4 Quantum Channels and Positive Maps

This chapter develops the theory of positive maps and quantum channels on matrix algebras, following [ Wol12 ] .

4.1 Positive and completely positive maps

Definition 4.1.1 Positive map

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A linear map $E : M_{n}(\mathbb {C}) \to M_{m}(\mathbb {C})$ is positive if $E(X) \ge 0$ whenever $X \ge 0$, where $X \ge 0$ means that $X$ is positive semidefinite.

Definition 4.1.2 Trace-preserving map

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A linear map $E : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is trace-preserving if $\operatorname{tr}(E(X)) = \operatorname{tr}(X)$ for all $X$.

Remark 4.1.3

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We use the Loewner order on matrices: $X \le Y$ means $Y - X \ge 0$.

Definition 4.1.4 Completely positive map

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A linear map $E : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is completely positive (CP) if it admits a Kraus representation: there exist operators $\{ K_i\} _{i=0}^{r-1}$ with $K_i \in M_{D}(\mathbb {C})$ such that, for every $X \in M_{D}(\mathbb {C})$,

\begin{align} E(X) & = \sum _{i=0}^{r-1} K_i X K_i^\dagger . \label{eq:channel_kraus_representation} \end{align}

The Kraus representation also gives entrywise positivity on every positive block matrix by Theorem 4.2.1, and hence the associated completely positive map between matrix $C^*$-algebras in Theorem 4.2.2.

4.2 Kraus representations and complete positivity

This section proves the two complete-positivity consequences stated after Definition 4.1.4: entrywise positivity on block matrices and the associated completely positive map between matrix $C^*$-algebras.

Theorem 4.2.1 Entrywise complete positivity from a Kraus representation

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Let $E : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ be completely positive in the Kraus sense. For every $k$ and every positive block matrix $M \in M_k(M_{D}(\mathbb {C}))$, the entrywise image $(E(M_{ab}))_{a,b}$ is again positive in $M_k(M_{D}(\mathbb {C}))$.

Proof ▶

Write $E$ in the Kraus form (??). For each $i$, let $d_i$ be the block-diagonal matrix with $K_i$ on every diagonal block. Then

\begin{align} (E(M_{ab}))_{a,b} & =\sum _i d_i M d_i^\dagger . \notag \end{align}

Each term on the right is positive because conjugation preserves positivity, and a finite sum of positive matrices is positive.

Theorem 4.2.2 Kraus maps as abstract completely positive maps

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Every Kraus-represented completely positive map $E : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ determines a completely positive map between the matrix $C^*$-algebras in the entrywise positivity sense.

Proof ▶

The defining entrywise positivity condition is exactly Theorem 4.2.1.

Thus the Kraus formulation of Definition 4.1.4 has the two consequences cited there: entrywise complete positivity and the associated abstract completely positive map.

Theorem 4.2.3 CP maps are positive

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Every completely positive map is positive.

Proof ▶

By the Kraus representation (??), if $X \ge 0$, then each summand $K_i X K_i^\dagger \ge 0$ because conjugation preserves positive semidefiniteness. Hence their sum is positive semidefinite.

4.3 Quantum channels

Definition 4.3.1 Quantum channel

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A linear map $E : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is a quantum channel if it is both completely positive and trace-preserving (CPTP), following [ Wol12 ] .

Theorem 4.3.2 Channels are positive

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Every quantum channel is positive.

Proof ▶

A quantum channel is completely positive by definition, so Theorem 4.2.3 applies.

4.4 Rectangular positive maps, trace bounds, and density matrices

Definition 4.4.1 Associated positive linear map

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A positive matrix map $E : M_{n}(\mathbb {C}) \to M_{m}(\mathbb {C})$ determines the same linear map regarded as a positive linear map: if $A,B\in M_{n}(\mathbb {C})$ and $A \le B$, then $E(A) \le E(B)$ in $M_{m}(\mathbb {C})$.

