entangled state yields a non-positive "density matrix".

Trace preservation on all inputs.If \(\operatorname{Tr}\mathcal{E}(\hat{\rho})\neq\operatorname{Tr}\hat{\rho}\) the completeness relation weakens to \(\sum_a\hat{K}_a^{\dagger}\hat{K}_a\le\hat{\mathbb{1}}\) (a trace-non-increasing, "selective" operation); probability is then not conserved and the Stinespring \(\hat{U}\) becomes a mere isometry into a larger output space.

Derivation

The engine is the Choi–Jamiołkowski isomorphism: a linear map on operators is completely encoded by the single operator it produces when fed one half of a maximally entangled state. Complete positivity becomes the ordinary positivity of that operator, to which the spectral theorem then applies.

\[ \lvert\Omega\rangle=\sum_{i=1}^{d}\lvert i\rangle_{A}\otimes\lvert i\rangle_{S}\in\mathcal{H}_A\otimes\mathcal{H}_S,\qquad \mathcal{H}_A\cong\mathcal{H}_S \]

Introduce an ancilla \(A\) that is a copy of the system and an unnormalised maximally entangled vector on \(A\otimes S\). Legal because \(\{\lvert i\rangle\}\) is an orthonormal basis of \(\mathcal{H}_S\). A

\[ \hat{J}(\mathcal{E})\;=\;(\hat{\mathbb{1}}_A\otimes\mathcal{E})\big(\lvert\Omega\rangle\langle\Omega\rvert\big)\;=\;\sum_{i,j=1}^{d}\lvert i\rangle\langle j\rvert_A\otimes\mathcal{E}\big(\lvert i\rangle\langle j\rvert_S\big) \]

Define the Choi operator by acting with \(\mathcal{E}\) on the \(S\)-half only. This is the Choi–Jamiołkowski isomorphism; it is a bijection between linear maps and operators on \(\mathcal{H}_A\otimes\mathcal{H}_S\) because the matrix units \(\lvert i\rangle\langle j\rvert\) span \(\mathcal{B}(\mathcal{H}_S)\). B

\[ \mathcal{E}\text{ completely positive}\;\Longleftrightarrow\;\hat{J}(\mathcal{E})\ge 0 \]

Complete positivity means \((\hat{\mathbb{1}}_A\otimes\mathcal{E})\) maps positive operators to positive operators for an ancilla of any dimension; taking the ancilla equal to the system and the positive input \(\lvert\Omega\rangle\langle\Omega\rvert\) gives \(\hat{J}\ge0\). The converse holds because \(\lvert\Omega\rangle\langle\Omega\rvert\) is (up to normalisation) the "hardest" positive input — this is Choi's theorem. C

\[ \hat{J}(\mathcal{E})=\sum_{a=1}^{r}\lambda_a\,\lvert v_a\rangle\langle v_a\rvert,\qquad \lambda_a>0,\quad \langle v_a\vert v_b\rangle=\delta_{ab},\quad r=\operatorname{rank}\hat{J}\le d^2 \]

Because \(\hat{J}\) is Hermitian and positive semidefinite it has a spectral decomposition (invokes the spectral theorem for Hermitian operators) with non-negative eigenvalues; keep only the \(r\) strictly positive ones. The space \(\mathcal{H}_A\otimes\mathcal{H}_S\) has dimension \(d^2\), bounding \(r\). B

\[ \sqrt{\lambda_a}\,\lvert v_a\rangle=\sum_{i,j}(\hat{K}_a)_{ji}\,\lvert i\rangle_A\otimes\lvert j\rangle_S \;\;\Longleftrightarrow\;\; \hat{K}_a:=\sqrt{\lambda_a}\sum_{i,j}\langle j\vert v_a^{(S)}\text{ for }\lvert i\rangle_A\rangle\,\lvert j\rangle\langle i\rvert \]

