Is there a class of Bell inequalities that are violated by certain mixtures of known orthogonal entangled states?

There exist a multitude of classes of Bell inequalities that can be used in conjunction with the corresponding experimental designs.

Bell inequalities from no-signalling distributions

All the Bell inequalities

We will consider the  following entangled states (Bell states):

$\Phi^+=\frac{1}{\sqrt{2}}(\vert 00\rangle+\vert 11\rangle)$

$\Phi^-=\frac{1}{\sqrt{2}}(\vert 00\rangle-\vert 11\rangle)$

$\Psi^+=\frac{1}{\sqrt{2}}(\vert 01\rangle+\vert 10\rangle)$

$\Psi^-=\frac{1}{\sqrt{2}}(\vert 01\rangle-\vert 10\rangle)$

Alice and Bob measure their photons using coincidence circuits and can perform a statistical analysis on the outcome of their measurements. They choose a class of Bell inequalities to analyze as well as the corresponding experimental setup (the Bell test experiment ) in order to assess whether the chosen inequalities are violated (in order to prove the presence of quantum entanglement ). 

Case 1. The source sends Alice and Bob only entangled pairs in the state $\Phi^+$ . In this case  it is known that there is a class of Bell inequalities and Bell test experiments that shows the violation of these inequalities.  That goes back to Bell's work. Similar results are valid if the source sends only entangled pairs in any of the other Bell states  $\Phi^-$$\Psi^+$ , or  $\Psi^-$ .

Case 2. The source sends Alice and Bob entangled pairs in the state $\Phi^+$  with probability $\alpha$, or in the state  $\Phi^-$  with probability  $\beta$ ,  or in the state $\Psi^+$  with probability $ \gamma$, or in the state  $\Psi^-$  with probability  $\delta$  , where $\alpha + \beta + \gamma + \delta  = 1$.  In this case  is there a class of Bell inequalities and Bell test experiments that shows the violation of these inequalities ?  Also for what range of the parameters $\alpha$ , $\beta$ , $\gamma$  , and  $\delta$  is this possible?

Question. In other words, can Alice and Bob prove the presence of quantum entanglement (correlations ) when they receive a mixture of known orthogonal entangled states? And for what range of the parameters?

Clearly for $\alpha = 1$  and $\beta = \gamma = \delta = 0$ this is possible  (this is case 1).


Justification for this question. In the following  I will describe the reason/motivation  behind this question.  This part is independent of the question itself but it is worth mentioning. It doesn't obscure the clarity of the question above.

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.

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.


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.

Secondary question. Could a reliable entanglement detection procedure (as described above, and the associated protocol ) be used for signalling in delayed choice entanglement swapping ?

  • 2
    $\begingroup$ Q1: Obviously this must still be true in some neighborhood of the pure state case due to continuity. $\endgroup$ Commented Mar 7, 2020 at 15:37
  • $\begingroup$ Thank you @NorbertSchuch . What about Q2? $\endgroup$ Commented Mar 8, 2020 at 1:08
  • 1
    $\begingroup$ There is no signaling. Never. $\endgroup$ Commented Mar 8, 2020 at 2:52
  • $\begingroup$ You're probably right, I probably have an error in my calculations or in my understanding of what's possible experimentally (specially the fact that in delayed choice entanglement swapping they use nonlinear SPDC crystals physics.stackexchange.com/q/533205/31339 ). Thank you. @NorbertSchuch $\endgroup$ Commented Mar 8, 2020 at 9:50
  • $\begingroup$ This has nothing to do with what is possible experimentally. $\endgroup$ Commented Mar 8, 2020 at 14:18

2 Answers 2


Figure 1 of https://iopscience.iop.org/article/10.1088/1751-8121/aa5dfd/meta has a nice visualization of all Wely states. All mixed states generated by randomly choosing a bell state are Wely states.

The figure shows the region which can violate the CHSH-Bell inequality and the region of separable states, where no Bell inequality can be violated.

I'm aware that this is not the full solution, but gives you at least some bounds on the full solution.


