In the second equality, we expanded the product and used that only terms with exactly one non-vacuum operator for each time step contribute. Note that the fact that there is no contribution from the terms with a control mismatch can be seen as a consequence of Lemma 2. This shows that the causal box is an extension of the QC-QC.

**Lemma 2 (No aborts).** *When composing the projective extension with local agents satisfying Assumption 1 and Assumption 2, the projective measurements yield the outcome accept with unit probability.*

*Proof.* We can show by induction that this is always the case. The global past is assumed to send a single message during the first time step and initialises the control correctly, thus the base case is trivial. Assume now the projective measurements {$P_{accept}^{I_k/O_k}, 1 - P_{accept}^{I_k/O_k}$} for $k < n'+1$ all yield accept. The internal operation thus acts effectively as $\bar{V}_{n+1} \circ P_{accept}^{O_n} = \bar{V}_{n+1}|_{H_{corr}^{O_n}}$, which outputs a state in $\mathcal{H}_{corr}^{I_{n+1}}$ by definition (Eq. (30)) and thus {$P_{accept}^{I_{n+1}}, 1 - P_{accept}^{I_{n+1}}$} yields accept. The local operations conserve the number of messages and cannot cause a control mismatch and so {$P_{accept}^{O_{n+1}}, 1 - P_{accept}^{O_{n+1}}$} must also give that outcome.

## E.2 Proofs for the appendix

**Lemma 4 (Isometry condition for internal operations).** For $K_{n-1}, \mathcal{L}_{n-1} \subset \mathcal{N}$ and $k_n \in \mathcal{N} \setminus K_{n-1}, l_n \in \mathcal{N} \setminus \mathcal{L}_{n-1}$ with $K_{n-1} \cup k_n = \mathcal{L}_{n-1} \cup l_n$, it holds that

$$ \sum_{k_{n+1}} V_{K_{n-1}, k_n}^{\rightarrow k_{n+1} \dagger} V_{\mathcal{L}_{n-1}, l_n}^{\rightarrow k_{n+1}} = \mathbb{1}_{A_{k_n}^O \alpha_n} \delta_{K_{n-1}, \mathcal{L}_{n-1}} \delta_{k_n, l_n}. \quad (91) $$

*Proof.* The lemma is a direct consequence of the maps $V_{n+1}: \mathcal{H}^{A^O \alpha_n C_n} \rightarrow \mathcal{H}^{A^I \alpha_{n+1} C_{n+1}}$ in the QC-QC picture being isometries,

$$ V_{n+1}^{\dagger} V_{n+1} = \mathbb{1}^{A^O \alpha_n C_n} = \sum_{K_{n-1}, k_n} \mathbb{1}_{A_{k_n}^O \alpha_n} \otimes |K_{n-1}, k_n\rangle \langle K_{n-1}, k_n|. \quad (92) $$

Using the definition of $V_{n+1}$ given in Eq. (52),

$$ \begin{aligned} V_{n+1}^{\dagger} V_{n+1} &= \sum_{K_{n-1}, \mathcal{L}_{n-1}, k_n, k_{n+1}} \sum_{l_n, l_{n+1}} V_{K_{n-1}, k_n}^{\rightarrow k_{n+1} \dagger} V_{\mathcal{L}_{n-1}, l_n}^{\rightarrow l_{n+1}} \otimes |K_{n-1}, k_n\rangle \langle K_{n-1} \cup k_n, k_{n+1}| \mathcal{L}_{n-1} \cup l_n, l_{n+1} \rangle \langle \mathcal{L}_{n-1}, l_n| \\ &= \sum_{K_n} \sum_{k_n, l_n \in K_n} \sum_{k_{n+1}} V_{K_n \setminus k_n, k_n}^{\rightarrow k_{n+1} \dagger} V_{K_n \setminus l_n, l_n}^{\rightarrow l_{n+1}} \otimes |K_n \setminus k_n, k_n\rangle \langle K_n \setminus l_n, l_n|. \end{aligned} \quad (93) $$

Comparing this to Eq. (92), we notice that terms with an off-diagonal projector over the control $|K_n \setminus k_n, k_n\rangle \langle K_n \setminus l_n, l_n|, k_n \neq l_n$, must vanish. This is only possible if $\sum_{k_{n+1}} V_{K_n \setminus k_n, k_n}^{\rightarrow k_{n+1} \dagger} V_{K_n \setminus l_n, l_n}^{\rightarrow l_{n+1}} = 0$ for $k_n \neq l_n$. Meanwhile, terms where $k_n = l_n$ must be the identity which implies

$$ \sum_{k_{n+1}} V_{K_n \setminus k_n, k_n}^{\rightarrow k_{n+1} \dagger} V_{K_n \setminus k_n, k_n}^{\rightarrow k_{n+1}} = \mathbb{1}_{A_{k_n}^O \alpha_n}. \quad (94) $$

Combining these, we obtain the statement of the lemma.