Every vector in a tensor product can be written \(\lvert v_a\rangle=\sum_{i,j}(M_a)_{ji}\lvert i\rangle_A\lvert j\rangle_S\); define \(\hat{K}_a=\sqrt{\lambda_a}\,M_a\) as the operator on \(\mathcal{H}_S\) whose matrix elements are \((\hat{K}_a)_{ji}\). This is just reading the coefficient array of \(\lvert v_a\rangle\) as a matrix ("vectorisation"). B

\[ \mathcal{E}\big(\lvert i\rangle\langle j\rvert\big)=\langle i\rvert_A\,\hat{J}(\mathcal{E})\,\lvert j\rangle_A =\sum_a \lambda_a\,\langle i\vert v_a\rangle_A\,\langle v_a\vert j\rangle_A =\sum_a \hat{K}_a\,\lvert i\rangle\langle j\rvert\,\hat{K}_a^{\dagger} \]

Recover the action of \(\mathcal{E}\) on a matrix unit by sandwiching the Choi operator between ancilla basis states — this inverts the isomorphism of step 2. Substituting the spectral form and the definition of \(\hat{K}_a\) collapses the ancilla indices into the operator-sum sandwich. C

\[ \hat{\rho}=\sum_{i,j}\rho_{ij}\,\lvert i\rangle\langle j\rvert \;\;\Longrightarrow\;\; \mathcal{E}(\hat{\rho})=\sum_{i,j}\rho_{ij}\sum_a \hat{K}_a\lvert i\rangle\langle j\rvert\hat{K}_a^{\dagger}=\sum_{a=1}^{r}\hat{K}_a\,\hat{\rho}\,\hat{K}_a^{\dagger} \]

Extend from matrix units to a general \(\hat{\rho}\) by linearity of \(\mathcal{E}\). This is the operator-sum (Kraus) form. A

\[ \operatorname{Tr}\mathcal{E}(\hat{\rho})=\operatorname{Tr}\!\Big(\sum_a\hat{K}_a^{\dagger}\hat{K}_a\,\hat{\rho}\Big)=\operatorname{Tr}\hat{\rho}\;\;\forall\hat{\rho} \;\;\Longleftrightarrow\;\; \sum_{a=1}^{r}\hat{K}_a^{\dagger}\hat{K}_a=\hat{\mathbb{1}}_S \]

Impose trace preservation. Using cyclicity of the trace, \(\operatorname{Tr}(\hat{K}_a\hat{\rho}\hat{K}_a^\dagger)=\operatorname{Tr}(\hat{K}_a^\dagger\hat{K}_a\hat{\rho})\); equality of traces for all \(\hat{\rho}\) forces the bracketed operator to be the identity. B

\[ \hat{V}\lvert\psi\rangle_S:=\sum_{a=1}^{r}\big(\hat{K}_a\lvert\psi\rangle_S\big)\otimes\lvert a\rangle_E,\qquad \{\lvert a\rangle_E\}\text{ orthonormal in }\mathcal{H}_E,\;\dim\mathcal{H}_E=r \]

Assemble the Kraus operators into a single map \(\hat{V}:\mathcal{H}_S\to\mathcal{H}_S\otimes\mathcal{H}_E\). Legal: the \(\lvert a\rangle_E\) are chosen orthonormal, so \(\hat{V}\) is well defined. B

\[ \hat{V}^{\dagger}\hat{V}=\sum_{a,b}\langle a\vert b\rangle_E\,\hat{K}_a^{\dagger}\hat{K}_b=\sum_a\hat{K}_a^{\dagger}\hat{K}_a=\hat{\mathbb{1}}_S \;\;\Longrightarrow\;\; \hat{V}\text{ is an isometry} \]

Compute \(\hat V^\dagger\hat V\) using orthonormality of the \(\lvert a\rangle_E\); the completeness relation of step 8 makes it the identity, so \(\hat{V}\) preserves inner products (isometry). An isometry into a space of equal-or-larger dimension extends to a unitary. C

\[ \hat{U}\big(\lvert\psi\rangle_S\otimes\lvert e_0\rangle_E\big):=\hat{V}\lvert\psi\rangle_S,\qquad \hat{U}\text{ extended to a unitary on }\mathcal{H}_S\otimes\mathcal{H}_E \]

