Version 1 · earlier version; the current one is v2 · External review, round 1: minor issues

# Detectability of λ through readings: unit, direction, common now, order

- **Subproject:** 04-ilang-time
- **Package:** 10-global-order
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-10

## Setup and assumptions

- **Setting.** Exactly as in question.md: $\mathcal H=\mathcal H_1\otimes\mathcal H_2$, all objects of different types, $C=C^\dagger$ independent of $\lambda$, start $\Psi_0$ at $\lambda=0$, only $\lambda\ge0$; schedules, re-timings $(a_1,a_2,C')$, equivalences and "detectable" as defined there.
- **Notation.** $h,h'$ are the joint places of $\mathcal P_1$, $r,r'$ those of $\mathcal P_2$, and $X_{hr,h'r'}:=\langle hr\vert X\vert h'r'\rangle$. A **place projector of $\mathcal P_1$** is $\Pi=\sum_{h\in S}\lvert h\rangle\langle h\rvert\otimes\mathbb 1$ for a set $S$ of places; of $\mathcal P_2$, $\Pi'=\mathbb 1\otimes\sum_{r\in S'}\lvert r\rangle\langle r\rvert$. These are exactly the class projectors of local readings (A4); each occurs in some reading; all are diagonal in the joint place basis and commute with each other. $X(\lambda):=e^{iC\lambda}Xe^{-iC\lambda}$, and $G_i:=C-a_iC'$.
- **Standard facts.** (F0) A self-adjoint $X$ with $\langle\Psi\vert X\vert\Psi\rangle=0$ for all unit vectors is zero. (F1) For orthogonal projectors with $\sum_s\Pi_s=\mathbb 1$: $X=\sum_s\Pi_sX\Pi_s$ iff $[X,\Pi_s]=0$ for all $s$.
- **Well-definedness (M6).** "Uncoupled", the split (10.5), the components (10.13) and every condition below are properties of the operator $C$; none refers to the list of contract terms or to the booking of single-object terms. $C_1,C_2$ are fixed by $C$ up to $(C_1+c\mathbb 1,\,C_2-c\mathbb 1)$. Adding $c\mathbb 1$ to a contract changes no $X(\lambda)$, hence no statistics.
- **Exactness.** Items 1–4 and 5(a) use no approximation: every statement is exact and proven for the whole stated class. Item 5(b) is an inequality valid for every start.
- **Inputs** from 07-causal-order@v2, as quoted in question.md: (7.1), (7.2), (7.11), (7.12), (7.15). Abbreviation: $\Sigma_r(\delta):=\sum_{k\ge r}(2\lVert J\rVert\delta)^k/k!$.

## Derivation

### Step 1. Statistics of a schedule

Number the events so that $\lambda_1\le\dots\le\lambda_K$; $\Pi^{(k)}_{s_k}$ is the class projector of the outcome $s_k$ of $R_k$. Applying (1.9) and (1.3) alternately and multiplying the conditional probabilities (the normalizations cancel):

$$
P(s_1,\dots,s_K)=\bigl\lVert\Pi^{(K)}_{s_K}(\lambda_K)\cdots\Pi^{(1)}_{s_1}(\lambda_1)\,\Psi_0\bigr\rVert^2 .
\tag{10.1}
$$

- **Equal $\lambda$.** Factors with the same $\lambda$ are $e^{iC\lambda}(\cdot)e^{-iC\lambda}$ of commuting place projectors, so they commute: the product, hence the statistics, does not depend on their order.
- **Comparison rule.** Two schedules, possibly under different contracts, have equal statistics for every start if every event has the same operator $\Pi^{(k)}_{s}(\lambda_k)$ in both and the two orders of the factors differ only by exchanges of commuting factors.

### Step 2. Unit and origin (item 1)

**(a)** For $a_1=a_2=a$, $C'=C/a$: $e^{-iC'(a\lambda)}=e^{-iC\lambda}$, so every factor of (10.1) is unchanged, and the order is unchanged because $a>0$. $C/a$ is a sum of pair terms. Hence the statistics of every schedule are identical, and

$$
(a,\,a,\,C/a)\ \text{is an equivalence of } C\quad\text{for every } C\text{ and every } a>0 .
\tag{10.2}
$$

Taking $a\neq1$: **the unit of $\lambda$ is not detectable, for any $C$** (no nondegeneracy condition; $C=0$ included).

