Entanglement swapping in the context of Indivisible Stochastic Process

[JonAce73]

TL;DR

What is the relative values of the transition matrix as defined in [2] during the polarization detection and before the entanglement swapping described in [1]?

I would highly appreciate it if you can answer my questions below on entanglement swapping described in [1] in the context of Indivisible Stochastic Processes explained in [2]. The symbols below should be interpreted as in Latex.

Suppose in figure 2 (attached) of [1], the two pairs of entangled photons (1,2) and (3,4) constitute the subject system $S$ while the polarization measuring devices named Alice and Bob, and the analyzer, which randomly performs the Bell-state measurement (BSM) or not, constitute the environment $E$ to which $S$ interacts. Suppose the photons are emitted at time zero. Then at time $t'$ photons 1 and 4 interact with Alice and Bob, respectively. Let the transition matrix, as defined in [2], from time zero to $t'$ be $\Gamma^S(t' \leftarrow 0)$ where at $t'$ BSM is not applied to photons 2 and 3 yet at the analyzer. Now, at time $t$ photons 2 and 3 undergo BSM thereby they are projected into one of the entangled Bell-states and the remaining photons 1 and 4 are projected into an entangled Bell-state as well. Let the transition matrix from time $t'$ to $t$ of $S$ be $\Gamma^S(t \leftarrow t')$. Thus, the transition matrix from time zero to $t$ is $\Gamma^S(t \leftarrow 0) = \Gamma^S(t \leftarrow t') \Gamma^S(t' \leftarrow 0)$, i.e., similar to Eq. 56 of [2].

Let $C'$ be the configuration having photons 1 and 2 entangled, and photons 3 and 4 entangled; and $C^t$ be the configuration having photons 1 and 4 in an entangled Bell-state, and photons 2 and 3 as well. Should $\Gamma^S(t' \leftarrow 0)$ (i.e., the transition matrix from time zero to the interaction time $t'$ at Alice and Bob) for configuration $C'$ be lesser than for configuration $C^t$ when in the future time $t$ BSM is applied (or not) on photons 2 and 3 at the analyzer?

My problem is, the relative values of the respective transition matrices for the two configurations cannot be determined prior to the random BSM application (or not) on photons 2 and 3. Perhaps, the matrices cannot be known until the interaction at the analyzer; or I misunderstood [2].

Thanks for entertaining my questions.

[1] D. Glick, “Timelike entanglement for delayed-choice entanglement swapping,” Studies in History and Philosophy of Science Part B: Studies in History and Philosophy of Modern Physics, vol. 68, pp. 16–22, 2019.
[2] J. A. Barandes, “The stochastic-quantum correspondence,” arXiv preprint arXiv:2302.10778, 2023.

 

[Morbert]

In this formalism, entanglement is not encoded in the configurations. It is encoded in the dynamics $\Gamma$. So e.g. considering the usual initial state $\ket{\psi^-}_{12}\ket{\psi^-}_{34}$ at time $0$, and tracing over $2$ and $3$, we have the transition matrix $\Gamma_{14}(t\leftarrow 0) = \Gamma_{1}(t\leftarrow 0)\otimes\Gamma_{4}(t\leftarrow 0)$ for all $t$. I.e. $1$ and $4$ are not entangled, where "not entangled" means the transition matrix that encodes all correlations will factorize.

You can select a subset of experimental runs such that the correlations between $1$ and $4$ in the subset are reproduced by transition matrices that don't factorize, and interpret this as entanglement swapping.

PS You can fix your LaTex by replacing you $s with $$ or ## where appropriate.

 

[PeterDonis]

the relative values of the respective transition matrices for the two configurations cannot be determined prior to the random BSM application (or not) on photons 2 and 3

This is just a consequence of the fact that, in general in quantum experiments, you have to know the entire experimental context in order to make predictions about the results. Here the "experimental context" includes whether or not the BSM application is made on a particular run. So until you know that, you can't make predictions about the results, which is equivalent to saying that you can't know the correct transition matrices for that run.