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 a fixed environment state independent of and initially uncorrelated with the input, apply a unitary, and trace out the environment:

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

Stinespring dilation proves the theorem. Here we use its construction to understand environment monitoring and repeated interactions.

Helpful background. Kraus representation describes the channel, and the partial trace removes an unobserved environment. Kraus, Choi, and Stinespring views compares the three representations.

A channel from a larger reversible evolution

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.

Building the isometry from Kraus operators

For a trace-preserving family $\sum_\alpha K_\alpha^\dagger K_\alpha=I$, introduce an orthonormal environment basis and set

$$V|\psi\rangle=\sum_\alpha K_\alpha|\psi\rangle\otimes|\alpha\rangle_E.$$

Orthogonality removes cross terms in $V^\dagger V$, giving $V^\dagger V=I$. It likewise gives

$$\operatorname{Tr}_E(V\rho V^\dagger)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger.$$

The dilation proof expands both steps. Notice that $V\rho V^\dagger$ itself retains cross terms between different environment labels; tracing or measuring the environment determines which terms remain relevant.

From an Isometry to a Unitary

An isometry preserves inner products but maps into a larger space. After embedding the input as a subspace of that larger space, complete an orthonormal basis on both sides. Mapping the completed bases to each other extends $V$ to a unitary; the extension away from the prepared input subspace is not fixed by the channel.

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 environment dimension in an isometric, or pure-initial-environment, representation 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 is a property of the operator map, not a unique count of microscopic noise mechanisms. If a mixed environment is allowed without counting its purification, a smaller apparatus can sometimes realize the same channel.

Examples:

Channel	Kraus rank	Qualification
Unitary channel	$1$	a one-dimensional environment suffices
Qubit dephasing with $\lvert c\rvert<1$	$2$	at $\lvert c\rvert=1$ the map is unitary
Qubit amplitude damping, $0<\gamma\le1$	$2$	at $\gamma=0$ it is the identity
Completely depolarizing qudit	$d^2$	this counts a pure dilation environment

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 $|c|\approx1$, coherence magnitude is nearly preserved, although its phase can rotate. For example, $c=i$ gives no loss of magnitude but rotates the input $|+x\rangle$ to $|-y\rangle$. 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 with $0\le\gamma\le1$, 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

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

For $\gamma>0$, amplitude damping is not a Pauli channel because it is nonunital: it pulls the center of the Bloch ball toward the ground state. Energy exchange alone is not the distinguishing criterion; a random bit flip can also change energy. See Amplitude-Damping Channel for the full channel analysis.

Reading the same environment in a different basis

For complete qubit dephasing, take $V|\psi\rangle=a|0\rangle|0\rangle_E+b|1\rangle|1\rangle_E$. Reading $E$ in its $z$ basis gives the usual projected system branches. Reading it instead in $|\pm\rangle_E=(|0\rangle_E\pm|1\rangle_E)/\sqrt2$ gives

$$K_+=I/\sqrt2,\qquad K_-=Z/\sqrt2.$$

Each result has probability $1/2$ for every input. The conditional states differ only by a known phase flip, so applying $Z$ after the minus result restores the input. Without the record and correction, averaging both branches still gives complete dephasing. Changing an unobserved environment measurement basis does not change the system channel.

Reusing the environment can restore coherence

The same one-step channel does not determine a many-step process. Let a system qubit control a NOT on a fresh environment qubit in $|0\rangle_E$. One interaction implements complete dephasing when $E$ is ignored. Two such interactions with independent fresh environments still give dephasing.

If the same coherent environment is retained, however, applying the controlled NOT twice gives $U^2=I$: the system returns to its input state and the environment to $|0\rangle_E$. For input $|+x\rangle$, the first reduced state is $I/2$, but the second is pure again. The intermediate correlations matter. A single Stinespring representation therefore establishes neither memoryless repeated dynamics nor a time-homogeneous master equation.

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.

References

• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Lecture Notes in Physics 190, Springer (1983). Publisher.
• W. F. Stinespring, “Positive functions on $C^*$-algebras,” Proceedings of the American Mathematical Society 6, 211–216 (1955). Article.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapter 2. Author’s text.