Version 2 · current · External review, round 1: minor issues

# Clock parts: self-driven evolution, ticking, ring clock, reading and resolution

- **Subproject:** 04-ilang-time
- **Package:** 02-clock
- **Version:** v2
- **Mode:** internal regeneration
- **Date:** 2026-10-09

## Changes from previous version

- **I1 (review.v1.md).** Step 6 now answers item 4(a) for every member with $\theta_m=0$, not only for the minimal member.
  - New results: the Fourier components $\chi_q$ of the single-reading distribution (2.22), and the necessary and sufficient criterion (2.23).
  - The sufficient condition (2.24) uses the point-mass set $\mathcal S$, and it is also necessary for $N=2$.
  - The example (2.25) shows that (2.24) is not necessary for $N=3$.
  - (2.17) is re-derived as the minimal-member special case of (2.23). Its content is unchanged, and the open issue on its "only if" is removed.
- **I2 (review.v1.md).** The Result now states that every result of item 4 rests on S2.
- **Minor correction in Step 6.** v1 reduced the minimal member to $c=0$ by saying that changing $c$ adds a constant to $C_K$. This is not exact. Changing $c$ conjugates $C_K$ by the place-diagonal unitary $Z^c$ and adds a constant, so $\ell_k\mapsto\ell_{k-c}+c$. All place statistics, and therefore the conclusions, are unaffected.
- Consistency check 2 now also covers (2.23)–(2.24) at $N=2$. All other steps, tags and meanings are as in v1.

## Setup and assumptions

The setting and the definitions are those of question.md: a clock part $K$ with companion $R\neq\emptyset$, $C=C_K\otimes\mathbb 1_R+\mathbb 1_K\otimes C_R+C_\partial$ with $C_\partial=\sum_nA_n\otimes B_n$, readings on $K$ with $p(r;\lambda)=\operatorname{Tr}(\Pi_rV_K(\lambda))$, and the notions self-driven, ticking, index, clock, uniform ring clock, $L_K$, $\Delta C_K$. There are no inputs from other packages.

- **S1.** $\mathcal H$ is finite-dimensional (A1) and the contracts do not depend on $\lambda$ (A5). The evolution keeps states admissible (A3, A6), so $C$ maps $\mathcal H_{\rm adm}$ into itself. "Every start" means every unit vector of $\mathcal H_{\rm adm}$ at $\lambda=0$. If no two objects have the same type, then $\mathcal H_{\rm adm}=\mathcal H$ by (1.11).
- **S2.** Items 1–3 and 5 use (1.9) without readings. In item 4 a place reading with outcome $m$ is applied at a stated $\lambda$ through (1.3) with projector $\Pi_m\otimes\mathbb 1_R$ (A4, A8), and (1.9) holds between readings. We assume the state after a reading is admissible, so that the self-driven property applies to it. This holds when no object of $R$ has the type of $k$; otherwise $\Pi_m\otimes\mathbb 1_R$ would violate (1.11).
- **Notation.** $V_0:=V_K(0)$. The imaginary unit is $i$, so the spectral index of $C_K$ is written $j,l$: $C_K=\sum_jE_jP_j$. For ring clocks we write $x:=\lambda/\lambda_0$, the shift $S\lvert m\rangle=\lvert m+1\rangle$, and the Fourier vectors $\lvert f_k\rangle=N^{-1/2}\sum_{m\in\mathbb Z_N}e^{2\pi ikm/N}\lvert m\rangle$ with $S\lvert f_k\rangle=e^{-2\pi ik/N}\lvert f_k\rangle$, $k\in\mathbb Z_N$.
- **Partial trace (standard).** $\operatorname{Tr}_R[(Y\otimes\mathbb 1)X]=Y\operatorname{Tr}_RX$, $\operatorname{Tr}_R[X(Y\otimes\mathbb 1)]=(\operatorname{Tr}_RX)Y$, $\operatorname{Tr}_R[(\mathbb 1\otimes W)X]=\operatorname{Tr}_R[X(\mathbb 1\otimes W)]$.
- **Units.** By (1.9), $C\lambda$ is dimensionless, so $C_K$, $E_j$ and $\Delta C_K$ carry the units of $1/\lambda$.

## Derivation

### Step 1. Self-driven parts (item 1)

Let $\rho(\lambda)=\lvert\Psi(\lambda)\rangle\langle\Psi(\lambda)\rvert$. By (1.9), $\dot\rho=-i[C,\rho]$. The partial-trace rules give $\operatorname{Tr}_R[\mathbb 1\otimes C_R,\rho]=0$ and $\operatorname{Tr}_R[C_K\otimes\mathbb 1,\rho]=[C_K,V_K]$, hence
$$
\frac{\mathrm dV_K}{\mathrm d\lambda}=-i[C_K,V_K]-i\operatorname{Tr}_R[C_\partial,\rho(\lambda)] .
$$

**Sufficient condition.** Suppose the cross terms act on $K$ as the identity:
$$
C_\partial=\mathbb 1_K\otimes B,\qquad B=B^\dagger\ \text{on}\ \mathcal H_R\quad(\text{in particular } C_\partial=0).
\tag{2.1}
$$
Then $\operatorname{Tr}_R[C_\partial,\rho]=0$ for every $\rho$, so $\dot V_K=-i[C_K,V_K]$. The unique solution of this linear equation is $e^{-iC_K\lambda}V_0e^{iC_K\lambda}$. Condition (2.1) does not depend on the pair term in which a single-object term of $R$ is booked (A5, M6). Shifting $C_K\to C_K+c\mathbb 1$ does not change self-drivenness.

