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

# The field as an object: sharp field states, reduced descriptions, conventions

- **Subproject:** 05-ilang-free-fall
- **Package:** 02-field-object
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-10

## Setup and assumptions

**Class.** Every description of the form stated in `question.md`: a body $b$, a field $f$ and a set $r$ of further objects (possibly empty), of pairwise different types, with $\mathcal H=\mathcal H_{br}\otimes\mathcal H_f$, $\mathcal H_{br}:=\mathcal H_b\otimes\mathcal H_r$, finite-dimensional (A1), and the contract

$$
C=C_{br}\otimes\mathbb 1_f+\sum_{j\in J}B_j\otimes\mathbb 1_r\otimes\Phi_j+\mathbb 1_{br}\otimes C_f \tag{2.1}
$$

for every self-adjoint $C_{br}$, $C_f$ and every finite family of self-adjoint $B_j$ (on $\mathcal H_b$) and $\Phi_j$ (on $\mathcal H_f$). The start is $\Psi_0=\Psi_{br}\otimes\varphi$ with $\Psi_{br}$ any unit vector of $\mathcal H_{br}$, and $\lambda\ge0$. The form (2.1), the product start and the field state $\varphi$ named in each step are assumptions (M8, A11). The results say what such descriptions give, not that the language enforces them (M10).

**Notation.** $B_j$ also denotes $B_j\otimes\mathbb 1_r$ on $\mathcal H_{br}$. $\langle O\rangle_\Psi:=\langle\Psi\vert O\vert\Psi\rangle$. $U(\lambda):=e^{-iC\lambda}$ and $\Psi(\lambda)=U(\lambda)\Psi_0$. $V(\lambda):=\operatorname{Tr}_f\lvert\Psi(\lambda)\rangle\langle\Psi(\lambda)\rvert$ is the view of $\{b\}\cup r$, and $V_f(\lambda)$ the view of $\{f\}$ (1.5).

**LI.** The hypothesis that $\mathbb 1_b$ and the $B_j$, $j\in J$, are linearly independent (over $\mathbb R$; for self-adjoint operators this is the same as over $\mathbb C$). Every statement below says whether it uses LI.

**Schedules.** A schedule consists of readings with projectors $\Pi_i\otimes\mathbb 1_f$ at $0\le\lambda_1\le\dots\le\lambda_k$, where $\Pi_i$ is diagonal in the joint places of $b$ and $r$ and belongs to the outcome $r_i$. By (1.3) and (1.9),

$$
p(r_1,\dots,r_k)=\bigl\lVert(\Pi_k\otimes\mathbb 1_f)\,U(\lambda_k-\lambda_{k-1})\cdots(\Pi_1\otimes\mathbb 1_f)\,U(\lambda_1)\,\Psi_0\bigr\rVert^2 . \tag{2.2}
$$

**Exact or estimate.** Every statement is exact, for all $\Psi_{br}$ and all $\lambda\ge0$, except (2.9), (2.10) and (2.14), which are expansions for $\lambda\to0$ at fixed operators and start, with the stated remainder.

**Input.** (17.15) of 03-ilang-space/17-motion@v2, as quoted in `question.md`: for $C=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$, start $\phi\otimes\chi$, record vectors $u_h=(K_h-\langle K_h\rangle)\chi$ and $K_{h_m}=mX$, one has $d_0(h_m,h_n)=\lvert m-n\rvert\,\sigma_X$ with $\sigma_X^2=\langle X^2\rangle-\langle X\rangle^2$. It is used only in Step 8.

## Derivation

### Step 1. Sharp and stationary field: state and views (item 1a)

Let $\Phi_j\varphi=n_j\varphi$ and $C_f\varphi=e_f\varphi$. For every $\Psi\in\mathcal H_{br}$, (2.1) gives

$$
C\,(\Psi\otimes\varphi)=\bigl((C_{\rm red}+e_f)\Psi\bigr)\otimes\varphi,\qquad C_{\rm red}:=C_{br}+\sum_{j\in J}n_j\,B_j\otimes\mathbb 1_r . \tag{2.3}
$$

So $\mathcal H_{br}\otimes\varphi$ is invariant under $C$, and (1.9) gives

$$
\Psi(\lambda)=e^{-ie_f\lambda}\,\Psi_{\rm red}(\lambda)\otimes\varphi,\qquad \Psi_{\rm red}(\lambda):=e^{-iC_{\rm red}\lambda}\Psi_{br}, \tag{2.4}
$$

$$
V(\lambda)=\lvert\Psi_{\rm red}(\lambda)\rangle\langle\Psi_{\rm red}(\lambda)\rvert,\qquad V_f(\lambda)=\lvert\varphi\rangle\langle\varphi\rvert . \tag{2.5}
$$

This is exact for every $\Psi_{br}$ (a product or not) and every $\lambda$, and LI is not used. Both views are pure and $V_f$ is constant: the field does not become entangled with $b$ or $r$. The factor $e^{-ie_f\lambda}$ is a common phase (1.2).

