Non maximally entangled states for QKD

Question

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?

Answer 1

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.

Answer 2

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.