**General criterion.** $K$ is self-driven if and only if
$$
\operatorname{Tr}_R[C_\partial,X]=0\quad\text{for every operator } X \text{ on } \mathcal H_{\rm adm}.
\tag{2.2}
$$
*Sufficiency.* $\rho(\lambda)$ stays on $\mathcal H_{\rm adm}$ (S1), and the argument above applies. *Necessity.* Differentiating the self-driven identity at $\lambda=0$ gives $\operatorname{Tr}_R[C_\partial,\lvert\Psi\rangle\langle\Psi\rvert]=0$ for every admissible unit $\Psi$. By polarization, the operators $\lvert\Psi\rangle\langle\Psi\rvert$ span all operators on $\mathcal H_{\rm adm}$.

**(2.1) is necessary when $\mathcal H_{\rm adm}=\mathcal H$.** Expand $C_\partial=\mathbb 1_K\otimes G_0+\sum_{\alpha\ge1}F_\alpha\otimes G_\alpha$, where the $F_\alpha$ form a basis of the traceless operators on $\mathcal H_K$. For $X=Y\otimes Z$, (2.2) reads $\bigl[\operatorname{Tr}(G_0Z)\mathbb 1+\sum_\alpha\operatorname{Tr}(G_\alpha Z)F_\alpha,\;Y\bigr]=0$ for all $Y$. The bracketed operator must therefore be a multiple of $\mathbb 1$, so $\operatorname{Tr}(G_\alpha Z)=0$ for all $Z$, i.e. $G_\alpha=0$. This gives (2.1) with $B=G_0$, which is self-adjoint because $C_\partial$ is.

**(2.1) is not necessary in general.** Take objects $1\in K$ and $2\in R$ of the same type $T$, with $d_T\ge2$ and either family $c_T$, and a single contract
$$
C=D\otimes\mathbb 1+\mathbb 1\otimes D+g\,P_{12},\qquad D=D^\dagger,\quad g\neq0,
\tag{2.3}
$$
with the single-object terms booked into it (A5), so that $C_K=C_R=D$ and $C_\partial=gP_{12}$. $C$ commutes with $P_{12}$, and $P_{12}=c_T$ on $\mathcal H_{\rm adm}$. Hence $[C_\partial,X]=0$ for every $X$ on $\mathcal H_{\rm adm}$, (2.2) holds, and $K$ is self-driven: $e^{-iC\lambda}\Psi=e^{-igc_T\lambda}(e^{-iD\lambda}\otimes e^{-iD\lambda})\Psi$. Yet $\operatorname{Tr}_2[P_{12},Y\otimes Z]=[Z,Y]\neq0$ for non-commuting $Y,Z$, so $gP_{12}\neq\mathbb 1\otimes B$. On admissible states the views of $\{1\}$ and $\{2\}$ coincide, so the view of $K$ does not depend on the labels (M6).

**Dependence on $R$.** For a self-driven $K$,
$$
p(r;\lambda)=\operatorname{Tr}\bigl(\Pi_r\,e^{-iC_K\lambda}V_0\,e^{iC_K\lambda}\bigr),
\tag{2.4}
$$
so the statistics depend on the state of $R$ only through $V_0$: **no** further dependence. Sequences of readings are treated in Step 6.

### Step 2. Statistics and ticking (item 2a)

Insert $e^{-iC_K\lambda}=\sum_je^{-iE_j\lambda}P_j$ into (2.4):
$$
V_K(\lambda)=\sum_{j,l}e^{-i(E_j-E_l)\lambda}P_jV_0P_l,\qquad
p(r;\lambda)=\sum_{j,l}e^{-i(E_j-E_l)\lambda}\operatorname{Tr}\bigl(\Pi_rP_jV_0P_l\bigr).
\tag{2.5}
$$
Group the terms by gap: $V_\omega:=\sum_{E_j-E_l=\omega}P_jV_0P_l$, so that $p(r;\lambda)=\sum_\omega e^{-i\omega\lambda}\operatorname{Tr}(\Pi_rV_\omega)$. The functions $e^{-i\omega\lambda}$ with distinct $\omega$ are linearly independent on $\mathbb R$, so $p(r;\cdot)$ is constant if and only if $\operatorname{Tr}(\Pi_rV_\omega)=0$ for every $\omega\neq0$. Every $\Pi_r$ is a sum of place projectors $\lvert m\rangle\langle m\rvert$, $m\in S_r$, and the place reading is itself a reading. Hence
$$
\text{some reading on } K \text{ ticks}\iff \exists\,m,\ \exists\,\omega\neq0:\ \sum_{E_j-E_l=\omega}\langle m\vert P_jV_0P_l\vert m\rangle\neq0 ,
\tag{2.6}
$$
and in that case the place reading ticks. Two conditions each exclude ticking: $[C_K,V_0]=0$, which gives $V_\omega=0$ for $\omega\neq0$; and $C_K$ diagonal in the place basis, since then each $\lvert m\rangle$ lies in one eigenspace and $\langle m\vert P_jV_0P_l\vert m\rangle=0$ for $j\neq l$.