### Step 2. Reduced description (item 1b)

By (2.3), $U(\ell)$ acts on $\mathcal H_{br}\otimes\varphi$ as $e^{-ie_f\ell}e^{-iC_{\rm red}\ell}\otimes\mathbb 1_f$, and $\Pi_i\otimes\mathbb 1_f$ maps this subspace into itself. Hence (2.2) becomes

$$
p(r_1,\dots,r_k)=\bigl\lVert\Pi_k\,e^{-iC_{\rm red}(\lambda_k-\lambda_{k-1})}\cdots\Pi_1\,e^{-iC_{\rm red}\lambda_1}\,\Psi_{br}\bigr\rVert^2 , \tag{2.6}
$$

which is the statistics of the same schedule in the reduced description with the objects $b$, $r$, the contract $C_{\rm red}$ of (2.3) and the start $\Psi_{br}$. The answer is yes, exactly, for every schedule and every $\Psi_{br}$.

*All contracts that do this.* A contract $C'$ on $\mathcal H_{br}$ gives the statistics (2.6) for every schedule and every $\Psi_{br}$ if and only if

$$
C'=C_{\rm red}+\Delta,\qquad \Delta\ \text{real and diagonal in the joint places of } b \text{ and } r,\qquad [\Delta,C_{\rm red}]=0 . \tag{2.7}
$$

*Proof.* ($\Leftarrow$) $e^{-iC'\ell}=e^{-i\Delta\ell}e^{-iC_{\rm red}\ell}$, and $e^{-i\Delta\ell}$ commutes with every $\Pi_i$ (both are diagonal in the joint places) and with every $e^{-iC_{\rm red}\ell'}$. All factors $e^{-i\Delta\ell}$ can be moved to the left in (2.6), where they do not change the norm. ($\Rightarrow$) Read all objects of $b$ and $r$ at one $\lambda$, with outcome the joint place $K$. Equal probabilities for every $\Psi_{br}$ mean $\lvert\langle K\vert e^{-iC'\lambda}\Psi_{br}\rangle\rvert=\lvert\langle K\vert e^{-iC_{\rm red}\lambda}\Psi_{br}\rangle\rvert$ for all $\Psi_{br}$, hence $e^{iC'\lambda}\lvert K\rangle=e^{i\theta_K(\lambda)}e^{iC_{\rm red}\lambda}\lvert K\rangle$. So $W(\lambda):=e^{-iC_{\rm red}\lambda}e^{iC'\lambda}$ is diagonal in the joint places for all $\lambda\ge0$, and so are its derivatives at $\lambda=0$: $W'(0)=i\Delta$ with $\Delta:=C'-C_{\rm red}$, and $W''(0)=[C_{\rm red},\Delta]-\Delta^2$. Thus $\Delta$ is diagonal (and real), and $[C_{\rm red},\Delta]$ is diagonal; its diagonal elements vanish identically, so it is zero. $\square$

A real constant $\Delta=c\,\mathbb 1_{br}$ always qualifies. Since $[\Delta,C_{\rm red}]_{KK'}=(\Delta_K-\Delta_{K'})(C_{\rm red})_{KK'}$, it is the only possibility when the nonzero off-diagonal elements of $C_{\rm red}$ connect all joint places.

### Step 3. Expansion of the purity for small $\lambda$ (order estimate)

For any unit $\varphi$ put $n_j:=\langle\varphi\vert\Phi_j\vert\varphi\rangle$ (the numbers of the definition when $\varphi$ is sharp), take $C_{\rm red}$ as in (2.3), and define

$$
\omega_j^{(k)}:=(\Phi_j-n_j)\,C_f^{\,k}\,\varphi,\qquad k=0,1,2,\dots \tag{2.8}
$$

$\varphi$ is sharp iff $\omega_j^{(0)}=0$ for all $j$. Suppose some $\omega_j^{(k)}\neq0$, and let $k_0$ be the smallest such $k$. Then $\langle\varphi\vert\omega_j^{(k_0)}\rangle=0$: for $k_0=0$ by the definition of $n_j$, for $k_0\ge1$ because $(\Phi_j-n_j)\varphi=0$.

1. Write $C=H_0+W$ with $H_0:=C_{\rm red}\otimes\mathbb 1_f+\mathbb 1_{br}\otimes C_f$ and $W:=\sum_jB_j\otimes(\Phi_j-n_j)$. The vector $\Psi_I(\lambda):=e^{iH_0\lambda}\Psi(\lambda)$ has the same $\operatorname{Tr}V^2$ as $\Psi(\lambda)$, because $e^{iH_0\lambda}$ is a product of a unitary on $\mathcal H_{br}$ and one on $\mathcal H_f$. It obeys $i\,\mathrm d\Psi_I/\mathrm d\lambda=W_I(\lambda)\Psi_I$ with $W_I(\mu):=e^{iH_0\mu}We^{-iH_0\mu}$.
2. By (2.8), $(\Phi_j-n_j)e^{-iC_f\mu}\varphi=\frac{(-i\mu)^{k_0}}{k_0!}\omega_j^{(k_0)}+O(\mu^{k_0+1})$, so $W_I(\mu)\Psi_0=\frac{(-i\mu)^{k_0}}{k_0!}\sum_jB_j\Psi_{br}\otimes\omega_j^{(k_0)}+O(\mu^{k_0+1})$. In the Dyson series the $n$-th term is therefore $O(\lambda^{k_0+n})$, and

$$
\Psi_I(\lambda)=\Psi_{br}\otimes\varphi+\frac{(-i\lambda)^{k_0+1}}{(k_0+1)!}\sum_{j}B_j\Psi_{br}\otimes\omega_j^{(k_0)}+O(\lambda^{k_0+2}) . \tag{2.9}
$$