This question was bumped by the Community bot, so I am supplying an answer. Better late than never?

  1. This is a clever question and has some good points mixed in. To discuss the main question:

Note: His reference "Experimental delayed-choice entanglement swapping" (Ma et al, 2012) is in fact one of the best on this fascinating subject. I highly would recommend this paper to anyone interested in entanglement. One of the co-authors (Zeilinger) won a Nobel in 2022 for this and other works.

a) Assuming either Alice (or Bob) individually could detect entanglement using the technique described by the paper of Pezze et al (2016): Alice (and Bob) would ALWAYS detect an entangled state regardless of what Victor does. So no signal can be sent by Victor.

The reason: All 4 photons in a delayed choice entanglement experiment start out entangled. Yes, Victor can choose to swap entanglement to 1 & 4, but otherwise 1 & 2 are entangled and 3 & 4 are entangled. Either way, Alice sees nothing but entangled photons.

b) Suppose we need Alice and Bob to collectively use the Pezze protocol. The hypothetical "statistical speed" velocity(delta theta)^2 - I'll call that V for simplicity - is based on timing of the arrival time (detection) of their respective photons (1 & 4) - specifically related to the difference in arrival times of each 1 & 4 pair as a function of V. Actually V - a very small number in the ideal case - becomes the standard deviation of a collection of trials of the difference between V and what they call Vmax. Let's call that V*, see for example their equations (9) and (10) to get a better idea, and Figure 2 as well. We are talking a microscopically small number.

But in fact, those arrival times do not necessarily mean anything at all. Normally, the arrival times of photons seen by Alice and Bob are within a coincidence time window of perhaps 5 nanoseconds. (Keep in mind that the window is adjusted for each photon's path length from source to detector.) Oh, and for an entanglement swap to operate, you need separated source entangled pairs - they cannot come from the same source at the same time or they will be in a GHZ state instead of the required Bell state.

Now, it is possible - actually required - that the newly created 1 & 2 pair and the 3 & 4 pair must be synchronized as to phase. Thus their time of creation must be a integer multiple N relative to the wavelength. That distance would be on the order of 400 nanometers, times N times c. And in fact in some experiments it has been possible to limit the difference in time of emission to a small pulse (translating to pairs emitted very close in time). But there will be a difference, sometimes with Alice's pair created first, and sometimes Bob's pair created first. The sequence and N are random. In principle, with enough iterations, you might get something close to a small average difference in creation times (let's call that CT). But it is almost certain to dwarf our V* number.

But even that doesn't help us much, no matter how small CT is. That's because we also have issues with detection times. Current photon detector technology is in the sub-nanosecond range, depending on a variety of factors. With one detector each for Alice and Bob, that's a lot of detection time (let's call that DT) difference to consider.

When we are considering our detection window, we only need both detection times to be close. But for consideration with the extremely small magnitude of our equation V*, the combined CT and DT differences need to be microscopically small relative to that. Looks to me as if those will will dwarf it instead.

In plain language: the margin of error in calculating the desired "statistical speed " formula will be much larger than the value itself.

c) Of course, the Pezze paper does not use the entanglement swapping protocol anyway. I don't really think it can be adapted for that, but that's just my best assessment given a minimal review. So that would be a problem of its own (finding a way to adapt to entanglement swapping).

d) Most importantly, we have assumed the theoretical basis of the 2016 Pezze paper as valid. Many papers go unpublished (this was not), and therefore there is no peer review process. Certainly, its validity - which goes against mainstream theory - can be questioned.

I had never heard of it, which alone is not proof of anything. Nor have I ever seen so much as a hint that swapped entanglement can be detected without using information from Victor to sort the data results based on the random Bell state which he records. That of course requires classical communication. That is the mainstream view.

But that paper has only been cited by 1 other work in the past 8 years. And that citation was in a paper co-authored by Pezze himself. Usually, I would expect a novel result to attract discussion from some of the thousands of papers written each month. You can judge for yourself what to make of that, and what weight to assign that.

  1. For the second question, I would answer a definitive NO for the totality of the reasons presented.

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