Stinespring Representation

A Stinespring representation writes a quantum channel as reversible dynamics on a larger Hilbert space followed by forgetting an environment. In finite dimensions, every completely positive trace-preserving map can be written this way.

For a channel

$$\Phi: \mathcal B(\mathcal H_{\mathrm{in}}) \to \mathcal B(\mathcal H_{\mathrm{out}}),$$

the channel-side form is

$$\Phi(\rho) = \operatorname{Tr}_E \!\left[ V\rho V^\dagger \right], \qquad V^\dagger V=I_{\mathrm{in}},$$

where

$$V: \mathcal H_{\mathrm{in}} \to \mathcal H_{\mathrm{out}}\otimes\mathcal H_E$$

is an isometry. Equivalently, when the input and output systems are embedded in a larger common space, one may prepare an environment state, apply a unitary, and trace out the environment:

$$\Phi(\rho) = \operatorname{Tr}_E \!\left[ U(\rho\otimes\eta_E)U^\dagger \right].$$

The proof of the theorem and its measurement-side interpretation live at Stinespring Dilation. This page is the quantum-channel working version: how to read the representation, build it from Kraus operators, and use it without overinterpreting the environment.

Kraus, Choi, and Stinespring Views compares minimal and padded Stinespring realizations with Kraus and Choi data and separates discarded from measured environment records; this page retains the channel-side construction, unitary extension, environment interpretation, minimality, and worked dilations.

Why This Representation Matters

The representation explains why completely positive maps are the right finite-time language for open quantum systems. A closed composite system can evolve unitarily, while a subsystem evolves nonunitarily because information, phase relations, or population have become correlated with degrees of freedom that are not retained.

The slogan is accurate but needs qualification:

$$\text{channel} \quad=\quad \text{unitary interaction} \quad+\quad \text{discarded environment}.$$

It is a representation theorem, not always a microscopic derivation. A mathematical environment in a minimal representation may encode only the noise alternatives needed to reproduce $\Phi$; it need not be the literal bath of a laboratory system.

For the reduced-dynamics route from a physical system-environment Hamiltonian, see Reduced Dynamics. For the operator-sum route, see Kraus Representation.

From Kraus Operators to an Isometry

Suppose a channel has Kraus representation

$$\Phi(\rho) = \sum_{\alpha=1}^r K_\alpha\rho K_\alpha^\dagger, \qquad \sum_{\alpha=1}^r K_\alpha^\dagger K_\alpha = I_{\mathrm{in}}.$$

Choose an environment with orthonormal basis

$$\{\lvert\alpha\rangle_E\}_{\alpha=1}^r.$$

Define

$$V\lvert\psi\rangle = \sum_{\alpha=1}^r K_\alpha\lvert\psi\rangle \otimes \lvert\alpha\rangle_E.$$

Then

$$\begin{aligned} V^\dagger V &= \sum_{\alpha=1}^r K_\alpha^\dagger K_\alpha\\ &= I_{\mathrm{in}}, \end{aligned}$$

so $V$ is an isometry. Tracing out the environment gives back the channel:

$$\begin{aligned} \operatorname{Tr}_E \!\left[ V\rho V^\dagger \right] &= \operatorname{Tr}_E \!\left[ \sum_{\alpha,\beta} K_\alpha\rho K_\beta^\dagger \otimes \lvert\alpha\rangle\langle\beta\rvert \right]\\ &= \sum_{\alpha} K_\alpha\rho K_\alpha^\dagger. \end{aligned}$$

This construction is often the fastest way to build a dilation for a finite-dimensional channel.

From an Isometry to a Unitary

An isometry preserves inner products but maps into a larger space. If the output-environment space has enough unused dimensions, the action of $V$ on the input subspace can be extended to a unitary on the whole larger space.