3. Decompose $\Psi_I=\Psi_a\otimes\varphi+\chi$ with $\chi\in\mathcal H_{br}\otimes\varphi^{\perp}$. Then the view is $\lvert\Psi_a\rangle\langle\Psi_a\rvert+V_\chi$ with $V_\chi:=\operatorname{Tr}_f\lvert\chi\rangle\langle\chi\rvert$ and $\lVert\Psi_a\rVert^2=1-\operatorname{Tr}V_\chi$, so exactly $1-\operatorname{Tr}V^2=2\operatorname{Tr}V_\chi-2\langle\Psi_a\vert V_\chi\vert\Psi_a\rangle-(\operatorname{Tr}V_\chi)^2-\operatorname{Tr}V_\chi^2$. By (2.9), $\chi$ is the second term of (2.9) up to $O(\lambda^{k_0+2})$, and $\Psi_a=\Psi_{br}+O(\lambda)$. Hence

$$
1-\operatorname{Tr}V(\lambda)^2=\frac{2\,\lambda^{2k_0+2}}{\bigl((k_0+1)!\bigr)^2}\,\Bigl\lVert\sum_{j}\bigl(B_j-\langle B_j\rangle_{\Psi_{br}}\bigr)\Psi_{br}\otimes\omega_j^{(k_0)}\Bigr\rVert^2+O(\lambda^{2k_0+3}) . \tag{2.10}
$$

The squared norm equals $\sum_{j,j'}\Gamma_{jj'}\,\langle\omega_j^{(k_0)}\vert\omega_{j'}^{(k_0)}\rangle$ with $\Gamma_{jj'}:=\langle B_jB_{j'}\rangle_{\Psi_{br}}-\langle B_j\rangle_{\Psi_{br}}\langle B_{j'}\rangle_{\Psi_{br}}$.

*Lemma (uses LI).* If $k_0$ exists, the coefficient in (2.10) is nonzero for some $\Psi_{br}$. *Proof.* Expand $\omega_j^{(k_0)}=\sum_\alpha w_{j\alpha}e_\alpha$ in an orthonormal basis of $\mathcal H_f$. If the coefficient vanished for every $\Psi_{br}$, every $\Psi_{br}$ would be an eigenvector of $M_\alpha\otimes\mathbb 1_r$, $M_\alpha:=\sum_jw_{j\alpha}B_j$, for each $\alpha$. Then $M_\alpha\in\mathbb C\,\mathbb 1_b$, and by LI all $w_{j\alpha}=0$, a contradiction. $\square$

### Step 4. Sharp field in general: the criterion (item 2a)

**Theorem.** Let $\varphi$ be sharp with the numbers $n_j$, and assume LI. The view $V(\lambda)$ is pure for every $\Psi_{br}$ and every $\lambda\ge0$ if and only if

$$
\Phi_j\,C_f^{\,k}\,\varphi=n_j\,C_f^{\,k}\,\varphi\qquad\text{for all } j\in J \text{ and all } k\ge0 . \tag{2.11}
$$

Equivalent forms of (2.11): each $\Phi_j$ acts as the number $n_j$ on $\mathcal K_\varphi:=\operatorname{span}\{C_f^{\,k}\varphi\}_{k\ge0}$ (so $k\le\dim\mathcal H_f-1$ suffices); the state $e^{-iC_f\lambda}\varphi$ is sharp with the same numbers for every $\lambda$; every nonzero component of $\varphi$ in an eigenspace of $C_f$ is sharp with the numbers $n_j$ (standard: $\mathcal K_\varphi$ is the span of these components). Item 1 is the case $\mathcal K_\varphi=\mathbb C\varphi$.

*Proof.* (Sufficient; no LI.) $\mathcal K_\varphi$ is invariant under $C_f$ and, by (2.11), under the $\Phi_j$. On $\mathcal H_{br}\otimes\mathcal K_\varphi$, (2.1) reads $C=C_{\rm red}\otimes\mathbb 1+\mathbb 1\otimes C_f$, so

$$
\Psi(\lambda)=\Psi_{\rm red}(\lambda)\otimes e^{-iC_f\lambda}\varphi,\qquad V(\lambda)=\lvert\Psi_{\rm red}(\lambda)\rangle\langle\Psi_{\rm red}(\lambda)\rvert,\qquad V_f(\lambda)=\lvert e^{-iC_f\lambda}\varphi\rangle\langle e^{-iC_f\lambda}\varphi\rvert . \tag{2.12}
$$

(Necessary; uses LI.) $1-\operatorname{Tr}V(\lambda)^2$ is analytic in $\lambda$. If it vanishes for all $\lambda\ge0$ and all $\Psi_{br}$, all its Taylor coefficients at $0$ vanish for all $\Psi_{br}$. If (2.11) failed, $k_0\ge1$ would exist and, by the Lemma, the coefficient of $\lambda^{2k_0+2}$ in (2.10) would be nonzero for some $\Psi_{br}$. $\square$

