Channel Composition and Fixed Points

Quantum channels are rarely used only once. A qubit may dephase during storage, relax during readout, and then pass through a detector model. A bosonic mode may lose photons in a fiber, pass through an amplifier, and then be measured. A Markovian noise model may be applied repeatedly in small time steps.

The channel-level questions are:

• how to compose two physical maps without losing complete positivity or trace preservation;
• when repeated applications are powers of one channel;
• which states or observables are unchanged by the noise;
• whether fixed states are actually attracting under repeated application.

This page treats the finite-dimensional channel picture first and then points to continuous-time semigroups. The full generator theory belongs to Quantum Dynamical Semigroups and Lindblad–GKSL Equation.

Serial Composition

Let

$$\Phi: \mathcal B(\mathcal H_A) \to \mathcal B(\mathcal H_B)$$

and

$$\Psi: \mathcal B(\mathcal H_B) \to \mathcal B(\mathcal H_C)$$

be quantum channels. Their serial composition is

$$(\Psi\circ\Phi)(\rho) = \Psi(\Phi(\rho)).$$

The rightmost channel acts first. This order convention is the same as ordinary function composition.

If

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

and

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

then the composite channel has Kraus operators

$$M_{\beta\alpha} = L_\beta K_\alpha,$$

because

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

Trace preservation is inherited:

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

Thus a serial composition of completely positive trace-preserving maps is again a channel. If one of the maps is trace nonincreasing, the composite is trace nonincreasing; it then describes a selected branch or filter rather than a deterministic channel.

Composition in Common Representations

Several representations make composition useful in different ways.

In Kraus form, composition multiplies Kraus operators as above. This is physically transparent when one channel follows another, but the number of Kraus operators can grow.

In a matrix representation of the linear map, vectorize operators as $\lvert X\rangle\rangle$. Then

$$\lvert\Phi(X)\rangle\rangle = S_\Phi\lvert X\rangle\rangle,$$

and serial composition becomes ordinary matrix multiplication:

$$S_{\Psi\circ\Phi} = S_\Psi S_\Phi.$$

This is the most convenient representation for eigenvalues, fixed points, and numerical iteration. The exact entries of $S_\Phi$ depend on the vectorization convention, so formulas with transposes must be used consistently.

In Choi form, composition is a contraction of the output leg of the first channel with the input leg of the second channel. This is often called a link product. Because tensor-ordering and transpose conventions differ across references, the safest route is to state the Choi convention first and then use the reconstruction formula from Choi Matrix.

Powers and Discrete-Time Channels

If the same channel is applied repeatedly, write

$$\Phi^n = \underbrace{\Phi\circ\Phi\circ\cdots\circ\Phi}_{n\ \mathrm{times}}.$$

This describes a discrete-time process with one fixed step. It is not automatically a continuous-time Markovian semigroup. A semigroup needs a family $\{\Phi_t:t\ge0\}$ satisfying $\Phi_{t+s}=\Phi_t\Phi_s$ for all nonnegative times, not only integer powers.

For common one-parameter channels, repeated application often composes through the physically natural survival or shrink factor:

Channel	One step	$n$ identical steps
depolarizing	traceless part $\tau\mapsto\lambda\tau$	$\tau\mapsto\lambda^n\tau$
dephasing	coherence $\rho_{ij}\mapsto\lambda\rho_{ij}$	$\rho_{ij}\mapsto\lambda^n\rho_{ij}$
amplitude damping	excited population survives by $1-\gamma$	decay probability $1-(1-\gamma)^n$
pure bosonic loss	transmissivity $\eta$	transmissivity $\eta^n$

These formulas assume the same basis, mode, and convention at every step. Interleaved Hamiltonian rotations or basis changes can turn an apparently simple composition into a different noise model.

Different Mechanisms Need Not Commute

Two channels generally do not commute:

$$\Psi\circ\Phi \ne \Phi\circ\Psi.$$

Order matters whenever the first channel changes the basis, population, coherence, or support on which the second channel acts. A dephasing channel in the $Z$ basis and a unitary rotation about the $X$ axis do not have the same effect in the two possible orders.