### Step 3. Recurrences and index (item 2b)

Let $\Omega:=\{E_j-E_l:\ P_jV_0P_l\neq0\}$ be the set of active gaps. It satisfies $\Omega=-\Omega$ because $(P_jV_0P_l)^\dagger=P_lV_0P_j$. The blocks $P_j(\cdot)P_l$ in (2.5) are independent: multiply by $P_j$ on the left and by $P_l$ on the right. Hence
$$
\{\lambda':V_K(\lambda')=V_K(\lambda)\}=\lambda+\mathcal T,\qquad
\mathcal T:=\{s\in\mathbb R:\ \omega s\in2\pi\mathbb Z\ \ \forall\,\omega\in\Omega\}.
\tag{2.7}
$$
$\mathcal T$ is a closed subgroup of $\mathbb R$ and does not depend on $\lambda$. There are three cases.
- $\Omega\subseteq\{0\}$, equivalently $[C_K,V_0]=0$. Then $\mathcal T=\mathbb R$ and $V_K$ is constant.
- $\Omega\neq\{0\}$ and all nonzero gaps are pairwise commensurate. The group $\mathbb Z\Omega$ generated by $\Omega$ lies in a cyclic group $(\omega_1/q)\mathbb Z$, so it is cyclic: $\mathbb Z\Omega=\omega_*\mathbb Z$ with $\omega_*>0$. Since $\omega_*$ is an integer combination of gaps and every gap is a multiple of $\omega_*$, $\mathcal T=\tau\mathbb Z$ with $\tau=2\pi/\omega_*$.
- Some $\omega_1,\omega_2\in\Omega\setminus\{0\}$ have $\omega_1/\omega_2\notin\mathbb Q$. A nonzero $s\in\mathcal T$ would give $\omega_1s=2\pi n_1$ and $\omega_2s=2\pi n_2$ with $n_2\neq0$, so $\omega_1/\omega_2=n_1/n_2$, a contradiction. Hence $\mathcal T=\{0\}$.

The view is injective on $\Lambda$ if and only if $(\Lambda-\Lambda)\cap\mathcal T=\{0\}$. Therefore
$$
V_K \text{ indexes } \lambda \text{ on all of } \mathbb R\iff \Omega \text{ contains two nonzero gaps with irrational ratio}.
\tag{2.8}
$$
If instead $\mathcal T=\tau\mathbb Z$, the maximal indexing intervals are the half-open intervals $[a,a+\tau)$ and $(a,a+\tau]$, $a\in\mathbb R$: any longer interval, or a closed one of length $\tau$, contains two points a distance $\tau$ apart. If $\mathcal T=\mathbb R$, no interval of positive length is indexed.

### Step 4. Uniform ring clocks (item 3a)

Let $U=e^{-iC_K\lambda_0}$, so $U\lvert m\rangle=e^{i\theta_m}\lvert m+1\rangle$, and let $\bar\theta:=N^{-1}\sum_m\theta_m$. Set $\alpha_0=0$ and $\alpha_{m+1}=\alpha_m+\theta_m-\bar\theta$. Summing gives $\alpha_{N-1}+\theta_{N-1}-\bar\theta=\alpha_0$, so with $\lvert\tilde m\rangle:=e^{i\alpha_m}\lvert m\rangle$ we have $U\lvert\tilde m\rangle=e^{i\bar\theta}\lvert\widetilde{m+1}\rangle$ for every $m\in\mathbb Z_N$. The orthonormal vectors $\lvert v_k\rangle=N^{-1/2}\sum_me^{2\pi ikm/N}e^{i\alpha_m}\lvert m\rangle$ satisfy $U\lvert v_k\rangle=e^{i\bar\theta}e^{-2\pi ik/N}\lvert v_k\rangle$, so $U$ has $N$ distinct eigenvalues. $C_K$ commutes with $U$ and therefore leaves each one-dimensional eigenspace invariant: $C_K=\sum_kE_k\lvert v_k\rangle\langle v_k\rvert$ with $e^{-iE_k\lambda_0}=e^{i\bar\theta-2\pi ik/N}$.

Conversely, take any data of the form below. Using $\lvert m\rangle=e^{-i\alpha_m}N^{-1/2}\sum_ke^{-2\pi ikm/N}\lvert v_k\rangle$, a direct computation gives $e^{-iC_K\lambda_0}\lvert m\rangle=e^{i(\alpha_{m+1}-\alpha_m-\varepsilon\lambda_0)}\lvert m+1\rangle$. Hence $K$ is a uniform ring clock with tick $\lambda_0$ if and only if
$$
C_K=\sum_{k\in\mathbb Z_N}E_k\lvert v_k\rangle\langle v_k\rvert,\quad
\lvert v_k\rangle=\frac{1}{\sqrt N}\sum_{m}e^{2\pi ikm/N}e^{i\alpha_m}\lvert m\rangle,\quad
E_k=\varepsilon+\frac{2\pi\ell_k}{N\lambda_0},\quad \ell_k\in\mathbb Z,\ \ell_k\equiv k\ (\mathrm{mod}\ N),
\tag{2.9}
$$
where the $\alpha_m$ and $\varepsilon$ are arbitrary reals. The phases are then $\theta_m=\alpha_{m+1}-\alpha_m-\varepsilon\lambda_0$ (mod $2\pi$, with $\alpha_N:=\alpha_0$). If $\theta_m=0$ for all $m$, then $e^{-iC_K\lambda_0}=S$. The same argument applied to $S$, whose eigenvalues $e^{-2\pi ik/N}$ on $\lvert f_k\rangle$ are distinct, gives exactly $C_K=\sum_k\frac{2\pi\ell_k}{N\lambda_0}\lvert f_k\rangle\langle f_k\rvert$ with $\ell_k\equiv k$. This $C_K$ commutes with $S$.