Fix a reference environment state \(\lvert e_0\rangle\). On the subspace \(\mathcal{H}_S\otimes\lvert e_0\rangle\) define \(\hat U\) to equal the isometry \(\hat V\); since \(\hat V\) maps an isometric copy of that subspace, Gram–Schmidt extends it to a full unitary on the whole tensor space. C

\[ \operatorname{Tr}_E\!\big[\hat{U}(\hat{\rho}\otimes\lvert e_0\rangle\langle e_0\rvert)\hat{U}^{\dagger}\big] =\operatorname{Tr}_E\!\big[\hat{V}\hat{\rho}\hat{V}^{\dagger}\big] =\sum_{a,b}\hat{K}_a\hat{\rho}\hat{K}_b^{\dagger}\,\langle b\vert a\rangle_E =\sum_a\hat{K}_a\hat{\rho}\hat{K}_a^{\dagger}=\mathcal{E}(\hat{\rho}) \]

Evolve the enlarged state unitarily and trace out the environment (uses reduced-density-matrix / partial-trace). Orthonormality of \(\{\lvert a\rangle_E\}\) collapses the double sum to the diagonal, reproducing the Kraus form — closing the equivalence. C

Result

\[ \boxed{\;\mathcal{E}(\hat{\rho})=\sum_{a=1}^{r}\hat{K}_a\hat{\rho}\hat{K}_a^{\dagger},\quad \sum_a\hat{K}_a^{\dagger}\hat{K}_a=\hat{\mathbb{1}}_S \;\;\Longleftrightarrow\;\; \mathcal{E}(\hat{\rho})=\operatorname{Tr}_E\!\big[\hat{U}(\hat{\rho}\otimes\lvert e_0\rangle\langle e_0\rvert)\hat{U}^{\dagger}\big]\;}\]

Reading. Any physical evolution of a quantum system is a stochastic sum of "branches" \(\hat{K}_a\), each a possible influence of the environment; the completeness relation says the branch probabilities sum to one. Equivalently, the system was never really open — it was one factor of a larger world evolving by a single unitary \(\hat{U}\), and the apparent randomness is our tracing away of the environment. The minimal number of Kraus operators, \(r=\operatorname{rank}\hat J\le d^2\), is the Kraus rank, the smallest environment that can generate the channel.

Units check. Density operators, Kraus operators, and \(\hat U\) are all dimensionless (they act on normalised state vectors). Each term \(\hat{K}_a\hat{\rho}\hat{K}_a^{\dagger}\) has the units of \(\hat{\rho}\), so the sum does too; \(\operatorname{Tr}\mathcal{E}(\hat\rho)=\operatorname{Tr}\hat\rho=1\) is a pure number, consistent with a probability. The completeness relation equates two dimensionless operators. All quantities are unit-free, as required for objects measured in probability.

Limiting cases

Single Kraus operator \(r=1\). Completeness forces \(\hat{K}_1^{\dagger}\hat{K}_1=\hat{\mathbb{1}}\Rightarrow\hat{K}_1=\hat{U}\) unitary; the channel reduces to closed reversible evolution \(\mathcal{E}(\hat\rho)=\hat U\hat\rho\hat U^\dagger\). The environment stays factored out and unentangled.
Unitary-mixture (random-unitary) channel. \(\hat{K}_a=\sqrt{p_a}\,\hat{U}_a\) with \(\sum_a p_a=1\): the system undergoes \(\hat{U}_a\) with classical probability \(p_a\). This is the intuitive "chance" picture; it is a strict subset of all channels for \(d\ge3\).
Measurement (selective) branch. Keeping a single outcome \(a\) gives the non-trace-preserving update \(\hat\rho\mapsto\hat{K}_a\hat\rho\hat{K}_a^{\dagger}/\operatorname{Tr}(\cdot)\), the generalised (POVM) measurement rule with effect \(\hat{E}_a=\hat{K}_a^{\dagger}\hat{K}_a\).
Complete decoherence. \(\hat{K}_a=\lvert a\rangle\langle a\rvert\) projects onto a basis; \(\mathcal{E}(\hat\rho)=\sum_a\lvert a\rangle\langle a\rvert\hat\rho\lvert a\rangle\langle a\rvert\) deletes all off-diagonal coherences, the classical limit.