Some special channels do commute with many others. The $d$-dimensional depolarizing channel

$$\mathcal D_\lambda(\rho) = \lambda\rho + (1-\lambda)\frac{I}{d}$$

commutes with unitary conjugations because it treats every traceless direction equally:

$$\mathcal D_\lambda(U\rho U^\dagger) = U\mathcal D_\lambda(\rho)U^\dagger.$$

Pure dephasing and amplitude damping in the same energy basis also compose simply on matrix elements: populations relax while coherences acquire both the dephasing factor and the damping factor. But this simplicity is basis dependent.

A common modeling mistake is to add finite-step probabilities directly. For independent amplitude-damping steps with probabilities $\gamma_1$ and $\gamma_2$, the total decay probability is

$$\gamma_{\mathrm{tot}} = 1-(1-\gamma_2)(1-\gamma_1),$$

not $\gamma_1+\gamma_2$ except to first order when both probabilities are small.

Fixed States

A fixed state of a channel $\Phi$ is a density operator $\rho_\ast$ satisfying

$$\Phi(\rho_\ast) = \rho_\ast.$$

Finite-dimensional quantum channels always have at least one fixed state. The reason is not specifically quantum: the state space is compact and convex, and a channel is a continuous map from that state space to itself. A fixed-point theorem then applies.

The fixed states form a convex set. If $\rho_1$ and $\rho_2$ are fixed, then any mixture

$$p\rho_1+(1-p)\rho_2, \qquad 0\le p\le1,$$

is also fixed.

Examples:

Channel	Fixed states
identity channel	every state
unitary channel $U\rho U^\dagger$	states commuting with $U$
projective dephasing	states block diagonal in the measured decomposition
depolarizing with $\lambda\ne1$	the maximally mixed state $I/d$
zero-temperature amplitude damping with $\gamma\gt0$	the ground state $\lvert0\rangle\langle0\rvert$
thermal oscillator damping	the thermal state of the oscillator
pure bosonic loss	the vacuum state

The existence statement above is finite dimensional. Infinite-dimensional channels need additional compactness, tightness, or energy-bound assumptions. Additive Gaussian noise, for example, keeps increasing covariance and has no normalizable finite-energy fixed state.

Fixed Observables

The adjoint channel $\Phi^\dagger$ acts on observables:

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

A fixed observable satisfies

$$\Phi^\dagger(A)=A.$$

This means that the expectation value of $A$ is conserved under the channel:

$$\operatorname{Tr}[A\Phi(\rho)] = \operatorname{Tr}[A\rho]$$

for all input states $\rho$.

Do not confuse fixed states with fixed observables. A dephasing channel fixes diagonal density matrices in the Schrödinger picture and fixes diagonal observables in the Heisenberg picture, but more general channels can have different-looking fixed-state and fixed-observable structures.

Attracting Fixed Points

A fixed state need not attract all initial states. Attraction is an asymptotic statement:

$$\lim_{n\to\infty} \Phi^n(\rho) = \rho_\ast.$$

In finite dimensions, represent $\Phi$ as a linear map on the vector space of operators. If

$$\Phi(X_\alpha) = \lambda_\alpha X_\alpha,$$

then $X_\alpha$ is an eigenoperator and $\lambda_\alpha$ is a channel eigenvalue. Fixed operators have $\lambda_\alpha=1$. Decaying modes have $|\lambda_\alpha|\lt1$.

For a completely positive trace-preserving map, the relevant spectrum lies in the unit disk. If the only eigenvalue on the unit circle is a simple eigenvalue $\lambda=1$ and the fixed-state space is one-dimensional, then repeated application converges to the unique fixed state. If other unit-modulus eigenvalues appear, the channel may preserve phases, support noiseless modes, or produce periodic behavior.