### Step 5. Ring-clock statistics and cost spread (item 3b)

Start at $\lvert0\rangle$, i.e. $V_0=\lvert0\rangle\langle0\rvert$. Then $V_K(\lambda)$ is the projector onto $e^{-iC_K\lambda}\lvert0\rangle$ by (2.4). We use $\lvert0\rangle=e^{-i\alpha_0}N^{-1/2}\sum_k\lvert v_k\rangle$, $\langle m\vert v_k\rangle=N^{-1/2}e^{2\pi ikm/N}e^{i\alpha_m}$, and $e^{2\pi ikm/N}=e^{2\pi i\ell_km/N}$. This gives $\langle m\vert e^{-iC_K\lambda}\vert0\rangle=e^{i(\alpha_m-\alpha_0)-i\varepsilon\lambda}G(m;\lambda)$, and hence
$$
p(m;\lambda)=\lvert G(m;\lambda)\rvert^2,\qquad G(m;\lambda):=\frac1N\sum_{k\in\mathbb Z_N}e^{2\pi i\ell_k(m-\lambda/\lambda_0)/N}.
\tag{2.10}
$$
This depends only on the $\ell_k$, not on the $\theta_m$. At $\lambda=j\lambda_0$, $p(m)=\delta_{m,\,j\bmod N}$.

$V_K(\lambda)$ is conjugated by a function of $C_K$, so $L_K$ and $\Delta C_K$ do not depend on $\lambda$. The weights are $\lvert\langle v_k\vert0\rangle\rvert^2=1/N$, so
$$
\Delta C_K=\frac{2\pi}{N\lambda_0}\Bigl[\frac1N\sum_k\ell_k^2-\Bigl(\frac1N\sum_k\ell_k\Bigr)^2\Bigr]^{1/2}.
\tag{2.11}
$$
The $\ell_k$ are $N$ distinct integers, since they are distinct mod $N$. Sort them as $\ell_{(1)}<\dots<\ell_{(N)}$. Their variance is $N^{-2}\sum_{a<b}(\ell_{(b)}-\ell_{(a)})^2\ge N^{-2}\sum_{a<b}(b-a)^2=(N^2-1)/12$, with equality if and only if all consecutive gaps equal $1$. $N$ consecutive integers form a complete residue system mod $N$, so the equality case lies in the family:
$$
\min\Delta C_K=\frac{2\pi}{\lambda_0}\sqrt{\frac{N^2-1}{12N^2}},\quad\text{attained iff } \{\ell_k\}=\{c,c+1,\dots,c+N-1\},\ c\in\mathbb Z .
\tag{2.12}
$$
Equivalently, the spectrum of $C_K$ consists of $N$ levels equally spaced by $2\pi/(N\lambda_0)$. For these minimal members, (2.10) is a geometric sum whose modulus does not depend on $c$:
$$
p(m;\lambda)=\frac{\sin^2(\pi\lambda/\lambda_0)}{N^2\sin^2\bigl(\pi(\lambda/\lambda_0-m)/N\bigr)}\qquad(\text{value }1\text{ where the denominator vanishes}).
\tag{2.13}
$$
For a minimal member every pair of eigenvectors is active and the gaps generate $(2\pi/(N\lambda_0))\mathbb Z$. By (2.7)–(2.8), $\tau=N\lambda_0$: the view indexes $\lambda$ on half-open intervals of length $N\lambda_0$.