**(b)** Increasing all $\lambda_k$ by $\mu\ge0$ replaces each factor of (10.1) by $e^{iC\mu}\Pi^{(k)}_{s_k}(\lambda_k)e^{-iC\mu}$, so for every start

$$
P_{\lambda+\mu}(s_1,\dots,s_K;\Psi_0)=P_{\lambda}(s_1,\dots,s_K;\,e^{-iC\mu}\Psi_0).
\tag{10.3}
$$

If $C\Psi_0=c\Psi_0$, then $e^{-iC\mu}\Psi_0=e^{-ic\mu}\Psi_0$ is the same state (1.2): **the statistics of every schedule are unchanged by a common shift**, and for a single event $p(s)=\lVert\Pi_s\Psi_0\rVert^2$ **does not depend on its $\lambda$**.

### Step 3. Direction (item 2)

$R$ on one part at $\lambda_1$ with projectors $\Pi_s$, $R'$ on the other part at $\lambda_2>\lambda_1$ with projectors $\Pi'_t$; $\Psi_1:=e^{-iC\lambda_1}\Psi_0$, $\delta:=\lambda_2-\lambda_1$, $Q_t:=\Pi'_t(\delta)$. By (10.1), $P(s,t)=\lVert Q_t\Pi_s\Psi_1\rVert^2$. Since the $Q_t$ are orthogonal projectors with $\sum_tQ_t=\mathbb 1$,

$$
\sum_tP(s,t)=\lVert\Pi_s\Psi_1\rVert^2,
\qquad
\sum_sP(s,t)-\lVert Q_t\Psi_1\rVert^2=\Delta_t:=\langle\Psi_1\vert\sum_s\Pi_s\,[Q_t,\Pi_s]\,\vert\Psi_1\rangle ,
\tag{10.4}
$$

where $\lVert\Pi_s\Psi_1\rVert^2$ and $\lVert Q_t\Psi_1\rVert^2$ are the distributions of the single-event schedules.

- **Marginal of $R$ (smaller $\lambda$):** it never depends on whether $R'$ is made — for every $C$, every start, every pair of readings. Exact.
- **Marginal of $R'$ (larger $\lambda$):** it changes by $\Delta_t$ when $R$ is made. For a given start it depends on $R$ iff $\Delta_t\neq0$ for some $t$. It is independent of $R$ for every start iff $[Q_t,\Pi_s]=0$ for all $s,t$ (F0, F1; $\Psi_1$ runs over all unit vectors with $\Psi_0$). This holds for all readings and all $\lambda_1<\lambda_2$ when $C$ is split (10.5), in particular for uncoupled parts, by (10.6). It fails in example (i) with the place readings, where $\Delta_t=\mp\tfrac12\sin(2g\delta)\,\langle\Psi_1\vert\sigma^x\otimes\sigma^y\vert\Psi_1\rangle$ (Check 3).
- **Implication.** The effect of a reading is one-sided: it can reach the statistics of a reading at larger $\lambda$, never at smaller $\lambda$. Whenever some pair of readings on the two parts has $\Delta\neq0$, the direction of increasing $\lambda$ is detectable: of two events, the one whose marginal changes when the other is omitted has the larger $\lambda$. If no pair of readings on the two parts has $\Delta\neq0$ (split $C$), two events on different parts do not show the direction.
- **What fixes the direction.** Not the evolution (1.9) (Laws 3, 4), which is reversible. The direction is fixed by the state update of (1.3) (Law 1: a reading is a write and is not reversible) as applied in A8 and in the schedule: the written state is the one that evolves on to larger $\lambda$, the start being prescribed at the smallest $\lambda$ and nothing at the largest. In (10.1) this is the fact that the sum over the outcomes of the last factor gives $\mathbb 1$, while the sum over an earlier factor does not.

### Step 4. Split contracts and uncoupled parts (item 3)

**Definition.** $C$ is **split** if there are self-adjoint $F_1,F_2$ with

$$
C=F_1+F_2,\qquad [F_1,F_2]=0,\qquad [F_1,\Pi']=0,\qquad [F_2,\Pi]=0
\tag{10.5}
$$

for every place projector $\Pi$ of $\mathcal P_1$ and $\Pi'$ of $\mathcal P_2$.

