# Question: 08-cell-medium

- **Subproject:** 03-ilang-space
- **Package:** 08-cell-medium
- **Equation tags:** (8.k)
- **Created:** 2026-10-08

## Goal

Determine how the effect of a body on a probe's witness geometry depends on where the body is recorded, when the medium consists of cells coupled by contracts. Specifically: when does the effect vanish, at which order in $\lambda$ does it first appear, and what is its first coefficient? The notation introduced below ($c_1,c_2,\dots$, $J$, $J_k$, $P_k$, $Q_k$, $r$, $c_4^{(\beta_0)}$) is fixed for later packages. "Probe", "body", "cell" and "link" are names only; no physical meaning is attached to them (M1, M4).

**Setting (two cells).**
- **Objects.** Four objects: the probe $p$, the body $b$, and the cells $c_1$ and $c_2$. They have pairwise different types, and each is the only instance of its type, so (1.11) imposes no constraint. The part is $A=\{p\}$, and its companion is $\{b,c_1,c_2\}$.
- **Contracts.** There are three pair contracts:
  $$C=\sum_h\lvert h\rangle\langle h\rvert_p\otimes K_h+\sum_\beta\lvert\beta\rangle\langle\beta\rvert_b\otimes B_\beta+J .$$
  - $K_h$ is self-adjoint on $\mathcal H_{c_1}$: the probe is recorded by $c_1$ only.
  - $B_\beta$ is self-adjoint on $\mathcal H_{c_2}$: the body is recorded by $c_2$ only.
  - $J$ is a self-adjoint operator on $\mathcal H_{c_1}\otimes\mathcal H_{c_2}$, the link.

  Each operator acts as the identity on the other objects. Write the link as a finite sum $J=\sum_kP_k\otimes Q_k$ with $P_k$ self-adjoint on $\mathcal H_{c_1}$ and $Q_k$ self-adjoint on $\mathcal H_{c_2}$; such a decomposition always exists.
- **Initial state.** A product: $\lvert\Psi(0)\rangle=\lvert\phi\rangle_p\otimes\lvert\beta_0\rangle_b\otimes\lvert\chi_1\rangle\otimes\lvert\chi_2\rangle$, with $\phi_h\neq0$ for every $h$. Write $\lvert\chi\rangle:=\lvert\chi_1\rangle\otimes\lvert\chi_2\rangle$.
- **Notation.** For a pair $h,h'$ of places of $p$, $D:=K_h-K_{h'}$ and $\tilde D:=D-\langle D\rangle$, as in 06-body-in-medium. Here $D$ acts on $c_1$ only.

**Use of 06.** The derivation of (6.6)–(6.12) uses only three facts:
- $K_h$, $K_{h'}$ and the operator common to both branches are self-adjoint;
- they act on a finite-dimensional medium;
- the medium starts in a unit vector.

Here the medium is $c_1c_2$, and both branches share $B_{\beta_0}+J$. State this, and then apply (6.6)–(6.12) with $\mathcal H_c\to\mathcal H_{c_1}\otimes\mathcal H_{c_2}$, $\lvert\chi\rangle\to\lvert\chi_1\rangle\otimes\lvert\chi_2\rangle$ and $B_{\beta_0}\to B_{\beta_0}+J$.

1. **Exact evolution.** Derive $\lvert\Psi(\lambda)\rangle$. Show that the body stays at $\beta_0$, and give the probe's witness data,
   $$W^{(\beta_0)}(h,h';\lambda)=\bigl\lvert\langle\chi\vert e^{i(K_{h'}+B_{\beta_0}+J)\lambda}e^{-i(K_h+B_{\beta_0}+J)\lambda}\vert\chi\rangle\bigr\rvert^2 .$$
   Define the body-dependent part of $W^{(\beta_0)}$ as $W^{(\beta_0)}$ minus its value at $B_{\beta_0}=0$.

2. **No link, no effect.** Show that for $J=0$ the body-dependent part vanishes for every $\lambda$.

3. **Delay through the link.** Expand
   $$W^{(\beta_0)}=1-\lambda^2\operatorname{Var}_\chi(D)+\lambda^3c_3^{(\beta_0)}+\lambda^4c_4^{(\beta_0)}+O(\lambda^5).$$
   - (a) Show that the body-dependent part of $c_3^{(\beta_0)}$ vanishes.
   - (b) Derive the body-dependent part of $c_4^{(\beta_0)}$ in closed form, first in operator form and then through the decomposition $J=\sum_kP_k\otimes Q_k$ in the product state.
   - (c) For $d_0(h,h')>0$, derive the corresponding body-dependent part of the witness angle at order $\lambda^3$.

4. **Light cone on a chain.** Let the medium be a chain of cells $c_1,\dots,c_L$ with pairwise different types, each the only instance of its type. Neighbouring cells are joined by links $J_k$, self-adjoint on $\mathcal H_{c_k}\otimes\mathcal H_{c_{k+1}}$, and there are no other medium contracts. The probe is recorded by $c_1$ ($K_h$ on $c_1$), and the body by $c_{r+1}$ ($B_\beta$ on $c_{r+1}$), with $0\le r\le L-1$. The initial state is a product over all objects. Show that the body-dependent part of $W^{(\beta_0)}(h,h';\lambda)$ is $O(\lambda^{3+r})$. For $r=0$, the first nonzero term can be the $\lambda^3$ term of 06.

