Contents

# Contents

## Idea

In quantum probability theory/open quantum systems, a unitary quantum channel is a quantum channel whose restriction to pure states acts by a unitary transformation just as a loss-less quantum gate does.

Concretely, in terms of Kraus decomposition, a quantum channel

$\array{ \mathscr{H}_1 \otimes \mathscr{H}_1^\ast & \longrightarrow & \mathscr{H}_2 \otimes \mathscr{H}_2^\ast \\ \rho &\mapsto& ch(\rho) }$

is unitary iff there exists a unitary operator $U \,\colon\, \mathscr{H}_1 \longrightarrow \mathscr{H}_2$ such that $ch$ is given by conjugation with this operator:

$ch(\rho) \;=\; U \cdot \rho \cdot U^\dagger \,.$

## Properties

### General

###### Proposition

(unitary channels are the reversible channels)
Unitary channels are clearly invertible. In fact, every invertible quantum channel must be a unitary quantum channel.

(eg. Preskill pp. 13)

### Environmental representation of quantum channels

The crux of dynamical quantum decoherence is that fundamentally the (time-)evolution of any quantum system $\mathscr{H}$ may be assumed unitary (say via a Schrödinger equation) when taking the whole evolution of its environment $\mathscr{B}$ (the “bath”, ultimately the whole observable universe) into account, too, in that the evolution of the total system $\mathscr{H} \otimes \mathscr{B}$ is given by a unitary operator

$\array{ \mathllap{ evolve \;\colon\; } \mathscr{H} \otimes \mathscr{B} &\longrightarrow& \mathscr{H} \otimes \mathscr{B} \\ \left\vert \psi, \beta \right\rangle &\mapsto& U_{tot} \left\vert \psi, \beta \right\rangle \mathrlap{\,,} }$

after understanding the mixed states $\rho \,\colon\, \mathscr{H} \otimes \mathscr{H}^\ast$ (density matrices) of the given quantum system as coupled to any given mixed state $env \,\colon\, \mathscr{B} \otimes \mathscr{B}^\ast$ of the bath (via tensor product)

$\array{ \mathllap{ couple \;\colon\; } \mathscr{H} \otimes \mathscr{H}^\ast & \longrightarrow & (\mathscr{H} \otimes \mathscr{B}) \otimes (\mathscr{H} \otimes \mathscr{B})^\ast \\ \rho &\mapsto& \rho \otimes env \mathrlap{\,;} }$

…the only catch being that one cannot — and in any case does not (want or need to) — keep track of the precise quantum state of the environment/bath, instead only of its average effect on the given quantum system, which by the rule of quantum probability is the mixed state that remains after the partial trace over the environment:

(1)$\array{ \mathllap{ average \;\colon\; } (\mathscr{H} \otimes \mathscr{B}) \otimes (\mathscr{H} \otimes \mathscr{B})^\ast &\longrightarrow& \mathscr{H} \otimes \mathscr{H}^\ast \\ \widehat{\rho} &\mapsto& Tr_{\mathscr{B}}\big(\widehat{\rho}\big) \mathrlap{\,.} }$

In summary this means for practical purposes that the probabilistic evolution of quantum systems $\mathscr{H}$ is always of the composite form

$\array{ \mathscr{H} \otimes \mathscr{H}^\ast & \xrightarrow{ \array{ \text{couple to} \\ \text{environment} } } & \left( \array{ \mathscr{H} \\ \otimes \\ \mathscr{B} } \right) \otimes \left( \array{ \mathscr{H} \\ \otimes \\ \mathscr{B} } \right)^\ast & \xrightarrow{ \array{ \text{total unitary} \\ \text{evolution} } } & \left( \array{ \mathscr{H} \\ \otimes \\ \mathscr{B} } \right) \otimes \left( \array{ \mathscr{H} \\ \otimes \\ \mathscr{B} } \right)^\ast & \xrightarrow{ \array{ \text{average over} \\ \text{environment} } } & \;\;\;\; \mathscr{H} \otimes \mathscr{H}^\ast \\ \rho &\mapsto& \rho \otimes env &\mapsto& \mathclap{ U_{tot} \cdot (\rho \otimes env) \cdot U_{tot}^\dagger } &\mapsto& \;\;\;\;\;\;\;\;\;\;\;\; \mathclap{ Tr_{\mathscr{B}} \big( U_{tot} \cdot (\rho \otimes env) \cdot U_{tot}^\dagger \big) } }$

This composite turns out to be a “quantum channel

The realization of a quantum channel in the form (2) is also called an environmental representation (eg. Życzkowski & Bengtsson 2004 (3.5)).

In fact all quantum channels on a fixed Hilbert space have such an evironmental representation:

###### Proposition

(environmental representation of quantum channels)

Every quantum channel

$chan \;\;\colon\;\; \mathscr{H} \otimes \mathscr{H}^\ast \longrightarrow \mathscr{H} \otimes \mathscr{H}^\ast$

may be written as

1. a unitary quantum channel, induced by a unitary operator $U_{tot} \,\colon\, \mathscr{H} \otimes \mathscr{B} \to \mathscr{H} \otimes \mathscr{B}$

2. on a compound system with some $\mathscr{B}$ (the “bath”), yielding a total system Hilbert space $\mathscr{H} \otimes \mathscr{B}$ (tensor product),

3. and acting on the given mixed state $\rho$ coupled (tensored) with any pure state of the bath system,

4. followed by partial trace (averaging) over $\mathscr{B}$ (leading to decoherence in the remaining state)

in that

(2)$chan(\rho) \;\;=\;\; Tr_{\mathscr{B}} \big( U_{tot} \cdot (\rho \otimes env) \cdot U_{tot}^\dagger \big) \,.$

Conversely, every operation of the form (2) is a quantum channel.