- **Sharpness is necessary too (LI).** The same argument with $k_0=0$ shows: for an arbitrary $\varphi$, the view is pure for every $\Psi_{br}$ and $\lambda\ge0$ iff $\varphi$ is sharp and (2.11) holds. Under LI these are exactly the field states in which the body and $r$ see definite numbers.
- **Reduced description.** When (2.11) holds, it exists, with the same contract $C_{\rm red}$ of (2.3): on $\mathcal H_{br}\otimes\mathcal K_\varphi$ every factor of (2.2) is a product, which gives (2.6). The class (2.7) is unchanged. The view of the field is pure but moves; there is no entanglement.
- **Without LI.** (2.11) is still sufficient and (2.12) holds, but (2.11) is no longer necessary. Choose a basis $\{\mathbb 1_b,A_\alpha\}$ of $\operatorname{span}\{\mathbb 1_b,B_j\}$ and write $B_j=m_{j0}\mathbb 1_b+\sum_\alpha m_{j\alpha}A_\alpha$. Then $C$ has the form (2.1) with the LI family $A_\alpha$, $\Phi'_\alpha:=\sum_jm_{j\alpha}\Phi_j$ and $C_f':=C_f+\sum_jm_{j0}\Phi_j$; $\varphi$ is sharp for the $\Phi'_\alpha$ with $n'_\alpha=\sum_jm_{j\alpha}n_j$. The necessary and sufficient condition is (2.11) for $(\varphi,C_f',\{\Phi'_\alpha\})$, and the reduced contract is $C_{br}+\sum_\alpha n'_\alpha A_\alpha=C_{\rm red}-\sum_jm_{j0}n_j\,\mathbb 1_{br}$. Example: $J=\{1\}$, $B_1=\mathbb 1_b$; the field is not coupled at all, and the view is pure for every $\varphi$ although (2.11) can fail.

### Step 5. Example and lowest power (item 2b)

**Example.** A field with two places $\lvert0\rangle,\lvert1\rangle$, one index, and $\gamma\neq0$:

$$
\Phi_1=\lvert1\rangle\langle1\rvert,\qquad C_f=\gamma\bigl(\lvert0\rangle\langle1\rvert+\lvert1\rangle\langle0\rvert\bigr),\qquad \varphi=\lvert0\rangle:\qquad n_1=0,\qquad \omega_1^{(1)}=\Phi_1C_f\varphi=\gamma\lvert1\rangle\neq0 . \tag{2.13}
$$

$\varphi$ is sharp and (2.11) fails at $k=1$.

**Lowest power.** For a sharp $\varphi$ that violates (2.11), $k_0\ge1$, so by (2.10) $1-\operatorname{Tr}V^2$ has no terms in $\lambda^2$ or $\lambda^3$, and

$$
1-\operatorname{Tr}V(\lambda)^2=\frac{\lambda^4}{2}\sum_{j,j'\in J}\Gamma_{jj'}\,\bigl\langle C_f\varphi\bigm\vert(\Phi_j-n_j)(\Phi_{j'}-n_{j'})\bigm\vert C_f\varphi\bigr\rangle+O(\lambda^5),\qquad \Gamma_{jj'}=\langle B_jB_{j'}\rangle_{\Psi_{br}}-\langle B_j\rangle_{\Psi_{br}}\langle B_{j'}\rangle_{\Psi_{br}} . \tag{2.14}
$$

The lowest power that can be nonzero is $\lambda^4$. Its coefficient is nonzero for some $\Psi_{br}$ iff some $(\Phi_j-n_j)C_f\varphi\neq0$ (Lemma, LI). If all these vectors vanish, the leading power is $\lambda^{2k_0+2}$ with the coefficient of (2.10). For comparison, a $\varphi$ that is not sharp has $k_0=0$ and starts at $\lambda^2$.

### Step 6. Static field that is not sharp (item 3)

Let the $\Phi_j$ commute with each other and with $C_f$; LI is not assumed. The $\Phi_j$ have joint spectral projectors $P_\nu$, labelled by the joint eigenvalues $\nu=(\nu_j)_{j\in J}\in\mathcal N$, with $P_\nu P_{\nu'}=\delta_{\nu\nu'}P_\nu$, $\sum_\nu P_\nu=\mathbb 1_f$ and $[P_\nu,C_f]=0$. Then

$$
\Phi_j=\sum_{\nu\in\mathcal N}\nu_j\,P_\nu,\qquad C=\sum_{\nu\in\mathcal N}C^{(\nu)}\otimes P_\nu+\mathbb 1_{br}\otimes C_f,\qquad C^{(\nu)}:=C_{br}+\sum_{j\in J}\nu_j\,B_j\otimes\mathbb 1_r . \tag{2.15}
$$

All summands commute, so $U(\lambda)=\sum_\nu e^{-iC^{(\nu)}\lambda}\otimes P_\nu e^{-iC_f\lambda}$. Let $w_\nu:=\lVert P_\nu\varphi\rVert^2$, $S:=\{\nu:w_\nu>0\}$, $\varphi_\nu:=P_\nu\varphi/\sqrt{w_\nu}$ and $\Psi^{(\nu)}(\lambda):=e^{-iC^{(\nu)}\lambda}\Psi_{br}$, the evolution in the reduced description $\mathcal R_\nu$ (objects $b$, $r$; contract $C^{(\nu)}$; start $\Psi_{br}$).

**(a)** The vectors $e^{-iC_f\lambda}\varphi_\nu$ are orthonormal, so

$$
\Psi(\lambda)=\sum_{\nu\in S}\sqrt{w_\nu}\;\Psi^{(\nu)}(\lambda)\otimes e^{-iC_f\lambda}\varphi_\nu,\qquad V(\lambda)=\sum_{\nu\in S}w_\nu\,\lvert\Psi^{(\nu)}(\lambda)\rangle\langle\Psi^{(\nu)}(\lambda)\rvert . \tag{2.16}
$$

The view is the mixture, with weights $w_\nu$, of the pure views of the reduced descriptions $\mathcal R_\nu$.

