Paper VII visual guide · Visual 33 of 40 · THE OPTIMAL SCALAR FOR TWO SPECTRAL POINTS

An approximate defect has a precise norm meaning

Traditional edition: Definition II.8.6, Proposition II.8.7.

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On a two-dimensional code, consider the overlap compression

\[ A=PE_a^*E_bP={\operatorname{diag}}(11/10,9/10). \]

For instance, within the code take \(E_a=I\) and \(E_b=A\). This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has

\[ \inf_{c\in{\mathbb C}}{\left\lVert A-cI\right\rVert}=\frac1{10},\qquad c=1\ \text{attains it}. \]

Indeed, for every complex \(c\), the triangle inequality gives

\[ \frac15=\left|\frac{11}{10}-\frac9{10}\right| \leq\left|\frac{11}{10}-c\right|+\left|\frac9{10}-c\right|. \]

Thus the larger of the two distances is at least \(1/10\), while both distances equal \(1/10\) at \(c=1\).

Three spectral points display scalar-approximation distances.
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

The marked scalar lies between two spectral endpoints, with equal distances shown above the line. The note below distinguishes the defect for one pair from the defect for the stated full operator family.

Drawing labels and data 6 labels

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  • \(9/10\)
  • \(1\)
  • \(11/10\)
  • \(1/10\)
  • \(1/10\)
  • one pair: \(1/10\), full family \(\{I,A\}\): \(1/5\)

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What one pair contributes to the family defect

For unit code vectors \(\psi,\phi\), this pair therefore satisfies

\[ \left|\langle E_a\psi,E_b\phi\rangle-\langle\psi,\phi\rangle\right| \leq\frac1{10}. \]

The bound is attained on either coordinate vector. For the whole declared family \(\{I,A\}\), the \(A^*A\) compression also enters the maximum: its diagonal entries are \(121/100,81/100\), giving pair defect \(1/5\). Hence the full family value is \(\varepsilon_{\rm KL}=1/5\), not \(1/10\).

This distinction prevents a tempting shortcut from one favorable overlap to an approximate correction claim. The quantity measures how close all overlap operators are to scalar action on the code. A recovery fidelity bound requires a declared normalized channel, a metric, and a separate continuity result.

What is checked exactly

For \(\varepsilon\in\{0,1/10,1/4,1\}\) the script checks \((A-I)^*(A-I)=\varepsilon^2I\) for \(A={\operatorname{diag}}(1+\varepsilon,1-\varepsilon)\) and several rational scalar controls. The triangle argument proves the lower bound for every complex scalar. Checking a finite list of candidates would not establish that conclusion. The family maximum above is also recorded separately.

Read this as

An overlap norm has a useful operational interpretation. Evaluate every error pair, then keep that statement separate from an unproved recovery fidelity estimate.

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