This is why uniqueness of a fixed state is stronger than existence but still must be paired with spectral information when one wants a convergence rate. The discrete-time spectral gap is controlled by the largest $|\lambda_\alpha|$ below one.

Solving for Fixed Points

In a finite-dimensional calculation, the fixed-point problem is linear before the positivity constraint is imposed:

$$(\Phi-I)(X)=0.$$

A practical workflow is:

1. choose a basis of operators or a superoperator matrix $S_\Phi$;
2. solve

$$(S_\Phi-I)\lvert X\rangle\rangle=0;$$

3. restrict to Hermitian $X$ with $\operatorname{Tr}X=1$;
4. keep the positive semidefinite solutions.

For a qubit affine Bloch-vector map

$$\mathbf r' = A\mathbf r+\mathbf c,$$

a fixed point satisfies

$$(I-A)\mathbf r_\ast = \mathbf c.$$

If $I-A$ is invertible, the fixed Bloch vector is unique. If $I-A$ is singular, there may be a continuum of fixed states or no physical solution inside the Bloch ball unless the channel constraints supply one.

Examples:

• depolarizing has $A=\lambda I_3$ and $\mathbf c=0$, so $\mathbf r_\ast=0$ when $\lambda\ne1$;
• dephasing has a whole fixed line along the dephasing axis;
• amplitude damping has a unique fixed point at the ground-state pole in the usual energy-basis convention.

For numerical examples of these affine maps, see Bloch Vector Noise Models. For general finite-dimensional channel matrices, see Simulating Quantum Channels.

Idempotent Channels

Some channels satisfy

$$\Phi^2=\Phi.$$

These are idempotent channels. After one application, the state is already in the fixed set.

The standard example is nonselective projective measurement:

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

Applying the same nonselective measurement again does nothing:

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

The fixed states are exactly those block diagonal in the measured decomposition:

$$\rho = \sum_a P_a\rho P_a.$$

This is the channel version of dephasing by unread measurement. It should not be confused with a selective measurement branch, which is trace nonincreasing before normalization.

Relation to Semigroups and Steady States

For a continuous-time Markovian semigroup

$$\Phi_t=e^{t\mathcal L},$$

a steady state satisfies

$$\mathcal L(\rho_{\mathrm{ss}})=0.$$

It is then fixed by every time-$t$ channel:

$$\Phi_t(\rho_{\mathrm{ss}}) = \rho_{\mathrm{ss}} \qquad t\ge0.$$

The converse needs care. A state fixed by one finite-time map $\Phi_T$ need not be stationary for all $t$. For example, a unitary rotation can return some coherences after one period even though they evolved nontrivially at intermediate times.

Thus:

• fixed by one channel is a discrete-time statement;
• stationary under a generator is a continuous-time statement;
• attracting under repeated application requires spectral information.

The generator-level treatment, including Liouvillian eigenvalues and relaxation modes, is developed in Quantum Dynamical Semigroups and Solving Lindblad Equations.

Physical Interpretation

Fixed points tell you what the environment is trying to stabilize or preserve. Depolarizing noise stabilizes the maximally mixed state. Thermal damping stabilizes a Gibbs or oscillator thermal state. Zero-temperature amplitude damping stabilizes the ground state. A projective dephasing channel preserves classical mixtures in the measurement basis.

The fixed set also tells you what information survives the noise. If the fixed set is large, some classical or quantum information may be protected. This is the entry point to Decoherence-Free Subspaces, Pointer States, and reservoir engineering, where the goal is to design a dissipative channel or generator whose desired state is the unique attracting fixed point.

The most useful diagnostic question is not only “what is fixed?” but also “what is attracted, how fast, and from which initial states?”

Common Mistakes

• Reading $\Psi\circ\Phi$ left to right instead of applying $\Phi$ first.
• Composing finite-step error probabilities by ordinary addition outside the small-probability limit.
• Treating tensor-product noise $\Phi_A\otimes\Phi_B$ as the same thing as serial composition.
• Assuming a fixed state is automatically unique or attracting.
• Applying finite-dimensional fixed-point existence arguments to infinite-dimensional channels without compactness or energy assumptions.
• Forgetting that a finite-time fixed point need not be a continuous-time steady state.
• Ignoring output spaces: channels can be composed only when the output space of the first map matches the input space of the next map.
• Confusing fixed density operators with conserved observables of the adjoint channel.

