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.
There is though another way based on statistics and a reliable entanglement witness.
When Victor entangles his photons 2 and 3, photons 1 and 4 are in a mixture of entangled states. We consider the transmitter (Victor) and the receiver (Alice and Bob) follow an agreed protocol. For each bit of information transferred (0/1), a certain number KN of pairs of photons are measured by both Victor and corespondingly by Alice/Bob. When he wants to send a 0, Victor does not entangle his photons. When he wants to send a 1, Victor entangles his photons. In order to decode the message Alice and Bob need a reliable procedure of entanglement detection . And they don't need to compare their records with Victor.
In the paper above it is discussed witnessing entanglement without entanglement witness operators. The method involves measuring the statistical response of a quantum system to an arbitrary nonlocal parametric evolution. The witness of entanglement is solely based on the visibility of an interference signal. If followed closely, this method never gives false positives.
In the protocol described , when Victor (the transmitter) and Alice and Bob (the receiver) measure N pairs of photons, then with probability $\frac{1}{4^N}$ all the N photon pairs measured by Alice and Bob will be in the same Bell state. So the transmitter and receiver can repeat measuring N pairs of photons (lets say K times) until the entanglement detection method described above will give a positive. At this point Alice and Bob know that Victor must be entangling his photons. When Victor does not entangle his photons, since the method of entanglement detection mentioned above does not give false positives, Alice and Bob will know that Victor does not entangle his photons for all the KN pairs of photons processed. For large N and K, the probability of error can be made arbitrarily small. Basically, without comparing records, Alice and Bob know what Victor is doing. That's signalling, and the no - signalling theorem can be circumvented due to the method of entanglement detection described above, which does not rely on witness operators.
In principle the problem seems to allow a solution. Reliable entanglement detection seems to circumvent the no - signalling theorem.
Question: Is back in time (classical) information transfer possible?
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, is this possible, in principle?
