Delayed choice entanglement swapping.
Two pairs of entangled photons are produced, and one photon from each pair is sent to a party called Victor. Of the two remaining photons, one photon is sent to the party Alice and one is sent to the party Bob. Victor can now choose between two kinds of measurements. If he decides to measure his two photons in a way such that they are forced to be in an entangled state, then also Alice's and Bob's photon pair becomes entangled.
If Victor chooses to measure his particles individually, Alice's and Bob's photon pair ends up in a separable state. Modern quantum optics technology allows to delay Victor's choice and measurement with respect to the measurements which Alice and Bob perform on their photons. Whether Alice's and Bob's photons are entangled and show quantum correlations or are separable and show classical correlations can be decided after they have been measured.
We follow the calculations in the reference.
$\Phi^+=\frac{1}{\sqrt{2}}(\vert 00\rangle+\vert 11\rangle)$
$\Phi^-=\frac{1}{\sqrt{2}}(\vert 00\rangle-\vert 11\rangle)$
$\Psi^+=\frac{1}{\sqrt{2}}(\vert 01\rangle+\vert 10\rangle)$
$\Psi^-=\frac{1}{\sqrt{2}}(\vert 01\rangle-\vert 10\rangle)$
$\vert 00\rangle=\frac{1}{\sqrt{2}}(\Phi^+ + \Phi^-)$
$\vert 11\rangle=\frac{1}{\sqrt{2}}(\Phi^+ - \Phi^-)$
$\vert 01\rangle=\frac{1}{\sqrt{2}}(\Psi^+ + \Psi^-)$
$\vert 10\rangle=\frac{1}{\sqrt{2}}(\Psi^+ - \Psi^-)$
Two pairs of entangled photons (1&2 and 3&4) are each produced in the antisymmetric polarization entangled Bell singlet state such that the total four photon state has the form:
$$\vert \Psi\rangle_{1234}=\vert \Psi^-\rangle_{12}\otimes\vert\Psi^-\rangle_{34}$$
In short, we write:
$$\vert \Psi\rangle_{1234}=\Psi^-_{12}\otimes\Psi^-_{34}$$
If Victor subjects his photons 2 and 3 to a Bell state measurement, they become entangled. Consequently photons 1 (Alice) and 4 (Bob) also become entangled, and entanglement swapping is achieved. This can be seen by writing $\vert \Psi\rangle_{1234}$ in the basis of Bell states of photons 2 and 3.
$$\vert\Psi\rangle_{1234}=\frac{1}{2}(\Psi^+_{14}\otimes\Psi^+_{23}-\Psi^-_{14}\otimes\Psi^-_{23}-\Phi^+_{14}\otimes\Phi^+_{23}+\Phi^-_{14}\otimes\Phi^-_{23})$$
This is relation (2) in the paper linked above.
In order to see the correlations between their particles, Alice and Bob must compare their coincidence records with Victor. Without comparing with Victor's records, they only see a perfect mixture of anti-correlated (the Ψ’s ) and correlated (the Φ’s ) photons, no pattern whatsoever.
Now we perform the same calculations, but we start with the general entangled state (the following calculations are not present in the paper linked above):
$\vert\Theta\rangle=\alpha\vert 01\rangle-\beta\vert 10\rangle$ with $\alpha^2+\beta^2=1$
The total four photon state has the form:
$\vert\Psi\rangle_{1234}=\vert\Theta\rangle_{12}\otimes\vert\Theta\rangle_{34}$
In short, we write:
$\vert\Psi\rangle_{1234}=\Theta_{12}\otimes\Theta_{34}$
Now we write $\vert\Psi\rangle_{1234}$ in the basis of Bell states of photons 2 and 3.
$$\vert\Psi\rangle_{1234}=\frac{1}{2}[(\alpha^2+\beta^2)(\Psi^+_{14}\otimes\Psi^+_{23}-\Psi^-_{14}\otimes\Psi^-_{23})+(\alpha^2-\beta^2)(-\Psi^+_{14}\otimes\Psi^-_{23}+\Psi^-_{14}\otimes\Psi^+_{23})+2\alpha\beta(-\Phi^+_{14}\otimes\Phi^+_{23}+\Phi^-_{14}\otimes\Phi^-_{23})]$$
If $|\alpha|^2$ is very close to $1$ and $|\beta|^2$ is very small , then Alice (photon 1) and Bob (photon 4) will see with high probability only anti-correlated photons (the $\Psi$'s) when Victor entangles his photons ( but the probabilities of Alice and Bob's anti-correlated photon states change in this mode of operation ) . They do not need to compare their results with Victor, in order to distinguish between the cases when Victor entangles his photons or measures them independently.
The sources of entangled particles send pairs of photons continuously at equal intervals of time (all the time), in a synchronized manner .
In order to send a bit of classical information, the transmitter (Victor ) entangles his two photons (that's a binary 1), or does not entangle his two photons (that's a binary 0). Groups of 2N entangled pairs of photons can be measured (by the transmitter and receiver ) for each bit of classical information transferred. Alice and Bob , the receiver, can use coincidence circuits only locally , and statistics, in order to decode the binary message sent by Victor. They do not need to compare their records with Victor.
In other words, Victor can send classical information back in time to Alice and Bob.
Question. Is back in time (classical ) information transfer possible? Is there an error in my calculations?
There are two operational modes. In one mode photons 1&2 also 3&4 are entangled. In the second mode of operation photons 2&3 also 1&4 are entangled. The coincidence probability distributions for Alice and Bob measurements (using coincidence circuits only locally ) are slightly different in these two operational modes, but only if $ \alpha$ is different than $\beta$ , and $\alpha$ is not 1. The no- signalling theorem does not cover this case when there is an entanglement redistribution between the four particles.
I am thinking about experiments that would validate Everett's many worlds interpretation of QM (or variants, because that's the only way to avoid the emerging logical paradoxes ). In fact, following Scott Aaronson (and others), computation with CTC's would have a great impact in the field. But first things first, are my calculations correct? Is this possible, in principle?