**Lemma.** Let $C$ be split, $\lambda,\lambda'\ge0$.
1. $[F_1,F_2]=0$ gives $e^{iC\lambda}=e^{iF_1\lambda}e^{iF_2\lambda}$; with $[F_2,\Pi]=0$ and $[F_1,\Pi']=0$: $\Pi(\lambda)=e^{iF_1\lambda}\Pi e^{-iF_1\lambda}$ and $\Pi'(\lambda')=e^{iF_2\lambda'}\Pi'e^{-iF_2\lambda'}$.
2. With $W:=e^{iF_1\lambda}e^{iF_2\lambda'}$, the same commutators give $W\Pi W^\dagger=\Pi(\lambda)$ and $W\Pi'W^\dagger=\Pi'(\lambda')$, so $[\Pi(\lambda),\Pi'(\lambda')]=W[\Pi,\Pi']W^\dagger=0$.
3. $C':=F_1/a_1+F_2/a_2$ is split with terms $F_i/a_i$; by item 1 an event on $\mathcal P_i$ at $a_i\lambda$ under $C'$ has the operator $e^{iF_i\lambda}(\cdot)e^{-iF_i\lambda}$, the same as at $\lambda$ under $C$.
4. A re-timing keeps the order of the events within each part ($a_i>0$) and can change only the relative order of events on different parts, whose operators commute (item 2). By the comparison rule of Step 1:

$$
[\Pi(\lambda),\Pi'(\lambda')]=0,\qquad\bigl(a_1,\,a_2,\,F_1/a_1+F_2/a_2\bigr)\ \text{is an equivalence of } C\ \text{ for all } a_1,a_2>0 ,
\tag{10.6}
$$

provided $F_1/a_1+F_2/a_2$ is a sum of pair terms (automatic when each part is one object).

**(a) Uncoupled parts** are split with $F_1=C_1\otimes\mathbb 1$, $F_2=\mathbb 1\otimes C_2$. Hence, for every start (product or not; the proof is an operator identity):

$$
\Bigl(a_1,\,a_2,\,\tfrac1{a_1}C_1\otimes\mathbb 1+\tfrac1{a_2}\mathbb 1\otimes C_2\Bigr)\ \text{is an equivalence for all } a_1,a_2>0 .
\tag{10.7}
$$

- $C'$ is a sum of pair terms: $C_1\otimes\mathbb 1=d_2^{-1}\operatorname{Tr}_2C\otimes\mathbb 1-\text{const}$ ($d_2=\dim\mathcal H_2$), and $\operatorname{Tr}_2$ maps each pair term of $C$ to a pair term inside $\mathcal P_1$, a single-object term or a constant (A5).
- M6: $(C_1,C_2)\to(C_1+c,C_2-c)$ changes $C'$ by $c\,(1/a_1-1/a_2)\mathbb 1$, which changes no statistics.

**(b)** (10.7) with $a_1\neq a_2$ exists for every uncoupled $C$: **the common now is not detectable for uncoupled parts.** For events $e_1$ on $\mathcal P_i$ and $e_2$ on $\mathcal P_j$, $j\neq i$, with $\lambda_1<\lambda_2$:

$$
a_i\lambda_1=a_j\lambda_2\iff\frac{a_i}{a_j}=\frac{\lambda_2}{\lambda_1},
\qquad
a_i\lambda_1>a_j\lambda_2\iff\frac{a_i}{a_j}>\frac{\lambda_2}{\lambda_1}.
\tag{10.8}
$$

- If $\lambda_1>0$: both are solvable, and (10.7) is an equivalence for every pair $(a_1,a_2)$. So an equivalence making the two values of $\lambda$ equal exists, and one reversing their order exists.
- If $\lambda_1=0$: $a_i\lambda_1=0<a_j\lambda_2$ for every re-timing, so no equivalence of the defined class does either. The cause is the fixed origin of a linear re-timing, not a detectable order: by (10.6) the statistics do not depend on the relative order of events on different parts.

### Step 5. Single-event conditions (item 4a)

1. A single event $(i,R,\lambda)$ has $p(s)=\langle\Psi_0\vert\Pi_s(\lambda)\vert\Psi_0\rangle$. Equality with the re-timed schedule under $C'$ for every start, every reading and every $\lambda\ge0$ is, by (F0), equivalent to

$$
e^{iC\lambda}\,\Pi\,e^{-iC\lambda}=e^{ia_iC'\lambda}\,\Pi\,e^{-ia_iC'\lambda}\qquad\text{for all }\lambda\ge0\text{ and all place projectors }\Pi\text{ of }\mathcal P_i,\ i=1,2 .
\tag{10.9}
$$