5. **Example.** Two qubit cells with Pauli matrices, $\lvert\chi_1\rangle=\lvert\chi_2\rangle=\lvert0\rangle$ where $\sigma_z\lvert0\rangle=\lvert0\rangle$.
   - The places of $p$ are $h_1,\dots,h_n$, with $K_{h_i}=i\,X_1$ and $X_1:=\sigma_x+\sigma_z$ on $c_1$.
   - The places of $b$ carry $B_\beta=\beta\,\sigma_y$ on $c_2$, with real $\beta$ labelling the places of $b$.
   - The link is $J=g\,\sigma_y\otimes\sigma_x$ with $g>0$.

   Derive $c_3^{(\beta_0)}(h_i,h_j)$, the body-dependent part of $c_4^{(\beta_0)}(h_i,h_j)$, and the body-dependent part of the witness angle at order $\lambda^3$. Contrast with the link $J=g\,\sigma_z\otimes\sigma_z$.

## Inputs

From 02-view-content@v1. Notation fixed there: $p_h:=\langle h\vert V_A\vert h\rangle$, $H_V:=\{h:p_h>0\}$, $\lvert\psi_h\rangle:=\bigl(\langle h\rvert\otimes\mathbb 1_{\bar A}\bigr)\lvert\Psi\rangle$, $G_{hh'}=\langle E_{h'}\vert E_h\rangle$.

Eq. (2.1):

$$
\lvert\Psi\rangle=\sum_h\lvert h\rangle\otimes\lvert\psi_h\rangle,
\qquad
\lvert\psi_h\rangle=\bigl(\langle h\rvert\otimes\mathbb 1_{\bar A}\bigr)\lvert\Psi\rangle=\sum_r\Psi_{hr}\lvert r\rangle,
\qquad
\sum_h\lVert\psi_h\rVert^2=\lVert\Psi\rVert^2=1 .
$$

Eq. (2.2):

$$
V_A=\sum_{h,h'}\langle\psi_{h'}\vert\psi_h\rangle\,\lvert h\rangle\langle h'\rvert,
\qquad\text{i.e.}\qquad
(V_A)_{hh'}=\langle\psi_{h'}\vert\psi_h\rangle,
\qquad
p_h=\lVert\psi_h\rVert^2 .
$$

Eq. (2.11):

$$
W(h)=1,\qquad W(h,h')=\lvert G_{hh'}\rvert^2=\frac{\lvert(V_A)_{hh'}\rvert^2}{p_hp_{h'}} .
$$

From 03-witness-distance@v1. Notation fixed there: $\alpha(x,x'):=\arccos\sqrt{W(x,x')}\in[0,\pi/2]$, and for places $\alpha(h,h'):=\alpha([h],[h'])$.

Eq. (3.5). Here $\alpha$ is a metric on $X_V$.

$$
\cos\alpha(x,x')=\sqrt{W(x,x')}=\frac{\lvert(V_A)_{hh'}\rvert}{\sqrt{p_h\,p_{h'}}}
=\frac{\lvert\langle h\vert V_A\vert h'\rangle\rvert}{\sqrt{\langle h\vert V_A\vert h\rangle\langle h'\vert V_A\vert h'\rangle}},\qquad h\in x,\ h'\in x' .
$$

From 05-recording-contract@v1. In the setting of 05, the leading-order distance is defined as follows.

Eq. (5.10):

$$
\lim_{\lambda\to0^+}\frac{\alpha(h,h';\lambda)}{\lambda}=d_0(h,h')=\bigl\lVert u_h-u_{h'}\bigr\rVert,
\qquad \lvert u_h\rangle=\bigl(K_h-\langle K_h\rangle\bigr)\lvert\chi\rangle .
$$

From 06-body-in-medium@v1. These hold in the setting of 06:
- objects: probe $p$, body $b$ at $\beta_0$, and a single medium object $c$;
- contracts: $K_h$ and $B_\beta$, both acting on $c$;
- initial state: $\lvert\phi\rangle\otimes\lvert\beta_0\rangle\otimes\lvert\chi\rangle$.

There, $D=K_h-K_{h'}$, $\tilde D=D-\langle D\rangle$, and $\mathrm i$ is the imaginary unit.