**(b)** $\Pi_i\otimes\mathbb 1_f$ commutes with $\mathbb 1_{br}\otimes P_\nu$. The vector in (2.2) is therefore $\sum_{\nu\in S}\sqrt{w_\nu}\,\bigl(\Pi_ke^{-iC^{(\nu)}(\lambda_k-\lambda_{k-1})}\cdots\Pi_1e^{-iC^{(\nu)}\lambda_1}\Psi_{br}\bigr)\otimes e^{-iC_f\lambda_k}\varphi_\nu$, with orthogonal summands, and

$$
p(r_1,\dots,r_k)=\sum_{\nu\in S}w_\nu\,p^{(\nu)}(r_1,\dots,r_k) , \tag{2.17}
$$

where $p^{(\nu)}$ is the statistics of the same schedule in $\mathcal R_\nu$. Any unit vector of the range of $P_\nu$ is sharp with the numbers $\nu$ and satisfies (2.11), because $[C_f,P_\nu]=0$; by Step 4 a start $\Psi_{br}\otimes(\text{such a vector})$ has the statistics $p^{(\nu)}$. Hence a start in which the field is sharp with numbers $\nu$ drawn once, with probability $w_\nu$, gives exactly (2.17). No schedule of readings on $b$ and $r$ distinguishes the two starts.

**(c)** From (2.16),

$$
\operatorname{Tr}V(\lambda)^2=\sum_{\nu,\nu'\in S}w_\nu w_{\nu'}\,\bigl\lvert\langle\Psi^{(\nu)}(\lambda)\vert\Psi^{(\nu')}(\lambda)\rangle\bigr\rvert^2=\sum_{\nu,\nu'\in S}w_\nu w_{\nu'}\,\bigl\lvert\langle\Psi_{br}\vert e^{iC^{(\nu)}\lambda}e^{-iC^{(\nu')}\lambda}\vert\Psi_{br}\rangle\bigr\rvert^2 . \tag{2.18}
$$

The view is pure for every $\Psi_{br}$ and every $\lambda\ge0$ if and only if

$$
\sum_{j\in J}(\nu_j-\nu'_j)\,B_j\in\mathbb R\,\mathbb 1_b\qquad\text{for all }\nu,\nu'\in S . \tag{2.19}
$$

*Proof.* Since $\sum_{\nu,\nu'}w_\nu w_{\nu'}=1$, purity means that every term of (2.18) has modulus one. ($\Leftarrow$) If $C^{(\nu)}-C^{(\nu')}=c\,\mathbb 1$, then $e^{iC^{(\nu)}\lambda}e^{-iC^{(\nu')}\lambda}=e^{ic\lambda}\mathbb 1$. ($\Rightarrow$) Modulus one for every unit $\Psi_{br}$ makes every $\Psi_{br}$ an eigenvector of the unitary $e^{iC^{(\nu)}\lambda}e^{-iC^{(\nu')}\lambda}$, which is then a multiple of $\mathbb 1$ for all $\lambda\ge0$; so is its derivative at $\lambda=0$, $i(C^{(\nu)}-C^{(\nu')})=i\sum_j(\nu_j-\nu'_j)B_j$. $\square$

Under LI, (2.19) holds iff $S$ has one element, that is, iff $\varphi$ is sharp. Without LI, (2.19) also holds for those non-sharp $\varphi$ whose $S$ lies in one coset of $\{a\in\mathbb R^J:\sum_ja_jB_j\in\mathbb R\,\mathbb 1_b\}$. When (2.19) fails, the purity is (2.18).

### Step 7. Conventions (item 4; LI assumed)

**(a)** *The replacements of $C_{br}$, the $\Phi_j$ and $C_f$ at fixed $B_j$ that leave $C$ unchanged are exactly*

$$
\Phi_j\to\Phi_j+a_j\,\mathbb 1_f,\qquad C_f\to C_f+c\,\mathbb 1_f,\qquad C_{br}\to C_{br}-\sum_{j\in J}a_j\,B_j\otimes\mathbb 1_r-c\,\mathbb 1_{br},\qquad a_j,c\in\mathbb R . \tag{2.20}
$$

*Proof.* (2.20) leaves (2.1) unchanged. Conversely, let $\delta_{br},\delta_j,\delta_f$ be self-adjoint changes with $\delta_{br}\otimes\mathbb 1_f+\sum_jB_j\otimes\delta_j+\mathbb 1_{br}\otimes\delta_f=0$. Multiply by $\mathbb 1_{br}\otimes F$ with $F$ self-adjoint and traceless, and take $\operatorname{Tr}_f$: $\sum_j\operatorname{Tr}(F\delta_j)\,B_j+\operatorname{Tr}(F\delta_f)\,\mathbb 1_{br}=0$. By LI, $\operatorname{Tr}(F\delta_j)=\operatorname{Tr}(F\delta_f)=0$ for all such $F$, so $\delta_j=a_j\mathbb 1_f$ and $\delta_f=c\,\mathbb 1_f$; then $\delta_{br}$ follows. $\square$

(2.20) preserves the sharpness and the stationarity of $\varphi$, and

$$
n_j\to n_j+a_j,\qquad e_f\to e_f+c,\qquad C_{\rm red}\to C_{\rm red}-c\,\mathbb 1_{br} . \tag{2.21}
$$

The $a_j$ are independent of each other. For a lapse-type term, $B_x=\lvert x\rangle\langle x\rvert\otimes G$ (LI holds iff $G\notin\mathbb R\,\mathbb 1_{\rm int}$), the numbers $a_x$ move the term $\sum_xa_x\lvert x\rangle\langle x\rvert\otimes G$, which is diagonal in the positions, between the pair term with the field and the body's own term: a place-dependent replacement.

