Kraus Map

Purpose

A Kraus representation writes a completely positive linear map as

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

where

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

The map is a deterministic quantum channel when

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

It is a trace-nonincreasing quantum operation when

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

The operator-sum form makes complete positivity automatic and turns trace conditions into elementary operator identities. The map $\Phi$, not a particular list of $K_\alpha$, is the physical channel.

The representation theorem and its proof are at Kraus Representation. This card collects the tests and constructions needed to use it.

At a glance

Task	Formula or condition
Apply a CP map	$\Phi(\rho)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger$
Trace preserving	$\sum_\alpha K_\alpha^\dagger K_\alpha=I_{\mathrm{in}}$
Trace nonincreasing	$\sum_\alpha K_\alpha^\dagger K_\alpha\le I_{\mathrm{in}}$
Unital	$\sum_\alpha K_\alpha K_\alpha^\dagger=I_{\mathrm{out}}$
Adjoint map	$\Phi^\dagger(A)=\sum_\alpha K_\alpha^\dagger A K_\alpha$
Compose two maps	Kraus operators $L_\beta K_\alpha$ for $\Psi\circ\Phi$
Tensor two maps	Kraus operators $K_\alpha\otimes L_\beta$
Outcome probability	$p(a)=\operatorname{Tr}\mathcal I_a(\rho)$
Conditional state	$\rho_a=\mathcal I_a(\rho)/p(a)$
POVM effect of an operation	$E_a=\sum_\mu M_{a\mu}^\dagger M_{a\mu}$
Minimal number of Kraus operators	$\operatorname{rank}J(\Phi)$

The rank identity is finite dimensional and uses the Choi matrix convention stated below.

Complete positivity

For every positive operator $X\ge0$,

$$K_\alpha X K_\alpha^\dagger\ge0,$$

so a Kraus map is positive. More importantly, for an arbitrary reference system $R$,

$$\begin{aligned} (\Phi\otimes I_R)(X) &= \sum_\alpha (K_\alpha\otimes I_R) X (K_\alpha^\dagger\otimes I_R) \\ &\ge0 \end{aligned}$$

whenever $X\ge0$ on $\mathcal H_{\mathrm{in}}\otimes\mathcal H_R$. Thus the map is completely positive: it remains positive when acting on one part of an entangled state.

In finite dimensions, the converse holds. Every completely positive map admits at least one Kraus representation. Positivity of $\Phi$ on isolated system states is weaker and does not by itself establish that $\Phi\otimes I_R$ is physically valid.

See Completely Positive Maps for the reference-system test and examples of positive but not completely positive maps.

Trace conditions

Using cyclicity of the trace,

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

Therefore

$$\operatorname{Tr}\Phi(\rho) = \operatorname{Tr}\rho$$

for every input exactly when

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

Similarly, the output trace cannot exceed the input trace for any $\rho\ge0$ exactly when

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

For a normalized input and a trace-nonincreasing operation,

$$p = \operatorname{Tr}\Phi(\rho) \in[0,1]$$

is the probability that the operation’s selected event occurs.

Unitality is different

A map is unital when

$$\Phi(I_{\mathrm{in}}) = I_{\mathrm{out}}.$$

In Kraus form,

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

Trace preservation uses the opposite operator order:

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

The two conditions coincide for a unitary channel but not for a general channel. Amplitude damping is trace preserving and nonunital. An unread projective measurement is both trace preserving and unital.

For equal input and output dimension, a unital channel preserves the maximally mixed state:

$$\Phi\left(\frac{I}{d}\right) = \frac{I}{d}.$$

This does not imply that every unital channel is unitary.

Adjoint map and observables

The Hilbert–Schmidt adjoint $\Phi^\dagger$ is defined by

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

For a Kraus map,

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

This is the Heisenberg-picture action of the channel on an observable. Trace preservation of $\Phi$ is equivalent to unitality of its adjoint:

$$\Phi^\dagger(I_{\mathrm{out}}) = I_{\mathrm{in}}.$$

Unitality of $\Phi$ is equivalent to trace preservation of $\Phi^\dagger$ with respect to the Hilbert–Schmidt trace pairing. Keeping the input and output identity operators labeled prevents dimension mistakes.

