Channel Composition and Fixed Points

Composing channels predicts successive state changes, provided each step really acts through the stated channel on the state it receives. Repeating a fixed channel gives a discrete-time evolution. Its fixed states need not attract other states: a channel can have one unique fixed state while other states keep oscillating.

This page develops composition and fixed points in finite dimensions, then compares Gaussian channels and continuous-time evolution. Infinite-dimensional fixed-point claims need additional assumptions.

Helpful background. Kraus operators let us multiply two channels explicitly. Quantum operations distinguishes deterministic channels from selected outcomes.

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 factor is completely positive and trace nonincreasing and the other is a channel, their composition remains completely positive and trace nonincreasing. Its trace gives the probability of the selected sequence; it is deterministic only if that trace equals one for every input state.

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 models a discrete-time process whose reduced step is the same channel for every input at every use. Fresh or reset independent environments provide one realization. Reusing an environment that retains correlations can invalidate the channel-power model; see the two-interaction example in Stinespring Representation. 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 $\tau\mapsto\lambda\tau$	$\tau\mapsto\lambda^n\tau$
qubit dephasing	upper coherence $\rho_{01}\mapsto\lambda\rho_{01}$	$\rho_{01}\mapsto\lambda^n\rho_{01}$
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 attenuation, $0\le\eta<1$	the finite-temperature bath state
pure bosonic loss, $0\le\eta<1$	the vacuum state

The existence statement above is finite dimensional. Infinite-dimensional channels need additional compactness, tightness, or energy-bound assumptions. Nonzero additive Gaussian noise, for example, keeps increasing at least one variance and has no fixed state with finite second moments. At $\eta=1$, the oscillator channels in the table are identities, so every state is fixed.

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

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

In finite dimensions (or for bounded observables in the infinite-dimensional setting), 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 finite-dimensional CPTP map, the whole spectrum lies in the unit disk. If $1$ is a simple eigenvalue and every other eigenvalue has modulus less than one, 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.

Uniqueness does not guarantee convergence. Take $K_0=|1\rangle\langle0|$ and $K_1=|0\rangle\langle1|$. Their channel deletes coherence and swaps the diagonal populations:

$$\Phi(\rho)=\begin{pmatrix}\rho_{11}&0\\0&\rho_{00}\end{pmatrix}.$$

Its only fixed density operator is $I/2$, but $|0\rangle$ and $|1\rangle$ alternate forever. The eigenoperator $Z$ has eigenvalue $-1$, so the stationary state is unique without being attracting.

For a mixing channel, the largest remaining eigenvalue modulus controls asymptotic exponential decay. Nontrivial Jordan blocks at eigenvalues inside the disk can add polynomial prefactors, so that modulus alone is not an exact finite-step error bound.

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, intersect the affine solution space with the Bloch ball. A valid qubit channel must have at least one physical solution; finding none signals a defective model or numerical calculation.

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 Lüders 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.

Gaussian composition: loss and amplification in different orders

The moment representation makes Gaussian composition explicit. If channel 1 has $(X_1,Y_1,d_1)$ and channel 2 has $(X_2,Y_2,d_2)$, applying 1 first gives

$$X=X_2X_1,\qquad Y=X_2Y_1X_2^{\mathsf T}+Y_2,\qquad d_0=X_2d_1+d_2.$$

The first channel’s added noise is transformed by the second channel as well as the signal.

For example, combine pure loss with $\eta=1/2$ and a quantum-limited amplifier with $G=2$, in the convention $V_{\mathrm{vac}}=I_2/2$. Both orders restore the first moments, because $\sqrt G\sqrt\eta=1$. But the noise differs:

$$\begin{aligned} \text{loss then gain:}\quad&V\mapsto V+I_2,\\ \text{gain then loss:}\quad&V\mapsto V+\tfrac12I_2. \end{aligned}$$

Neither restores the input state. Amplifying a signal after loss cannot undo the quantum noise introduced during attenuation. Gaussian Channels derives the noise floors used here.

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.

References

• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018). Author’s text.
• C. Weedbrook, S. Pirandola, R. García-Patrón, N. J. Cerf, T. C. Ralph, J. H. Shapiro, and S. Lloyd, “Gaussian quantum information,” Reviews of Modern Physics 84, 621–669 (2012). DOI.
• M. M. Wolf, Quantum Channels & Operations: Guided Tour, lecture notes (5 July 2012), Sections 6.4–6.5 and 8.7. Author’s notes.