*Without LI:* every relation $\sum_jz_jB_j=z_0\mathbb 1_b$ adds the replacements $\Phi_j\to\Phi_j+z_jE$, $C_f\to C_f-z_0E$, with $E$ any self-adjoint operator on $\mathcal H_f$. If $E\varphi\notin\mathbb C\varphi$ they destroy the sharpness for the family, so that "sharp" itself is then not well defined.

**(b)** From (2.21):
- The numbers $n_j$ are **not** well defined.
- The differences $n_j-n_{j'}$, $j\neq j'$, are **not** well defined: they change by $a_j-a_{j'}$, which is arbitrary. For a lapse-type term, neither the profile $n_x$ nor its difference between two positions is well defined.
- $C_{\rm red}$ is well defined up to a real multiple of $\mathbb 1_{br}$, which changes the common phase (1.2) in (2.4) and nothing in (2.5), (2.6); this is the case $\Delta=c\,\mathbb 1$ of (2.7). Exactly well defined is

$$
\hat C:=C_{\rm red}+e_f\,\mathbb 1_{br}=(\mathbb 1_{br}\otimes\langle\varphi\rvert)\,C\,(\mathbb 1_{br}\otimes\lvert\varphi\rangle) , \tag{2.22}
$$

  which depends only on $C$ and $\varphi$, and is therefore the same in every listing of the terms of $C$, at fixed $B_j$ or not.
- The hypothesis of item 1 is well defined: under LI, $\varphi$ is sharp and stationary iff $C$ maps $\mathcal H_{br}\otimes\varphi$ into itself. ($\Rightarrow$) is (2.3). ($\Leftarrow$) With $Q:=\mathbb 1_f-\lvert\varphi\rangle\langle\varphi\rvert$, $\sum_jB_j\Psi\otimes Q\Phi_j\varphi+\Psi\otimes QC_f\varphi=0$ for all $\Psi$; taking components in a basis of $\mathcal H_f$ and using LI gives $Q\Phi_j\varphi=0=QC_f\varphi$.
- Invariant under (2.20) are also the differences between two field states, $n_j(\varphi)-n_j(\varphi')$, and in Step 6 the weights $w_\nu$, the differences $\nu_j-\nu'_j$ and $C^{(\nu)}-C^{(\nu')}$.

**On-site blocks.** Let $\mathcal H_b=\mathcal H_{\rm pos}\otimes\mathcal H_{\rm int}$ (positions $x$, internal labels), $B_x=\lvert x\rangle\langle x\rvert\otimes G$, and $C_{br}=C_b\otimes\mathbb 1_r+\sum_x\lvert x\rangle\langle x\rvert\otimes\mathbb 1_{\rm int}\otimes K_x+R$, where $C_b$ is the body's own term, $K_x$ acts on the medium and $R$ collects the remaining terms. For an operator $O$ on $\mathcal H_{br}$ let $O_{xx}:=(\langle x\rvert\otimes\mathbb 1)\,O\,(\lvert x\rangle\otimes\mathbb 1)$, an operator on $\mathcal H_{\rm int}\otimes\mathcal H_r$. The on-site block of $C_{\rm red}$ is

$$
\Omega_x:=(C_{\rm red})_{xx}=(C_b)_{xx}\otimes\mathbb 1_r+\mathbb 1_{\rm int}\otimes K_x+n_x\,G\otimes\mathbb 1_r+R_{xx}=\hat C_{xx}-e_f . \tag{2.23}
$$

In every listing of $C$ in which $\varphi$ is sharp and stationary, (2.3) gives $C_{\rm red}=\hat C-e_f\mathbb 1_{br}$ with the number $e_f$ of that listing; this uses no LI. Hence:
- $\Omega_x$ is well defined up to a real constant that is the same at all positions;
- $\Omega_x-\Omega_{x'}=\hat C_{xx}-\hat C_{x'x'}$ is well defined exactly;
- the three place-dependent parts of $\Omega_x$ are **not** separately well defined. With arbitrary real $a_x$, $k_x$, the replacements
  1. body and field: $\Phi_x\to\Phi_x+a_x\mathbb 1_f$, $C_b\to C_b-\sum_xa_x\lvert x\rangle\langle x\rvert\otimes G$, a case of (2.20);
  2. body and medium: $K_x\to K_x+k_x\mathbb 1$, $C_b\to C_b-\sum_xk_x\lvert x\rangle\langle x\rvert\otimes\mathbb 1_{\rm int}$;
  3. medium and field: $K_x\to K_x+k_x\mathbb 1$, and the pair term with the field enlarged by the members $B'_x=\lvert x\rangle\langle x\rvert\otimes\mathbb 1_{\rm int}$, $\Phi'_x=-k_x\mathbb 1_f$ (not at fixed $B_j$; $\varphi$ stays sharp)

  leave $C$, $\hat C$ and every $\Omega_x$ unchanged, and change $n_x$, $(C_b)_{xx}$ and $K_x$ by arbitrary functions of $x$. In $\Omega_x-\Omega_{x'}$ only the sum of the three contributions is well defined.