Definition 4.4.2 Positive map between matrix algebras

A linear map $T : M_{D}(\mathbb {C}) \to M_{D'}(\mathbb {C})$ between matrix algebras of possibly different dimensions is positive if $T(X) \ge 0$ whenever $X \ge 0$, where $X \ge 0$ means that $X$ is positive semidefinite. For $D' = D$ this is the notion of Definition 4.1.1.

Definition 4.4.3 Trace-preserving map between matrix algebras

A linear map $T : M_{D}(\mathbb {C}) \to M_{D'}(\mathbb {C})$ between matrix algebras of possibly different dimensions is trace-preserving if $\operatorname{tr}(T(X)) = \operatorname{tr}(X)$ for all $X$. For $D' = D$ this is the notion of Definition 4.1.2.

Theorem 4.4.4 The trace bounds a positive matrix

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Let $D\geq 1$ and let $P\in M_{D}(\mathbb {C})$ be positive semidefinite. Then

\begin{align} P & \preceq \operatorname{tr}(P)\mathbb {1}. \label{eq:channel_psd_trace_bound} \end{align}

Proof ▶

Diagonalize $P$ with non-negative eigenvalues $\lambda _1,\ldots ,\lambda _D$. Each is at most their sum $\operatorname{tr}(P)$, so every eigenvalue of $\operatorname{tr}(P)\mathbb {1}-P$ is non-negative.

Theorem 4.4.5 Positive maps preserve Hermiticity

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If $E:M_{n}(\mathbb {C})\to M_{m}(\mathbb {C})$ is positive and $X\in M_{n}(\mathbb {C})$ satisfies $X = X^\dagger $, then $E(X) = E(X)^\dagger $ in $M_{m}(\mathbb {C})$.

Proof ▶

Regard $E$ as the associated positive linear map (Definition 4.4.1). For positive linear maps between matrix $C^*$-algebras, $E(Y^\ast )=E(Y)^\ast $. Since $X=X^\dagger $ means $X=X^\ast $, one obtains $E(X)^\dagger =E(X)^\ast =E(X^\ast )=E(X)$.

Definition 4.4.6 Density matrices

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The set of density matrices in $M_{D}(\mathbb {C})$ is

\begin{align} \mathcal{D}_D & =\{ \rho \in M_{D}(\mathbb {C}) : \rho \ge 0,\ \operatorname{tr}(\rho ) = 1\} . \label{eq:channel_density_matrices} \end{align}

4.5 Maximally entangled states and Choi matrices

The maximally entangled state and the flip operator give the basic matrices used by the Choi–Jamiołkowski correspondence.

Definition 4.5.1 Maximally entangled state

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For every $d\in \mathbb {N}$, define $|\Omega \rangle \! \langle \Omega |$ on $\mathbb {C}^d\otimes \mathbb {C}^d$ using

\begin{align} |\Omega \rangle & = \frac{1}{\sqrt{d}} \sum _{j=0}^{d-1} |j,j\rangle . \label{eq:representations_max_entangled} \end{align}

When $d\geq 1$, this is the maximally entangled state: a rank-one projector with $(|\Omega \rangle \! \langle \Omega |)_{(i_1,i_2),(j_1,j_2)} = \frac{1}{d}\delta _{i_1 i_2}\delta _{j_1 j_2}$ and $\operatorname{tr}(|\Omega \rangle \! \langle \Omega |)=1$.

Definition 4.5.2 Flip operator

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The flip operator $F$ on $\mathbb {C}^d \otimes \mathbb {C}^d$ is

\begin{align} F & = \sum _{i,j=0}^{d-1} |ij\rangle \! \langle ji|. \label{eq:representations_flip} \end{align}

Its matrix entries are

\begin{align} F_{(i_1,i_2),(j_1,j_2)} & = \begin{cases} 1, & i_1=j_2 \text{ and } i_2=j_1,\\ 0, & \text{otherwise}. \end{cases} \notag \end{align}

Definition 4.5.3 Two-dimensional antisymmetric vector

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In $\mathbb {C}^2 \otimes \mathbb {C}^2$, let $v = |01\rangle - |10\rangle $. Equivalently, $v_{(0,1)}=1$, $v_{(1,0)}=-1$, and all other coefficients are zero.