Breaks when

The map is positive but not completely positive. The partial transpose \(\hat\rho\mapsto\hat\rho^{T}\) sends valid single-system states to valid states, yet \(\hat J\) has a negative eigenvalue, so no real \(\hat{K}_a\) exist. Applied to half of a Bell pair it produces an operator with a negative eigenvalue — not a physical state. No Kraus/Stinespring form exists.
Infinite-dimensional systems without normality/trace-class control. For bosonic modes the Choi "operator" \(\lvert\Omega\rangle\langle\Omega\rvert\) is not trace class and step 4's discrete spectral sum can diverge; a naive finite Kraus decomposition fails and one must use the general (continuous) Stinespring dilation.
Non-linear or time-non-local dynamics. Mean-field or feedback evolutions where \(\mathcal{E}\) depends on \(\hat\rho\) itself have no fixed Choi operator; the operator-sum form does not apply and, if forced, permits acausal signalling.
Initial system–environment correlations. The derivation assumes a fixed product input \(\hat\rho\otimes\lvert e_0\rangle\langle e_0\rvert\). If the system is already correlated with its environment before the unitary, the reduced dynamics need not even be a positive map, let alone CPTP.

Failure modes

Forgetting completeness. Writing \(\sum_a\hat{K}_a\hat{K}_a^{\dagger}=\hat{\mathbb{1}}\) (wrong order). Trace preservation constrains \(\hat{K}_a^{\dagger}\hat{K}_a\); \(\sum_a\hat{K}_a\hat{K}_a^{\dagger}=\hat{\mathbb{1}}\) is instead the condition for the channel to be unital (fixes the maximally mixed state), a different property.
Believing Kraus operators are unique. They are fixed only up to a right multiplication by an isometry: \(\hat{K}'_b=\sum_a W_{ba}\hat{K}_a\) with \(W^\dagger W=\mathbb{1}\) gives the same channel. Students who "read off" physical outcomes from one particular \(\{\hat K_a\}\) forget this gauge freedom.
Confusing complete positivity with positivity. Checking \(\mathcal{E}(\hat\rho)\ge0\) only for single-system inputs and concluding the map is a channel; the transpose passes this test but is not CP.
Treating the environment dimension as physical. Assuming the real bath must have dimension \(r\); \(r\) is only the minimal mathematical environment. The true bath may be far larger, and any dilation with \(\dim\mathcal{H}_E\ge r\) reproduces the same channel.
Normalisation slip in \(\lvert\Omega\rangle\). Using the normalised \(\tfrac{1}{\sqrt d}\sum_i\lvert ii\rangle\) and then dropping the compensating factor of \(d\) in the recovery step 6, yielding \(\mathcal{E}\) that is off by \(1/d\).

Discussion

The theorem unifies three notions that look distinct in elementary quantum mechanics: unitary evolution, measurement, and noise. All three are single points in the same convex set of CPTP maps. Unitary evolution is the extreme case of Kraus rank one; a projective measurement (unread) is a channel with orthogonal projectors as Kraus operators; decoherence is a channel that shrinks off-diagonal terms. The Stinespring dilation is what glues them: it says every one of these is a unitary on a bigger space followed by a partial trace, so the "collapse" and "noise" that seem to violate reversibility are bookkeeping for entanglement we declined to track.

The freedom in the decomposition is physically important, not a blemish. Two Kraus sets related by \(\hat{K}'_b=\sum_a W_{ba}\hat{K}_a\) (an isometry \(W\)) describe the same channel but correspond to different measurements the environment could perform. This "unravelling" freedom is the mathematical root of quantum trajectory theory and of the fact that an eavesdropper's information about a channel is basis-dependent — central to quantum key distribution and to the theory of complementary channels.