### Step 6. Reading the clock (item 4a)

A place reading with outcome $m$ acts by $\Pi_m\otimes\mathbb 1_R$ (1.3). The partial-trace rules give the new view
$$
V_K\ \longmapsto\ \frac{\Pi_mV_K\Pi_m}{\operatorname{Tr}(\Pi_mV_K)}=\lvert m\rangle\langle m\rvert .
\tag{2.14}
$$
By S2 and (2.4), the statistics of a sequence of readings on a self-driven $K$ are fixed by $C_K$ and $V_0$, and after outcome $m$ the clock restarts from $\lvert m\rangle$. For $\theta_m=0$, $C_K=\sum_kE_k\lvert f_k\rangle\langle f_k\rvert$ with $E_k=2\pi\ell_k/(N\lambda_0)$ commutes with $S$ (Step 4), so $\langle m'\vert e^{-iC_K\lambda}\vert m\rangle=G(m'-m;\lambda)$ with $G$ from (2.10) and no phase prefactor. Write $\Delta:=\lambda_2-\lambda_1$, $x_1:=\lambda_1/\lambda_0$, $x_\Delta:=\Delta/\lambda_0$. The joint distribution of the two outcomes is
$$
P(m_1,m_2)=p(m_1;\lambda_1)\,p(m_2-m_1;\Delta).
\tag{2.15}
$$
Without the first reading, $p(m_2;\lambda_2)=\bigl\lvert\sum_{m_1}G(m_2-m_1;\Delta)G(m_1;\lambda_1)\bigr\rvert^2$, because $e^{-iC_K\lambda_2}=e^{-iC_K\Delta}e^{-iC_K\lambda_1}$. With the first reading applied but not recorded,
$$
P_2(m_2)=\sum_{m_1}p(m_1;\lambda_1)\,p(m_2-m_1;\Delta),\qquad
p(m_2;\lambda_2)-P_2(m_2)=\sum_{m_1\neq m_1'}G(m_2-m_1;\Delta)G(m_1;\lambda_1)\,G(m_2-m_1';\Delta)^*G(m_1';\lambda_1)^* .
\tag{2.16}
$$
The unrecorded reading removes the interference terms. The first reading changes the distribution of the second outcome if and only if $D(m):=p(m;\lambda_2)-P_2(m)$ is not identically zero.

**General criterion (every member with $\theta_m=0$).** Let $Z:=\sum_me^{2\pi im/N}\lvert m\rangle\langle m\rvert$, so that $Z^{-1}\lvert f_k\rangle=\lvert f_{k-1}\rangle$, and $\psi(\lambda):=e^{-iC_K\lambda}\lvert0\rangle=N^{-1/2}\sum_ke^{-iE_k\lambda}\lvert f_k\rangle$. The discrete Fourier components of the single-reading distribution are
$$
\chi_q(\lambda):=\sum_{m\in\mathbb Z_N}e^{-2\pi iqm/N}p(m;\lambda)=\langle\psi(\lambda)\vert Z^{-q}\vert\psi(\lambda)\rangle=\frac1N\sum_{k\in\mathbb Z_N}e^{-i(E_k-E_{k-q})\lambda},\qquad q\in\mathbb Z_N .
\tag{2.22}
$$
By (2.16), $P_2$ is the convolution on $\mathbb Z_N$ of $p(\cdot\,;\lambda_1)$ and $p(\cdot\,;\Delta)$, so its components are $\chi_q(\lambda_1)\chi_q(\Delta)$. The discrete Fourier transform on $\mathbb Z_N$ is invertible. The component $q=0$ of $D$ vanishes because both distributions are normalized, and the component $N-q$ is the complex conjugate of the component $q$ because $D$ is real. Hence the first reading leaves the distribution of the second outcome unchanged if and only if
$$
\hat D(q):=\sum_me^{-2\pi iqm/N}D(m)=\chi_q(\lambda_2)-\chi_q(\lambda_1)\,\chi_q(\lambda_2-\lambda_1)=0\qquad\text{for } q=1,\dots,\lfloor N/2\rfloor .
\tag{2.23}
$$
Since $e^{-iC_K\lambda_0}=S$, shifting $\lambda_1$ or $\Delta$ by $\lambda_0$ shifts $p(\cdot\,;\lambda_2)$, $P_2$ and hence $D$ by one place. So (2.23) depends on $\lambda_1$ and $\Delta$ only modulo $\lambda_0$.

**A sufficient condition.** Let $\mathcal S$ be the set of $s$ at which $p(\cdot\,;s)$ is a point mass. If $e^{-iC_Ks}\lvert0\rangle=e^{i\varphi}\lvert j\rangle$, then $e^{-iC_Ks}\lvert m\rangle=S^me^{-iC_Ks}\lvert0\rangle=e^{i\varphi}\lvert m+j\rangle$, i.e. $e^{-iC_Ks}=e^{i\varphi}S^j$.
- If $\lambda_1\in\mathcal S$, the state at $\lambda_1$ is $\lvert j\rangle$ up to phase, so $p(m;\lambda_2)=p(m-j;\Delta)=P_2(m)$.
- If $\Delta\in\mathcal S$, then $\psi(\lambda_2)=e^{i\varphi}S^j\psi(\lambda_1)$, so $p(m;\lambda_2)=p(m-j;\lambda_1)=P_2(m)$.

The phases $e^{-2\pi im/N}$, $m\in\mathbb Z_N$, are distinct. By the triangle inequality, $\lvert\chi_1(s)\rvert=1$ therefore holds if and only if $p(\cdot\,;s)$ is a point mass. For $N=2$, $G:=\ell_1-\ell_0$ is odd and $\chi_1(\lambda)=\cos(\pi G\lambda/\lambda_0)$. Then $\cos(a+b)-\cos a\cos b=-\sin a\sin b$ turns (2.23) into the second line of
$$
\begin{aligned}
&\mathcal S=\{s:\ \lvert\chi_1(s)\rvert=1\}\supseteq\lambda_0\mathbb Z,\qquad \lambda_1\in\mathcal S\ \text{ or }\ \lambda_2-\lambda_1\in\mathcal S\ \Longrightarrow\ D\equiv0;\\
&N=2:\quad \hat D(1)=-\sin\frac{\pi G\lambda_1}{\lambda_0}\,\sin\frac{\pi G(\lambda_2-\lambda_1)}{\lambda_0},\qquad \mathcal S=\frac{\lambda_0}{\lvert G\rvert}\mathbb Z .
\end{aligned}
\tag{2.24}
$$
So for $N=2$ the sufficient condition is also necessary. For example, $\ell=(0,3)$ gives no change for every $\Delta\in(\lambda_0/3)\mathbb Z$.