Quantum instruments and selected outcomes

A quantum instrument associates a completely positive trace-nonincreasing map $\mathcal I_a$ with each reported outcome $a$:

$$\mathcal I_a(\rho) = \sum_\mu M_{a\mu} \rho M_{a\mu}^\dagger.$$

The outcome probability is

$$\begin{aligned} p(a) &= \operatorname{Tr}\mathcal I_a(\rho) \\ &= \operatorname{Tr} \left( E_a\rho \right), \end{aligned}$$

where

$$E_a = \sum_\mu M_{a\mu}^\dagger M_{a\mu}$$

is the associated POVM effect. If $p(a)>0$, the conditional output is

$$\rho_a = \frac{ \mathcal I_a(\rho) }{ p(a) }.$$

The operation $\mathcal I_a$ is linear. The normalized conditional update is nonlinear because its denominator depends on $\rho$.

The nonselective map is

$$\Phi(\rho) = \sum_a \mathcal I_a(\rho).$$

It is trace preserving exactly when

$$\sum_aE_a=I.$$

A POVM $\{E_a\}$ determines outcome probabilities but not post-measurement states. Different sets of $M_{a\mu}$ can have the same effects and different backaction. See Quantum Instruments for that distinction.

Composition and tensor products

Suppose

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

and

$$\Psi(\sigma) = \sum_\beta L_\beta\sigma L_\beta^\dagger.$$

Then

$$\begin{aligned} (\Psi\circ\Phi)(\rho) &= \sum_{\alpha,\beta} L_\beta K_\alpha \rho K_\alpha^\dagger L_\beta^\dagger. \end{aligned}$$

Thus $\{L_\beta K_\alpha\}$ is a Kraus list for the composition. The order is physical: $K_\alpha$ acts first.

For independent maps on systems $A$ and $B$,

$$(\Phi_A\otimes\Psi_B)(\rho_{AB}) = \sum_{\alpha,\beta} (K_\alpha\otimes L_\beta) \rho_{AB} (K_\alpha^\dagger\otimes L_\beta^\dagger).$$

If the component maps are trace preserving, their composition and tensor product are trace preserving. The displayed lists need not be minimal even when the input lists are minimal.

Kraus-representation freedom

Kraus operators are not unique. If

$$L_\beta = \sum_\alpha u_{\beta\alpha}K_\alpha$$

and

$$u^\dagger u=I$$

on the span of the original Kraus labels, then

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

For two lists of the same length, $u$ can be unitary. Lists of different length can be padded with zero operators and related by a unitary, or a minimal list can be embedded into a longer one by an isometry.

This freedom has several consequences:

• an individual Kraus operator is generally not an invariant property of a channel;
• Kraus probabilities depend on the chosen environment measurement or unraveling;
• the number of operators in an arbitrary list is not the channel’s minimal Kraus rank;
• two very different-looking lists can implement exactly the same map.

Physical meaning can be assigned to a Kraus label only after a monitoring scheme or environment record has been specified.

Choi matrix and minimal rank

Fix an input basis $\{\lvert i\rangle\}$ and use the output–input convention

$$J(\Phi) = \sum_{i,j} \Phi\left( \lvert i\rangle\langle j\rvert \right) \otimes \lvert i\rangle\langle j\rvert.$$

With column-stacking vectorization,

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

the Kraus form gives

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

Hence $J(\Phi)\ge0$. Conversely, diagonalize

$$J(\Phi) = \sum_{r:\lambda_r>0} \lambda_r \lvert v_r\rangle\langle v_r\rvert.$$

Unvectorizing

$$\sqrt{\lambda_r}\, \lvert v_r\rangle$$

produces a Kraus operator $K_r$. In finite dimensions,

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

The same convention gives the trace tests

$$\operatorname{Tr}_{\mathrm{out}}J(\Phi) = I_{\mathrm{in}}$$

for trace preservation and

$$\operatorname{Tr}_{\mathrm{in}}J(\Phi) = I_{\mathrm{out}}$$

for unitality. Swapping the tensor-factor convention swaps which partial trace appears, so always state the Choi ordering.

See Choi Matrix for normalization choices and channel-state duality.

Stinespring construction

Choose orthonormal environment labels $\{\lvert\alpha\rangle_E\}$ and define

$$V = \sum_\alpha K_\alpha \otimes \lvert\alpha\rangle_E.$$

Equivalently,

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

Then

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

For a trace-preserving channel, $V^\dagger V=I$, so $V$ is an isometry. The channel is recovered by discarding the environment:

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

If the environment begins in a fixed pure state $\lvert e_0\rangle$ and the joint evolution is unitary $U$, then

$$K_\alpha = {}_E\langle\alpha\rvert U \lvert e_0\rangle_E.$$

Changing the environment basis mixes the Kraus operators while leaving $\Phi$ unchanged. The minimal environment dimension equals the Choi rank for a finite-dimensional minimal dilation.

