Stinespring Dilation

Stinespring dilation is the channel-side counterpart of the indirect-measurement idea. For a channel, it represents the state transformation as an isometry into a larger space followed by ignoring an environment. A general completely positive map has a similar representation, but its embedding operator need not preserve norm.

The proof below explains the theorem. Stinespring representation applies it to channel models and environment measurements.

Helpful background. Kraus operators provide the construction, while the partial trace removes an unobserved subsystem.

For a quantum channel, the operational slogan is:

$$\text{channel} \quad=\quad \text{unitary system-environment dynamics} \quad+\quad \text{discarded environment}.$$

This theorem is one of the main bridges between generalized measurements, quantum channels, decoherence, and open-system dynamics.

Stinespring representation in finite dimensions

Let

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

be a completely positive map. In finite dimensions, there exists an auxiliary Hilbert space $\mathcal H_E$ and a linear operator

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

such that

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

If $\Phi$ is trace preserving, then $V$ is an isometry:

$$V^\dagger V=I_{\mathrm{in}}.$$

If $\Phi$ is trace nonincreasing, then

$$V^\dagger V\le I_{\mathrm{in}}.$$

The trace-preserving case is a quantum channel. A trace-nonincreasing map can describe a selected outcome; its unitary realization also requires selection, not merely discarding an environment. Complete positivity alone allows trace increase and does not by itself make a map a physical operation.

Construction from Kraus Operators

Suppose

$$\Phi(\rho) = \sum_{\alpha=1}^r K_\alpha\rho K_\alpha^\dagger,$$

where

$$K_\alpha: \mathcal H_{\mathrm{in}} \to \mathcal H_{\mathrm{out}}.$$

Let the environment have 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

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

Therefore, if $\Phi$ is trace preserving,

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

and $V$ is an isometry.

Tracing out the environment recovers the Kraus representation:

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

This construction is the simplest finite-dimensional proof of the theorem.

Unitary Form

When the map is trace preserving, the isometry can be realized by a unitary on a larger space. In the common case $\mathcal H_{\mathrm{in}}=\mathcal H_{\mathrm{out}}=\mathcal H_S$, prepare an environment in a fixed state $\lvert0\rangle_E$ and choose a unitary $U$ such that

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

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

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

This is the standard reduced-dynamics formula for a system initially uncorrelated with its environment. The theorem says that every channel has such a representation, although the environment in a representation need not be a literal microscopic bath.

Physical Interpretation

If one actually measures the dilation environment in the specified basis, $\alpha$ labels a possible recorded outcome:

$$K_\alpha \quad\leftrightarrow\quad \text{environment record }\alpha.$$

If the environment is ignored, the alternatives add incoherently:

$$\Phi(\rho) = \sum_\alpha K_\alpha\rho K_\alpha^\dagger.$$

If the environment is measured, one obtains an outcome-resolved instrument:

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

Changing that environment measurement basis mixes the Kraus operators and generally gives a different instrument with the same unread channel. The representation does not guarantee experimental access to this environment or basis.

Minimal Environment Dimension

The construction above uses one environment basis state per Kraus operator. A nonminimal Kraus representation therefore gives a nonminimal dilation.

In finite dimensions, the smallest environment dimension in the isometric (equivalently, pure-initial-environment) representation equals the Kraus rank:

$$\dim\mathcal H_E^{\min} = \operatorname{rank}J_\Phi,$$

where $J_\Phi$ is the Choi matrix of the channel. This number is independent of the chosen Kraus representation. It does not give the minimum size of a mixed-state apparatus in every realization: for example, swapping a $d$-level system with a $d$-level environment in $I/d$ implements complete depolarization, whereas its pure-environment dilation has dimension $d^2$. Purifying that mixed apparatus supplies the extra dimension.

For examples:

• a unitary channel needs a one-dimensional environment;
• a qubit amplitude-damping channel with $0\lt p\lt1$ has Kraus rank $2$;
• a fully depolarizing qudit channel has Kraus rank $d^2$.

Example: Dephasing from an Environment Record

Let a qubit interact with an environment so that

$$\lvert0\rangle\lvert e_0\rangle \mapsto \lvert0\rangle\lvert e_0'\rangle, \qquad \lvert1\rangle\lvert e_0\rangle \mapsto \lvert1\rangle\lvert e_1'\rangle.$$

For an input density matrix

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

tracing out the environment gives

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

where

$$\gamma = \langle e_1'|e_0'\rangle.$$

When the environment states are orthogonal, $\gamma=0$, and the environment has a perfect record of the qubit basis state. When $|\gamma|\approx1$, the two pointer rays are nearly the same: coherence magnitude is nearly preserved. The phase of $\gamma$ can still rotate the relative system phase, even when $|\gamma|=1$.

This is the same structural calculation behind decoherence.

