Kraus Representation

A Kraus representation writes a completely positive map as a sum of ordinary operator conjugations. It is the most common working form of a finite-dimensional quantum channel:

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

The representation is useful because complete positivity is automatic, trace conditions are easy to check, and many physical models of noise or unread measurement lead directly to Kraus operators.

The main warning is that Kraus operators are not unique. A channel is the map $\Phi$; a Kraus representation is one way to write it.

Simulating quantum channels shows how to calculate outputs and check complete positivity numerically.

Helpful background. Complete positivity explains the condition being represented. Kraus, Choi, and Stinespring views compares the equivalent descriptions.

Operator-Sum Form

Let the input Hilbert space be $\mathcal H_{\mathrm{in}}$ and the output Hilbert space be $\mathcal H_{\mathrm{out}}$. A Kraus operator is a linear map

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

The corresponding operator-sum map is

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

For any positive operator $X\ge0$, each term

$$K_\alpha X K_\alpha^\dagger$$

is positive. The same remains true after tensoring with an idle reference system, so every Kraus map is completely positive.

In finite dimensions, the converse also holds: every completely positive map has at least one Kraus representation.

Trace Preservation

The trace of the output is

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

Therefore $\Phi$ is trace preserving if and only if

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

It is trace nonincreasing if and only if

$$\sum_\alpha K_\alpha^\dagger K_\alpha\le I_{\mathrm{in}}.$$

These are normalization conditions, not complete-positivity conditions. Complete positivity is already built into the operator-sum form.

Unitality Is Different

For a map from a Hilbert space to itself, $\Phi$ is unital if

$$\Phi(I)=I.$$

In Kraus form this is

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

This condition is often confused with trace preservation, but it is not the same equation. Trace preservation uses $K_\alpha^\dagger K_\alpha$; unitality uses $K_\alpha K_\alpha^\dagger$.

A unitary channel is both trace preserving and unital. Amplitude damping is trace preserving but not unital. A selected measurement outcome is usually neither trace preserving nor unital before normalization.

Representation Theorem

For finite-dimensional matrix algebras, the following statements are equivalent:

1. $\Phi$ is completely positive.
2. $\Phi$ has a Kraus representation.
3. The Choi matrix $J(\Phi)$ is positive.

One way to see the connection is to use vectorization. For an operator

$$K:\mathcal H_{\mathrm{in}}\to\mathcal H_{\mathrm{out}},$$

define

$$|K\rangle\rangle = \sum_{a,i} K_{ai}|a\rangle_{\mathrm{out}}\otimes |i\rangle_{\mathrm{in}}.$$

If

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

then the Choi matrix can be written as

$$J(\Phi) = \sum_\alpha |K_\alpha\rangle\rangle \langle\langle K_\alpha|.$$

Thus $J(\Phi)\ge0$. Conversely, if

$$J(\Phi) = \sum_{\ell=1}^r \lambda_\ell |v_\ell\rangle\langle v_\ell|, \qquad \lambda_\ell>0,$$

then unvectorizing

$$\sqrt{\lambda_\ell}\,|v_\ell\rangle$$

gives a set of Kraus operators for $\Phi$, up to the chosen vectorization convention.

Minimal Kraus Rank

The smallest possible number of Kraus operators in a representation of $\Phi$ is the Kraus rank. In finite dimensions,

$$\text{Kraus rank of }\Phi = \operatorname{rank}J(\Phi).$$

This rank is representation independent. It counts the minimum number of operator-sum terms, or the minimum environment dimension in a pure-environment dilation. It does not count uniquely identifiable physical noise mechanisms.

Examples:

• a unitary channel $\Phi(\rho)=U\rho U^\dagger$ has Kraus rank $1$;
• a qubit phase-flip channel $(1-p)\rho+pZ\rho Z$ has rank $2$ for $0<p<1$, and rank $1$ at the unitary endpoints;
• a completely depolarizing channel on a $d$-dimensional system has Kraus rank $d^2$ in a minimal full-rank Choi representation.

The number of Kraus operators in a nonminimal representation may be larger than the Kraus rank.

Unitary Freedom

Kraus representations are not unique. If

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

and $u_{\beta\alpha}$ is a unitary matrix, then

$$L_\beta = \sum_{\alpha=1}^r u_{\beta\alpha}K_\alpha$$

gives another representation of the same map:

$$\Phi(\rho) = \sum_{\beta=1}^r L_\beta\rho L_\beta^\dagger.$$

If two representations have different numbers of operators, one can pad the shorter list with zero operators and then use a unitary mixing on the padded list. Equivalently, a nonminimal representation can be obtained from a minimal one by an isometry on the Kraus-label space.

This freedom is why an individual Kraus operator should not automatically be assigned direct physical meaning.

Environmental Interpretation

A trace-preserving Kraus map can be represented by an isometry

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

defined by

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

The trace-preserving condition gives

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

The channel is recovered by tracing out the environment:

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

If the environment is measured in the basis $\{|\alpha\rangle_E\}$, the Kraus label can become an observed record. If the environment is ignored, the same formula is just an unread noise channel.

This is the conceptual bridge between channels, quantum instruments, and reduced dynamics. The channel-side environment construction is developed in Stinespring Representation.

