Tensor Network Theory: A formalization blueprint

B Perron–Frobenius Theory for Channels and Transfer Maps: Supporting Results

This appendix supplies the compactness, gauge, similarity, and finite-sum calculations used in Chapter 6.

B.1 Density matrices, Brouwer’s theorem, and Cesàro limits

Write \(\mathcal{D}_D\) for the density matrices of Definition 4.4.6. The next results provide the compact convex domain and the limit argument used for channel fixed points.

Theorem B.1.1 Density matrices are compact
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\(\mathcal{D}_D\) is compact in the entrywise topology on \(M_{D}(\mathbb {C})\).

Proof

The positive semidefinite cone is closed, being the preimage of the closed non-negative cone under continuous quadratic forms. If \(\rho \ge 0\) with \(\operatorname{tr}(\rho ) = 1\), then each entry satisfies \(|\rho _{ij}|^2 \le \rho _{ii}\rho _{jj} \le 1\): the diagonal entries are non-negative and sum to \(1\), and the off-diagonal bound is the positive semidefinite Cauchy–Schwarz inequality. Hence the matrix entries are uniformly bounded, and Heine–Borel gives compactness.

Theorem B.1.2 Density matrices are convex
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\(\mathcal{D}_D\) is convex.

Proof

A convex combination of positive semidefinite matrices is positive semidefinite, and the trace is linear.

Theorem B.1.3 Density matrices are nonempty
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For \(D \ge 1\), \(\mathcal{D}_D \neq \varnothing \).

Proof

\(\frac{1}{D}\mathbb {1}_D\) is a density matrix.

Theorem B.1.4 Channels preserve density matrices

If \(E\) is a quantum channel, then \(E(\mathcal{D}_D) \subseteq \mathcal{D}_D\).

Proof

Complete positivity implies positivity by Theorem 4.2.3, so \(E(\rho ) \ge 0\); trace preservation gives \(\operatorname{tr}(E(\rho )) = \operatorname{tr}(\rho ) = 1\).

Let \(D {\gt} 0\). Every continuous map from the compact convex set of density matrices in \(M_{D}(\mathbb {C})\) to itself has a fixed point.

Proof

The proof starts from Brouwer’s theorem for products of simplices, transfers it to closed cubes, and then to compact retracts of finite-dimensional real normed spaces. The density-matrix set is such a compact retract, using the explicit density retraction.

Lemma B.1.6 Cesàro means remain density matrices

Let \(E\) be a quantum channel and \(\rho \in \mathcal{D}_D\). Then, for every \(N\geq 0\),

\begin{align} \frac{1}{N+1}\sum _{n=0}^{N}E^n(\rho ) & \in \mathcal{D}_D. \label{eq:qpf_cesaro_density} \end{align}
Proof

Every iterate \(E^n(\rho )\) belongs to \(\mathcal{D}_D\) by Theorem B.1.4. Equation (??) is their convex average: positivity is preserved by sums and multiplication by \((N+1)^{-1}\), while its trace is \((N+1)^{-1}\sum _{n=0}^{N}1=1\).

Lemma B.1.7 Subsequential Cesàro limits are fixed points

Let \(E\) be a quantum channel, \(\rho \in \mathcal{D}_D\), and \(\psi :\mathbb {N}\to \mathbb {N}\) satisfy \(\psi (k)\to \infty \). If

\begin{align} \frac{1}{\psi (k)+1}\sum _{n=0}^{\psi (k)}E^n(\rho ) & \longrightarrow \sigma , \notag \end{align}

then \(\sigma \in \mathcal{D}_D\) and \(E(\sigma )=\sigma \).

Proof

Lemma B.1.6 and closedness of the compact set \(\mathcal{D}_D\) give \(\sigma \in \mathcal{D}_D\). The telescope identity yields

\begin{align} E(\sigma _{\psi (k)+1})-\sigma _{\psi (k)+1} & =\frac{E^{\psi (k)+1}(\rho )-\rho }{\psi (k)+1}. \notag \end{align}

Both \(E^{\psi (k)+1}(\rho )\) and \(\rho \) lie in the compact set \(\mathcal{D}_D\), so their difference is uniformly bounded. The right-hand side tends to zero. Continuity of \(E\) and convergence of the subsequence therefore give \(E(\sigma )-\sigma =0\).

B.2 Canonical-gauge algebra

For the right-canonical gauge in Theorem 6.5.1, each summand becomes

\begin{align} (S^{-1}A^iS)(S^{-1}A^iS)^\dagger & =S^{-1}A^iSS^\dagger (A^i)^\dagger (S^\dagger )^{-1} \notag \\ & =S^{-1}A^i\rho (A^i)^\dagger (S^\dagger )^{-1}. \notag \end{align}

Hence

\begin{align} \sum _iA’^i(A’^i)^\dagger & =S^{-1}\mathcal{E}_A(\rho )(S^\dagger )^{-1} =S^{-1}\rho (S^\dagger )^{-1} =\mathbb {1}. \notag \end{align}

For the left-canonical gauge in Theorem 6.5.3,

\begin{align} (SA^iS^{-1})^\dagger (SA^iS^{-1}) & =(S^\dagger )^{-1}(A^i)^\dagger S^\dagger SA^iS^{-1} \notag \\ & =(S^\dagger )^{-1}(A^i)^\dagger \sigma A^iS^{-1}, \notag \end{align}

and therefore

\begin{align} \sum _i(A’^i)^\dagger A’^i & =(S^\dagger )^{-1} \left(\sum _i(A^i)^\dagger \sigma A^i\right)S^{-1} =(S^\dagger )^{-1}\sigma S^{-1} =\mathbb {1}. \notag \end{align}