**Proposition 3 (Sequence representation for QC-QCs).** *The maps $\bar{V}_1, \dots, \bar{V}_{N+1}$ as defined by Eq. (63) are isometries.*

*Proof.* We use the convention of Eq. (40) for the symmetric tensor product. The proof relies on the fact that the inner product is invariant if one replaces vectors of the form
$$ \bigotimes_k |\psi_k\rangle^{A_k} \qquad (95) $$

with vectors

$$ \bigodot_k |\psi_k\rangle^{A_k}. \qquad (96) $$

This follows from the fact that the different spaces associated to different agents are orthogonal. We can also show this by explicitly calculating the two inner products (using Eq. (41)).

$$ \begin{aligned} \bigodot_k \langle \psi_k |^{A_k} \bigodot_l | \phi_l \rangle^{A_l} &= \sum_{\pi} \prod_k \langle \psi_k |^{A_k} | \phi_l \rangle^{A_l(l)} \\ &= \sum_{\pi} \prod_k \langle \psi_k | \phi_k \delta_{k, \pi(k)} \\ &= \prod_k \langle \psi_k | \phi_k \\ &= \bigotimes_k \langle \psi_k |^{A_k} \bigotimes_l | \psi_l \rangle^{A_l} \end{aligned} \qquad (97) $$

where we have used the orthogonality of the different Hilbert spaces.

Due to Lemma 3, it suffices to consider the inner products of elementary tensors with a fixed but arbitrary number of messages $M$

$$ \bigotimes_{k_n} \bigodot_{i=i_{k_n}}^{i_{k_n+1}-1} |\psi^i\rangle^{\bar{A}_{k_n}} |\mathcal{K}_{n-1}^i, k_n\rangle \qquad (98) $$

where $1 = i_1 < i_2 < ... < i_{N+1} = M+1$ and we dropped the time stamps to reduce clutter. This is because the subspaces corresponding to different numbers of messages are orthogonal to each other and the operators conserve the number of messages, allowing us to invoke condition 4 of Lemma 3, while the elementary tensors span the whole space, therefore containing in particular a basis (condition 3).

Replacing the tensor product in Eq. (98) with a symmetric tensor product and using Eq. (41), the inner product of two such tensor products is

$$ \sum_{\pi \in S^M} \prod_i \langle \psi^i | \phi^{\pi(i)} \rangle \langle \mathcal{K}_{n-1}^i, k_n^i | \mathcal{L}_{n-1}^i, l_n^{\pi(i)} \rangle \qquad (99) $$

where we added indexes $i$ to $k_n$ and $l_n$ so that there is only one product over $i$ instead of one over the same and one over the agents' labels.

On the other hand, applying $\tilde{V}_{n+1}$ to Eq. (98), we obtain

$$ \mathrm{symm}(\bigotimes_{k_n} \bigotimes_{i=i_{k_n}}^{i_{k_n+1}-1} \sum_{k_{n+1}^i} \bar{V}_{\mathcal{K}_{n-1}^i, k_n}^{\to k_{n+1}^i} |\psi^i\rangle^{\bar{A}_{k_n}} |\mathcal{K}_{n-1}^i \cup k_n, k_{n+1}^i\rangle). \qquad (100) $$

