Paper VII visual guide · Visual 18 of 40 · WHY COMPLETE POSITIVITY IS THE RELEVANT CONDITION

An ancilla tests more than positivity

Traditional edition: Classical input II.7.7.

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Ordinary transpose \(\mathsf T\) sends a positive matrix on one system to another positive matrix and preserves its trace. Those properties are insufficient for a quantum channel on a subsystem of an entangled state. The test must include \({\mathrm{id}}\otimes\mathsf T\) on the larger system.

A partially transposed Bell matrix has a negative test direction.
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

The left matrix is the displayed scaled partial-transpose output. The right side supplies a vector and its negative quadratic form. The accompanying note distinguishes positivity on one system from complete positivity.

Drawing labels and data 24 labels

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\[2({\mathrm{id}}\otimes{\mathsf T})(\rho_{\rm Bell})\]
Drawing grid 1 · rows top to bottom, columns left to right
Row / columnColumn 1Column 2Column 3Column 4
00\(1\)\(0\)\(0\)\(0\)
01\(0\)\(0\)\(1\)\(0\)
10\(0\)\(1\)\(0\)\(0\)
11\(0\)\(0\)\(0\)\(1\)
  • \(w=({\left|01\right\rangle}-{\left|10\right\rangle})/\sqrt2\)

\(w^*({\mathrm{id}}\otimes{\mathsf T})(\rho)w=-1/2\)

Transpose is positive on one system.
It is not completely positive.

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Apply this extended map to the Bell density matrix. The result is

\[ ({\mathrm{id}}\otimes\mathsf T)(\rho_{\rm Bell}) =\frac12{\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}}. \]

Its middle block has one negative eigenvalue. More concretely, the normalized vector \(w=({\left|01\right\rangle}-{\left|10\right\rangle})/\sqrt2\) satisfies

\[ w^*({\mathrm{id}}\otimes\mathsf T)(\rho_{\rm Bell})w=-\frac12. \]

This exact negative value of the quadratic form certifies failure of positivity on the composite system. No numerical eigenvalue tolerance is needed.

Why the Kraus form avoids this problem

For every auxiliary dimension and positive \(X\), each term \((I\otimes K_a)X(I\otimes K_a)^*\) is positive. Their sum is positive as well. This argument explains the word completely. The conclusion holds for every auxiliary dimension, which a test of one input density matrix on the original system cannot establish.

The transpose counterexample is an explanatory finite calculation, not a new addition to the formal theorem count. It makes visible the hypothesis that the manuscript's Kraus result already supplies. The exact check constructs the partial transpose with the stated tensor indexing and evaluates its negative witness directly.

Read this as

A map can look harmless on an isolated system and fail on an entangled extension. Complete positivity retains compatibility with every auxiliary system.

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