In the common same-system case $\mathcal H_{\mathrm{in}}=\mathcal H_{\mathrm{out}}=\mathcal H_S$, choose a fixed environment state $\lvert0\rangle_E$ and define a unitary $U$ satisfying

$$U \left( \lvert\psi\rangle_S\otimes\lvert0\rangle_E \right) = V\lvert\psi\rangle_S$$

for every $\lvert\psi\rangle_S$. Then

$$\Phi(\rho) = \operatorname{Tr}_E \!\left[ U \left( \rho\otimes\lvert0\rangle\langle0\rvert_E \right) U^\dagger \right].$$

If the environment begins in a mixed state

$$\eta_E=\sum_j q_j\lvert j\rangle\langle j\rvert,$$

one can purify $\eta_E$ by adding another reference system, or equivalently absorb the classical label $j$ into a larger environment. This is why a pure environment input is sufficient for the representation theorem, even when a microscopic model uses a thermal environment.

Environment Records and Kraus Freedom

The environment basis used above turns Kraus labels into orthogonal records. If the environment is ignored, those alternatives add incoherently and produce the channel.

If instead the environment is measured in the basis $\{\lvert\alpha\rangle_E\}$, the branches are

$$\mathcal I_\alpha(\rho) = K_\alpha\rho K_\alpha^\dagger,$$

with branch probabilities

$$p_\alpha(\rho) = \operatorname{Tr} \!\left( K_\alpha\rho K_\alpha^\dagger \right).$$

Changing the environment measurement basis mixes the Kraus operators by a unitary matrix. Thus different Kraus representations of the same channel can correspond to different ways of monitoring the same environment, or merely to different mathematical decompositions.

The channel $\Phi$ is the invariant object. Individual Kraus operators acquire direct physical meaning only after a concrete measurement record, environment basis, or experimental unraveling has been specified. This point is essential in quantum instruments and in trajectory descriptions.

Minimal Environment Dimension

The smallest possible environment dimension in a Stinespring representation of a finite-dimensional channel equals the minimal number of Kraus operators. Equivalently,

$$d_E^{\min} = \operatorname{rank}J_\Phi,$$

where $J_\Phi$ is the Choi matrix.

This rank is sometimes called the Choi rank or Kraus rank. It measures how many independent noise alternatives are needed to represent the channel.

Examples:

channel	typical Kraus rank	minimal environment reading
unitary channel	$1$	no irreducible environment record
generic qubit dephasing	$2$	two phase alternatives
qubit amplitude damping with $0\lt\gamma\lt1$	$2$	no-jump and jump alternatives
completely depolarizing qudit	$d^2$	full operator-basis randomization

The minimal environment is representation efficient, not necessarily physically complete. A real bath may have many more degrees of freedom than $d_E^{\min}$.

Example: Unitary Channel

For a unitary channel

$$\Phi_U(\rho) = U_S\rho U_S^\dagger,$$

one may take a one-dimensional environment and

$$V=U_S.$$

There is no unavoidable discarded record. The channel is reversible, has Kraus rank $1$, and maps pure states to pure states. If a larger environment is included but factorizes again at the end, the reduced channel can still be unitary.

Example: Dephasing from an Environment Record

Let a qubit system interact with an environment initially in $\lvert e_0\rangle$ by

$$\begin{aligned} \lvert0\rangle\lvert e_0\rangle &\longmapsto \lvert0\rangle\lvert e_0'\rangle,\\ \lvert1\rangle\lvert e_0\rangle &\longmapsto \lvert1\rangle\lvert e_1'\rangle. \end{aligned}$$

For an input density matrix

$$\rho = \begin{pmatrix} \rho_{00}&\rho_{01}\\ \rho_{10}&\rho_{11} \end{pmatrix},$$

tracing out the environment gives

$$\rho \longmapsto \begin{pmatrix} \rho_{00} & c\,\rho_{01}\\ c^*\,\rho_{10} & \rho_{11} \end{pmatrix}, \qquad c=\langle e_1'|e_0'\rangle.$$

If the two environment states are nearly identical, $|c|\approx1$ and little dephasing occurs. If they are orthogonal, $c=0$ and the environment carries perfect which-alternative information. This is the channel-level bridge to decoherence and the canonical dephasing channel.