From a Unitary System-Environment Model

Suppose an environment starts in a fixed pure state $|e_0\rangle$, independent of and uncorrelated with the system input, the joint system evolves with a unitary $U$, and the environment is later ignored. Choose an environment basis $\{|\alpha\rangle\}$. Define

$$K_\alpha = \langle\alpha|U|e_0\rangle.$$

Here the bracket is taken only in the environment Hilbert space, leaving an operator from system input to system output. Then

$$\Phi(\rho) = \operatorname{Tr}_E \left[ U(\rho\otimes |e_0\rangle\langle e_0|)U^\dagger \right] = \sum_\alpha K_\alpha\rho K_\alpha^\dagger.$$

Changing the environment basis changes the Kraus representation without changing the channel. This is the physical origin of unitary freedom in many examples.

Basic Examples

Unitary Channel

A closed-system unitary evolution has one Kraus operator:

$$K_1=U, \qquad \Phi(\rho)=U\rho U^\dagger.$$

The trace-preserving condition is $U^\dagger U=I$.

Nonselective Projective Measurement

An unread Lüders measurement with projectors $\{P_a\}$ has

$$\Phi(\rho) = \sum_a P_a\rho P_a.$$

The Kraus operators are $K_a=P_a$, and

$$\sum_a P_a^\dagger P_a = \sum_a P_a =I.$$

This channel removes coherences between the measured eigenspaces while preserving each diagonal block. The same PVM can also have other state updates.

Random Unitary Channel

If unitary $U_j$ is applied with classical probability $p_j$, then

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

A Kraus representation is

$$K_j=\sqrt{p_j}\,U_j.$$

Random unitary channels are trace preserving and unital. Not every channel is random unitary.

A rank-one channel that changes dimension

A single Kraus operator of a channel satisfies $K^\dagger K=I_{\mathrm{in}}$: it is an isometry. It is unitary only when the finite input and output dimensions agree. For example,

$$K=\begin{pmatrix}1&0\\0&0\\0&i\end{pmatrix}$$

embeds a qubit into a three-level output. It sends $|+\rangle$ to $(|0\rangle+i|2\rangle)/\sqrt2$, preserving purity and inner products. With the output-input vectorization defined above,

$$|K\rangle\rangle=(1,0,0,0,0,i)^{\mathsf T}, \qquad J_\Phi=|K\rangle\rangle\langle\langle K|.$$

The Choi matrix has rank one and trace two, while $KK^\dagger=\operatorname{diag}(1,0,1)\ne I_3$. This illustrates why a channel can have minimal Kraus rank one without being a unitary operation on a fixed system.

Channel Kraus Operators Versus Measurement Kraus Operators

The same algebraic formula appears in two related settings:

Setting	Object	Meaning of Kraus labels
channel	$\Phi(\rho)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$	unobserved alternatives in a nonselective map
instrument	$\mathcal I_m(\rho)=\sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger$	hidden alternatives inside reported outcome $m$

For measurement outcomes, see Kraus Operators. For channels, the Kraus label usually represents an environment basis, a noise alternative, or a mathematical decomposition. It becomes an observed result only after a specific monitoring scheme is chosen.

Infinite-Dimensional Caution

The finite-dimensional Kraus theorem is the right default for qubits, finite spin systems, finite-level atoms, and finite-dimensional quantum information models. In infinite-dimensional Hilbert spaces, analogous operator-sum or integral representations exist under additional technical hypotheses, but domains, convergence, normality, and trace-class continuity matter.

For most quantum-mechanics applications, one first builds intuition with the finite-dimensional theorem and then checks the functional-analytic assumptions when working with unbounded operators or continuous-variable systems.

Exercises

1. Unitary mixing. Suppose $L_\beta=\sum_\alpha u_{\beta\alpha}K_\alpha$, where $u$ is unitary. Show that $\sum_\beta L_\beta\rho L_\beta^\dagger=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$.

Solution

Substitute the definition:

$$\sum_\beta L_\beta\rho L_\beta^\dagger = \sum_{\beta,\alpha,\gamma} u_{\beta\alpha}u_{\beta\gamma}^* K_\alpha\rho K_\gamma^\dagger.$$

Unitarity gives

$$\sum_\beta u_{\beta\alpha}u_{\beta\gamma}^* = \delta_{\alpha\gamma}.$$

Therefore

$$\sum_\beta L_\beta\rho L_\beta^\dagger = \sum_\alpha K_\alpha\rho K_\alpha^\dagger.$$

2. Isometry from Kraus operators. Let $\Phi$ be trace preserving, and define $V|\psi\rangle=\sum_\alpha K_\alpha|\psi\rangle\otimes|\alpha\rangle_E$. Show that $V$ is an isometry.

Solution

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

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

Trace preservation gives $\sum_\alpha K_\alpha^\dagger K_\alpha=I$. Hence

$$\langle\phi|V^\dagger V|\psi\rangle = \langle\phi|\psi\rangle$$

for all vectors, so $V^\dagger V=I$. Thus $V$ preserves inner products and is an isometry.

3. Unitality check. Let $\Phi(\rho)=\sum_j p_j U_j\rho U_j^\dagger$, where $\sum_jp_j=1$ and each $U_j$ is unitary. Show that $\Phi$ is both trace preserving and unital.

Solution

Use Kraus operators $K_j=\sqrt{p_j}U_j$. Then

$$\sum_jK_j^\dagger K_j = \sum_j p_j U_j^\dagger U_j = \sum_jp_j I =I,$$

so the channel is trace preserving. Also,

$$\Phi(I) = \sum_jp_jU_jIU_j^\dagger = \sum_jp_jI =I,$$

so the channel is unital.

References

• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285–290 (1975). Article.
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Lecture Notes in Physics 190, Springer (1983). Publisher.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapter 2. Author’s text.