Definition 4.5.4 Tensor product of a map with identity

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Given a linear map $T : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$, the tensor extension $T \otimes \operatorname{id}$ acts on bipartite matrices $X \in M_{D \times D}(\mathbb {C})$ by applying $T$ to each “slice”:

\begin{align} (T \otimes \operatorname{id})(X)_{(i_1,i_2),(j_1,j_2)} & = (T(X^{(i_2,j_2)}))_{i_1 j_1}, \label{eq:representations_tensor_map_id} \end{align}

where $X^{(i_2,j_2)}_{ab} = X_{(a,i_2),(b,j_2)}$ is the bipartite slice.

The Choi–Jamiołkowski isomorphism bends the input legs of the channel $T$ around the maximally entangled pair $|\Omega \rangle $, presenting the map as the matrix $\tau =(T\otimes \operatorname{id})|\Omega \rangle \! \langle \Omega |$ on the doubled space:

$\begin{tenkz}[ rows={wire,wire}, east={cup=$\ketbra{\Omega}{\Omega}$}, west label=$\tau$ ] \tnghost{} & \tnX[wires=2]{T\otimes\id} & \tnghost{} \\ \tnghost{} & & \tnghost{} \end{tenkz}$

Definition 4.5.5 Choi matrix

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The Choi matrix of a linear map $T : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is

\begin{align} \tau & = (T \otimes \operatorname{id})(|\Omega \rangle \! \langle \Omega |) \in M_{D \times D}(\mathbb {C}). \label{eq:representations_choi} \end{align}

Concretely, $\tau _{(i_1,i_2),(j_1,j_2)} = \frac{1}{D}\, (T(E_{i_2 j_2}))_{i_1 j_1}$ where $E_{i_2 j_2}$ is the matrix unit. This is the Choi–Jamiol\- kowski convention of [ Wol12 , Proposition 2.1 ] .

Theorem 4.5.6 Choi matrix of the identity map

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The identity channel has Choi matrix $|\Omega \rangle \! \langle \Omega |$.

Proof ▶

Apply the identity map in the definition of the Choi matrix.

Theorem 4.5.7 Choi matrix of transposition

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Let $\theta (X)=X^T$ be matrix transposition on $M_{D}(\mathbb {C})$, with $D \ge 1$. Then

\begin{align} (\theta \otimes \operatorname{id})(|\Omega \rangle \! \langle \Omega |) & = \frac{1}{D}F. \label{eq:representations_transpose_choi} \end{align}

This is [ Wol12 , Equation 3.1 ] .

Proof ▶

The $(i_2,j_2)$ slice of $|\Omega \rangle \! \langle \Omega |$ is $D^{-1}E_{i_2j_2}$. Transposition sends this matrix unit to $D^{-1}E_{j_2i_2}$, whose $(i_1,j_1)$ entry is $D^{-1}$ exactly when $i_1=j_2$ and $i_2=j_1$.

Theorem 4.5.8 Complete positivity iff the Choi matrix is positive semidefinite

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Let $D\geq 1$. A linear map $T : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is completely positive (in the Kraus sense) if and only if its Choi matrix $\tau \ge 0$.

Proof ▶

($\Rightarrow $) If $T(X) = \sum _i K_i X K_i^\dagger $, then $\tau = \sum _i |v_i\rangle \! \langle v_i|$ where $(v_i)_{(a,b)} = c\, K_i(a,b)$ and $c = 1/\sqrt{D}$. This is a sum of rank-one PSD matrices.

($\Leftarrow $) If $\tau \ge 0$, write $\tau = \sum _m |v_m\rangle \! \langle v_m|$ by spectral decomposition, define $K_m(a,b) = v_m(a,b)/c$, and recover $T(X) = \sum _m K_m X K_m^\dagger $ by linearity on matrix units.

Unitary changes of Kraus operators preserve the corresponding completely positive map.

