Researchers demonstrated violation of Bell's inequality on frequency-bin entangled photon pairs

May 8, 2017, Hong Kong University of Science and Technology
Quantum entanglement. Credit: Physics Department, HKUST

Quantum entanglement, one of the most intriguing features of multi-particle quantum systems, has become a fundamental building block in both quantum information processing and quantum computation. If two particles are entangled, no matter how far away they are separated, quantum mechanics predicts that measurement of one particle leads to instantaneous wave-function collapse of the other particle.

Such "spooky action at a distance" is non-intuitive, and in 1935, Einstein attempted to use entanglement to criticize to suggest that the quantum description of physical reality is incomplete. Einstein believed that no information could travel faster than light, and suggested that there might be some local hidden variable theories that could explain the world in a deterministic way, if and only if they obey realism and locality. In 1964, J. S. Bell showed that the debate can be experimentally resolved by testing an ; by measuring correlations between entangled parties, the result calculated from local hidden variable theories should be constrained by the Bell inequality, which, on the other hand, can be violated in the predictions of quantum mechanics.

By reducing the velocity of light dramatically, researchers at the Hong Kong University of Science and Technology implemented a Bell Test and were able to generate frequency-bin entangled narrowband biphotons from spontaneous four-wave mixing (SFWM) in cold atoms with a double-path configuration, where the phase difference between the two spatial paths can be controlled independently and nonlocally.

Their findings were published in the journal Optica on April 15, 2017.

"We tested the CHSH Bell inequality and registered |S|=2.52±0.48|S|=2.52±0.48, which violates the Bell inequality |S|≤2," said Shengwang Du, professor of Physics at HKUST and the leader of the research team. "We have unambiguously demonstrated the generation of frequency-bin entangled narrowband (about 1 MHz) biphotons, which can efficiently interact with stationary atomic quantum nodes in an atom-photon quantum network. Because of their narrow bandwidth, these biphotons can be stored and retrieved from a quantum memory with high efficiency."

"Our result, for the first time, tests the Bell inequality in a nonlocal temporal correlation of frequency-bin entangled narrowband biphotons with time-resolved detection," said Xianxin Guo, a co-author of the paper. "This will have applications in involving time-frequency entanglement."

The study revealed temporal details that agree well with theory calculations based on mechanics, and implies the possibility of encoding and decoding qubit information in the time domain.

"Our narrowband frequency-bin entangled biphoton source in this work can be ideally implemented to produce pure heralded single photons in a two-color qubit state with a tunable phase, making use of entanglement, linear optics, and time-resolved detection," said Guo.

Explore further: All quantum communication involves nonlocality

More information: Xianxin Guo et al, Testing the Bell inequality on frequency-bin entangled photon pairs using time-resolved detection, Optica (2017). DOI: 10.1364/OPTICA.4.000388

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AmritSorli
May 08, 2017
This comment has been removed by a moderator.
AmritSorli
1 / 5 (1) May 08, 2017
entanglement runs over quantum vacuum http://article.sc....13.html
Da Schneib
5 / 5 (1) May 08, 2017
This is entanglement on the time-energy Heisenberg uncertainty. I'll have a close look at this; off the top of my head, it's a little less definitive than using spins, because both time and energy are continuous symmetries, not discrete like spin. This could have effects on accuracy of imposition of the values of the qubits being transmitted on the receiving quantum system.
Ryan1981
not rated yet May 09, 2017
Researchers (...) were able to generate frequency-bin entangled narrowband biphotons from spontaneous four-wave mixing (SFWM) in cold atoms with a double-path configuration, where the phase difference between the two spatial paths can be controlled independently and non-locally.


For me this is too complex a description to understand what is going on.
antialias_physorg
5 / 5 (3) May 09, 2017
For me this is too complex a description to understand what is going on.

I have the same problem. Looking around the nety I'll try to give what I found. Maybe DaSchneib can correct me on the points that I misrepresent:

1) Biphotons
A 'biphoton' are two photons that are entangled.

2) Frequency entanglement:
Entanglement can be had from many properties. The one most are 'familiar' with is probably polarization, but you can entangle spin, position, momentum....

A 'frequency entanglement' happens when you take a very narrowband source (e.g. a monochromatic pump laser) and create two entangled photons from one 'pump' photon. This gives you two photons where the SUM of the frequencies is well defined (ferquency is directly related to energy of a photon - and you know the energy of the initial photon), but the frequencies of EACH ONE is uncertain.
The frequencies of the two photons are entangled (measure one and you know the other).
antialias_physorg
5 / 5 (3) May 09, 2017
3) Binning:
This is a function of the detectors. The detectors can detect photons in a certain energy range - not exact energies. If two photons are within the energy range of a detector then they are said to lie in the same 'bin'. If they are in the same bin then they are not distinguishable. (If they come to lie in different bins then they are distinguishable)

4) Four wave mixing:
In some non-linear materials you can augment the frequency of n-1 wavelengths (photons) when you input n wavelengths (photons). In this case - since they are creating photon pairs - The input should be three photons and you get two up-converted, entangled photons (not tooooo sure if I interpret this correctly)

How exactly the rest tests for Bell inequality I haven't figured yout yet. Maybe tomorrow.
Seeker2
5 / 5 (1) May 09, 2017
@AP
2) Frequency entanglement:
Entanglement can be had from many properties. The one most are 'familiar' with is probably polarization, but you can entangle spin, position, momentum....