**Minimal member.** Changing $c$ in (2.12) replaces $C_K$ by $Z^cC_KZ^{-c}+2\pi c/(N\lambda_0)$, i.e. $\ell_k\mapsto\ell_{k-c}+c$. This permutes the differences $\ell_k-\ell_{k-q}$ over $k$, so $\chi_q$ is unchanged. Take $c=0$, $\ell_k=k$ for $k=0,\dots,N-1$. Then $E_k-E_{k-q}=2\pi q/(N\lambda_0)$ for $k\ge q$ and $2\pi(q-N)/(N\lambda_0)$ for $k<q$, so
$$
\chi_q(\lambda)=\tfrac1Ne^{-2\pi iqx/N}\bigl(N-q+q\,e^{2\pi ix}\bigr),\qquad
\hat D(q)=\tfrac{q(N-q)}{N^2}\,e^{-2\pi iq(x_1+x_\Delta)/N}(e^{2\pi ix_1}-1)(e^{2\pi ix_\Delta}-1).
$$
Its component $q=1$ is the result of v1:
$$
\hat D(1)=\frac{N-1}{N^2}\,e^{-2\pi i(x_1+x_\Delta)/N}\,(e^{2\pi ix_1}-1)(e^{2\pi ix_\Delta}-1),\qquad
\lvert\hat D(1)\rvert=\frac{4(N-1)}{N^2}\,\lvert\sin\pi x_1\,\sin\pi x_\Delta\rvert .
\tag{2.17}
$$
Since $q(N-q)\neq0$ for $1\le q\le N-1$, for the minimal member the first reading changes the distribution of the second outcome **if and only if** $\lambda_1\notin\lambda_0\mathbb Z$ and $\lambda_2-\lambda_1\notin\lambda_0\mathbb Z$. Here $\lvert\chi_1(s)\rvert=1$ iff $e^{2\pi is/\lambda_0}=1$, so $\mathcal S=\lambda_0\mathbb Z$ and (2.24) is again necessary.

**(2.24) is not necessary in general.** For $N=3$ only $q=1$ enters (2.23). Take $(\ell_0,\ell_1,\ell_2)=(0,1,5)$, so $E_k-E_{k-1}=2\pi g_k/(3\lambda_0)$ with $(g_0,g_1,g_2)=(-5,1,4)$, and $\lambda_1=3\lambda_0/8$, $\lambda_2=5\lambda_0/4$, $\Delta=7\lambda_0/8$. Summing the three phases $e^{-2\pi ig_k\lambda/(3\lambda_0)}$ in (2.22) gives
$$
\chi_1(\lambda_1)=-\frac{1+i\sqrt2}{3},\qquad
\chi_1(\Delta)=-\frac{e^{i\pi/6}(\sqrt2+i)}{3},\qquad
\chi_1(\lambda_2)=\frac{e^{2\pi i/3}}{3}=\chi_1(\lambda_1)\,\chi_1(\Delta).
\tag{2.25}
$$
The last equality uses $(1+i\sqrt2)(\sqrt2+i)=3i$. The first reading therefore leaves the second distribution unchanged, although $\lvert\chi_1(\lambda_1)\rvert=\lvert\chi_1(\Delta)\rvert=1/\sqrt3<1$, i.e. $\lambda_1,\Delta\notin\mathcal S$.

### Step 7. Frequent reading (item 4b)

By (2.14), each outcome $0$ restarts the clock at $\lvert0\rangle$. Given outcome $0$ at the previous reading (or the start at $\lambda=0$), outcome $0$ at the next reading has probability $p(0;\lambda/n)=\lvert\langle0\vert e^{-iC_K\lambda/n}\vert0\rangle\rvert^2$. Hence
$$
P_n=\Bigl\lvert\frac1N\sum_ke^{-2\pi i\ell_k\lambda/(nN\lambda_0)}\Bigr\rvert^{2n}
\ \overset{\text{min.}}{=}\ \Bigl[\frac{\sin^2(\pi\lambda/(n\lambda_0))}{N^2\sin^2(\pi\lambda/(nN\lambda_0))}\Bigr]^{n},
\qquad P_n=e^{-\lambda^2\Delta C_K^2/n+O(n^{-3})}\xrightarrow[n\to\infty]{}1 .
\tag{2.18}
$$
For the expansion: $\lvert\langle0\vert e^{-iC_K\delta}\vert0\rangle\rvert^2=1-\delta^2\Delta C_K^2+O(\delta^4)$, which is even in $\delta$, so $n\ln p(0;\lambda/n)=-\lambda^2\Delta C_K^2/n+O(n^{-3})$. Frequent reading freezes the clock at $\lvert0\rangle$ (Law 1).