2. Both sides are entire functions of $\lambda$, so (10.9) is equivalent to the equality of all Taylor coefficients at $\lambda=0$. At first and second order:

$$
[C,\Pi]=a_i\,[C',\Pi]\quad\Longleftrightarrow\quad[\,C-a_iC',\,\Pi\,]=0 ,
\tag{10.10}
$$

$$
[C,[C,\Pi]]=a_i^{2}\,[C',[C',\Pi]] .
\tag{10.11}
$$

3. Given (10.10), the right side of (10.11) is $a_i[C',[C,\Pi]]=[C-G_i,[C,\Pi]]$, so (10.11) $\iff[G_i,[C,\Pi]]=0$. By the Jacobi identity and (10.10), $[G_i,[C,\Pi]]=[[G_i,C],\Pi]+[C,[G_i,\Pi]]=a_i[[C,C'],\Pi]$. Hence, given (10.10):

$$
\text{(10.11)}\iff[[C,C'],\Pi]=0\ \ \text{for all place projectors of }\mathcal P_i;\qquad\text{for both parts together}\iff[C,C']=0 .
\tag{10.12}
$$

   Proof of the last step: commuting with all $\lvert h\rangle\langle h\rvert\otimes\lvert r\rangle\langle r\rvert$ means that $[C,C']$ is diagonal; and its diagonal vanishes by item 4 below.

4. **Matrix form.** Decompose every operator by the kind of transition, $X=X^{\rm d}+X^{(1)}+X^{(2)}+X^{(12)}$: the elements $X_{hr,h'r'}$ with ($h=h'$, $r=r'$), ($h\neq h'$, $r=r'$), ($h=h'$, $r\neq r'$), ($h\neq h'$, $r\neq r'$). Since $[G_1,\Pi]=0$ for all $\Pi$ of $\mathcal P_1$ iff $(G_1)_{hr,h'r'}=0$ for $h\neq h'$, and likewise for $G_2$:

$$
\text{(10.10) for both parts}\iff C'^{(1)}=\frac{C^{(1)}}{a_1},\quad C'^{(2)}=\frac{C^{(2)}}{a_2},\quad C'^{(12)}=\frac{C^{(12)}}{a_1}=\frac{C^{(12)}}{a_2};\qquad C'^{\rm d}\ \text{free (real diagonal)}.
\tag{10.13}
$$

   So off the diagonal $C_{hr,h'r'}=\alpha\,C'_{hr,h'r'}$ and $C_{h'r',hr}=\alpha\,C'_{h'r',hr}$ with the same real $\alpha\in\{a_1,a_2\}$, and $\langle hr\vert[C,C']\vert hr\rangle=\sum_{h'r'}\bigl(C_{hr,h'r'}C'_{h'r',hr}-C'_{hr,h'r'}C_{h'r',hr}\bigr)=0$.

### Step 6. All equivalences and the two sufficient conditions (item 4b)

**Theorem (exact, every $C$).**

$$
(a_1,a_2,C')\ \text{is an equivalence of } C\iff[\,C-a_iC',\Pi\,]=0\ \text{for all place projectors }\Pi\text{ of }\mathcal P_i\ (i=1,2)\ \text{ and }\ [C,C']=0 .
\tag{10.14}
$$

- *Necessity:* (10.10) and (10.12), from single-event schedules.
- *Sufficiency:* $[C,C']=0$ gives $[G_i,C]=0$, so $e^{ia_iC'\lambda}=e^{iC\lambda}e^{-iG_i\lambda}$, and $[G_i,\Pi]=0$ gives (10.9): every event has the same operator in both schedules. If $a_1=a_2$ the order of the events is unchanged. If $a_1\neq a_2$, put $F_1:=\frac{a_1}{a_1-a_2}G_2$, $F_2:=\frac{a_2}{a_2-a_1}G_1$. Then $F_1+F_2=C$, $F_1/a_1+F_2/a_2=C'$, $[F_1,F_2]\propto[G_2,G_1]=(a_2-a_1)[C,C']=0$, and (10.10) gives the last two relations of (10.5). So $C$ is split, and by (10.6) the operators of events on different parts commute. The comparison rule of Step 1 applies in both cases.

**Consequences.**
- For $a_1=a_2=a$, (10.10) for both parts says that $C-aC'$ is diagonal:

$$
(a,a,C')\ \text{is an equivalence of } C\iff C'=C/a+D,\ \ D\ \text{real diagonal in the joint places},\ \ [C,D]=0 .
\tag{10.15}
$$

- An equivalence with $a_1\neq a_2$ exists iff $C$ is split (10.5) with $F_2$ a sum of pair terms ($C'=C/a_1+(1/a_2-1/a_1)F_2$). So the two conditions below are sufficient, and "not split" is the exact criterion for a detectable common now.