### Step 8. One object as recording medium and field (item 5)

Here $\mathcal H=\mathcal H_b\otimes\mathcal H_c$ with $\mathcal H_b=\mathcal H_{\rm pos}\otimes\mathcal H_{\rm int}$, the start is $\phi\otimes\chi$, readings are made on $b$ only, and

$$
C=C_b\otimes\mathbb 1_c+\sum_x\lvert x\rangle\langle x\rvert\otimes\Bigl(\mathbb 1_{\rm int}\otimes\kappa\,x\,X+G\otimes(\mathbb 1_c+g\,x\,Y)\Bigr) . \tag{2.24}
$$

This is (2.1) with $r=\emptyset$, $f=c$, $C_f=0$ and the commuting field operators $\kappa xX$ and $\mathbb 1_c+gxY$, so Step 6 applies (it uses no LI). Let $P_{y\xi}$ be the joint spectral projectors of $(Y,X)$, $P_y:=\sum_\xi P_{y\xi}$, $w_{y\xi}:=\lVert P_{y\xi}\chi\rVert^2$ and $w_y:=\sum_\xi w_{y\xi}$.

**(a)** Since $g\neq0$ and a position $x\neq0$ exists, $\chi$ is sharp for the $\Phi_x$ iff $Y\chi=y\chi$ for a real $y$. Then $(\Phi_x-\langle\Phi_x\rangle)\chi=0$, so the lapse-type term adds nothing to the record vectors, $u_x=\kappa x\,(X-\langle X\rangle)\chi$, and (17.15) with $X\to\kappa X$ gives $d_0(x,x')=\kappa\lvert x-x'\rvert\,\sigma_X$. The condition is

$$
Y\chi=y\,\chi\qquad\text{and}\qquad\sigma_X^2=\langle\chi\vert X^2\vert\chi\rangle-\langle\chi\vert X\vert\chi\rangle^2>0 : \tag{2.25}
$$

$\chi$ lies in one eigenspace $\mathcal E_y$ of $Y$ and has components in at least two eigenspaces of $X$. Such a $\chi$ exists iff $X$ is not a function of $Y$; in particular none exists for $Y=X$. Under (2.25), $\mathcal H_b\otimes\mathcal E_y$ contains the start and is invariant under $C$ and under the readings on $b$, and on it

$$
C=\Bigl(C_b+\sum_xn_x\,\lvert x\rangle\langle x\rvert\otimes G\Bigr)\otimes\mathbb 1_c+\sum_x\lvert x\rangle\langle x\rvert\otimes\mathbb 1_{\rm int}\otimes\kappa\,x\,X,\qquad n_x=1+g\,y\,x . \tag{2.26}
$$

Exactly, for every $\phi$ and every schedule of readings on $b$: the factor $\Phi_x$ amounts to the number $n_x$, a profile linear in $x$ with slope $g\,y$ (flat if $y=0$). The body behaves as a body with the own term $C_b+\sum_xn_x\lvert x\rangle\langle x\rvert\otimes G$ that is recorded by $c$ through $K_x$ alone, a description of the form of 17-motion. By Step 7 this term cannot be told from a position-dependent own term of the body's type (A10).

**(b)** If $\chi$ is not an eigenvector of $Y$, (2.16) with $\nu=(y,\xi)$ gives

$$
V_b(\lambda)=\sum_{y,\xi}w_{y\xi}\,\lvert\phi^{(y,\xi)}(\lambda)\rangle\langle\phi^{(y,\xi)}(\lambda)\rvert,\qquad \phi^{(y,\xi)}(\lambda)=e^{-iC_b^{(y,\xi)}\lambda}\phi,\qquad C_b^{(y,\xi)}:=C_b+\sum_x\lvert x\rangle\langle x\rvert\otimes\bigl((1+g\,y\,x)\,G+\kappa\,\xi\,x\,\mathbb 1_{\rm int}\bigr) . \tag{2.27}
$$

The view of the body is the mixture, with weights $w_{y\xi}$, of the pure views of reduced descriptions of the body alone. The reduced description $(y,\xi)$ has the profile $n_x=1+g\,y\,x$ and the record gradient $\kappa\xi$ (the record term reduced to the number $\kappa\xi x$). A profile and a gradient occur together exactly for the pairs with $w_{y\xi}>0$, with the joint distribution $w_{y\xi}$ of the commuting operators $Y$ and $X$ in $\chi$. Summed over $\xi$ at fixed $y$: $V_b=\sum_yw_yV_b^{[y]}$, where $V_b^{[y]}$ is the view of the body in the description (2.26) started in $\phi\otimes\chi_y$, $\chi_y:=P_y\chi/\sqrt{w_y}$, whose record distances are $\kappa\lvert x-x'\rvert\,\sigma_X^{[y]}$ with $\sigma_X^{[y]}$ the standard deviation of $X$ in $\chi_y$. If $X=h(Y)$, for example $Y=X$, then $\xi=h(y)$: each profile occurs with exactly one gradient, and $\sigma_X^{[y]}=0$ for every $y$.