Example: Amplitude Damping

For the zero-temperature qubit amplitude-damping channel, one Stinespring isometry is

$$\begin{aligned} \lvert0\rangle\lvert0\rangle_E &\longmapsto \lvert0\rangle\lvert0\rangle_E,\\ \lvert1\rangle\lvert0\rangle_E &\longmapsto \sqrt{1-\gamma}\, \lvert1\rangle\lvert0\rangle_E + \sqrt{\gamma}\, \lvert0\rangle\lvert1\rangle_E. \end{aligned}$$

The environment state $\lvert1\rangle_E$ records that a decay quantum was emitted. Tracing out the environment yields Kraus operators

$$K_0 = \begin{pmatrix} 1&0\\ 0&\sqrt{1-\gamma} \end{pmatrix}, \qquad K_1 = \begin{pmatrix} 0&\sqrt{\gamma}\\ 0&0 \end{pmatrix}.$$

This representation makes clear why amplitude damping is not a Pauli channel: the process exchanges energy and pulls the Bloch ball toward the ground state. See Amplitude-Damping Channel for the full channel analysis.

Common Mistakes

• Treating the Stinespring environment as uniquely determined by the channel. Minimal dilations are unique only up to an appropriate isometry; nonminimal environments can be much larger.
• Giving physical meaning to a Kraus label without specifying how the environment or apparatus is monitored.
• Assuming every open-system derivation starts from a pure environment state. A pure environment is enough for a representation, while realistic baths are often mixed or thermal.
• Confusing a channel representation with a Markovian master equation. A single channel need not belong to a time-homogeneous semigroup.
• Forgetting the initial-correlation assumption in microscopic reduced dynamics. A fixed channel on the system alone generally requires a fixed environment state independent of the system input.

References

• W. F. Stinespring, “Positive functions on C*-algebras,” Proceedings of the American Mathematical Society 6, 211-216 (1955).
• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285-290 (1975).
• K. Kraus, States, Effects, and Operations, Springer (1983).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press (2010).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).

Exercises

1. Recover the Kraus form. Starting from $V\lvert\psi\rangle=\sum_\alpha K_\alpha\lvert\psi\rangle\otimes\lvert\alpha\rangle_E$, show explicitly that $\operatorname{Tr}_E[V\rho V^\dagger]=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$.

Solution

For an arbitrary density operator $\rho$,

$$V\rho V^\dagger = \sum_{\alpha,\beta} K_\alpha\rho K_\beta^\dagger \otimes \lvert\alpha\rangle\langle\beta\rvert.$$

Using $\operatorname{Tr}_E(\lvert\alpha\rangle\langle\beta\rvert)=\delta_{\alpha\beta}$ gives

$$\operatorname{Tr}_E[V\rho V^\dagger] = \sum_{\alpha,\beta} K_\alpha\rho K_\beta^\dagger \delta_{\alpha\beta} = \sum_\alpha K_\alpha\rho K_\alpha^\dagger.$$

2. Trace preservation and isometry. Show that the map built from $V\lvert\psi\rangle=\sum_\alpha K_\alpha\lvert\psi\rangle\otimes\lvert\alpha\rangle_E$ is trace preserving exactly when $V^\dagger V=I_{\mathrm{in}}$.

Solution

The adjoint product is

$$V^\dagger V = \sum_\alpha K_\alpha^\dagger K_\alpha.$$

For the corresponding channel,

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

This equals $\operatorname{Tr}\rho$ for every $\rho$ exactly when

$$\sum_\alpha K_\alpha^\dagger K_\alpha=I_{\mathrm{in}},$$

which is the same condition as $V^\dagger V=I_{\mathrm{in}}$.

3. Environment overlap and dephasing. In the dephasing example, compute the output of the input state $(\lvert0\rangle+\lvert1\rangle)/\sqrt2$ when $\langle e_1'|e_0'\rangle=c$.

Solution

The input density matrix is

$$\rho = \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.$$

After tracing out the environment,

$$\rho' = \frac12 \begin{pmatrix} 1&c\\ c^*&1 \end{pmatrix}.$$

The populations remain $1/2$, while the coherence is multiplied by $c$. Orthogonal environment records give $c=0$ and complete dephasing in this basis.