For the inner product, we can once again use tensors of the form

$$ \bigodot_i \sum_{k_{n+1}^i} \bar{V}_{\mathcal{K}_{n-1}^i, k_n}^{\to k_{n+1}^i} |\psi^i\rangle^{\bar{A}_{k_n}} |\mathcal{K}_{n-1}^i \cup k_n, k_{n+1}^i\rangle. \qquad (101) $$

which yields
$$
\begin{align}
& \sum_{\pi \in S^M} \prod_i \sum_{k_{n+1}^i, l_{n+1}^{\pi(i)}} \langle \psi^i | \bar{V}_{\mathcal{K}_{n-1}^i, k_n^i}^{\rightarrow k_{n+1}^i \dagger} \bar{V}_{\mathcal{L}_{n-1}^{\pi(i)}, l_n^{\pi(i)}}^{\rightarrow l_{n+1}^{\pi(i)}} | \phi^{\pi(i)} \rangle \langle \mathcal{K}_{n-1}^i \cup k_n^i, k_{n+1}^i | \mathcal{L}_{n-1}^{\pi(i)} \cup l_n^{\pi(i)}, l_{n+1}^{\pi(i)} \rangle \nonumber \\
& = \sum_{\pi \in S^M} \prod_i \sum_{k_{n+1}^i} \langle \psi^i | \bar{V}_{\mathcal{K}_{n-1}^i, k_n^i}^{\rightarrow k_{n+1}^i \dagger} \bar{V}_{\mathcal{L}_{n-1}^{\pi(i)}, l_n^{\pi(i)}}^{\rightarrow k_{n+1}^i} | \phi^{\pi(i)} \rangle \langle \mathcal{K}_{n-1}^i \cup k_n^i, k_{n+1}^i | \mathcal{L}_{n-1}^{\pi(i)} \cup l_n^{\pi(i)}, k_{n+1}^i \rangle \nonumber \\
& = \sum_{\pi \in S^M} \prod_i \sum_{k_{n+1}^i} \langle \psi^i | \bar{V}_{\mathcal{K}_{n-1}^i, k_n^i}^{\rightarrow k_{n+1}^i \dagger} \bar{V}_{\mathcal{L}_{n-1}^{\pi(i)}, l_n^{\pi(i)}}^{\rightarrow k_{n+1}^i} | \phi^{\pi(i)} \rangle \delta_{\mathcal{K}_{n-1}^i \cup k_n^i, \mathcal{L}_{n-1}^{\pi(i)} \cup l_n^{\pi(i)}} \tag{102}
\end{align}
$$

$$
\begin{align*}
& \stackrel{\text{Lemma 4}}{=} \sum_{\pi \in S^M} \prod_i \langle \psi^i | \phi^{\pi(i)} \rangle \delta_{\mathcal{K}_{n-1}^i, \mathcal{L}_{n-1}^{\pi(i)}} \delta_{k_n^i, l_n^{\pi(i)}} \delta_{\mathcal{K}_{n-1}^i \cup k_n^i, \mathcal{L}_{n-1}^{\pi(i)} \cup l_n^{\pi(i)}} \\
& = \sum_{\pi \in S^M} \prod_i \langle \psi^i | \phi^{\pi(i)} \rangle \langle \mathcal{K}_{n-1}^i, k_n^i | \mathcal{L}_{n-1}^i, l_n^{\pi(i)} \rangle
\end{align*}
$$

where in the second equality we used that if the target spaces are different (i.e. $k_{n+1}^i \neq l_{n+1}^{\pi(i)}$),
the states are orthogonal. $\square$

**Proposition 4** (Causal box description of QC-QCs). The causal box with sequence representation given by the isometries $\bar{V}_1, \dots, \bar{V}_{N+1}$ from Eq. (63) is an extension of the QC-QC described by the process vector $|w_{N,F}\rangle = \sum_{\{k_1, \dots, k_N\}} |V_{\emptyset, \emptyset}^{\rightarrow k_1}\rangle * \dots * |V_{\{k_1, \dots, k_{N-2}\}, k_N-1}^{\rightarrow k_N}\rangle * |V_{N \setminus \{k_N, k_N\}}^{\rightarrow F}\rangle$.

