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. $\endgroup$ Mar 7, 2020 at 15:37
  • $\begingroup$ Thank you @NorbertSchuch . What about Q2? $\endgroup$ Mar 8, 2020 at 1:08
  • 1
    $\begingroup$ There is no signaling. Never. $\endgroup$ 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 $\endgroup$ Mar 8, 2020 at 9:50
  • $\begingroup$ This has nothing to do with what is possible experimentally. $\endgroup$ Mar 8, 2020 at 14:18

1 Answer 1


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.


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