Theorem 4.5.9 Unitary freedom in Kraus operators

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Let $U$ be a unitary $r \times r$ matrix ($U^\dagger U = \mathbb {1}$) and suppose $K_j = \sum _\ell U_{j\ell } \tilde{K}_\ell $. Then $\{ K_j\} $ and $\{ \tilde{K}_\ell \} $ define the same Kraus map: for every $X \in M_{D}(\mathbb {C})$,

\begin{align} \sum _j K_j X K_j^\dagger & = \sum _\ell \tilde{K}_\ell X \tilde{K}_\ell ^\dagger . \notag \end{align}

Proof ▶

Substitute the combination formula, expand the triple sum over $j,\ell ,\ell '$, swap summation order, and use the entry-wise form of $U^\dagger U = \mathbb {1}$ to collapse to $\ell = \ell '$ diagonal terms.

Theorem 4.5.10 Rectangular Kraus freedom

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If $\sum _\alpha B_\alpha X B_\alpha ^\dagger = \sum _j A_j X A_j^\dagger $ for all $X$ and $r_2 \le r_1$ (where $\{ B_\alpha \} _{\alpha =0}^{r_1-1}$ is the larger family and $\{ A_j\} _{j=0}^{r_2-1}$ the smaller), then there exists a rectangular isometry $V$ ($r_1 \times r_2$, $V^\dagger V = \mathbb {1}_{r_2}$) such that $B_\alpha = \sum _j V_{\alpha j}\, A_j$.

Proof ▶

Pad the smaller family by zero operators to a family $A'$ of size $r_1$. Evaluating equality of the two maps on matrix units and taking matrix entries gives equality of the Gram matrices of the Stinespring vectors, $M_B^\dagger M_B=M_{A'}^\dagger M_{A'}$. Hence the assignment from each column of $M_{A'}$ to the corresponding column of $M_B$ is well-defined and isometric on their common span. Extend this isometry to the ambient space $\mathbb {C}^{r_1}$ and let $W$ be its unitary matrix. Restricting $W$ to the original $r_2$ coordinates gives a rectangular matrix $V$ with $V^\dagger V=\mathbb {1}_{r_2}$. Reading the column identity entrywise yields $B_\alpha =\sum _jV_{\alpha j}A_j$.

4.6 Irreducibility

Definition 4.6.1 Orthogonal projection

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A matrix $P \in M_{D}(\mathbb {C})$ is an orthogonal projection if $P = P^\dagger $ and $P^2 = P$.

Definition 4.6.2 Irreducible map

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A linear map $E : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is irreducible if whenever $P$ is an orthogonal projection satisfying $E(P M_{D}(\mathbb {C}) P) \subseteq P M_{D}(\mathbb {C}) P$, then $P = 0$ or $P = \mathbb {1}$. This is [ Wol12 , Theorem 6.2(1) ] . The definition applies to any linear map; complete positivity is not required.

Theorem 4.6.3 Orthogonal projections and involutive-algebra projections

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For a complex square matrix, the two descriptions $P=P^\dagger $, $P^2=P$ and $P^*=P$, $P^2=P$ are equivalent.

Proof ▶

Since the involution on $M_{D}(\mathbb {C})$ is conjugate transpose, $P^*=P^\dagger $. Therefore $P^\dagger =P$, $P^2=P$ if and only if $P^*=P$, $P^2=P$.

Theorem 4.6.4 Coisometric transport of a star projection

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Let $Q$ be a self-adjoint idempotent on a finite-dimensional space, and let $U$ be a coisometry onto that space, so that $UU^\dagger =\mathbb {1}$. Then $U^\dagger Q U$ is self-adjoint and idempotent.

Proof ▶

Self-adjointness follows from $(U^\dagger Q U)^\dagger =U^\dagger Q^\dagger U=U^\dagger Q U$. Idempotence follows from

\begin{align} (U^\dagger Q U)^2 & =U^\dagger Q(UU^\dagger )QU =U^\dagger Q^2U =U^\dagger Q U. \notag \end{align}

Lemma 4.6.5 Pairwise orthogonality from a projection resolution

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Let $(Q_k)_k$ be a finite family of orthogonal projections satisfying $\sum _k Q_k=\mathbb {1}$. If $k\ne \ell $, then $Q_kQ_\ell =0$.