References

• M. M. Wolf, Quantum Channels & Operations: Guided Tour (lecture notes, 2012).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 10th anniversary ed. (2010).
• A. Arias, A. Gheondea, and S. Gudder, “Fixed points of quantum operations,” Journal of Mathematical Physics 43, 5872-5881 (2002).
• B. Baumgartner and H. Narnhofer, “Analysis of quantum semigroups with GKS-Lindblad generators: II. General,” Journal of Physics A: Mathematical and Theoretical 41, 395303 (2008).
• R. Alicki and K. Lendi, Quantum Dynamical Semigroups and Applications, Springer, 2nd ed. (2007).

Exercises

1. Composite Kraus operators. Suppose $\Phi$ has Kraus operators $\{K_\alpha\}$ and $\Psi$ has Kraus operators $\{L_\beta\}$. Prove that $\{L_\beta K_\alpha\}_{\beta,\alpha}$ is a Kraus representation of $\Psi\circ\Phi$.

Solution

Insert the Kraus form of $\Phi$ into $\Psi$:

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

Thus the composite Kraus operators are $M_{\beta\alpha}=L_\beta K_\alpha$. If both channels are trace preserving, then

$$\sum_{\beta,\alpha} M_{\beta\alpha}^\dagger M_{\beta\alpha} = I,$$

as shown in the main text.

2. Depolarizing powers. Let $\mathcal D_\lambda(\rho)=I/d+\lambda(\rho-I/d)$. Show that $\mathcal D_\lambda^n=\mathcal D_{\lambda^n}$.

Solution

Write $\rho=I/d+\tau$ with $\operatorname{Tr}\tau=0$. Then

$$\mathcal D_\lambda(\rho) = \frac{I}{d} + \lambda\tau.$$

After $n$ applications,

$$\tau \mapsto \lambda^n\tau.$$

Therefore

$$\mathcal D_\lambda^n(\rho) = \frac{I}{d} + \lambda^n \left( \rho-\frac{I}{d} \right) = \mathcal D_{\lambda^n}(\rho).$$

3. Amplitude-damping probabilities. Two amplitude-damping channels in the same basis have decay probabilities $\gamma_1$ and $\gamma_2$. Find the total decay probability.

Solution

The excited population survives the first channel with probability $1-\gamma_1$ and the second with probability $1-\gamma_2$. The total survival probability is

$$(1-\gamma_2)(1-\gamma_1).$$

Therefore the total decay probability is

$$\gamma_{\mathrm{tot}} = 1-(1-\gamma_2)(1-\gamma_1).$$

For small probabilities, this is approximately $\gamma_1+\gamma_2$, but the exact expression includes the product term.

4. Fixed states of dephasing. Let $\Phi(\rho)=\sum_a P_a\rho P_a$ for orthogonal projectors $\{P_a\}$ with $\sum_aP_a=I$. Show that $\rho$ is fixed exactly when it is block diagonal in this decomposition.

Solution

Decompose the state into blocks:

$$\rho = \sum_{a,b} P_a\rho P_b.$$

The channel removes the off-block terms:

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

Thus $\Phi(\rho)=\rho$ exactly when

$$P_a\rho P_b=0 \qquad a\ne b.$$

That is the statement that $\rho$ is block diagonal in the projective decomposition.

5. Fixed does not mean attracting. Give an example of a channel with many fixed states that is not mixing to one state.

Solution

The identity channel is the simplest example:

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

Every state is fixed, so there is no unique attracting state. A less trivial example is a unitary channel $\Phi(\rho)=U\rho U^\dagger$. States that commute with $U$ are fixed, while coherences between different eigenvalue sectors generally rotate rather than decay.