Complete positivity — not mere positivity — is the crux, and it is subtle enough to be worth dwelling on. A map can preserve the positivity of every state you feed it in isolation and still be unphysical, because in a quantum world your system may be entangled with a spectator it never interacts with. The transpose map is the standard warning: it is a perfectly good operation on classical-like correlations but turns a Bell pair into a pseudo-state with negative probability. Complete positivity is precisely the demand that the map stay physical no matter what the system is secretly entangled with, and Choi's theorem compresses that infinite family of tests into the single condition \(\hat J\ge0\).

At a deeper level the Choi–Jamiołkowski isomorphism is a manifestation of channel–state duality: the space of channels \(S\to S\) is isomorphic (as a convex set, preserving the CP cone) to the space of bipartite states on \(A\otimes S\) with fixed reduced state on \(A\). This turns dynamical questions into static entanglement questions — channel capacities become entanglement measures of the Choi state, and the structure of the CP cone becomes the structure of separable-versus-entangled states. The Stinespring \(\hat U\) is unique up to a unitary on the environment (Stinespring's uniqueness theorem), which is exactly the isometric freedom \(W\) seen in the Kraus picture; the minimal dilation, of dimension \(r=\operatorname{rank}\hat J\), is the analogue of the GNS construction for completely positive maps on operator algebras.

Common misconceptions. (i) "The environment must be measured for the channel to act" — no; tracing out, i.e. simply ignoring the environment, already produces the full channel. (ii) "The Kraus operators are the physical measurement outcomes" — only up to isometric mixing; the outcomes are not uniquely defined by the channel. (iii) "A channel is just a probabilistic mixture of unitaries" — true only for the random-unitary subclass; for \(d\ge3\) most channels (e.g. amplitude damping) are not unitary mixtures.

Worked examples

Example 1 — Amplitude damping (spontaneous emission of a qubit) and its dilation.

\[ \hat{K}_0=\begin{pmatrix}1&0\\[2pt]0&\sqrt{1-\gamma}\end{pmatrix},\qquad \hat{K}_1=\begin{pmatrix}0&\sqrt{\gamma}\\[2pt]0&0\end{pmatrix},\qquad \gamma\in[0,1] \]

Model a two-level atom decaying \(\lvert1\rangle\to\lvert0\rangle\) with probability \(\gamma\); \(\hat K_1\) is the "a photon was emitted" branch, \(\hat K_0\) the "no emission" branch (which still deforms the state). A

\[ \hat{K}_0^{\dagger}\hat{K}_0+\hat{K}_1^{\dagger}\hat{K}_1=\begin{pmatrix}1&0\\0&1-\gamma\end{pmatrix}+\begin{pmatrix}0&0\\0&\gamma\end{pmatrix}=\hat{\mathbb{1}} \]

Verify the completeness relation (step 8) — the channel is trace-preserving. A

\[ \hat\rho=\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\;\Longrightarrow\; \mathcal{E}(\hat\rho)=\begin{pmatrix}\rho_{00}+\gamma\rho_{11}&\sqrt{1-\gamma}\,\rho_{01}\\[2pt]\sqrt{1-\gamma}\,\rho_{10}&(1-\gamma)\rho_{11}\end{pmatrix} \]

Apply \(\mathcal{E}(\hat\rho)=\hat K_0\hat\rho\hat K_0^\dagger+\hat K_1\hat\rho\hat K_1^\dagger\). Population flows \(\rho_{11}\to\rho_{00}\); coherences shrink by \(\sqrt{1-\gamma}\). B

\[ \hat U\lvert 0\rangle_S\lvert0\rangle_E=\lvert0\rangle_S\lvert0\rangle_E,\qquad \hat U\lvert1\rangle_S\lvert0\rangle_E=\sqrt{1-\gamma}\,\lvert1\rangle_S\lvert0\rangle_E+\sqrt{\gamma}\,\lvert0\rangle_S\lvert1\rangle_E \]

