Manuscript I-B@1 · I-B-v1.tex · proposition · 5357688f

Proposition: Complete history-conditioning identity

What destructive clock reading does and does not do › A nontrivial finite history readout · lines 959–988 · proof included or linked

I-B@1#prop:history, I-B@1#eq:history-channel

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For every operator $Z$ on $H_{X}$ ,

M_{C} (J Z J^{†}) = \frac{1}{D} k = 0 \sum^{D - 1} ∣ k ⟩ ⟨ k ∣_{R} \otimes U_{k} Z U_{k}^{†} .

This equality holds with every untouched reference. A record-controlled application of $U_{k}^{†}$ returns the original matter input, including its reference correlations. It does not invert $M_{C}$ on the full clock–matter input space.

Proof. Orthogonality gives

J^{†} J = \frac{1}{D} k \sum U_{k}^{†} U_{k} = I_{X}, D_{k} J = D^{- 1/2} U_{k} .

Substitution into (eq:instrument) proves (eq:history-channel). Controlled inversion gives $D^{- 1} \sum_{k} ∣ k ⟩ ⟨ k ∣ \otimes Z$ . These are operator identities and remain true with a reference. The inverse acts only on the encoded matter task; the phase-erasure counterexample of Proposition prop:no-return still applies to the full input space.

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I-B@1#eq:history-channel · line 962
$M_{C} (J Z J^{†}) = \frac{1}{D} k = 0 \sum^{D - 1} ∣ k ⟩ ⟨ k ∣_{R} \otimes U_{k} Z U_{k}^{†} .$
unlabeled · line 977
$J^{†} J = \frac{1}{D} k \sum U_{k}^{†} U_{k} = I_{X}, D_{k} J = D^{- 1/2} U_{k} .$

Proposition: Complete history-conditioning identity

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