## Result

- **Item 1.** (a) State (2.4) and views (2.5): a product for all $\lambda$, both views pure, no entanglement; exact, no LI. (b) Yes: every schedule has the statistics (2.6) of the reduced description with the contract $C_{\rm red}$ of (2.3). The contracts that do this are exactly those of (2.7).
- **Item 2.** (a) Under LI the view is pure for every $\Psi_{br}$, $\lambda\ge0$ iff (2.11); then the state is (2.12) and the reduced description exists with the same $C_{\rm red}$ (2.3). Without LI, (2.11) is sufficient but not necessary; the criterion is (2.11) for the reduced family of Step 4. (b) Example (2.13). The lowest power is $\lambda^4$ with the coefficient of (2.14); in general $\lambda^{2k_0+2}$ with the coefficient of (2.10) (order estimates).
- **Item 3.** (a) Mixture (2.16) of the reduced descriptions with contracts (2.15). (b) Statistics (2.17); readings on $b$ and $r$ cannot distinguish the start from a sharp field with numbers drawn with probabilities $w_\nu$. (c) Pure iff (2.19), which under LI means $\varphi$ sharp; otherwise the purity is (2.18).
- **Item 4.** (a) Replacements (2.20), transformations (2.21). (b) Not well defined: $n_j$, and $n_j-n_{j'}$. Well defined: $C_{\rm red}$ up to a real multiple of $\mathbb 1_{br}$, and $\hat C$ (2.22) exactly. The on-site block (2.23) is well defined up to a constant common to all positions, the differences of blocks exactly; the split of a block into body, medium and field parts is not well defined.
- **Item 5.** (a) Condition (2.25); the factor amounts to the linear profile $n_x=1+g\,y\,x$ of (2.26). (b) Mixture (2.27), with profile and record gradient distributed jointly as $Y$ and $X$ in $\chi$.
- **Controls (M9).** (2.13)–(2.14) and (2.16)–(2.18) are the descriptions in which (2.4)–(2.6) fail: a sharp field that does not stay sharp, and a static field that is not sharp.

## Consistency checks

1. **Dimensions.** A change of unit $\lambda\to\lambda/s$ takes $C\to sC$, so each product $B_j\otimes\Phi_j$ and $C_f$ scale with $s$. With dimensionless $\Phi_j$: $n_j$, $w_\nu$ and $\operatorname{Tr}V^2$ are dimensionless, (2.3), (2.15), (2.22) are homogeneous, and the coefficient of (2.10) scales as $s^{2k_0+2}$ (two factors $B_j$, $2k_0$ factors $C_f$), which compensates $\lambda^{2k_0+2}$. Passed.
2. **$\Phi_j=\mathbb 1_f$ for all $j$ (this includes a field with one place).** Every $\varphi$ is sharp with $n_j=1$, (2.11) holds for every $\varphi$ and $C_f$, all $\omega_j^{(k)}=0$, (2.16) has the single sector $\nu=(1,\dots,1)$, and $C_{\rm red}=C_{br}+\sum_jB_j$. Directly: $C=(C_{br}+\sum_jB_j)\otimes\mathbb 1_f+\mathbb 1_{br}\otimes C_f$, the field decouples. Passed.
3. **Two-place field, one index.** Body with two places, $B_1=\beta\sigma_3$, $C_{br}=0$, $r=\emptyset$, $\Psi_{br}=(\lvert0\rangle+\lvert1\rangle)/\sqrt2$, so $\Gamma_{11}=\beta^2$. (i) The field of (2.13): with $E_\pm(\lambda)=e^{-i(\pm\beta\Phi_1+C_f)\lambda}\lvert0\rangle$ one finds $E_+-E_-=-\lambda^2\beta\gamma\lvert1\rangle+O(\lambda^3)$ and, by (1.7), $1-\operatorname{Tr}V^2=\tfrac12\bigl(1-\lvert\langle E_-\vert E_+\rangle\rvert^2\bigr)=\tfrac12\beta^2\gamma^2\lambda^4+O(\lambda^5)$, equal to (2.14). (ii) Static: $C_f=0$, $\varphi=(\lvert0\rangle+\lvert1\rangle)/\sqrt2$; (2.18) gives $1-\operatorname{Tr}V^2=\tfrac12\sin^2\beta\lambda$, and (2.10) with $k_0=0$, $\lVert\omega_1^{(0)}\rVert^2=\tfrac14$, gives $\tfrac12\beta^2\lambda^2$. Passed.

## Open issues

- (2.10) and (2.14) are expansions for $\lambda\to0$; no bound on the remainder, and no range of $\lambda$ on which the leading term dominates, is derived here.
- A field whose $\Phi_j$ do not commute with $C_f$ is treated only through the criterion (2.11) and the expansion (2.10); its view at finite $\lambda$ is left to a later package.
- Starts in which the field is entangled with $b$ or $r$, and readings on the field, are outside the question. In Step 8, readings on the medium are not treated.
- Which operator $G$ the field multiplies, and what the well-defined differences $\Omega_x-\Omega_{x'}$ of (2.23) mean for a clock and for motion, belong to 03-clock-and-fall. A later package that uses a profile $n_x$ must fix the listing, or work with (2.22) and (2.23).

## Methods used

- Invariant subspaces of a self-adjoint operator
- Partial trace, purity of a view
- Interaction picture and Dyson series
- Krylov subspace, analyticity in $\lambda$
- Joint spectral decomposition of commuting self-adjoint operators
- Linear independence in tensor products of operator spaces

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