*Proof.* W.l.o.g., we assume the global past to be trivial. If we have a QC-QC with non-trivial global past, compose this QC-QC with some arbitrary state $\rho^P \in \mathcal{L}(\mathcal{H}^P)$ to obtain a QC-QC with trivial global past and compose the causal box we obtain from the QC-QC with $\rho^P \otimes |t=1\rangle\langle t=1|$. Note that the behaviour of the original QC-QC and causal box are thus completely determined by QC-QCs and causal boxes with trivial past since $\rho^P$ is arbitrary. Thus, it is enough to prove the statement in the case of trivial global past. Further instead of considering the Choi vector, we can consider its projection to the space

$$
\text{Span}[\{ \bigotimes_{n=1}^{N} |i_{k_n}, 2n\rangle^{\bar{A}_{k_n}^I} |\Omega, (2n)^c\rangle^{\bar{A}_{k_n}^I} |j_{k_n}, 2n+1\rangle^{\bar{A}_{k_n}^O} |\Omega, (2n+1)^c\rangle^{\bar{A}_{k_n}^O} |i_F, 2N+2\rangle^{\bar{F}} \} \begin{array}{c} i_{k_1} \neq \Omega; \\ i_{k_n}, j_{k_n}; \\ i_{k_n} = \Omega \\ \Rightarrow j_{k_n} = \Omega \end{array} . \tag{103}
$$

This space is obtained by imposing the following constraints: firstly, at any time there is a message on at most a single wire. Secondly, at $t=2$, there cannot be vacuum on all input wires. Thirdly, if the input wire of an agent during one time step carries the vacuum, so does their output wire during the next time step. The first and second constraints follow from the fact that the causal boxes as we defined them conserve the number of messages. Additionally, since we assumed the global past to be trivial, we have that $V_1$ is simply a state (potentially in superposition on different wires), i.e. the number of messages is exactly 1. The third constraint is justified by the fact that we assume that the local agents only output a message at $t=2n+1$ if they received a message at $t=2n$. Thus, terms that do not fulfill this constraint simply vanish when composing with agents.

Let us now consider a basis state from Eq. (103) and see what happens when we compose it with the pure local operations from Def. 3
$$
\begin{align}
& \bigotimes_{k=1}^{N} |\bar{A}_k\rangle * \bigotimes_{n=1}^{N} |i_{k_n}, t=2n\rangle^{\bar{A}_{k_n}^I} |\Omega, (2n)^c\rangle^{\bar{A}_{k_n}^I} |j_{k_n}, t=2n+1\rangle^{\bar{A}_{k_n}^O} |\Omega, (2n+1)^c\rangle^{\bar{A}_{k_n}^O} |i_F, t=2N+2\rangle^{\bar{F}} \nonumber \\
& = \left( \bigotimes_{n=1}^{N} \bigotimes_{t \in T^I} \left( |A_n\rangle + |\Omega\rangle^{\bar{A}_{k_n}^I} |\Omega\rangle^{\bar{A}_{k_n}^O} \right) \otimes |t+1\rangle |t\rangle \right) * \nonumber \\
& \bigotimes_{n=1}^{N} |i_{k_n}, t=2n\rangle^{\bar{A}_{k_n}^I} |\Omega, (2n)^c\rangle^{\bar{A}_{k_n}^I} |j_{k_n}, t=2n+1\rangle^{\bar{A}_{k_n}^O} |\Omega, (2n+1)^c\rangle^{\bar{A}_{k_n}^O} |i_F, t=2N+2\rangle^{\bar{F}} \nonumber \\
& = \prod_{n=1}^{N} (|A_{k_n}\rangle + |\Omega\rangle^{\bar{A}_{k_n}^I} |\Omega\rangle^{\bar{A}_{k_n}^O}) \nonumber \\
& \quad * \left( |i_{k_n}\rangle^{\bar{A}_{k_n}^I} |j_{k_n}\rangle^{\bar{A}_{k_n}^O} \right) \left( \left( |\Omega\rangle^{\bar{A}_{k_n}^I} |\Omega\rangle^{\bar{A}_{k_n}^O} \right) * \left( |\Omega\rangle^{\bar{A}_{k_n}^I} |\Omega\rangle^{\bar{A}_{k_n}^O} \right) \right)^3 |i_F, t=2N+2\rangle \nonumber \\
& = \begin{cases}
\prod_{n=1}^{N} |A_{k_n}\rangle * (|i_{k_n}\rangle^{\bar{A}_{k_n}^I} |j_{k_n}\rangle^{\bar{A}_{k_n}^O}) |i_F, t=2N+2\rangle, & \text{if } i_{k_n}, j_{k_n} \neq \Omega, \forall n \\
0, & \text{else}
\end{cases} \tag{104}
\end{align}
$$

In the first equality, we used the definition of extensions of local operations. In the second equality, we contracted the time stamps and also used that $|A_{k_n}\rangle * (|\Omega\rangle^{\bar{A}_{k_n}^I} \otimes |\Omega\rangle^{\bar{A}_{k_n}^O}) = 0$ for all $k_n$. Finally, we used that both the local and internal operations map non-vacuum states to non-vacuum states. The contraction thus vanishes unless all $i_{k_n}, j_{k_n}$ are either the vacuum or the non-vacuum. However, as we stated earlier, $i_{k_1} \neq \Omega$ and so only the latter is actually possible.