Read the Stinespring dilation off the Kraus operators via \(\hat V\lvert\psi\rangle=\sum_a(\hat K_a\lvert\psi\rangle)\lvert a\rangle_E\) (steps 9–11): \(\lvert1\rangle_E\) is "the environment holds the emitted photon." Extend to a full \(2\times2\) block unitary on the \(\{\lvert1,0\rangle,\lvert0,1\rangle\}\) subspace. C

\[ \gamma=\tfrac12,\;\hat\rho=\lvert1\rangle\langle1\rvert:\quad \mathcal{E}(\hat\rho)=\begin{pmatrix}0.5&0\\0&0.5\end{pmatrix};\qquad \gamma=0.1,\;\hat\rho=\lvert+\rangle\langle+\rvert:\quad \mathcal{E}=\begin{pmatrix}0.55&0.474\\0.474&0.45\end{pmatrix} \]

Plug numbers. For an excited atom with \(\gamma=0.5\) the output is maximally mixed. For \(\lvert+\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle)\) with \(\gamma=0.1\): \(\rho_{00}=0.5+0.1(0.5)=0.55\), off-diagonal \(=0.5\sqrt{0.9}=0.474\). B

\[ \mathcal{E}(\lvert+\rangle\langle+\rvert)\big|_{\gamma=0.1}=\begin{pmatrix}0.55&0.4743\\0.4743&0.45\end{pmatrix},\quad \operatorname{Tr}=1 \]

Reading. A weak damping of \(\gamma=0.1\) leaves populations nearly balanced but has already begun both to bias the atom toward \(\lvert0\rangle\) and to erode the coherence (\(0.5\to0.474\)). The purity \(\operatorname{Tr}\hat\rho^2=0.55^2+0.45^2+2(0.4743)^2=0.955<1\): the state is no longer pure because information has leaked into the environment photon.

Example 2 — Single-qubit bit-flip channel: minimal dilation and unitary form.

\[ \mathcal{E}(\hat\rho)=(1-p)\,\hat\rho+p\,\hat X\hat\rho\hat X,\qquad \hat X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\;p\in[0,1] \]

A qubit is flipped with probability \(p\). This is a random-unitary channel, so the Kraus operators are proportional to unitaries. A

\[ \hat{K}_0=\sqrt{1-p}\,\hat{\mathbb{1}},\qquad \hat{K}_1=\sqrt{p}\,\hat X,\qquad \hat K_0^\dagger\hat K_0+\hat K_1^\dagger\hat K_1=(1-p)\hat{\mathbb{1}}+p\hat{\mathbb{1}}=\hat{\mathbb{1}} \]

Identify the Kraus operators and confirm completeness. Kraus rank \(r=2\Rightarrow\dim\mathcal{H}_E=2\), one environment qubit. A

\[ \hat U(\lvert\psi\rangle_S\otimes\lvert0\rangle_E)=\sqrt{1-p}\,\lvert\psi\rangle_S\lvert0\rangle_E+\sqrt{p}\,(\hat X\lvert\psi\rangle_S)\lvert1\rangle_E \]

Build the isometry \(\hat V\) (step 9). This is a controlled-\(X\) with the control on the environment prepared in \(\sqrt{1-p}\lvert0\rangle+\sqrt p\lvert1\rangle\) — indeed \(\hat U=\text{CNOT}_{E\to S}\big(\hat{\mathbb1}\otimes\hat R\big)\) with \(\hat R\lvert0\rangle=\sqrt{1-p}\lvert0\rangle+\sqrt p\lvert1\rangle\). C

\[ p=0.2,\;\hat\rho=\lvert+\rangle\langle+\rvert:\quad \hat X\lvert+\rangle=\lvert+\rangle\;\Rightarrow\;\mathcal{E}(\lvert+\rangle\langle+\rvert)=\lvert+\rangle\langle+\rvert \]