**Sufficient condition for $a_1=a_2$ in every equivalence.**

$$
C^{(12)}\neq0\quad\text{or}\quad[C^{(1)},C^{(2)}]\neq0\qquad\Longrightarrow\qquad\text{every equivalence has } a_1=a_2 :\ \text{the common now is detectable.}
\tag{10.16}
$$

- If $C^{(12)}\neq0$: (10.13) gives $(a_1-a_2)\,C'^{(12)}=0$ with $C'^{(12)}\neq0$.
- If $C^{(12)}=0$: by (10.13), $C'=C'^{\rm d}+C^{(1)}/a_1+C^{(2)}/a_2$, and
  $[C,C']=[C^{(1)},C'^{\rm d}-C^{\rm d}/a_1]+[C^{(2)},C'^{\rm d}-C^{\rm d}/a_2]+(1/a_2-1/a_1)[C^{(1)},C^{(2)}]$. The three terms contain only transitions of kind (1), (2), (12) respectively, so by (10.14) each vanishes; the third gives $a_1=a_2$.
- In words: the contract changes the places of both parts in one step, or its place-changing terms of the two parts do not commute. The condition is sufficient, not necessary.

**Sufficient condition for an equivalence with $a_1\neq a_2$ although the parts are coupled.** $C$ is split (10.5) and coupled, with $F_2$ a sum of pair terms: then (10.6). A class that is always of this kind: $C$ commutes with every place projector of one part, say $\mathcal P_2$ (the places of $\mathcal P_2$ are conserved), $C=\sum_rC_r\otimes\lvert r\rangle\langle r\rvert$. Then $F_1=C$, $F_2=0$, and $(a_1,a_2,C/a_1)$ is an equivalence for all $a_1,a_2>0$. The parts are coupled iff $C_r-C_{r'}\notin\mathbb R\,\mathbb 1$ for some $r,r'$ (taking the $(r,r)$ block of $C=C_1\otimes\mathbb 1+\mathbb 1\otimes C_2$ gives $C_r=C_1+(C_2)_{rr}\mathbb 1$, and conversely).

### Step 7. The three examples (item 4c)

Each part is one object with places $\lvert1\rangle,\lvert2\rangle$, $\sigma^z=\lvert1\rangle\langle1\rvert-\lvert2\rangle\langle2\rvert$, $\sigma^x=\lvert1\rangle\langle2\rvert+\lvert2\rangle\langle1\rvert$; every self-adjoint $C'$ is one pair term. $D=\sum d_{hr}\lvert hr\rangle\langle hr\rvert$ is real diagonal, and $[C,D]_{hr,h'r'}=C_{hr,h'r'}(d_{h'r'}-d_{hr})$. In (ii), $h$ is the coupling constant. All three sets are exact, by (10.14).

**(i) $C=g\,\sigma^x\otimes\sigma^x$.** $C=C^{(12)}\neq0$, so $a_1=a_2=a$ by (10.16). By (10.15), $C'=C/a+D$ with $[C,D]=0$, i.e. $d_{11}=d_{22}$ and $d_{12}=d_{21}$:

$$
\mathcal E_{\rm(i)}=\bigl\{(a,\,a,\,C/a+c\,\mathbb 1+\mu\,\sigma^z\otimes\sigma^z):\ a>0,\ c,\mu\in\mathbb R\bigr\};\qquad\text{common now detectable.}
\tag{10.17}
$$

**(ii) $C=g\,\sigma^z\otimes\sigma^z+h\,\sigma^x\otimes\mathbb 1$.** $C^{\rm d}=g\,\sigma^z\otimes\sigma^z$, $C^{(1)}=h\,\sigma^x\otimes\mathbb 1$, $C^{(2)}=C^{(12)}=0$. By (10.13), $C'=C^{(1)}/a_1+C'^{\rm d}$, with no condition on $a_2$. Then $[C,C']=[C^{(1)},E]$ with $E:=C'^{\rm d}-C^{\rm d}/a_1$, which vanishes iff $e_{1r}=e_{2r}$, i.e. $E=c\,\mathbb 1+\mu\,\mathbb 1\otimes\sigma^z$:

$$
\mathcal E_{\rm(ii)}=\bigl\{(a_1,\,a_2,\,C/a_1+c\,\mathbb 1+\mu\,\mathbb 1\otimes\sigma^z):\ a_1,a_2>0,\ c,\mu\in\mathbb R\bigr\};\qquad\text{common now not detectable.}
\tag{10.18}
$$

The parts are coupled ($g\neq0$), and the places of $\mathcal P_2$ are conserved: the second sufficient condition of Step 6.

