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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?


Cross-posted on physics.SE

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  • $\begingroup$ Latex (actually) MathJax) support is built into this web site; it doesn't matter what input device you use to type. $\endgroup$ – tparker Feb 21 at 13:57
  • $\begingroup$ The question is suitable for both sites, even if now I know that cross-posting is not encouraged, even if in some cases is allowed. $\endgroup$ – Cristian Dumitrescu Feb 21 at 17:16
  • $\begingroup$ Hi Christian, welcome to QCSE. I've taken the liberty of formatting your question into latex, I hope I didn't miss anything. You can see how I've edited it by reviewing the edit button. As to the question "Is back in time (classical ) information transfer possible?" the answer is "no", but it's not entirely clear from the question what you are asking of Victor, Alice, and Bob. It seems like your approach must invalidate the monogamy of entanglement somehow. When Victor entangles his photons, Alice and Bob don't care. $\endgroup$ – Mark S Feb 22 at 0:23
  • $\begingroup$ Thank you @MarkS for latex formatting my question. Before the Bell-state measurement particles 1&2 also 3&4 are entangled. After Victor performs his Bell-state measurement on particles 2 & 3, they become entangled, also particles 1&4 become entangled. Entanglement is redistributed among the 4 particles. This scheme does not contradict the monogamy of entanglement. Please see the reference . $\endgroup$ – Cristian Dumitrescu Feb 22 at 6:32
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Any scheme of back-in-time information transfer based on entangled particles runs up against the no-signalling theorem.

Initially it's not clear how you conclude that "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."

For example, in your scheme Victor could just as easily be light-years away from Alice and Bob. Rather than "back in time" information transfer, if Victor's actions could send signal to Alice/Bob, then he could transfer a bit to them faster than light, violating the no-signalling theorem.

Taking it a step further, consider $\alpha^2=1$. Then up to global phase your system is:

$$\vert\Psi\rangle_{1234}=\vert 0101\rangle,$$

Alice and Bob's qubits are:

$$\vert\Psi\rangle_{14}=\vert 01\rangle,$$

But Victor's qubits are:

$$\vert\Psi\rangle_{23}=\vert 10\rangle,$$

Victor can do anything he wants with his two qubits that are in the state $\vert 10\rangle$, but Alice/Bob's are already fixed to $\vert 01\rangle$.

Your Hilbert space only has dimension $2^4 = 16$ , which is not so bad to work out the amplitudes of each basis directly. Thus I recommend picking a Pythagorean triple for $\alpha$ and $\beta$, and working out for yourself all $16$ amplitudes for your state. From letting, say, $\alpha=\frac{12}{13}$ and $\beta=\frac{5}{13}$, you can calculate each amplitude and you should see that local operations on $\vert\Psi\rangle_{23}$ do not impact amplitudes on $\vert\Psi\rangle_{14}$.

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    $\begingroup$ You want Victor to make a local operation on his two qubits $2$ and $3$, in such a manner that two other parties Alice and Bob can perform local operations on their qubits $1$ and $4$ and compare statistics among each other, such that Victor's local operations alone are sufficient to send a signal to Alice and Bob such that their statistics will vary based on Victor's local operations. That is, you want Victor to send a signal to the system of (Alice and Bob). I don't think you can do what you want to do locally. $\endgroup$ – Mark S Feb 22 at 20:53
  • $\begingroup$ I've extended a pretty big hand by Latex'ing your question, and have provided an answer based on my understanding of your question. I've suggested to you that you do the calculations yourself based on $\alpha=\frac{12}{13}$. I don't know how you conclude "in the second mode of operation (when photons 1&4 become entangled), these probabilities are equal." Again you are asking for Victor to violate the no-signalling theorem. I am not motivated to work on this question further. If you want a further answer you might post a bounty on your question. $\endgroup$ – Mark S Feb 25 at 12:59
  • $\begingroup$ But once again, thank you for your help @MarkS I do appreciate it. $\endgroup$ – Cristian Dumitrescu Feb 25 at 16:55
  • $\begingroup$ I deleted my irrelevant comments and I edited the question for clarity. In other words, I gave it some thought. Feedback appreciated. $\endgroup$ – Cristian Dumitrescu Mar 11 at 6:46
  • $\begingroup$ I edited my question the second time @MarkS I rely on a different entanglement detection method. Feedback appreciated. $\endgroup$ – Cristian Dumitrescu Mar 20 at 6:34

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