3
$\begingroup$

I am currently reading this paper about Quantum Key Distribution Protocols which use Franson Interferometers to secure against eavesdroppers. I am having trouble understanding how the interferometers implement the CHSH game in order to show the correlation between the photons in the experiment has not been tampered with.

I do not come from an experimentalist background and understanding the workings of the Franson interferometer vs other interferometers is proving difficult. I know that in the CHSH game both parties (Alice and Bob) will choose between two measurements to maximize the violation of Bells Inequality. With Franson interferometers what is the analog for these measurements? Is it just the counts of how many photons choose the long branch vs the short? I know some experiments use phase differences in the branch paths and Alice and Bob choose their own phase delays. Does this not destroy the temporal correlation between the two photons?

I know this is a long slew of questions so any help is greatly appreciated!

$\endgroup$

1 Answer 1

1
$\begingroup$

What a Franson interferometer does is to superpose two-photon wavepackets generated at different times (within the coherence time of the pump, supposing we are generating a pair of entangled photons by e.g. SPDC). This superposition is realized by an unbalanced MZI where the time delay between short and long arm should be in principle orders of magnitude smaller than the coherence time of the pump.

The CHSH plays a role here because we can say that overpassing a certain visibility value (~77%) means no local hidden variable theory can support this non-local effect. Therefore the violation of Bell´s inequality follows can be seen as the two-photon (Franson) interference. This visibility can be seen in the post processing of the arrival time of the individual photons in Alice and Bob sides and when we just look at the central peak (we reduce the coincidence window to only the central peak) we see the entangled state |phi> = |S>|S> + exp(phi(alice)+(phi(bob)) |L>|L> where as you can see this is a non-local state because depends on both Alice and Bob Phases. (Strictly speaking use post-selection gives rise to loopholes on the system but this doesnt matter to understand the general picture)

Recall that because the generated individual photons (e.g. signal or idler) have a small coherence time and therefore when they go trough the interferometer they individually act as a particle, but still the overall entangled quantum state has a coherence time larger (given by the pump) than the imbalance of the unbalanced MZI so therefore two-photon interference is still there.

When someone tries to tamper this two-photon interference the non-local correlation gets lost or (or the visibility decreases) depending on the attack to the quantum channel so therefore also by looking to the visibility of the interferometer you can "secure" the communication protocol (altough maybe one Franson interferometer won´t be enough to secure it).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.