### Step 8. Rate and orthogonality bound (item 5a)

Up to phases, $\langle\psi(\lambda)\vert\psi(\lambda+s)\rangle=\langle\psi(\lambda)\vert e^{-iC_Ks}\vert\psi(\lambda)\rangle=1-is\langle C_K\rangle-\tfrac{s^2}{2}\langle C_K^2\rangle+\tfrac{is^3}{6}\langle C_K^3\rangle+O(s^4)$. Therefore $\cos^2\theta=1-s^2\Delta C_K^2+O(s^4)$, where $\Delta C_K$ is computed from $V_K(\lambda)$ and does not depend on $\lambda$ (Step 5). Hence $\sin^2\theta=s^2\Delta C_K^2+O(s^4)$ and
$$
\lim_{\lambda'\to\lambda}\frac{\theta(\lambda,\lambda')}{\lvert\lambda'-\lambda\rvert}=\Delta C_K .
\tag{2.19}
$$
We use the standard fact that $\theta$ is the Fubini–Study distance on unit rays, which is a metric. Split $[\lambda,\lambda']$ into $M$ equal pieces. The triangle inequality and (2.19) give $\theta(\lambda,\lambda')\le\Delta C_K\lvert\lambda'-\lambda\rvert+O(1/M)$, because the remainder is uniform: $\lvert\langle C_K^n\rangle\rvert\le\lVert C_K\rVert^n$. Letting $M\to\infty$ and using $\theta=\pi/2$ at orthogonality:
$$
\theta(\lambda,\lambda')\le\Delta C_K\lvert\lambda'-\lambda\rvert,\qquad
\psi(\lambda')\perp\psi(\lambda)\ \Longrightarrow\ \lvert\lambda'-\lambda\rvert\ge\frac{\pi}{2\Delta C_K}.
\tag{2.20}
$$
This is the Mandelstam–Tamm inequality for the unitary group $e^{-iC_K\lambda}$. If $\Delta C_K=0$, $\psi_0$ is an eigenvector of $C_K$ and orthogonality never occurs.

### Step 9. Ring clock at the bound (item 5b)

Take $\theta_m=0$, the minimal member, and the start $\lvert0\rangle$. Then $\langle\psi(0)\vert\psi(\lambda)\rangle=G(0;\lambda)=N^{-1}\sum_{j=0}^{N-1}e^{-2\pi i(c+j)x/N}$. This vanishes if and only if $e^{-2\pi ix}=1$ and $e^{-2\pi ix/N}\neq1$, i.e. $x\in\mathbb Z\setminus N\mathbb Z$. The smallest such $\lambda>0$ is $\lambda_\perp=\lambda_0$ (for $N\ge2$). Inserting (2.12) into (2.20):
$$
\lambda_\perp=\lambda_0,\qquad \frac{\pi}{2\min\Delta C_K}=\frac{\sqrt3\,N\lambda_0}{2\sqrt{N^2-1}},\qquad
r_N:=\frac{\lambda_\perp}{\pi/(2\min\Delta C_K)}=\frac{2}{\sqrt3}\sqrt{1-\frac1{N^2}}\ \xrightarrow[N\to\infty]{}\ \frac{2}{\sqrt3}.
\tag{2.21}
$$
$r_N$ is strictly increasing in $N$ and $r_2=1$, so the bound is attained if and only if $N=2$.

