Manuscript En07-C@1 · En07-C.tex · proposition · 3f7e9ebb
Proposition: Selected entropy and spectra do not determine recovery
Entropy, energy, erasure cost, and leakage › Selected entropy and spectral data · lines 1256–1306 · proof included or linked
En07-C@1#prop:entropy, En07-C@1#eq:entropy-good, En07-C@1#eq:entropy-bad
Choose any finite state , a purification , and a logical register . Let
Define two complete isometries, with tensor factors ordered as and , by
Their accessible channels are
For every pure logical input, the two accessible outputs have the same eigenvalues and the same entropy . For the input , the entire accessible density operators are equal. Nevertheless
If , every energy-measurement distribution is also equal in the two models for all logical inputs.
Proof. Both maps preserve logical inner products. The partial traces give the displayed channels. For a pure , both accessible states are a rank-one logical state tensored with ; therefore their nonzero spectra and entropies agree. The chosen input makes the density operators identical. Tracing exactly decodes the return model. The retain model is a replacement channel, so Lemma lem:constant-error gives its error. A spectral effect of acts as the identity on the logical factor and has the same expectation in both outputs.
Displayed formulas in this passage
En07-C@1#eq:entropy-good · line 1266
En07-C@1#eq:entropy-bad · line 1266
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