The full representation theorem and its uniqueness properties are at Stinespring Representation.

Example: amplitude damping

For decay probability $0\le\gamma\le1$, take

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

The completeness relation is

$$K_0^\dagger K_0 + K_1^\dagger K_1 = I.$$

For

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

the channel gives

$$\Phi_\gamma(\rho) = \begin{pmatrix} \rho_{00}+\gamma\rho_{11} & \sqrt{1-\gamma}\,\rho_{01} \\ \sqrt{1-\gamma}\,\rho_{10} & (1-\gamma)\rho_{11} \end{pmatrix}.$$

It is not unital for $\gamma>0$:

$$K_0K_0^\dagger + K_1K_1^\dagger = \begin{pmatrix} 1+\gamma&0\\ 0&1-\gamma \end{pmatrix} \ne I.$$

If the environment record is monitored, the jump probability is

$$p_1 = \operatorname{Tr} \left( K_1\rho K_1^\dagger \right) = \gamma\rho_{11}.$$

If the record is ignored, the two branches are summed and the map is deterministic. The physical model and its $T_1$ limit are at Amplitude-Damping Channel.

Other common forms

Unitary channel

One Kraus operator is enough:

$$\Phi_U(\rho)=U\rho U^\dagger, \qquad U^\dagger U=I.$$

This channel has Choi rank one.

Random-unitary channel

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

$$\Phi(\rho) = \sum_j p_jU_j\rho U_j^\dagger$$

has Kraus operators

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

It is trace preserving and unital. Not every channel is random unitary.

Unread projective measurement

For orthogonal projectors satisfying $P_aP_b=\delta_{ab}P_a$ and $\sum_aP_a=I$,

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

This channel preserves block-diagonal populations and removes coherence between different projected sectors.

Finite map versus continuous-time dynamics

A single CPTP map describes a finite input–output operation. It need not be:

• invertible;
• part of a time-homogeneous semigroup;
• divisible into CPTP maps at every intermediate time;
• generated by a time-independent Lindblad operator;
• associated with a unique environment trajectory.

To obtain a Markovian master equation, one needs a family $\{\Phi_t\}_{t\ge0}$ with additional continuity and composition properties. Do not infer a Lindblad generator from one isolated Kraus list without checking whether an appropriate logarithm and CPTP interpolation exist.

Scope and initial correlations

The standard system-only channel construction assumes a fixed environment state independent of the system input:

$$\rho_{SE}(0) = \rho_S\otimes\rho_E.$$

Initial system–environment correlations can prevent one linear CP map from describing arbitrary hypothetical system inputs. A reduced evolution may still be well defined on a restricted compatibility domain, but the usual all-input Kraus interpretation then requires care.

In infinite dimensions, operator sums may be countably infinite or replaced by integrals. Normality, trace-class continuity, convergence topology, and domains of unbounded operators matter. The finite-dimensional Choi-rank and matrix algorithms should not be transferred without checking those hypotheses.

Calculation workflow

1. State the input and output Hilbert spaces and write every Kraus matrix with a fixed basis ordering.
2. Check dimensions of $K_\alpha\rho K_\alpha^\dagger$.
3. Verify complete positivity from the Kraus form or, for a supplied superoperator, from Choi positivity.
4. Compute $\sum_\alpha K_\alpha^\dagger K_\alpha$ to classify the map as trace preserving or trace nonincreasing.
5. Separately compute $\sum_\alpha K_\alpha K_\alpha^\dagger$ if unitality matters.
6. For an outcome operation, calculate its probability before normalizing the conditional state.
7. Test Hermiticity, positivity, and trace of representative outputs.
8. When comparing two Kraus lists, compare their induced maps or Choi matrices, not operators label by label.
9. Distinguish a finite channel from a continuous-time dynamical model.

