Why aren't non maximally entangled states produced and used in quantum key distribution schemes? What would be the advantage/disadvantage to use such states rather than maximally entangled ones?
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$\begingroup$ If I understand things correctly, BB84 does not use any entanglement, while E91 uses maximally entangled states. Are you wondering about advantages/disadvantages of a hybrid between the two? $\endgroup$– Mark SpinelliAug 10, 2019 at 12:59
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1$\begingroup$ one advantage of not using entanglement is that it's difficult to create entanglement between distant parties $\endgroup$– user2723984Aug 10, 2019 at 13:43
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$\begingroup$ @MarkS I don't get you. Yes your understanding is right about BB84 and E91. $\endgroup$– Sujan VijayarajAug 11, 2019 at 18:59
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$\begingroup$ By hybrid I mean a protocol that includes some features of BB84 and some of E91. Is this what you are asking about? $\endgroup$– Mark SpinelliAug 12, 2019 at 11:52
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2 Answers
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In schemes like E91, the idea behind using an entangled state is that:
- in a particular measurement basis (for both parties), the measurement outcomes are perfectly correlated but completely random (50:50 outcomes).
- you can perform a Bell test on the state to verify its nature.
Using a maximally entangled state gives you the property of the 50:50 outcomes (which you can see from the reduced density matrix having two eigenvalues of 1/2). You could use a non-maximally entangled state and still get perfectly correlated outcomes, just with an imbalance in the probabilities. You could compensate for this later (e.g. privacy amplification) but it reduces the length of the key you manage to produce, so best avoided if you have a choice.
Also, violating the bound on some Bell test is tricky. You want to give yourself the maximum chance of violating a Bell inequality. You can easily prove (Tsirelson's bound) that the maximally entangled state is the state that gives maximum violation of the CHSH inequality. This gives you the most overhead to work with given the potential presence of noise in the experiment. Again, provided your state does give violation of Bell inequality, it doesn't have to be maximal, but it will cost you in the rate of key generation.
answered Aug 12, 2019 at 8:24
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$\begingroup$ Thanks. I was actually wondering about using non maximally entangled states in E91 to reduce the information leakage to Eve before Alice and Bob compare their basis to potentially know Eve's presence. With your explanation that seems plausible. $\endgroup$ Aug 13, 2019 at 12:21
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$\begingroup$ @SujanVijayaraj I would guess (but don't know) that the reduction in information leakage is compensated for by the reduction in information that Alice and Bob can share,so that it doesn't help. $\endgroup$ Aug 13, 2019 at 12:35
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They did use them in the Vienna loophole-free Bell test, which is a pre-condition for implementing DIQKD. Their benefit is that you can do a loophole-free violation of the CHSH inequality efficiency down to $2/3$, whereas if you use maximally entangled states the required efficiency is much higher, $2\sqrt2-2$.
Therefore I fully expect them to be used for DIQKD based on photons. Note that the Delft loophole-free Bell test used maximally entangled states, though, as they were using NV centres instead.