Eq. (6.6):

$$
W^{(\beta_0)}(h,h';\lambda)=\frac{\lvert(V_A)_{hh'}\rvert^2}{p_hp_{h'}}=\bigl\lvert\langle\chi\vert e^{\mathrm i(K_{h'}+B_{\beta_0})\lambda}e^{-\mathrm i(K_h+B_{\beta_0})\lambda}\vert\chi\rangle\bigr\rvert^2 .
$$

Eq. (6.7). This holds if $[B_{\beta_0},K_h]=[B_{\beta_0},K_{h'}]=0$.

$$
G^{(\beta_0)}_{hh'}(\lambda)=\langle\chi\vert e^{\mathrm iK_{h'}\lambda}e^{-\mathrm iK_h\lambda}\vert\chi\rangle,\qquad
W^{(\beta_0)}(h,h';\lambda)=W(h,h';\lambda)\ \text{of (5.6)}\quad\text{for every }\lambda\ge0 .
$$

Eq. (6.9):

$$
W^{(\beta_0)}(h,h';\lambda)=1-\lambda^2\operatorname{Var}_\chi(D)+\lambda^3c_3^{(\beta_0)}(h,h')+O(\lambda^4),
$$

Eq. (6.10):

$$
c_3^{(\beta_0)}(h,h')=\operatorname{Im}\langle\chi\vert\,(K_{h'}+B_{\beta_0})\,\tilde D^2\vert\chi\rangle ,
$$

Eq. (6.11):

$$
c_3^{(\beta_0)}=c_3^{(0)}+\operatorname{Im}\langle\chi\vert B_{\beta_0}\tilde D^2\vert\chi\rangle,\qquad
c_3^{(0)}(h,h')=\operatorname{Im}\langle\chi\vert K_{h'}\tilde D^2\vert\chi\rangle,\qquad
\operatorname{Im}\langle\chi\vert B_{\beta_0}\tilde D^2\vert\chi\rangle=\frac{\langle\chi\vert[B_{\beta_0},\tilde D^2]\vert\chi\rangle}{2\mathrm i}.
$$

Eq. (6.12):

$$
\alpha^{(\beta_0)}(h,h';\lambda)=\lambda d_0(h,h')+\lambda^2\alpha_2^{(\beta_0)}(h,h')+O(\lambda^3),\qquad
\alpha_2^{(\beta_0)}=-\frac{c_3^{(\beta_0)}}{2d_0},
$$

## Assumptions

- The settings stated under "Goal". All contracts are independent of $\lambda$ (A5), and $\lambda\ge0$.
- Items 3–5 are expansions for $\lambda\to0^+$. Items 1 and 2 hold for every $\lambda$.

## Scope

- In scope:
  - the probe's witness data with the body recorded by a different cell;
  - the order and the first coefficient of the body-dependent part;
  - the light-cone bound on a chain;
  - the qubit example.
- Out of scope:
  - medium states that are not products;
  - identical-type objects (Law 5);
  - large $\lambda$;
  - the dependence of the effect on a distance in the derived geometry, as opposed to the number of links;
  - the relation to the cost $L$ (1.8), and motion;
  - any reading of $\lambda$ as time (M2), and any reading of the effect as gravity.

## Depth

- Item 1: short argument.
- Item 2: short argument.
- Item 3: derive. The Baker–Campbell–Hausdorff formula to third order and the cumulant expansion of $\langle e^{-i\lambda H}\rangle$ to fourth order may be used without proof.
- Item 4: short argument. A counting of operator supports and commutators suffices.
- Item 5: derive.

## Expected result

The main model expects the following. Derive each statement independently, and report any discrepancy in a sign or a factor.

- Item 2: an exact equality for every $\lambda$.
- Item 3:
  - the body-dependent part of $c_3^{(\beta_0)}$ is $\operatorname{Im}\langle\chi\vert B_{\beta_0}\tilde D^2\vert\chi\rangle=0$ in the product state;
  - the body-dependent part of $c_4^{(\beta_0)}$ is
    $$\tfrac16\langle\chi\vert\{\tilde D,[B_{\beta_0},[J,D]]\}\vert\chi\rangle=-\tfrac23\sum_k\operatorname{Im}\langle\chi_1\vert P_k\tilde D^2\vert\chi_1\rangle\,\operatorname{Im}\langle\chi_2\vert B_{\beta_0}Q_k\vert\chi_2\rangle ;$$
  - the body-dependent part of $\alpha$ at order $\lambda^3$ is $-\lambda^3$ times this quantity divided by $2d_0$.
- Item 4: the bound $O(\lambda^{3+r})$.
- Item 5:
  - $c_3^{(\beta_0)}(h_i,h_j)=0$;
  - the body-dependent part of $c_4^{(\beta_0)}$ is $\tfrac43\,g\,\beta_0\,(i-j)^2$;
  - the body-dependent part of $\alpha$ is $-\tfrac23\,g\,\beta_0\,\lvert i-j\rvert\,\lambda^3$;
  - for $J=g\,\sigma_z\otimes\sigma_z$ the body-dependent part of $c_4^{(\beta_0)}$ vanishes.

Give every main result a tag $(8.k)$.

Consistency checks, at most three, chosen from:
- with $J=0$ the body-dependent part vanishes, in agreement with (6.7);
- with $B_{\beta_0}=\beta_0\mathbb 1$ there is no effect;
- with the body recorded by $c_1$ instead of $c_2$ ($r=0$), the $\lambda^3$ term of (6.11) returns.

## Code

None.