Position only for bosons only, obviously? Would there be a restriction on the phase angles of the two entangled photons?
antialias_physorg
5 / 5 (1) May 10, 2017
Position only for bosons only, obviously?

I guess you could position-entangle fermions (like electrons), too. However the associated frequency would be very high because the wavelike properties are described by lambda = h / p
where lambda is the frequency, h is the planck number and p is the momentum. Since the momentum of an electron is far greater than that of your run-of-the-mill photon you get a very short effective wavelength.
https://en.wikipe...ter_wave

Reminds me of a question on a physics test in school: Teacher with mass 80kg walks into the room through a 80cm slit (door) at 6km/h: calculate his diffraction pattern.

Would there be a restriction on the phase angles of the two entangled photons?

If the only condition of these is 'monochromatic' then there is no restriction on phase. If they are monochromatic and phase correlated then there *might* be a restriction on phase - but that also depends on the material used for the up-conversion
Da Schneib
not rated yet May 10, 2017
@Seeker, why would there be only entanglement on position for bosons? Not sure I understand your reasoning here.
johnqsmith
5 / 5 (1) May 10, 2017
Methinks the original post is directed to physics colleagues who are fairly familiar with the subject matter. There seem to be a larger number of professional physicists here who can dig out what the OP is talking about. Then there's me who took college and grad school physics courses but I'm not a physicist. I'm not sure what the moderators of this forum prefer, but I'd be much obliged if posters translated their thoughts to the amateur physicist level.

I appreciate the efforts by antialias and others to help, and I could understand their translations.

In the article it states: "In 1964, J. S. Bell showed that the debate can be experimentally resolved by testing an inequality; by measuring correlations between entangled parties, the result calculated from local hidden variable theories should be constrained by the Bell inequality, which, on the other hand, can be violated in the predictions of quantum mechanics."

What does this mean, please? to a non-professional.
Seeker2
5 / 5 (1) May 11, 2017
@DS
@Seeker, why would there be only entanglement on position for bosons? Not sure I understand your reasoning here.
I was thinking fermions could not be entangled at the same position because of the Pauli Exclusion principle. Presuming that's what entanglement on position means. Probably misunderstood the Pauli Exclusion principle though.
Da Schneib
5 / 5 (1) May 11, 2017
@Seeker, Pauli Exclusion doesn't prohibit two fermions being emitted from the same particle (fermion or boson) at the same time. For example, a decaying neutron emits an electron and an electron antineutrino, both fermions. More complex examples involve two fermions of identical nature emerging from the interaction, with momenta that add to the original momentum of the original particle; and these two fermions would be entangled on position/momentum as well as energy/time (or as in this article, frequency/time) degrees of freedom. Measuring one would tell you information not available under Heisenberg uncertainty about the other, the standard measure of entanglement.
Da Schneib
3 / 5 (2) May 11, 2017
@johnq, the best explanation I've seen is in Greene's The Elegant Universe. What it boils down to is, if classical correlations only are considered, the correlation can only be at maximum 2; but if quantum correlations are present, it can be greater than that. Bell's Inequality precisely quantifies this. So if we run an experiment, and the correlations are 2 or less, then both realism and locality are preserved; if the correlations are greater, then either locality or realism must be false.

The details are rather complex; I can gather them and comment if you're really interested. It's a good question, so I gave you a 5.
Da Schneib
not rated yet May 11, 2017
@Seeker, an additional detail that may not be clear: two particles emerging from a single interaction are necessarily entangled, on many different parameters which are conserved, and uncertain in the Heisenberg sense. It is these two properties, conservation and uncertainty, that cause these interactions to generate entangled particles. While it is common to measure spins since they are discrete, it is possible if one has sufficient accuracy to measure continuous symmetries as well, and that is what the experiment reported in this article and its associated paper are showing.