Additionally, note that there is an obvious isomorphism between the global future state in the above equation and the global future in the QC-QC, namely $|i_F, t = 2N + 2\rangle \cong |i_F\rangle$. We will thus drop the time stamp and the vacuum states of the global future from here on out.

The coefficient of the basis state in Eq. (104) can be found using the definition of the Choi vector. The Choi vector of some operator $T: \mathcal{H}^X \to \mathcal{H}^Y$ is $T|1\rangle\rangle^{XX} = \sum_i |i\rangle^X T|i\rangle^X = \sum_{ij} |i\rangle^X |j\rangle^Y \langle j|Y T|i\rangle^X = \sum_{ij} \langle j|T|i\rangle^X |j\rangle^Y$. Applied to our basis state and with the help of Eq. (61), we find the coefficient to be

$$ \langle i_F | V_{N \setminus k_N, k_N}^{F} | j_{k_N} \rangle * \star_{n=1}^{N-1} \langle i_{k_{n+1}} | V_{k_{n-1}, k_n}^{k_{n+1}} | j_{k_n} \rangle * \langle i_{k_1} | V_{\emptyset, \emptyset}^{k_1} \quad (105) $$

where we used the notation $\star_{n=1}^N A_n = A_1 * A_2 * \dots * A_N$.

With Eq. (104) and Eq. (105), we can now calculate the contraction $\bigotimes_{k=1}^N |\bar{A}_k\rangle * |\bar{V}\rangle$.

$$
\begin{align}
& \bigotimes_{k=1}^{N} |\bar{A}_k\rangle * |\bar{V}\rangle \nonumber \\
& \cong \sum_{(k_1, \dots, k_N)} \sum_{\substack{i_1, \dots, i_N \\ j_1, \dots, j_N, t_F}} \langle i_F | V_{N \setminus k_N, k_N}^{F} | j_{k_N} \rangle * \star_{n=1}^{N-1} \langle i_{k_{n+1}} | V_{k_{n-1}, k_n}^{k_{n+1}} | j_{k_n} \rangle * \langle i_{k_1} | V_{\emptyset, \emptyset}^{k_1} \nonumber \\
& \qquad \prod_{n=1}^{N} |A_{k_n}\rangle * (|i_{k_n}\rangle^{\bar{A}_{k_n}^I} |j_{k_n}\rangle^{\bar{A}_{k_n}^O}) |i_F\rangle \tag{106} \\
& = \sum_{(k_1, \dots, k_N)} \star_{n=1}^{N} (|A_{k_n}\rangle * \sum_{\substack{i_{k_{n+1}}, j_{k_n} \\ i_{k_{n+1}}}} \langle i_{k_{n+1}} | V_{k_{n-1}, k_n}^{k_{n+1}} | j_{k_n} \rangle (|i_{k_n}\rangle^{\bar{A}_{k_n}^I} |j_{k_n}\rangle^{\bar{A}_{k_n}^O})) \nonumber \\
& = \sum_{(k_1, \dots, k_N)} \star_{n=1}^{N} |A_{k_n}\rangle * |V_{k_{n-1}, k_n}^{k_{n+1}}\rangle \nonumber \\
& = \bigotimes_{k=1}^{N} |A_k\rangle * |w\rangle . \nonumber
\end{align}
$$
In the above equation, one should interpret $k_{N+1} = F$ and $|j_{k_0}\rangle = 1$. In the third equality, we used again that $|T\rangle\langle T| = \sum_{ij} \langle i|T|j\rangle \langle j|^{X}|j\rangle^{Y}$. In the last equality, we used commutativity of the link product and the definition of the process vector of the QC-QC.

This shows that we have found a causal box extension of the QC-QC. ☐

<figure><img src="image_21.png" /></figure>