Proof ▶

Multiplying the resolution of the identity on both sides by $Q_k$ gives $\sum _j Q_kQ_jQ_k=Q_k$. Each summand is positive semidefinite because $Q_kQ_jQ_k=(Q_jQ_k)^*(Q_jQ_k)$, while the $j=k$ summand is already $Q_k$. Subtracting that term and taking traces gives

\begin{align} 0 & =\sum _{j\ne k}\operatorname{tr}((Q_jQ_k)^*(Q_jQ_k)) =\sum _{j\ne k}\lVert Q_jQ_k\rVert _{\mathrm F}^{2}. \notag \end{align}

Hence $Q_jQ_k=0$ for $j\ne k$, and taking adjoints gives $Q_kQ_j=0$.

Theorem 4.6.6 Complement of an orthogonal projection

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If $P \in M_{D}(\mathbb {C})$ is an orthogonal projection, then $\mathbb {1}-P$ is an orthogonal projection.

Proof ▶

In the involutive algebra $M_{D}(\mathbb {C})$, $P$ satisfies $P^\dagger =P$ and $P^2=P$. Hence $(\mathbb {1}-P)^\dagger =\mathbb {1}-P$ and $(\mathbb {1}-P)^2=\mathbb {1}-2P+P^2=\mathbb {1}-P$. Thus $\mathbb {1}-P$ is an orthogonal projection.

Theorem 4.6.7 Injectivity implies irreducibility

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If $A$ is an injective MPS tensor, then its transfer map $\mathcal{E}_A$ is irreducible.

Proof ▶

If $P$ is an invariant projection for $\mathcal{E}_A$, then $(\mathbb {1}-P)A^iP=0$ for all $i$. Since the $\{ A^i\} $ span $M_{D}(\mathbb {C})$, $(\mathbb {1}-P)MP=0$ for all $M$, so $P=0$ or $P=\mathbb {1}$.

4.7 Transfer maps

Theorem 4.7.1 Transfer map is CP

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The transfer map $\mathcal{E}_A(X)=\sum _i A^iX(A^i)^\dagger $ is completely positive.

Proof ▶

It is already written in the Kraus form (??), with Kraus operators $\{ A^i\} $.

Theorem 4.7.2 Transfer map is a channel

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If $\sum _i(A^i)^\dagger A^i=\mathbb {1}$, then the transfer map $\mathcal{E}_A(X)=\sum _i A^iX(A^i)^\dagger $ is a quantum channel (CPTP).

Proof ▶

Complete positivity is Theorem 4.7.1. Trace preservation follows by cyclicity and the normalization hypothesis:

\begin{align} \operatorname{tr}(\mathcal{E}_A(X)) & =\sum _i\operatorname{tr}(A^iX(A^i)^\dagger ) =\operatorname{tr}\! \left(\left(\sum _i(A^i)^\dagger A^i\right)X\right) =\operatorname{tr}(X). \notag \end{align}

4.8 Peripheral spectrum and primitivity

Definition 4.8.1 Peripheral spectrum

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The peripheral spectrum of a bounded linear operator $T$ is the set of spectral values $\mu $ with $|\mu |$ equal to the spectral radius of $T$.

Definition 4.8.2 Peripheral eigenvalues

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The peripheral eigenvalues of a linear map $E$ on a finite-dimensional space are the eigenvalues $\lambda $ on the unit circle: $|\lambda | = 1$. For a channel, the spectral radius is $1$ [ Wol12 , Proposition 6.1 ] , so these coincide with the eigenvalues of maximal modulus.

The condition $|\lambda | = 1$ is appropriate for channels, since the spectral radius of a channel is $1$; for a general linear map with spectral radius $r \neq 1$ the condition would generalise to $|\lambda | = r$.

Remark 4.8.3 Peripheral conventions for channels

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The spectral-radius convention of Definition 4.8.1 is the usual bounded-operator notion. For channels the spectral radius is $1$, so the channel arguments below use the unit-circle set of Definition 4.8.2.