Common mistakes

• Treating positivity on the system as sufficient without checking complete positivity.
• Reversing the trace-preserving product and testing $\sum_\alpha K_\alpha K_\alpha^\dagger=I$.
• Assuming trace preservation implies unitality.
• Normalizing each Kraus branch before adding an unread channel.
• Forgetting that a selected operation is trace nonincreasing before conditioning.
• Using a POVM effect to infer post-measurement backaction.
• Treating Kraus operators or their branch probabilities as unique.
• Reading the number of operators in a nonminimal list as the Kraus rank.
• Reversing $L_\beta K_\alpha$ in a composed channel.
• Mixing Choi tensor-order or vectorization conventions.
• Assuming one CPTP map automatically has a Lindblad generator.
• Ignoring initial system–environment correlations or infinite-dimensional convergence issues.

Exercises

1. Verify trace preservation and nonunitality of the amplitude-damping Kraus operators.

Solution

First,

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

Their sum is $I$, so the map is trace preserving. In the other order,

$$K_0K_0^\dagger + K_1K_1^\dagger = \begin{pmatrix} 1+\gamma&0\\ 0&1-\gamma \end{pmatrix}.$$

This equals $I$ only at $\gamma=0$, so nontrivial amplitude damping is not unital.

2. Let $\Phi$ have Kraus operators $K_\alpha$ and $\Psi$ have Kraus operators $L_\beta$. Prove that $\{L_\beta K_\alpha\}$ represents $\Psi\circ\Phi$ and preserves trace when both input maps do.

Solution

Substitution gives

$$\begin{aligned} \Psi(\Phi(\rho)) &= \sum_\beta L_\beta \left( \sum_\alpha K_\alpha\rho K_\alpha^\dagger \right) L_\beta^\dagger \\ &= \sum_{\alpha,\beta} (L_\beta K_\alpha) \rho (L_\beta K_\alpha)^\dagger. \end{aligned}$$

If both maps are trace preserving, then

$$\begin{aligned} \sum_{\alpha,\beta} (L_\beta K_\alpha)^\dagger (L_\beta K_\alpha) &= \sum_\alpha K_\alpha^\dagger \left( \sum_\beta L_\beta^\dagger L_\beta \right) K_\alpha \\ &= \sum_\alpha K_\alpha^\dagger K_\alpha = I. \end{aligned}$$

3. Suppose $L_\beta=\sum_\alpha u_{\beta\alpha}K_\alpha$ with $u^\dagger u=I$. Show that the $L_\beta$ and $K_\alpha$ lists define the same map.

Solution

Expand the new operator sum:

$$\begin{aligned} \sum_\beta L_\beta\rho L_\beta^\dagger &= \sum_{\beta,\alpha,\gamma} u_{\beta\alpha} u_{\beta\gamma}^* K_\alpha\rho K_\gamma^\dagger \\ &= \sum_{\alpha,\gamma} (u^\dagger u)_{\gamma\alpha} K_\alpha\rho K_\gamma^\dagger \\ &= \sum_\alpha K_\alpha\rho K_\alpha^\dagger. \end{aligned}$$

The individual branches changed, but the nonselective map did not.

4. An instrument has one Kraus operator $M_a$ for each outcome and satisfies $\sum_aM_a^\dagger M_a=I$. Derive its outcome probabilities and prove that they sum to one.

Solution

The probability of outcome $a$ is

$$p(a) = \operatorname{Tr} \left( M_a\rho M_a^\dagger \right) = \operatorname{Tr} \left( M_a^\dagger M_a\rho \right).$$

Therefore

$$\begin{aligned} \sum_a p(a) &= \operatorname{Tr} \left[ \left( \sum_aM_a^\dagger M_a \right) \rho \right] \\ &= \operatorname{Tr}\rho = 1. \end{aligned}$$

For $p(a)>0$, the conditional state is $M_a\rho M_a^\dagger/p(a)$.

Canonical links

• Kraus Representation
• Completely Positive Maps
• Trace-Preserving Maps
• Choi Matrix
• Stinespring Representation
• Quantum Instruments
• Amplitude-Damping Channel

References

• W. F. Stinespring, “Positive Functions on C*-Algebras,” Proceedings of the American Mathematical Society 6, 211–216 (1955).
• E. C. G. Sudarshan, P. M. Mathews, and J. Rau, “Stochastic Dynamics of Quantum-Mechanical Systems,” Physical Review 121, 920–924 (1961).
• K. Kraus, “General State Changes in Quantum Theory,” Annals of Physics 64, 311–335 (1971).
• 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: Fundamental Notions of Quantum Theory, Springer, 1983.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010, Ch. 8.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, Chs. 2–3.