javisot said:
In this case, it's not from Zeilinger, but I would like to see the following translated into Barandés' terms:
https://www.nature.com/articles/ncomms2076
I would like to see the translation of, for example, the following works by Zeilinger:
https://arxiv.org/abs/2507.07756 ,
https://arxiv.org/abs/quant-ph/0201134
The latter Zeilinger paper maps nicely onto section VIII of Barandes's
new prospects paper. We have 7 subsystems: particles 1,2,3,4 (which we will relabel P, Q, R, S) and Alice, Bob, and Charles (A, B, and C respectively). The initial state, before any preparation into Bell states, is$$\rho_\mathrm{All}(0) = \ket{p_0,q_0,r_0,s_0,a_0,b_0,c_0}\bra{p_0,q_0,r_0,s_0,a_0,b_0,c_0}$$This corresponds to Barandes's equation 66. Preparing subsystems PQ and RS each in the usual Bell state ##\psi^-## at time ##t'## gets us to$$\begin{eqnarray*}
\rho_\mathrm{All}(t') &=& U_\mathrm{All}(t')\rho_\mathrm{All}U^\dagger_\mathrm{All}(t')\\
&=& \ket{\psi^-_{PQ},\psi^-_{RS}, a_0, b_0, c_0}\bra{\psi^-_{PQ},\psi^-_{RS}, a_0, b_0, c_0}
\end{eqnarray*}$$This corresponds to Barandes's equation 67, 68. Next, let's get the reduced density matrix for Alice, Bob, and their particles.$$\begin{align*}
\rho_{PSAB}(t) &=
\operatorname{tr}_{QRC}\Bigl(
(U_{PSAB}(t\leftarrow t')\otimes U_{QRC}(t\leftarrow t')) \\
&\qquad \rho_{\mathrm{All}}(t')\,
(U_{PSAB}(t\leftarrow t')\otimes U_{QRC}(t\leftarrow t'))^\dagger
\Bigr) \\
&=\operatorname{tr}_{QR}\Bigl(
(U_{PSAB}(t\leftarrow t')\otimes I_{QR}) \\
&\qquad \ket{\psi^-_{PQ},\psi^-_{RS}, a_0, b_0}\bra{\psi^-_{PQ},\psi^-_{RS}, a_0, b_0}
(U_{PSAB}(t\leftarrow t')\otimes I_{QR})^\dagger
\Bigr)
\end{align*}$$This corresponds to Barandes's equation 70. Notice that all dependencies on ##c_0## have disappeared. Following Barandes's equations 71, 72:$$\begin{align*}
p((a_t,b_t), t | (p_0, q_0, r_0, s_0, a_0, b_0, c_0), 0) &=
\sum_{p_t,q_t,r_t,s_t,c_t}p((p_t,q_t,r_t,s_t,a_t,b_t), t | (p_0, q_0, r_0, s_0, a_0, b_0, c_0), 0) \\
&=p(a_t,b_t, t | (p_0, q_0, r_0, s_0, a_0, b_0), 0)
\end{align*}$$where$$\begin{align*}
p(a_t,b_t, t | (p_0, q_0, r_0, s_0, a_0, b_0), 0) &=\bra{a_t,b_t}\operatorname{tr}_{QR}\Bigl(
(U_{PSAB}(t\leftarrow t')\otimes I_{QR})\\
&\qquad\ket{\psi^-_{PQ},\psi^-_{RS}, a_0, b_0}\bra{\psi^-_{PQ},\psi^-_{RS}, a_0, b_0}
(U_{PSAB}(t\leftarrow t')\otimes I_{QR})^\dagger
\Bigr)\ket{a_t,b_t}\end{align*}$$The probabilities for Alice's and Bob's final states are not conditioned on Charles's initial state.
In section VIII, in the standard EPR scenario, Barandes shows Bob has no influence on Alice, as the computed probabilities for Alice turn out to not be conditioned on Bob. (He elaborates on the relation between causal relations and probabilities in section VI). Analogously, in the entanglement swapping scenario, we Charles has no causal influence on Alice or Bob, as their probabilities are not conditioned on Charles. This hinges on the factorizeability of ##U_\mathrm{All}(t\leftarrow t') = U_{PSAB}(t\leftarrow t')\otimes U_{QRC}(t\leftarrow t') ##