**(iii) $C=t\,\sigma^x\otimes\mathbb 1+\kappa\,\lvert2\rangle\langle2\rvert\otimes\sigma^x$.** $C^{(1)}=t\,\sigma^x\otimes\mathbb 1$, $C^{(2)}=\kappa\,\lvert2\rangle\langle2\rvert\otimes\sigma^x$, $C^{\rm d}=C^{(12)}=0$, and $[C^{(1)},C^{(2)}]=t\kappa\,(\lvert1\rangle\langle2\rvert-\lvert2\rangle\langle1\rvert)\otimes\sigma^x\neq0$, so $a_1=a_2=a$ by (10.16). By (10.15), $C'=C/a+D$ with $[C,D]=0$: $d_{1r}=d_{2r}$ (from $t$) and $d_{21}=d_{22}$ (from $\kappa$), so $D=c\,\mathbb 1$:

$$
\mathcal E_{\rm(iii)}=\bigl\{(a,\,a,\,C/a+c\,\mathbb 1):\ a>0,\ c\in\mathbb R\bigr\};\qquad\text{common now detectable.}
\tag{10.19}
$$

**Without $g,h,t,\kappa>0$.** If a constant vanishes, the contract is diagonal ((i) $g=0$, (ii) $h=0$), uncoupled ((ii) $g=0$, (iii) $\kappa=0$) or conserves the places of $\mathcal P_1$ ((iii) $t=0$). Each of these is split, and the common now is not detectable.

### Step 8. The order of two events on a chain (item 5)

With $c_{st}:=[Q_t,\Pi_s]$: $P(s,t)=\langle\Psi_1\vert\Pi_sQ_t\Pi_s\vert\Psi_1\rangle$ and $\tilde P(s,t)=\langle\Psi_1\vert Q_t\Pi_sQ_t\vert\Psi_1\rangle$.

**(a)** By (F0) ($\Psi_1$ runs over all unit vectors), $P=\tilde P$ for every start iff $\Pi_sQ_t\Pi_s=Q_t\Pi_sQ_t=:T$ for all $s,t$. Then $T=TQ_t=T\Pi_s$, so $c_{st}^\dagger c_{st}=-(Q_t\Pi_s-\Pi_sQ_t)^2=-(TQ_t-\Pi_sQ_t\Pi_s-Q_t\Pi_sQ_t+T\Pi_s)=0$; the converse is evident. By (7.1) with (F0), (F1), $R_i$ influences $R_j$ for no start iff $\mathcal M_i(Q_t)=Q_t$ for all $t$ iff $[Q_t,\Pi_s]=0$ for all $s,t$. Hence

$$
P=\tilde P\ \text{for every start}\iff[Q_t,\Pi_s]=0\ \text{for all }s,t\iff R_i\text{ influences }R_j\text{ for no start.}
\tag{10.20}
$$

For a single start only one direction holds: $\sum_s\tilde P(s,t)=\lVert Q_t\Psi_1\rVert^2=P_j(t)$ and $\sum_sP(s,t)=P_j(t\mid R_i)$, so

$$
\sum_s\bigl(P(s,t)-\tilde P(s,t)\bigr)=\Delta_t,\qquad\delta P\le\sum_{s,t}\bigl\lvert P(s,t)-\tilde P(s,t)\bigr\rvert .
\tag{10.21}
$$

$P=\tilde P$ implies no influence for that start; the converse fails (Check 3).

**(b)**
1. $Q_t\Pi_s=\Pi_sQ_t+c_{st}$ gives $\Pi_sQ_t\Pi_s-Q_t\Pi_sQ_t=\Pi_sc_{st}-c_{st}Q_t$, so $\lvert P-\tilde P\rvert(s,t)\le\lVert c_{st}\rVert\,\bigl(\lVert\Pi_s\Psi_1\rVert+\lVert Q_t\Psi_1\rVert\bigr)$.
2. $c_{st}=[A(\delta),B]$ with $A=\Pi'_t-\tfrac12\mathbb 1$ on cell $j$ and $B=\Pi_s-\tfrac12\mathbb 1$ on cell $i$, $\lVert A\rVert,\lVert B\rVert\le\tfrac12$; by (7.11), $\lVert c_{st}\rVert\le\tfrac12\Sigma_r(\delta)$.
3. $\sum_s\lVert\Pi_s\Psi_1\rVert^2=1$ and the Cauchy–Schwarz inequality give $\sum_s\lVert\Pi_s\Psi_1\rVert\le\sqrt m$; likewise $\sum_t\lVert Q_t\Psi_1\rVert\le\sqrt{m'}$, with $m'$ the number of classes of $R_j$.
4. Summing over the $m\,m'$ pairs, and using the last bound of (7.12): for every start, $\delta\ge0$, $r\ge1$,