Finally, for the square-root gauge (??), self-adjointness of \(\sigma ^{1/2}\) gives

\begin{align} (B^i)^\dagger B^i & =\sigma ^{-1/2}(A^i)^\dagger \sigma ^{1/2}\sigma ^{1/2}A^i\sigma ^{-1/2} \notag \\ & =\sigma ^{-1/2}(A^i)^\dagger \sigma A^i\sigma ^{-1/2}. \notag \end{align}

Summing, substituting the adjoint fixed-point equation, and cancelling the outer square-root factors yields

\begin{align} \sum _i(B^i)^\dagger B^i & =\sigma ^{-1/2} \left(\sum _i(A^i)^\dagger \sigma A^i\right)\sigma ^{-1/2} =\sigma ^{-1/2}\sigma \sigma ^{-1/2} =\mathbb {1}. \notag \end{align}
Lemma B.2.1 Square-root trace-preserving gauge is a gauge equivalence
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If \(\sigma \) is positive definite and \(B\) is defined by (??), then \(A\) and \(B\) are gauge equivalent.

Proof

The gauge matrix is \(\sigma ^{1/2}\), which is invertible because \(\sigma \) is positive definite. Thus (??) has the form of the gauge relation (??).

B.3 Similarity bookkeeping

Lemma B.3.1 Similarity is an involution
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For any invertible \(C \in M_{D}(\mathbb {C})\) and any linear map \(E\) on \(M_{D}(\mathbb {C})\), write \(S_C(E)(X)=C^{-1}E(CXC^\dagger )(C^\dagger )^{-1}\) for the similarity transform by \(C\). The transforms by \(C\) and \(C^{-1}\) compose to the identity:

\begin{align} S_{C^{-1}}(S_C(E))(X) & =C\left[C^{-1}E\! \left(C(C^{-1}X(C^\dagger )^{-1})C^\dagger \right) (C^\dagger )^{-1}\right]C^\dagger \notag \\ & =E(X). \label{eq:qpf_similarity_involution} \end{align}
Proof

Direct computation: the inner conjugation \(C(C^{-1}X(C^\dagger )^{-1})C^\dagger \) collapses to \(X\) using \(CC^{-1} = \mathbb {1}\) and \((C^\dagger )^{-1}C^\dagger = \mathbb {1}\), and the outer conjugation by \(C(\, \cdot \, )C^\dagger \) then cancels the inner \(C^{-1}(\, \cdot \, )(C^\dagger )^{-1}\) on \(E(X)\) for the same reason.

Lemma B.3.2 Irreducibility is invariant under similarity

For any invertible \(C \in M_{D}(\mathbb {C})\) and any linear map \(E\) on \(M_{D}(\mathbb {C})\), the similarity transform \(X \mapsto C^{-1}E(CXC^\dagger )(C^\dagger )^{-1}\) is irreducible if and only if \(E\) is.

Proof

The forward direction follows from Lemma 6.6.1 applied with the inverse \(C^{-1}\), whose determinant is also nonzero: if the similarity transform of \(E\) by \(C\) is irreducible, then so is its similarity transform by \(C^{-1}\), which equals \(E\) by (??). The reverse direction is Lemma 6.6.1 itself with \(c = 1\).

B.4 Auxiliary Perron reductions

Theorem B.4.1 Adjoint transfer map is nonzero

If some Kraus operator \(A^i \neq 0\), then the adjoint transfer map \(\mathcal{E}_A^\dagger (X) = \sum _i (A^i)^\dagger X A^i\) is nonzero.

Proof

If \(\mathcal{E}_A^\dagger = 0\) then \(\mathcal{E}_A^\dagger (\mathbb {1}) = 0\), so \(\sum _i (A^i)^\dagger A^i = 0\). Since each summand \((A^i)^\dagger A^i\) is positive semidefinite, each \((A^i)^\dagger A^i = 0\) and hence each \(A^i = 0\).

Theorem B.4.2 Real spectral-radius identity

Under the hypotheses of Theorem 6.10.1, the real-valued spectral radius of \(E\) is also equal to \(r\).

Proof

This is the real-valued reformulation of Theorem 6.10.1.

B.5 Exponential truncation and scalar reformulations

Theorem B.5.1 Exponential truncation is positive definite

Let \(E\) be an irreducible completely positive map on \(M_{D}(\mathbb {C})\), let \(A \ge 0\) be nonzero, and let \(t {\gt} 0\). Then the finite exponential truncation satisfies

\begin{align} \sum _{k=0}^{D-1}\frac{t^k}{k!}E^k(A) & {\gt}0. \label{eq:qpf_exp_truncation} \end{align}

This is the finite-sum core of the completely positive specialization of [ Wol12 , Theorem 6.2(3) ] .

Proof

If a nonzero vector \(v\) annihilated the quadratic form in (??), positivity of every summand would make \(E^k(A)v=0\) for \(0\leq k\leq D-1\). Expanding \((\mathbb {1}+E)^{D-1}(A)\) by the binomial theorem would then make it annihilate \(v\), contrary to Theorem 6.2.1.