## Result

- **(2.1)** Sufficient condition for a self-driven part: $C_\partial=\mathbb 1_K\otimes B$ (the cross terms act trivially on $K$).
- **(2.2)** Necessary and sufficient condition: $\operatorname{Tr}_R[C_\partial,X]=0$ for all $X$ on $\mathcal H_{\rm adm}$. It reduces to (2.1) when $\mathcal H_{\rm adm}=\mathcal H$, for example when no two objects share a type.
- **(2.3)** Counterexample to the necessity of (2.1): two identical-type objects coupled by a swap contract $gP_{12}$.
- **(2.4)** For a self-driven $K$, the statistics depend on $R$ only through $V_K(0)$.
- **(2.5)** Closed form of $V_K(\lambda)$ and $p(r;\lambda)$. **(2.6)** Ticking criterion: some active gap $\omega\neq0$ has a nonzero place-diagonal component.
- **(2.7)** The recurrence set is $\lambda+\mathcal T$, with period group $\mathcal T$ fixed by the active gaps $\Omega$. **(2.8)** The view indexes $\lambda$ on all of $\mathbb R$ if and only if $\Omega$ contains two nonzero gaps with irrational ratio. Otherwise the maximal indexing intervals are half-open of length $\tau=2\pi/\omega_*$.
- **(2.9)** All uniform ring clocks: phase-twisted Fourier eigenvectors with eigenvalues $\varepsilon+2\pi\ell_k/(N\lambda_0)$, $\ell_k\equiv k$ (mod $N$).
- **(2.10)** $p(m;\lambda)$ for every member. **(2.11)** $\Delta C_K$ over the family. **(2.12)** Its minimum, attained for equally spaced levels. **(2.13)** $p(m;\lambda)$ for minimal members.
- **Item 4 rests on S2.** (2.14)–(2.18) and (2.23)–(2.25) assume that the state after each reading is admissible. This holds when no object of $R$ has the type of the clock object $k$.
- **(2.14)** A reading resets the view of $K$ to $\lvert m\rangle\langle m\rvert$. **(2.15)** Joint distribution of two readings. **(2.16)** The unrecorded first reading removes interference.
- **(2.22)** Fourier components $\chi_q(\lambda)=N^{-1}\sum_ke^{-i(E_k-E_{k-q})\lambda}$ of the single-reading distribution.
- **(2.23)** For every member with $\theta_m=0$, the first reading leaves the distribution of the second outcome unchanged if and only if $\chi_q(\lambda_2)=\chi_q(\lambda_1)\chi_q(\lambda_2-\lambda_1)$ for $q=1,\dots,\lfloor N/2\rfloor$. Otherwise it changes it. The condition depends on $\lambda_1$ and $\lambda_2-\lambda_1$ only modulo $\lambda_0$.
- **(2.24)** No change if $\lambda_1$ or $\lambda_2-\lambda_1$ lies in the point-mass set $\mathcal S\supseteq\lambda_0\mathbb Z$. For $N=2$ this is also necessary, with $\mathcal S=(\lambda_0/\lvert\ell_1-\ell_0\rvert)\mathbb Z$.
- **(2.17)** Minimal member, special case of (2.23): the first reading changes the second distribution if and only if $\lambda_1,\lambda_2-\lambda_1\notin\lambda_0\mathbb Z$ ($=\mathcal S$).
- **(2.25)** (2.24) is not necessary in general. For $N=3$, $\ell=(0,1,5)$, $\lambda_1=3\lambda_0/8$, $\lambda_2=5\lambda_0/4$, nothing changes, although $\lambda_1,\lambda_2-\lambda_1\notin\mathcal S$.
- **(2.18)** All-zero probability under $n$ readings, $P_n\to1$ (frequent reading stops the clock).
- **(2.19)** Angular rate $=\Delta C_K$. **(2.20)** Orthogonality needs $\lvert\lambda'-\lambda\rvert\ge\pi/(2\Delta C_K)$.
- **(2.21)** Ring clock: $\lambda_\perp=\lambda_0$, $r_N=\frac{2}{\sqrt3}\sqrt{1-N^{-2}}\to2/\sqrt3$, and the bound is attained only for $N=2$.

## Consistency checks

1. **Units and normalization.**
   - $\Delta C_K\propto1/\lambda_0$ and the bound $\propto\lambda_0$, so $r_N$ is dimensionless.
   - In (2.10), $\sum_m p(m;\lambda)=1$ because $\sum_me^{2\pi i(\ell_k-\ell_{k'})m/N}=N\delta_{kk'}$ (as $\ell_k\equiv k$).
   - (2.15) sums to $1$, $\sum_mD(m)=0$ in (2.16), and $\chi_0=1$ in (2.22), so $\hat D(0)=0$.

   Passed.
2. **$N=2$.** Take $C_K=(\pi/\lambda_0)\lvert f_1\rangle\langle f_1\rvert$, i.e. $G=1$.
   - Directly, $\langle0\vert\psi(\lambda)\rangle=(1+e^{-i\pi x})/2$, so $p(0;\lambda)=\cos^2(\pi x/2)$, which agrees with (2.13). Also $\Delta C_K=\pi/(2\lambda_0)$, which agrees with (2.12).
   - A direct computation gives $D(0)=-\tfrac12\sin\pi x_1\sin\pi x_\Delta$. Since $D(1)=-D(0)$, this equals $\hat D(1)/2$ from both (2.24) and (2.17).
   - Finally $\lambda_\perp=\lambda_0=\pi/(2\Delta C_K)$.

   Passed.
3. **Start commuting with $C_K$** ($V_0=\lvert v_k\rangle\langle v_k\rvert$ or $\mathbb 1/d_K$). Here $\Omega\subseteq\{0\}$: by (2.5)–(2.8), nothing ticks, $\mathcal T=\mathbb R$ and no interval is indexed. For $\lvert v_k\rangle$, $\Delta C_K=0$, and (2.20) correctly excludes orthogonality. Passed.

## Open issues

- The necessity of (2.1) is shown only for $\mathcal H_{\rm adm}=\mathcal H$. With identical-type objects on both sides of the cut, only (2.2) and the counterexample (2.3) are given. How such parts are specified without labels (A7, M6) is not treated.
- Item 4 assumes that post-reading states are admissible, which holds when no object of $R$ has the type of $k$ (S2).
- Indexing on all of $\mathbb R$ in (2.8) is exact injectivity only. The quasi-periodic near-recurrences that limit a finite-resolution reading are not quantified.
- Out of scope: clock time versus other parts (criterion 2), comparison of clocks (criterion 3), readings represented by recording contracts (criterion 7), and clocks that are not self-driven.

## Methods used

- partial trace, Heisenberg-type equation for the view
- spectral decomposition, linear independence of exponentials
- closed subgroups of $\mathbb R$, commensurability
- circulant operators, discrete Fourier transform on $\mathbb Z_N$, convolution theorem, geometric sums
- variance of distinct integers
- projective readings (1.3), interference and decoherence of the view
- Taylor expansion, frequent-reading (Zeno-type) limit
- Fubini–Study metric, Mandelstam–Tamm inequality

Author: Ivan Honis · ivan.honis@ndot.io · License: CC BY 4.0 · Human–AI note