$$
\sum_{s,t}\bigl\lvert P(s,t)-\tilde P(s,t)\bigr\rvert\le\Theta_{mm'}\,\Sigma_r(\delta)\le\Theta_{mm'}\,\frac{(2\lVert J\rVert\delta)^r}{r!}\,e^{2\lVert J\rVert\delta},
\qquad\Theta_{mm'}:=\tfrac12\sqrt{m\,m'}\,\bigl(\sqrt m+\sqrt{m'}\bigr).
\tag{10.22}
$$

5. The set on which this bound is below $\epsilon$, in the notation of (7.15):

$$
\tilde{\mathcal S}_\epsilon:=\bigl\{(r,\delta):\ r\ge1,\ \delta\ge0,\ \Theta_{mm'}\Sigma_r(\delta)<\epsilon\bigr\}=\mathcal S_{2\epsilon/\Theta_{mm'}}=\bigl\{(r,\delta):\ 0\le\delta<\delta_{2\epsilon/\Theta_{mm'}}(r)\bigr\}
\tag{10.23}
$$

   (last form for $\lVert J\rVert>0$; for $\lVert J\rVert=0$ it is all $(r,\delta)$ and $P=\tilde P$).
6. Comparison with (7.15): the same family of regions, with $\epsilon$ replaced by $2\epsilon/\Theta_{mm'}$. For $m,m'\ge2$, $\Theta_{mm'}\ge2\sqrt2>2$, so $\tilde{\mathcal S}_\epsilon\subseteq\mathcal S_\epsilon$, as (10.21) requires. If $m=1$ or $m'=1$, a projector is $\mathbb 1$ and $P=\tilde P$.
7. An explicit part of the set: for $2\lVert J\rVert\delta\le r/e^2$, the last bound of (7.12) and $r!\ge(r/e)^r$ give $\Sigma_r(\delta)\le e^{-r}e^{r/e^2}$, so

$$
\delta\le\frac{r}{2e^2\lVert J\rVert}\ \Longrightarrow\ \sum_{s,t}\lvert P-\tilde P\rvert\le\Theta_{mm'}\,e^{-(1-e^{-2})\,r};\qquad(r,\delta)\in\tilde{\mathcal S}_\epsilon\ \text{ if also }\ r>\frac{\ln(\Theta_{mm'}/\epsilon)}{1-e^{-2}} .
\tag{10.24}
$$

**(c)**
- **Equal $\lambda$** ($\delta=0$): $Q_t=\Pi'_t$ commutes with $\Pi_s$, so $P=\tilde P$ exactly, for every start and every pair of readings on different cells (consistent with (7.2)). The order of two events at equal $\lambda$ is not detectable.
- **Small delay over a large distance** ($(r,\delta)\in\tilde{\mathcal S}_\epsilon$, e.g. (10.24)): the statistics differ from the order-exchanged ones by less than $\epsilon$ for every start. The order is undetectable to accuracy $\epsilon$, but for $\delta>0$ in general not exactly.
- **In general:** by (10.20) the order of the two events is detectable for some start iff $R_i$ can influence $R_j$, and its statistical trace is at most (10.22). The order given by $\lambda$ is therefore detectable only where an influence is possible; elsewhere it is carried by the labels $\lambda_k$ alone. The boundary $\delta_{2\epsilon/\Theta}(r)$ depends on $\epsilon$: the detectable order is not sharp.