Definition 4.8.4 Primitive map

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A linear map is primitive if its only peripheral eigenvalue is $\lambda = 1$, i.e. $\mathrm{peripheral}(E) = \{ 1\} $. For channels this corresponds to [ Wol12 , Theorem 6.7 ] . This definition applies to any linear endomorphism, not only to channels.

4.9 Fixed-point projection

Definition 4.9.1 Fixed-point projection

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Given a linear map $E:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})$ and a fixed point $\rho $ with $E(\rho )=\rho $ and $\operatorname{tr}(\rho )\neq 0$, the fixed-point projection is the rank-one map

\begin{align} P(X) & =\frac{\operatorname{tr}(X)}{\operatorname{tr}(\rho )}\rho . \label{eq:channel_fixed_point_proj} \end{align}

This projects onto the span of $\rho $ along the kernel of the trace functional. When the fixed point is unique up to scaling, this span is the full fixed-point space.

Theorem 4.9.2 Power decomposition

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If $E$ is trace-preserving and $E(\rho )=\rho $, then, for $n\geq 1$,

\begin{align} E^n & =P+(E-P)^n, \label{eq:channel_power_decomp} \end{align}

where $P$ is the fixed-point projection.

Proof ▶

By (??), $P$ is idempotent. The fixed-point relation $E(\rho )=\rho $ gives $E\circ P=P$, and trace preservation gives $P\circ E=P$. These imply $(E-P)\circ P=0$ and $P\circ (E-P)=0$, so in the binomial expansion of $(P+(E-P))^n$ all cross terms vanish, yielding (??).

4.10 Peripheral eigenvalues and powering

Theorem 4.10.1 $1$ is a peripheral eigenvalue

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If $E$ has a nonzero fixed point $\rho \neq 0$ with $E(\rho )=\rho $, then $1$ is a peripheral eigenvalue of $E$.

Proof ▶

$\rho $ is an eigenvector with eigenvalue $1$, and $|1|=1$.

Theorem 4.10.2 Finiteness of peripheral eigenvalues

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On a finite-dimensional space, the set of peripheral eigenvalues is finite.

Proof ▶

Peripheral eigenvalues are a subset of all eigenvalues, which is a finite set in finite dimensions.

Theorem 4.10.3 Power-stable peripheral eigenvalues are roots of unity

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Let $E$ be a linear endomorphism on a finite-dimensional space. If $\mu $ is a peripheral eigenvalue of $E$ and $\mu ^n$ is an eigenvalue of $E$ for every $n\geq 1$, then $\mu $ is a root of unity.

Proof ▶

Since $\mu ^n$ is an eigenvalue for every positive $n$, the pigeonhole principle on the finite eigenvalue set gives $\mu ^a=\mu ^b$ for some $a\neq b$, hence $\mu ^{|a-b|}=1$.

Theorem 4.10.4 Peripheral powers imply root of unity

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If the set of peripheral eigenvalues of a linear endomorphism $E$ on a finite-dimensional space is closed under powers ($\mu \in \mathrm{peripheral}(E)$ and $n\geq 1$ imply $\mu ^n\in \mathrm{peripheral}(E)$), then every peripheral eigenvalue is a root of unity.

Proof ▶

The closure hypothesis provides $\mu ^n\in \mathrm{peripheral}(E)$ for all positive $n$, which in particular means $\mu ^n$ is an eigenvalue. Theorem 4.10.3 then gives $\mu ^p=1$ for some $p{\gt}0$.

4.10.1 Periodicity removal by powering

Remark 4.10.1.1

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The results in this subsection are purely spectral: they use only finiteness of a set of complex numbers and the spectral mapping theorem for powers. In particular, they do not rely on complete positivity. For transfer maps, replacing a tensor by its $p$-blocked tensor replaces $\mathcal{E}_A$ by $\mathcal{E}_A^p$, so this powering step is the operator-theoretic form of blocking.

Lemma 4.10.1.2 Common exponent for a finite set of roots of unity

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Let $s\subseteq \mathbb {C}$ be a finite set. Assume that for every $\mu \in s$ there exists an exponent $p_\mu {\gt}0$ with $\mu ^{p_\mu }=1$. Then there exists $p{\gt}0$ such that $\mu ^p=1$ for all $\mu \in s$.

