Paper VIIReading edition

Kraus transfer

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I.7.2 Kraus transfer

A quantum channel can transfer both populations and coherences between packet addresses. A Kraus representation describes this transfer by summing the contributions of its individual branches. The normalization of those branches ensures that the total trace is preserved.

Classical input I.7.2: Finite Kraus channel

Let \(K_1,\ldots,K_r\) satisfy

\[ \sum_aK_a^*K_a=I. \]

Then

\[ \boxed{ \Phi(\rho)=\sum_aK_a\rho K_a^* } \]

is completely positive and trace preserving.

Proof of Classical input I.7.2

Trace preservation follows from

\[ {\operatorname{Tr}}\Phi(\rho) ={\operatorname{Tr}}\left(\rho\sum_aK_a^*K_a\right) ={\operatorname{Tr}}\rho. \]

For any auxiliary dimension and positive matrix \(X\), each term

\[ (I\otimes K_a)X(I\otimes K_a)^* \]

is positive, so their sum is positive. This is complete positivity.

Packet superoperator blocks are defined by

\[ \Phi_{\alpha\beta;\gamma\delta}(X) = P_\alpha \Phi(P_\gamma XP_\delta) P_\beta. \]

Summing all four packet indices reconstructs \(\Phi(X)\) exactly.

For amplitude damping with a real parameter \(0\le\gamma\le1\),

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

The excited state becomes

\[ \Phi({\left|1\right\rangle}{\left\langle1\right|}) = \gamma{\left|0\right\rangle}{\left\langle0\right|}+(1-\gamma){\left|1\right\rangle}{\left\langle1\right|}. \]
Amplitude damping gives a diagonal mixture of ground and excited populations.
Amplitude damping transfers population from the excited packet to the ground packet while preserving trace and positivity.
Text description

A two-by-two density-matrix heatmap shows positive entries on the diagonal and zero off-diagonal entries for the displayed excited-state input. The ground population receives the damping contribution while the excited population retains the remainder.

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Numerical certificate I.7.3: Channel reconstruction

For \(\gamma=0.23\), the residuals for Kraus completeness and the output trace were \(0\). The computed output from \({\left|1\right\rangle}{\left\langle1\right|}\) was \({\operatorname{diag}}(0.23,0.77)\). In the Bell state test, packet dephasing left every measurement probability in the computational basis unchanged.

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