## Result

- **Item 1.** (10.2): $(a,a,C/a)$ is an equivalence for every $C$; the unit of $\lambda$ is never detectable. (10.3): for an eigenvector start, a common shift changes no statistics, and a single-event schedule does not depend on its $\lambda$.
- **Item 2.** (10.4): the marginal at the smaller $\lambda$ never depends on the later reading; the marginal at the larger $\lambda$ changes by $\Delta_t$, which vanishes for every start iff $[Q_t,\Pi_s]=0$. The direction is detectable wherever $\Delta\neq0$; it is fixed by the state update (1.3) (Law 1) as applied in A8, not by (1.9).
- **Item 3.** (10.5)–(10.7): the family $(a_1,a_2,C_1/a_1\otimes\mathbb 1+\mathbb 1\otimes C_2/a_2)$ for every start; the common now is not detectable; by (10.8) equal values and a reversed order are reachable iff $\lambda_1>0$.
- **Item 4.** (10.9)–(10.13): the single-event conditions; (10.14): all equivalences of a general $C$; (10.15): those with $a_1=a_2$; (10.16) and the split condition (10.5): the two sufficient conditions; (10.17)–(10.19): the examples, with the common now detectable in (i) and (iii), not in (ii).
- **Item 5.** (10.20), (10.21): $P=\tilde P$ for every start iff all $[Q_t,\Pi_s]=0$ iff no influence for any start; (10.22): the bound; (10.23), (10.24): the set; item (c) in Step 8.

## Consistency checks

1. **Dimensions and $C=0$.** $C\lambda$ is dimensionless, $a_i$ are pure numbers, so $C/a$, $F_i/a_i$ and $\lVert J\rVert\delta$ have the right dimensions. For $C=0$, (10.14) gives $[C',\Pi]=0$ for all place projectors: the equivalences are all $(a_1,a_2,D)$ with $D$ real diagonal. Directly: under a diagonal contract every place projector is constant, so the statistics do not depend on any $\lambda_k$. Passed.
2. **Contract diagonal in the places**, e.g. $C=g\,\sigma^z\otimes\sigma^z$ (coupled). (10.14) again gives all $(a_1,a_2,D)$: the common now is not detectable although the parts are coupled, in agreement with the split $F_1=C$, $F_2=0$ and with (10.1), where all factors are constant and commute. Also $c_{st}=0$ while (10.22) is positive: the bound is not tight, as allowed. Passed.
3. **Example (i) against the exact evolution of two cells** ($n=2$, $r=1$, $\lVert J\rVert=g$, place readings). $C$ anticommutes with $\mathbb 1\otimes\sigma^z$, so $(\mathbb 1\otimes\sigma^z)(\delta)=\cos(2g\delta)\,\mathbb 1\otimes\sigma^z+\sin(2g\delta)\,\sigma^x\otimes\sigma^y$.
   - $\mathcal M(Q_\pm)-Q_\pm=\mp\tfrac12\sin(2g\delta)\,\sigma^x\otimes\sigma^y$: the value of $\Delta_t$ used in Step 3; $\lVert c_{st}\rVert=\tfrac12\lvert\sin2g\delta\rvert\le\tfrac12(e^{2g\delta}-1)$, as in Step 8(b).
   - $\Psi_1=\lvert11\rangle$: $P=(\cos^2,\sin^2,0,0)$ and $\tilde P=(\cos^4,\sin^4,\sin^2\cos^2,\sin^2\cos^2)$ for $(s,t)=(1,1),(1,2),(2,1),(2,2)$, argument $g\delta$. So $\sum\lvert P-\tilde P\rvert=\sin^2(2g\delta)\le2\sqrt2\,(e^{2g\delta}-1)$, and $\Delta_t=0$ although $P\neq\tilde P$.
   - Under $C'=C/a+\mu\,\sigma^z\otimes\sigma^z$ at $a\delta$ the same operator results, because $\sigma^z\otimes\sigma^z$ commutes with $C$ and with $\mathbb 1\otimes\sigma^z$: agreement with (10.17). Passed.

## Open issues

- "Not split" is an exact criterion for a detectable common now, but it is not turned into an explicit classification of the coupled contracts (out of scope). For more than two objects the proviso "$F_2$ is a sum of pair terms" is part of the criterion.
- Item 2 treats two events on different parts. Whether the direction can be detected when no reading on one part influences a reading on the other (split $C$) was not examined; (10.4) itself holds for any two readings.
- $\tilde P$ is the comparator defined in the question, not the statistics of a schedule; item 5(c) refers to this notion. The constant $\Theta_{mm'}$ in (10.22) is not optimized.
- Re-timings that are not linear, more than two parts and readings represented by records are out of scope.

## Methods used

- Heisenberg form of sequential reading statistics
- Operator identities from quadratic forms (polarization)
- Taylor expansion of conjugations, nested commutators, Jacobi identity
- Decomposition of an operator by transition type in the place basis
- Commuting splits of a self-adjoint operator
- Pauli algebra for two two-place objects
- Commutator bound (7.11) with Cauchy–Schwarz inequality

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