This is originally due to Lindblad 1975 (see top of p. 149 and inside the proof of Lem. 5). For exposition and review see: Nielsen & Chuang 2000 §8.2.2-8.2.3. An account of the infinite-dimensional case is in Attal, Thm. 6.5 & 6.7. These authors focus on the case that the environment is in a pure state, the (parital) generalization to mixed environment states is discussed in Bengtsson & Życzkowski 2006 pp. 258.

###### Proof

We spell out the proof assuming finite-dimensional Hilbert spaces. (The general case follows the same idea, supplemented by arguments that the following sums converge.)

Now given a completely positive map:

$chan \,\colon\, \mathscr{H} \otimes \mathscr{H}^\ast \longrightarrow \mathscr{H} \otimes \mathscr{H}^\ast \,,$

then by operator-sum decomposition there exists a set (finite, under our assumptions) inhabited by at least one element

$s_{ini} \,\colon\, S \,,$

and an $S$-indexed set of linear operators

(3)$s \,\colon\, S \;\;\; \vdash \;\;\; E_s \;\colon\; \mathscr{H} \longrightarrow \mathscr{H} \,,\;\;\;\; \text{with} \;\;\;\; \underset{s}{\sum} E_s^\dagger \cdot E_s \,=\, Id \mathrlap{\,,}$

such that

$chan(\rho) \;=\; \underset{s}{\sum} \, E_s \cdot \rho \cdot E_s^\dagger \,.$

Now take

$\mathscr{B} \,\equiv\, \underset{S}{\oplus} \mathbb{C}$

with its canonical Hermitian inner product-structure with orthonormal linear basis $\big(\left\vert s \right\rangle\big)_{s \colon S}$ and consider the linear map

$\array{ \mathllap{ V \;\colon\;\; } \mathscr{H} &\longrightarrow& \mathscr{H} \otimes \mathscr{B} \\ \left\vert \psi \right\rangle &\mapsto& \underset{s}{\sum} \, E_s \left\vert \psi \right\rangle \otimes \left\vert s \right\rangle \mathrlap{\,.} }$

Observe that this is a linear isometry

$\begin{array}{ll} \left\langle \psi \right\vert V^\dagger V \left\vert \psi \right\rangle \\ \;=\; \underset{s,s'}{\sum} \left\langle \psi \right\vert E_{s'}^\dagger E_s \left\vert \psi \right\rangle \underset{ \delta_s^{s'} }{ \underbrace{ \left\langle s' \vert s \right\rangle } } \\ \;=\; \left\langle \psi \right\vert \underset{ Id }{ \underbrace{ \underset{s}{\sum} E_{s}^\dagger E_s } } \left\vert \psi \right\rangle \\ \;=\; \left\langle \psi \vert \psi \right\rangle \mathrlap{\,.} \end{array}$

This implies that $V$ is injective so that we have a direct sum-decomposition of its codomain into its image and its cokernel orthogonal complement, which is unitarily isomorphic to $dim(\mathscr{B})-1$ summands of $\mathscr{H}$ that we may identify as follows:

$\mathscr{H} \otimes \mathscr{B} \;\simeq\; V\big( \mathscr{H} \big) \oplus \Big( \mathscr{H} \otimes \big( \mathscr{B} \ominus \mathbb{C}\left\vert s_0 \right\rangle \big) \Big) \,.$

In total this yields a unitary operator

$U \;\colon\; \mathscr{H} \otimes \mathscr{B} \,\simeq\, \mathscr{H} \oplus \Big( \mathscr{H} \otimes \big( \mathscr{B} \ominus \mathbb{C}\left\vert s_{ini} \right\rangle \big) \Big) \underoverset{}{}{\longrightarrow} V\big( \mathscr{H} \big) \oplus \Big( \mathscr{H} \otimes \big( \mathscr{B} \ominus \mathbb{C}\left\vert s_{ini} \right\rangle \big) \Big) \;\simeq\; \mathscr{H} \otimes \mathscr{B}$

and we claim that this has the desired action on couplings of the $\mathscr{H}$-system to the pure bath state $\left\vert s_{ini} \right\rangle$:

$\begin{array}{l} trace^{\mathscr{B}} \Big( U \big( \left\vert s_{ini} \right\rangle \rho \left\langle s_{ini} \right\vert \big) U^\dagger \Big) \\ \;=\; \underset{s,s'}{\sum} trace^{\mathscr{B}} \big( \left\vert s \right\rangle E_s \cdot \rho \cdot E_{s'}^\dagger \left\langle s' \right\vert \big) \\ \;=\; \underset{s,s'}{\sum} \underset{ \delta_{s}^{s'} }{ \underbrace{ \left\langle s' \vert s \right\rangle } } E_s \cdot \rho \cdot E_{s'}^\dagger \\ \;=\; \underset{s}{\sum} E_s \cdot \rho \cdot E_{s}^\dagger \\ \;=\; chan(\rho) \,. \end{array}$

This concludes the construction of an environmental representation where the environment is in a pure state.

###### Remark

The above theorem is often phrased as “… and the environment can be assumed to be in a pure stated”. But in fact the proof crucially uses the assumption that the environment is in a pure state. It is not clear that there is a proof that works more generally.

In fact, if the environment is taken to be in the maximally mixed state, then the resulting quantum channels are called noisy operations or unistochastic quantum channels and are not expected to exhaust all quantum channels.

## References

(For more see the references at quantum channel.)

Last revised on September 24, 2023 at 10:49:21. See the history of this page for a list of all contributions to it.