Proof ▶

Take $p$ to be the least common multiple of the exponents $p_\mu $.

Lemma 4.10.1.3 Powering collapses peripheral eigenvalues

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Let $E:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})$ be a linear map, and assume that $E$ has a nonzero fixed point $\rho \neq 0$. Let $p{\gt}0$. If every peripheral eigenvalue $\mu $ of $E$ satisfies $\mu ^p=1$, then $\mathrm{peripheral}(E^p)=\{ 1\} $.

Proof ▶

The fixed point $\rho $ gives $1\in \mathrm{peripheral}(E^p)$. For the reverse inclusion, let $\nu \in \mathrm{peripheral}(E^p)$. Spectral mapping gives $\nu =\mu ^p$ for some $\mu \in \operatorname{spec}(E)$. From $|\nu |=1$ and $p{\gt}0$ we obtain $|\mu |=1$, hence $\mu \in \mathrm{peripheral}(E)$. The hypothesis $\mu ^p=1$ then forces $\nu =1$.

For a normalized MPS tensor, the transfer map is naturally trace-preserving but need not be unital. In general one should not expect a single gauge choice to make it both unital and trace-preserving.

Remark 4.10.1.4 Bi-canonical special case

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When a Kraus map happens to be both unital and trace-preserving (the “bi-canonical” case), the Kadison–Schwarz equality at peripheral eigenvectors (Theorem 5.2.3) shows directly that the peripheral eigenvalues are closed under powers, and hence are roots of unity by Theorem 4.10.4. It requires both normalizations simultaneously. The more general adjoint-fixed-point formulation below covers the case where only unitality and a positive definite adjoint fixed point are available.

4.10.2 Peripheral closure via adjoint fixed point

The Kraus map, its adjoint, and the unital normalization are those of Definitions 5.1.1, 5.1.2, and 5.1.3. The following results combine the canonical Kadison–Schwarz and multiplicative-domain theorems of Chapter 5 with a positive definite fixed point of the adjoint map, replacing trace preservation by the weighted-trace argument in [ CPGSV16 , Appendix A ] .

Lemma 4.10.2.1 Peripheral closure via adjoint fixed point

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Let $E(X)=\sum _iK_iXK_i^\dagger $ be a Kraus map that is unital ($\sum _iK_iK_i^\dagger =\mathbb {1}$), and assume:

the adjoint Kraus map $E^*(X)=\sum _iK_i^\dagger XK_i$ has a positive definite fixed point $\rho {\gt}0$, and
$E$ is irreducible.

Then $\mathrm{peripheral}(E)$ is closed under powers: if $\mu \in \mathrm{peripheral}(E)$, then $\mu ^n\in \mathrm{peripheral}(E)$ for every $n\in \mathbb {N}$.

Proof ▶

Let $E(X)=\mu X$ with $|\mu |=1$, and set $G:=E(X^\dagger X)-E(X)^\dagger E(X)$. By (??), $G\geq 0$. Since $E^*(\rho )=\rho $,

\begin{align} \operatorname{tr}(\rho G) & =\operatorname{tr}(\rho E(X^\dagger X)) -\operatorname{tr}(\rho E(X)^\dagger E(X)) \notag \\ & =\operatorname{tr}(E^*(\rho )X^\dagger X) -\operatorname{tr}(\rho E(X)^\dagger E(X)) \notag \\ & =\operatorname{tr}(\rho X^\dagger X) -|\mu |^2\operatorname{tr}(\rho X^\dagger X) =0. \end{align}

Because $\rho {\gt}0$ and $G\geq 0$, this forces $G=0$, so $E(X^\dagger X)=E(X)^\dagger E(X)=X^\dagger X$. Hence $X^\dagger X$ is a positive semidefinite fixed point, which is positive definite by irreducibility (Theorem 6.2.4). Thus $X$ is invertible. Theorem 5.2.2 gives the corresponding intertwining relation with each Kraus operator. Iterating the multiplicative identity of Theorem 5.2.5 gives $E(X^n)=\mu ^nX^n$, so $\mu ^n$ is peripheral.