Incidentally, this paper is open access from the journal of record, and everyone should go read it. This is an important result, extending entanglement from the discrete to the continuous realm. Not unexpected; but this is another of those experiments that's important because we didn't discover anything new.
Seeker2
not rated yet May 12, 2017
@DS
@Seeker, Pauli Exclusion doesn't prohibit two fermions being emitted from the same particle (fermion or boson) at the same time. For example, a decaying neutron emits an electron and an electron antineutrino, both fermions. More complex examples involve two fermions of identical nature emerging from the interaction, with momenta that add to the original momentum of the original particle; and these two fermions would be entangled on position/momentum as well as energy/time (or as in this article, frequency/time) degrees of freedom.
By position I assume you mean the position at emission or positon of the entangled fermions then and thereafter? That is I'm thinking the two fermions go off to different positions after emission.
Da Schneib
not rated yet May 12, 2017
Until they decohere, yes. They have to split the momentum, and that means they are entangled on the position/momentum degree of freedom.
Seeker2
not rated yet May 12, 2017
They have to split the momentum, and that means they are entangled on the position/momentum degree of freedom.
So does that mean their momentum has equal magnitude but opposite direction until they decohere?
image
1 / 5 (1) May 12, 2017
Bell's definition of locality is unphysical in the sense that it excludes usual classical correlation. This holds for the CHSH inequalities, too. Hence, the measurment of violations of those inequalities are null statements. Another read about that can be found here:
physics.stackexchange.com/a/171607/75518
Seeker2
not rated yet May 12, 2017
Does, or could, gravity decohere entangled particles?
Da Schneib
not rated yet May 12, 2017
They have to split the momentum, and that means they are entangled on the position/momentum degree of freedom.
So does that mean their momentum has equal magnitude but opposite direction until they decohere?
Not necessarily, they might not split it 50/50, but you can think of it that way and not be far wrong. Note that if their masses are not equal they will have different vectors, but they'll still have to vector add to the original momentum of the progenitor particle.

Does, or could, gravity decohere entangled particles?
Good question. Unfortunately without a theory of quantum gravity we can't tell.
Da Schneib
5 / 5 (1) May 12, 2017
@image, bah, Bell's definition cannot be unphysical since it is confirmed by experiment. A 1 for you. See Afshar and follow-ons.
Seeker2
not rated yet May 14, 2017
Does, or could, gravity decohere entangled particles?
Good question. Unfortunately without a theory of quantum gravity we can't tell.
If gravity warps spacetime wouldn't it similarly affect the wave fundtion?
Dingbone
May 14, 2017
This comment has been removed by a moderator.
Da Schneib
not rated yet May 14, 2017
We know this much: under Earth's gravity entangled photons can go 143 km without being decohered: http://www.pnas.o...202.full
Seeker2
not rated yet May 14, 2017
I'm guessing virtual particle pairs are entangled. Also that entangled particles have the same wave function. You can probably see where I'm going with this if my premises are correct.
Da Schneib
not rated yet May 14, 2017
It's probably best if you don't think of virtual particles as having real existence. They are a calculational tool that helps calculate the interaction of real particles with fields.
johnqsmith
5 / 5 (1) May 14, 2017
Da Schneib, thanks for the explanation of the significance of Bell's inequality, and I now have a better sense of it. I would be interested in more details should you be willing.
Seeker2
not rated yet May 14, 2017
I was thinking virtual particles have something to do with Hawking radiation?
Da Schneib
not rated yet May 14, 2017
@johnq, sure! :D

"Locality" means that local effects have local causes. "Realism" means that even though we cannot measure some parameters due to Heisenberg uncertainty, they still have exact values.

Bell's Theorem is a provable theorem of mathematics that shows that if both locality and realism are true, then measurements of Heisenberg uncertain parameters should obey a certain rule that limits their correlation to a value of 2; but if one or the other fails, then their correlations should obey rules of quantum mechanics, which make the correlation greater than 2.

Experiments, called "Bell Test experiments," have all shown correlation values greater than 2. Experiment, as always, trumps theory; any hypothesis that doesn't predict a value greater than 2 for the correlation is therefore untenable.
[contd]
Da Schneib
not rated yet May 14, 2017
[contd]
Classical physics predicts a value of 2. Therefore, classical physics does not correctly predict quantum physics. Note that this does not mean quantum physics does not correctly predict classical physics. This is because in the limits of size and time, the value approaches 2. For further details, you should examine the proof of the Fluctuation Theorem, and experimental results pertaining to it.

I will not restate Greene's discussion here; I recommend you read the book. Note carefully that Bell's Theorem is a provable (and proven) theorem of mathematics; this is not a physical theory subject to possible future disproof.
Da Schneib
not rated yet May 14, 2017
I was thinking virtual particles have something to do with Hawking radiation?
Hawking radiation is a process by which virtual interactions are manifested as real particles by extracting energy from a black hole's gravity field at the event horizon. They are then no longer virtual.
Seeker2
not rated yet May 15, 2017
So these real particles are not entangled?
johnqsmith
not rated yet May 15, 2017
Got it. Thanks. I found the Wiki articles "Bell's Theorem" and "Counterfactual Defiteness" also helpful.
Da Schneib
not rated yet May 16, 2017
So these real particles are not entangled?
With what? If something is inside a black hole you can never check for entanglement. You can't measure it.
Seeker2
not rated yet May 16, 2017
So I guess real particles produced from virtual particle pairs outside black holes are entangled?
Seeker2
not rated yet Jul 02, 2017
entanglement runs over quantum vacuum http://article.sc....13.html
Per your reference time t means numerical order of material changes. I don't see it. Changes yes. Material? The material really doesn't have to change, it's the spatial configuration of the materials. Or, I would say, the ratio between changes in spatial configurations. To be meaningful you need to have a reference - days, seconds, or whatever. At any rate a ratio is not a dimension. This ratio is not even a ratio of dimensions. Just had to get that off my chest. Thanks.

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