Numbers: \(\lvert+\rangle\) is the \(+1\) eigenstate of \(\hat X\), so the bit-flip channel leaves it invariant. B

\[ p=0.2,\;\hat\rho=\lvert0\rangle\langle0\rvert:\quad \mathcal{E}=0.8\lvert0\rangle\langle0\rvert+0.2\lvert1\rangle\langle1\rvert=\begin{pmatrix}0.8&0\\0&0.2\end{pmatrix} \]

A computational-basis state is instead mixed: 80% survives, 20% flips. A

\[ \mathcal{E}(\lvert0\rangle\langle0\rvert)\big|_{p=0.2}=\begin{pmatrix}0.8&0\\0&0.2\end{pmatrix},\qquad \mathcal{E}(\lvert+\rangle\langle+\rvert)=\lvert+\rangle\langle+\rvert \]

Reading. The channel is diagonal in the \(\hat X\)-eigenbasis: states aligned with the flip axis (\(\lvert\pm\rangle\)) are untouched, while states orthogonal to it (\(\lvert0\rangle,\lvert1\rangle\)) decohere toward the mixed state. The Stinespring picture makes the mechanism explicit — the flip axis is exactly the basis in which system and environment do not become entangled.

Problems

Show that the depolarising channel \(\mathcal{E}(\hat\rho)=(1-p)\hat\rho+\frac{p}{3}(\hat X\hat\rho\hat X+\hat Y\hat\rho\hat Y+\hat Z\hat\rho\hat Z)\) has Kraus operators \(\sqrt{1-p}\,\hat{\mathbb1},\ \sqrt{p/3}\,\hat X,\ \sqrt{p/3}\,\hat Y,\ \sqrt{p/3}\,\hat Z\), and verify completeness. What is its Kraus rank?
Solution
Writing the map as \(\sum_a\hat K_a\hat\rho\hat K_a^\dagger\) with \(\hat K_0=\sqrt{1-p}\,\hat{\mathbb1}\) and \(\hat K_{1,2,3}=\sqrt{p/3}\,\{\hat X,\hat Y,\hat Z\}\) reproduces each term since the Paulis are Hermitian, \(\hat X\hat\rho\hat X^\dagger=\hat X\hat\rho\hat X\). Completeness: \(\sum_a\hat K_a^\dagger\hat K_a=(1-p)\hat{\mathbb1}+\frac{p}{3}(\hat X^2+\hat Y^2+\hat Z^2)=(1-p)\hat{\mathbb1}+\frac{p}{3}(3\hat{\mathbb1})=\hat{\mathbb1}\), using \(\hat X^2=\hat Y^2=\hat Z^2=\hat{\mathbb1}\). Four Kraus operators, all linearly independent (the Paulis plus identity span \(\mathcal{B}(\mathbb{C}^2)\)), so Kraus rank \(r=4=d^2\) — the depolarising channel is maximal-rank.
The Choi operator of a channel on a qubit is \(\hat J=\lvert\Phi^+\rangle\langle\Phi^+\rvert\) with \(\lvert\Phi^+\rangle=\lvert00\rangle+\lvert11\rangle\) (unnormalised). Identify the channel and its Kraus rank.
Solution
\(\hat J\) is rank one and positive, so Kraus rank \(r=1\): the channel is unitary. Vectorising \(\lvert\Phi^+\rangle=\sum_{i}\lvert i\rangle_A\lvert i\rangle_S\) gives the coefficient matrix \((\hat K_1)_{ji}=\delta_{ji}\), i.e. \(\hat K_1=\hat{\mathbb1}\). Hence \(\mathcal{E}(\hat\rho)=\hat\rho\): it is the identity channel. (One checks completeness \(\hat K_1^\dagger\hat K_1=\hat{\mathbb1}\).) A single positive eigenvalue of \(\hat J\) always signals reversible, closed evolution.
Prove the unitary-freedom theorem in one direction: if \(\hat K'_b=\sum_a W_{ba}\hat K_a\) with \(\sum_b W^*_{ba}W_{bc}=\delta_{ac}\), then \(\{\hat K'_b\}\) generates the same channel and satisfies completeness.