Theorem 4.10.2.2 Roots of unity via adjoint fixed point

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Let $E(X)=\sum _iK_iXK_i^\dagger $ be a Kraus map that is unital, and assume that the adjoint Kraus map $E^*(X)=\sum _iK_i^\dagger XK_i$ has a positive definite fixed point and that $E$ is irreducible. Then every peripheral eigenvalue of $E$ is a root of unity.

Proof ▶

Combine the closure-under-powers result (Lemma 4.10.2.1) with Theorem 4.10.4.

Remark 4.10.2.3

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Lemma 4.10.2.1 and Theorem 4.10.2.2 cover the bi-canonical (unital + TP) case as well: trace preservation gives $E^*(\mathbb {1})=\mathbb {1}$, which is a positive definite fixed point of the adjoint. The adjoint-fixed-point formulation matches the argument in [ CPGSV16 , Appendix A ] , which uses a faithful invariant state rather than bi-canonicality. Theorem 4.10.2.2 establishes the root-of-unity conclusion in this adjoint-fixed-point formulation. In the same adjoint-fixed-point formulation, Chapter 7 proves the corresponding cyclic structure: the cyclic group of peripheral eigenvalues is Theorem 7.1.1, and the cyclic projection decomposition is Theorem 7.9.1.1.

Definition 4.10.2.4 Channel period

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The period of a channel $E$ is the number of peripheral eigenvalues, counted without multiplicity: $p(E):=|\mathrm{peripheral}(E)|$.

4.11 Primitivity and the complementary transfer-map gap

Theorem 4.11.1 Primitivity equals period one

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A linear map with a nonzero fixed point is primitive if and only if its period is $1$.

Proof ▶

If $\mathrm{peripheral}(E)=\{ 1\} $, then $|\mathrm{peripheral}(E)|=1$. Conversely, if $|\mathrm{peripheral}(E)|=1$, then $1\in \mathrm{peripheral}(E)$ by Theorem 4.10.1, so the only element is $1$.

Remark 4.11.2 Period-one characterization

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The period $p(E)$ counts peripheral eigenvalues without multiplicity. Theorem 4.11.1 is therefore the period-one characterization of primitivity. The spectral estimates below are stated directly with Definition 4.8.4, since they use the absence of peripheral eigenvalues other than $1$, not a separate counting hypothesis.

Theorem 4.11.3 Complementary transfer-map gap for primitive maps

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Let $E:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})$ be trace-preserving, and let $\rho \neq 0$ satisfy $E(\rho )=\rho $ and $\operatorname{tr}(\rho )\neq 0$. Let $P$ be the fixed-point projection associated to $\rho $, as defined in (??). Assume that:

$E$ is primitive;
every eigenvalue $\mu $ of $E$ satisfies $|\mu |\le 1$;
whenever $E(X)=X$ and $\operatorname{tr}(X)=0$, one has $X=0$.

Then every eigenvalue $\nu $ of $E-P$ satisfies $|\nu |{\lt}1$.

Proof ▶

Let $(E-P)(X)=\nu X$ with $X\neq 0$. Then $E(X)=\nu X+P(X)$. If $\operatorname{tr}(X)=0$, then $P(X)=0$, so $E(X)=\nu X$ and hence $\nu $ is an eigenvalue of $E$ with $|\nu |\le 1$. If $|\nu |=1$, primitivity forces $\nu =1$, so $X$ is a trace-zero fixed point of $E$; by assumption this implies $X=0$, a contradiction. Thus $|\nu |{\lt}1$.

If instead $\operatorname{tr}(X)\neq 0$, trace preservation and $\operatorname{tr}(P(X))=\operatorname{tr}(X)$, which follows from (??), give

\begin{align} \operatorname{tr}(X) & =\operatorname{tr}(E(X)) =\nu \operatorname{tr}(X)+\operatorname{tr}(P(X)) =\nu \operatorname{tr}(X)+\operatorname{tr}(X). \notag \end{align}

Hence $\nu \operatorname{tr}(X)=0$ and therefore $\nu =0$. In either case, $|\nu |{\lt}1$.