Example: Amplitude Damping

For a two-level zero-temperature decay channel with $0\le p\le1$, use an environment with states $\lvert0\rangle_E$ and $\lvert1\rangle_E$ and define

$$\begin{aligned} U\lvert0\rangle_S\lvert0\rangle_E &= \lvert0\rangle_S\lvert0\rangle_E,\\ U\lvert1\rangle_S\lvert0\rangle_E &= \sqrt{1-p}\, \lvert1\rangle_S\lvert0\rangle_E + \sqrt p\, \lvert0\rangle_S\lvert1\rangle_E. \end{aligned}$$

Tracing out $E$ gives the channel with Kraus operators

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

The environment state $\lvert1\rangle_E$ records that a quantum of energy was emitted. If that record is ignored, the system undergoes amplitude damping. If it is monitored, the same dilation supports a quantum-jump description.

Example: Random Unitary Noise

A random unitary channel has the form

$$\Phi(\rho) = \sum_j p_j U_j\rho U_j^\dagger.$$

A Stinespring isometry is

$$V\lvert\psi\rangle = \sum_j \sqrt{p_j}\, U_j\lvert\psi\rangle \otimes \lvert j\rangle_E.$$

In this pure-environment representation the register need not be classical before it is measured or dephased. For example, take $p_0=p_1=1/2$, $U_0=I$, $U_1=\sigma_z$, and input $|+\rangle$. The joint output is

$$V|+\rangle=\frac{|+\rangle|0\rangle_E+|-\rangle|1\rangle_E}{\sqrt2},$$

an entangled state, whose system marginal is $I/2$. The same channel can alternatively be implemented by a classical random choice of $I$ or $\sigma_z$. A channel does not uniquely specify which environment was used.

Not every channel is random unitary: any random mixture of unitaries satisfies $\Phi(I)=I$, whereas amplitude damping with $p>0$ gives $\Phi(I)=\operatorname{diag}(1+p,1-p)$. Entanglement in a chosen dilation is therefore not the criterion separating these classes.

Relation to Naimark Dilation

Naimark dilation starts with a POVM and represents outcome probabilities as a projective measurement on a larger Hilbert space:

$$F_m=V^\dagger\Pi_mV.$$

Stinespring dilation starts with a completely positive map and represents the state transformation using a linear operator $V$ followed by a partial trace:

$$\Phi(\rho) = \operatorname{Tr}_E[V\rho V^\dagger].$$

Here $V$ is an isometry precisely when the map is trace preserving. For a selected outcome it need only obey $V^\dagger V\le I$; $\operatorname{Tr}(V\rho V^\dagger)$ is then the outcome probability.

In a full measurement model, both appear. A system couples unitarily to an apparatus and environment; a pointer may be projectively measured; ignoring or reading different parts produces channels, instruments, and POVMs.

Relation to Reduced Dynamics

The reduced-dynamics formula

$$\rho_S(t) = \operatorname{Tr}_E \!\left[ U_{SE}(t) (\rho_S(0)\otimes\rho_E) U_{SE}^\dagger(t) \right]$$

defines a channel on arbitrary system inputs when $\rho_E$ is fixed independently of the input and the initial state factorizes. If $\rho_E$ is mixed,

$$\rho_E=\sum_\beta q_\beta \lvert e_\beta\rangle\langle e_\beta\rvert,$$

one obtains Kraus operators

$$K_{\alpha\beta}(t) = \sqrt{q_\beta}\, \langle e_\alpha|U_{SE}(t)|e_\beta\rangle_E.$$

If the initial system-environment state is correlated, the simple channel representation for arbitrary system inputs may fail. That caveat belongs to the physics of reduced dynamics, not to a failure of Stinespring’s theorem.

See Reduced Dynamics for the system-environment setting.

Exercises

Verify the isometry condition

Given $V\lvert\psi\rangle=\sum_\alpha K_\alpha\lvert\psi\rangle\otimes\lvert\alpha\rangle$, show that $V^\dagger V=\sum_\alpha K_\alpha^\dagger K_\alpha$.

Solution

For arbitrary $\lvert\phi\rangle$ and $\lvert\psi\rangle$,

$$\langle\phi\rvert V^\dagger V\lvert\psi\rangle = \sum_{\alpha,\beta} \langle\phi\rvert K_\alpha^\dagger K_\beta\lvert\psi\rangle \langle\alpha|\beta\rangle.$$

Since $\langle\alpha|\beta\rangle=\delta_{\alpha\beta}$,

$$\langle\phi\rvert V^\dagger V\lvert\psi\rangle = \sum_\alpha \langle\phi\rvert K_\alpha^\dagger K_\alpha\lvert\psi\rangle.$$

Thus

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

Recover the Kraus map

Show that tracing out the environment in the same construction gives $\Phi(\rho)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$.

Solution

First write

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

Using

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

we obtain

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

Amplitude-damping probabilities

For the amplitude-damping unitary above, what is the probability that the environment is found in $\lvert1\rangle_E$ for input state $\rho$?

Solution

The environment outcome $\lvert1\rangle_E$ corresponds to the Kraus operator

$$K_1 = \begin{pmatrix} 0 & \sqrt p\\ 0 & 0 \end{pmatrix}.$$

Thus

$$p(1) = \operatorname{Tr}(K_1\rho K_1^\dagger) = \operatorname{Tr}(K_1^\dagger K_1\rho).$$

Since

$$K_1^\dagger K_1 = p\lvert1\rangle\langle1\rvert,$$

the probability is

$$p(1)=p\,\rho_{11}.$$

The environment clicks only if the system had excited-state population.

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.