Solution
Same channel: \(\sum_b\hat K'_b\hat\rho\hat K'^\dagger_b=\sum_b\sum_{a,c}W_{ba}W^*_{bc}\hat K_a\hat\rho\hat K_c^\dagger=\sum_{a,c}\big(\sum_b W^*_{bc}W_{ba}\big)\hat K_a\hat\rho\hat K_c^\dagger=\sum_{a,c}\delta_{ca}\hat K_a\hat\rho\hat K_c^\dagger=\sum_a\hat K_a\hat\rho\hat K_a^\dagger=\mathcal{E}(\hat\rho)\), using \(W^\dagger W=\mathbb1\). Completeness: \(\sum_b\hat K'^\dagger_b\hat K'_b=\sum_{a,c}\big(\sum_b W^*_{ba}W_{bc}\big)\hat K_a^\dagger\hat K_c=\sum_{a,c}\delta_{ac}\hat K_a^\dagger\hat K_c=\sum_a\hat K_a^\dagger\hat K_a=\hat{\mathbb1}\). Both properties are preserved, proving the Kraus set is fixed only up to an isometry \(W\) (padding with zero operators lets \(W\) be rectangular).
For the amplitude-damping channel of Example 1, compute the purity \(\operatorname{Tr}\mathcal{E}(\hat\rho)^2\) for input \(\hat\rho=\lvert1\rangle\langle1\rvert\) as a function of \(\gamma\), and find the \(\gamma\) that minimises it.
Solution
From Example 1, \(\mathcal{E}(\lvert1\rangle\langle1\rvert)=\operatorname{diag}(\gamma,1-\gamma)\) (a diagonal state: \(\rho_{00}=\gamma,\ \rho_{11}=1-\gamma,\ \rho_{01}=0\)). Purity \(=\gamma^2+(1-\gamma)^2=1-2\gamma(1-\gamma)\). Minimise: \(\frac{d}{d\gamma}[1-2\gamma+2\gamma^2]=-2+4\gamma=0\Rightarrow\gamma=\tfrac12\), giving purity \(=\tfrac14+\tfrac14=\tfrac12\), the maximally mixed value for a qubit. So an initially excited atom loses the most information (about itself) at half-decay, when the emitted-photon environment is maximally entangled with it.
A channel has Kraus operators \(\hat K_0=\begin{pmatrix}1&0\\0&\cos\theta\end{pmatrix}\), \(\hat K_1=\begin{pmatrix}0&0\\0&\sin\theta\end{pmatrix}\). Is it trace-preserving? Is it unital (does it fix \(\hat{\mathbb1}/2\))? Give the output for \(\hat\rho=\hat{\mathbb1}/2\) at \(\theta=\pi/3\).
Solution
Trace-preserving: \(\hat K_0^\dagger\hat K_0+\hat K_1^\dagger\hat K_1=\operatorname{diag}(1,\cos^2\theta)+\operatorname{diag}(0,\sin^2\theta)=\operatorname{diag}(1,1)=\hat{\mathbb1}\). Yes. Unital: \(\hat K_0\hat K_0^\dagger+\hat K_1\hat K_1^\dagger=\operatorname{diag}(1,\cos^2\theta)+\operatorname{diag}(0,\sin^2\theta)=\hat{\mathbb1}\)? That equals \(\operatorname{diag}(1,1)=\hat{\mathbb1}\) only because the operators are diagonal here — compute: \(\hat K_0\hat K_0^\dagger=\operatorname{diag}(1,\cos^2\theta)\), \(\hat K_1\hat K_1^\dagger=\operatorname{diag}(0,\sin^2\theta)\), sum \(=\hat{\mathbb1}\). So it is unital and fixes \(\hat{\mathbb1}/2\). Check directly at \(\theta=\pi/3\): \(\mathcal{E}(\hat{\mathbb1}/2)=\tfrac12[\hat K_0\hat K_0^\dagger+\hat K_1\hat K_1^\dagger]=\tfrac12\hat{\mathbb1}=\operatorname{diag}(0.5,0.5)\). The maximally mixed state is unchanged, confirming unitality.